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Notes for 4H Galois Theory 2003–4
Andrew Baker
[29/05/2004]
Department of Mathematics, University of Glasgow.
Email address: a.baker@maths.gla.ac.uk
URL: http://www.maths.gla.ac.uk/∼ajb
Introduction: What is Galois Theory?
Much of early algebra centred around the search for explicit formulæ for roots of polynomial
equations in one or more unknowns. The solution of linear and quadratic equations in a single
unknown was well understood in antiquity, while formulæ for the roots of general real cubics
and quartics was solved by the 16th century. These solutions involved complex numbers rather
than just real numbers. By the early 19th century no general solution of a general polynomial
equation ‘by radicals’ (i.e., by repeatedly taking nth roots for various n) was found despite
considerable effort by many outstanding mathematicians. Eventually, the work of Abel and
Galois led to a satisfactory framework for fully understanding this problem and the realization
that the general polynomial equation of degree at least 5 could not always be solved by radicals. At a more profound level, the algebraic structure of Galois extensions is mirrored in the
subgroups of their Galois groups, which allows the application of group theoretic ideas to the
study of fields. This Galois Correspondence is a powerful idea which can be generalized to apply
to such diverse topics as ring theory, algebraic number theory, algebraic geometry, differential
equations and algebraic topology. Because of this, Galois theory in its many manifestations is
a central topic in modern mathematics.
In this course we will focus on the following topics.
• The solution of polynomial equations over a field, including relationships between roots,
methods of solutions and location of roots.
• The structure of finite and algebraic extensions of fields and their automorphisms.
We will study these in detail, building up a theory of algebraic extensions of fields and their
automorphism groups and applying it to solve questions about roots of polynomial equations.
The techniques we will meet can also be applied to study the following some of which may be
met by people studying more advanced courses.
• Classic topics such as squaring the circle, duplication of the cube, constructible numbers
and constructible polygons.
• Applications of Galois theoretic ideas in Number Theory, the study of differential
equations and Algebraic Geometry.
There are many good introductory books on Galois Theory, some of which are listed in the
Bibliography. In particular, [2, 3, 7] are all excellent sources and have many similarities to the
present approach to the material.
Some suggestions on using these notes. These notes cover more than the content of
the course and should be used in parallel with the lectures. The problem sets contain samples
of the kind of problems likely to occur in the final examination and should be attempted as an
integral part of learning the subject.
♠
The symbol ♥ ♦ indicates that the adjacent portion of the notes (proof, stated result,
♣
example,. . .) is not deemed to be examinable.
c A. J. Baker (2004)
Contents
Introduction: What is Galois Theory?
i
Chapter 1. Integral domains, fields and polynomial rings
Basic notions, convention, etc
1.1. Recollections on integral domains and fields
1.2. Polynomial rings
1.3. Identifying irreducible polynomials
1.4. Finding roots of complex polynomials of small degree
1.5. Automorphisms of rings and fields
Exercises on Chapter 1
1
1
1
5
10
14
17
21
Chapter 2. Fields and their extensions
2.1. Fields and subfields
2.2. Simple and finitely generated extensions
Exercises on Chapter 2
25
25
27
30
Chapter 3. Algebraic extensions of fields
3.1. Algebraic extensions
3.2. Splitting fields and Kronecker’s Theorem
3.3. Monomorphisms between extensions
3.4. Algebraic closures
3.5. Multiplicity of roots and separability
3.6. The Primitive Element Theorem
3.7. Normal extensions and splitting fields
Exercises on Chapter 3
33
33
36
39
41
44
47
49
50
Chapter 4. Galois extensions and the Galois Correspondence
4.1. Galois extensions
4.2. Working with Galois groups
4.3. Subgroups of Galois groups and their fixed fields
4.4. Subfields of Galois extensions and relative Galois groups
4.5. The Galois Correspondence and the Main Theorem of Galois Theory
4.6. Galois extensions inside the complex numbers and complex conjugation
4.7. Galois groups of even and odd permutations
4.8. Kaplansky’s Theorem
Exercises on Chapter 4
51
51
52
54
55
56
57
58
60
63
Chapter 5. Galois extensions for fields of positive characteristic
5.1. Finite fields
5.2. Galois groups of finite fields and Frobenius mappings
5.3. The trace and norm mappings
Exercises on Chapter 5
65
65
68
69
71
Chapter 6. A Galois Miscellany
6.1. A proof of the Fundamental Theorem of Algebra
6.2. Cyclotomic extensions
73
73
73
iii
6.3. Artin’s Theorem on linear independence of characters
6.4. Simple radical extensions
6.5. Solvability and radical extensions
6.6. Symmetric functions
Exercises on Chapter 6
Bibliography
Solutions
Chapter
Chapter
Chapter
Chapter
Chapter
Chapter
77
79
81
83
84
87
89
89
95
97
99
101
103
1
2
3
4
5
6
iv
CHAPTER 1
Integral domains, fields and polynomial rings
Basic notions, convention, etc
In these notes, a ring will always be a ring with unity 1 = 0. Most of the rings encountered
will also be commutative. An ideal I R will always mean a twosided ideal. An ideal I R in
a ring R is proper if I = R, or equivalently if I R. Under a ring homomorphism ϕ : R −→ S,
1 ∈ R is sent to 1 ∈ S, i.e., ϕ(1) = 1.
1.1. Definition. Let ϕ : R −→ S be a ring homomorphism.
• ϕ is a monomorphism if it is injective, i.e., if for r1 , r2 ∈ R,
ϕ(r1 ) = ϕ(r2 )
=⇒
r 1 = r2 ,
or equivalently if ker ϕ = {0}.
• ϕ is an epimorphism if it is surjective, i.e., if for every s ∈ S there is an r ∈ R with
ϕ(r) = s.
• ϕ is an isomorphism if it is both a monomorphism and an epimorphism, i.e., if it is
invertible (in which case its inverse is also an isomorphism).
1.1. Recollections on integral domains and fields
The material in this section is standard and most of it should be familiar. Details may be
found in [3, 4] or other books containing introductory ring theory.
1.2. Definition. A commutative ring R in which there are no zerodivisors is called an
integral domain or an entire ring. This means that for u, v ∈ R,
uv = 0
=⇒
u = 0 or v = 0.
1.3. Example. The following rings are integral domains.
(i) The ring of integers, Z.
(ii) If p is a prime, the ring of integers modulo p, Fp = Z/p = Z/(p).
(iii) The rings of rational numbers, Q, real numbers, R, and complex numbers, C.
(iv) The polynomial ring R[X], where R is an integral domain; in particular, the polynomial
rings Z[X], Q[X], R[X] and C[X] are all integral domains.
1.4. Definition. Let I R be a proper ideal in a commutative ring R.
• I is a prime ideal if for u, v ∈ R,
uv ∈ I
u ∈ I or v ∈ I.
=⇒
• I is a maximal ideal R if whenever J R is a proper ideal and I ⊆ J then J = I.
• I R is principal if
I = (p) = {rp : r ∈ R}
for some p ∈ R. Notice that if p, q ∈ R, then (q) = (p) if and only if q = up for some
unit u ∈ R. We also write p  x if x ∈ (p).
• p ∈ R is prime if (p) R is a prime ideal; this is equivalent to the requirement that
whenever p  xy with x, y ∈ R then p  x or p  y.
• R is a principal ideal domain if it is an integral domain and every ideal I R is principal.
1
1.5. Example. Every ideal I Z is principal, so I = (n) for some n ∈ Z which we can
always take to be nonnegative, i.e., n 0. Hence Z is a principal ideal domain.
1.6. Proposition. Let R be a commutative ring and I R an ideal.
(i) The quotient ring R/I is an integral domain if and only if I is a prime ideal.
(ii) The quotient ring R/I is a field if and only if I is a maximal ideal.
1.7. Example. If n
if n is a prime.
0, the quotient ring Z/n = Z/(n) is an integral domain if and only
For any (not necessarily commutative) ring with unity there is an important ring homomorphism η : Z −→ R called the unit or characteristic homomorphism which is defined by
1 + · · · + 1 if n > 0,
n
η(n) = n1 = −(1 + · · · + 1) if n < 0,
−n
0
if n = 0.
Since 1 ∈ R is nonzero, ker η Z is a proper ideal and using the Isomorphism Theorems we
see that there is a quotient monomorphism η : Z/ ker η −→ R which allows us to identify the
quotient ring Z/ ker η with the image ηZ ⊆ R as a subring of R. By Example 1.5, there is a
unique nonnegative integer p 0 such that ker η = (p); this p is called the characteristic of R
and denoted char R.
1.8. Lemma. If R is an integral domain, its characteristic char R is a prime.
Proof. Consider p = char R. If p = 0 we are done. So suppose that p > 0. The quotient
monomorphism η : Z/ ker η −→ R identifies Z/ ker η with the subring im η = im η of the integral
domain R. But every subring of an integral domain is itself an integral domain, hence Z/ ker η is
an integral domain. Now by Proposition 1.6(i), ker η = (p) is prime ideal and so by Example 1.7,
p is a prime.
1.9. Remark. When discussing a ring with unit R, we can consider it as containing as a
subring of the form Z/(char R) since the quotient homomorphism η : Z/(char R) −→ R gives
an isomorphism Z/(char R) −→ im η, allowing us to identify these rings. In particular, every
integral domain contains as a subring either Z = Z/(0) (if char R = 0) or Z/(p) if p = char R > 0
is a nonzero prime. This subring is sometimes called the characteristic subring of R. The rings
Z and Z/n = Z/(n) for n > 0 are often called core rings. When considering integral domains,
the rings Z and Fp = Z/p = Z/(p) for p > 0 a prime are called prime rings.
Here is a useful and important fact about rings which contain a finite prime ring Fp .
1.10. Theorem (Idiot’s Binomial Theorem). Let R be a commutative ring with unit containing Fp for some prime p > 0. If u, v ∈ R, then
(u + v)p = up + v p .
Proof. We have p1 = 0 in R, hence pt = 0 for any t ∈ R. The Binomial Expansion yields
p p−1
p p−2 2
p
(1.1)
(u + v)p = up +
u v+
u v + ··· +
uv p−1 + v p .
1
2
p−1
Now suppose that 1 j p − 1. Then we have
(p − 1)!
p
p (p − 1)!
=p×
.
=
j! (p − j)!
j! (p − j)!
j
There are no factors of p appearing in (p − 1)!, j! or (p − j)!, so since this number is an integer
it must be divisible by p, i.e.,
p
(1.2a)
p
,
j
2
or equivalently
p
j
(1.2b)
≡0
(mod p).
Hence in R we have
p
1 = 0.
j
Combining the divisibility conditions of (1.2) with the expansion of (1.1), we obtain the
required equation in R,
(u + v)p = up + v p .
1.11. Definition. Let R be a ring. An element u ∈ R is a unit if it is invertible, i.e., there
is and element v ∈ R for which
uv = 1 = vu.
We usually write u−1 for this element v, which is necessarily unique and is called the (multiplicative) inverse of u in R. We will denote the set of all invertible elements of R by R× and
note that it always forms a group under multiplication.
1.12. Definition. A commutative ring k is a field if every nonzero element u ∈ k is a unit.
This is equivalent to requiring that k× = k − {0}.
The familiar rings Q, R and C are all fields.
1.13. Example. If n
1, the quotient ring Z/n is a field if and only if n is a prime.
1.14. Proposition. Every field is an integral domain.
Proof. Let k be a field. Suppose that u, v ∈ k and uv = 0. If u = 0, we can multiply by
u−1 to obtain
v = u−1 uv = 0,
hence v = 0. So at least one of u, v must be 0.
A nonzero element p ∈ R is irreducible if for u, v ∈ R,
p = uv
=⇒
u or v is a unit.
1.15. Lemma. Let R be an integral domain. If p ∈ R is a nonzero prime then it is an
irreducible.
Proof. Suppose that p = uv for some u, v ∈ R. Then p  u or p  v, and we might as well
assume that u = tp for some t ∈ R. Then (1 − tv)p = 0 and so tv = 1, showing that v is a unit
with inverse t.
Now let D be an integral domain. A natural question to ask is whether D is isomorphic to
a subring of a field. This is certainly true for the integers Z which are contained in the field of
rational numbers Q, and for a prime p > 0, the prime ring Fp is itself a field.
1.16. Definition. The fields Q and Fp where p > 0 is a prime are the prime fields.
Of course, we can view Z as a subring of any subfield of the complex numbers so an answer
to this question may not be unique! However, there is always a ‘smallest’ such field which is
unique up to an isomorphism.
1.17. Theorem. Let D be an integral domain.
(i) There is a field of fractions of D, Fr(D), which contains D as a subring.
(ii) If ϕ : D −→ F is a ring monomorphism into a field F , there is a unique homomorphism
ϕ : Fr(D) −→ F such that ϕ(t) = ϕ(t) for all t ∈ D ⊆ Fr(D).
3
ϕ
D
inc
/
0. Then there are unique polynomials q(X), r(X) ∈ k[X] for
which
f (X) = q(X)d(X) + r(X)
and either deg r(X) < deg d(X) or r(X) = 0.
In the situation discussed in this result, the following names are often used. We refer to the
process of finding q(X) and r(X) as long division of f (X) by d(X). Also,
f (X) = the dividend ,
d(X) = the divisor ,
q(X) = the quotient,
r(X) = the remainder .
1.27. Example. For k = Q, find the quotient and remainder when f (X) = 6X 4 − 6X 3 +
3X 2 − 3X + 1 is divided by d(X) = 2X 2 + 1.
Solution. In the usual notation we have the following calculation.
3X 2 − 3X
2X 2 + 1  6X 4 − 6X 3 + 3X 2 − 3X + 1
6X 4 + 0X 3 + 3X 2 + 0X + 0
− 6X 3 + 0X 2 − 3X + 1
− 6X 3 + 0X 2 − 3X + 0
1
7
Hence
6X 4 − 6X 3 + 3X 2 − 3X + 1 = (3X 2 − 3X)(2X 2 + 1) + 1,
giving q(X) = 3X 2 − 3X and r(X) = 1.
1.28. Example. For k = F5 , find the quotient and remainder when f (X) = 10X 5 + 6X 4 −
+ 3X 2 − 3X + 1 is divided by d(X) = 2X 2 + 1.
6X 3
Solution. First notice that working modulo 5 we have
f (X) = 10X 5 + 6X 4 − 6X 3 + 3X 2 − 3X + 1 ≡ X 4 + 4X 3 + 3X 2 + 2X + 1
(mod 5).
Notice also following multiplicative inverses in F5 :
2−1 ≡ 3
(mod 5),
3−1 ≡ 2
(mod 5),
4−1 ≡ 4
(mod 5).
We have the following calculation.
3X 2 + 2X
2X 2 + 1 6X 4 + 4X 3 + 3X 2 + 2X + 1
6X 4 + 0X 3 + 3X 2 + 0X + 0
4X 3 + 0X 2 + 2X + 1
4X 3 + 0X 2 + 2X + 0
1
Hence
6X 4 − 6X 3 + 3X 2 − 3X + 1 ≡ (3X 2 + 2X)(2X 2 + 1) + 1
giving q(X) = 3X 2 + 2X and r(X) = 1.
(mod 5),
An important consequence of Theorem 1.26 is the following which makes use of the Euclidean
Algorithm.
1.29. Corollary. Let k be a field and X an indeterminate. Let f (X), g(X) ∈ k[X] be
nonzero. Then there are a(X), b(X) ∈ k[X] such that
a(X)f (X) + b(X)g(X) = gcd(f (X), g(X)).
Here the greatest common divisor gcd(f (X), g(X)) of f (X), g(X) is the monic polynomial
of greatest degree which divides both of f (X), g(X).
1.30. Proposition. Let k be a field and X an indeterminate. Then a nonconstant polynomial p(X) ∈ k[X] is an irreducible if and only if it is a prime.
♠
♥ ♦
♣
Proof. By Lemma 1.15 we already know that p(X) is irreducible if it is prime. So
suppose that p(X) is irreducible and that p(X)  u(X)v(X) for u(X), v(X) ∈ k[X]. Then by
Corollary 1.29, there are a(X), b(X) ∈ k[X] such that
a(X)p(X) + b(X)u(X) = gcd(p(X), u(X)).
But since p(X) is irreducible, gcd(p(X), u(X)) = 1, hence
a(X)p(X) + b(X)u(X) = 1.
Multiplying through by v(X) gives
a(X)p(X)v(X) + b(X)u(X)v(X) = v(X)
and so p(X)  v(X). This shows that p(X)  u(X) or p(X)  v(X) and so p(X) is prime.
1.31. Theorem. Let k be a field and X an indeterminate.
(i) Every ideal I k[X] is principal, i.e., I = (h(X)) for some h(X) ∈ k[X].
(ii) The ideal (p(X)) k[X] is prime if and only if p(X) = 0 or p(X) is irreducible in k[X].
8
(iii) The quotient ring k[X]/(p(X)) is an integral domain if and only if p(X) = 0 or p(X)
is irreducible in k[X].
(iv) The quotient ring k[X]/(p(X)) is a field if and only if p(X) is an irreducible in k[X].
♠
♥ ♦
♣
Proof. (i) Let I
k[X] and assume that I = (0). Then there must be at least
one element of I with positive degree and so we can choose h(X) ∈ I of minimal degree, say
d = deg h(X).
Now let p(X) ∈ I. By Long Division, there are q(X), r(X) ∈ k[X] such that
p(X) = q(X)h(X) + r(X)
and deg r(X) < d or r(X) = 0.
Since p(X) and h(X) are in the ideal I, we also have
r(X) = p(X) − q(X)h(X) ∈ I.
If r(X) = 0, this would contradict the minimality of d, so we must have r(X) = 0, showing that
p(X) = q(X)h(X). Thus I ⊆ (p(X)) ⊆ I and therefore I = (p(X)).
(ii) This follows from Proposition 1.30.
(iii) This follows from Proposition 1.6(i).
(iv) Since k[X] is an integral domain and not a field, it is follows that if k[X]/(p(X)) is a field
then because it is an integral domain, p(X) is an irreducible by (iii).
Suppose that p(X) is irreducible (and hence is nonzero). Then for any q(X) ∈ k[X] with
q(X) ∈
/ (p(X)), by Corollary 1.29 we can find suitable a(X), b(X) ∈ k[X] for which
a(X)p(X) + b(X)q(X) = gcd(p(X), q(X)).
But gcd(p(X), q(X)) = 1 since p(X) is irreducible, so
a(X)p(X) + b(X)q(X) = 1.
This shows that in the quotient ring k[X]/(p(X)) the residue class of q(X) has the residue class
of b(X) as its inverse.
1.32. Remark. In connection with Theorem 1.31(i), notice that if p(X) ∈ k[X], then
provided d = deg p(X) > 0, we have for some pd = 0,
p(X) = p0 + p1 X + · · · + pd X d = pd q(X),
where
−1
−1
d−1
+ X d.
q(X) = p−1
d p0 + pd p1 X + · · · + pd pd−1 X
This easily implies that as ideals of k[X], (p(X)) = (q(X)). So we can always find a monic
polynomial as the generator of a given ideal, and this monic polynomial is unique.
1.33. Proposition (Unique Factorization Property). Every nonconstant polynomial f (x) ∈
k[X] has a factorization
f (x) = cp1 (X) · · · pk (X),
where c ∈ k, and p1 (X), . . . , pk (X) ∈ k[X] are irreducible monic polynomials. Moreover, c is
unique and the sequence of polynomials p1 (X), . . . , pk (X) is unique apart from the order of the
terms.
♠
♥ ♦
♣
Proof. (Sketch)
Existence is proved by induction on the degree of f (X) and begins with the obvious case
deg f (X) = 1. If deg f (X) > 1, then either f (X) is already irreducible, or f (X) = f1 (X)f2 (X)
with both factors of positive degree, and therefore deg fj (X) < deg f (X). This gives the
inductive step.
To prove uniqueness, suppose that
p1 (X) · · · pk (X) = q1 (X) · · · q (X)
9
where pi (X), qj (X) ∈ k[X] are irreducible monic polynomials. Then by Proposition 1.30, each
pi (X) is prime, hence divides one of the qj (X), hence must equal it. By reordering we can
assume that pi (X) = qi (X) and k
. After cancelling common factors we obtain
qk+1 (X) · · · q (X) = 1,
and so we see that k = .
1.34. Corollary. Suppose that f (X) ∈ k[X] factors into linear factors
f (X) = c(X − u1 ) · · · (X − ud ),
where u1 , . . . , ud ∈ k. Then the sequence of roots u1 , . . . , ud is unique apart from the order. In
particular, if v1 , . . . , vr are the distinct roots, then
f (X) = c(X − v1 )m1 · · · (X − vr )mr ,
where mi > 0 and this factorization is unique apart from the order of the pairs (vi , mi ).
1.35. Corollary. The number of distinct roots of a nonconstant polynomial f (X) ∈ k[X]
is at most deg f (X).
1.36. Definition. If k is a field and X an indeterminate, then the field of fractions of k[X]
is the field of rational functions, k(X). The elements of k(X) are fractions of the form
a0 + a1 X + · · · + am X m
b0 + b1 X + · · · + bn X n
with ai , bj ∈ k and b0 + b1 X + · · · + bn X n = 0.
1.3. Identifying irreducible polynomials
When k is a field, we will need some effective methods for deciding when a polynomial in
k[X] is irreducible.
Let us consider factorisation of polynomials over Q. If f (X) ∈ Z[X] then we can also consider
f (X) as an element of Q[X]. If R = Z or Q, we say that f (X) has a proper factorisation over
R if f (X) = g(X)h(X) for some g(X), h(X) ∈ R[X] with deg g(X) > 0 and deg h(X) > 0.
1.37. Proposition (Gauss’s Lemma). Let f (X) ∈ Z[X]. Then f (X) has a proper factorisation over Z if and only it has a proper factorisation over Q.
So to find factors of f (X) it is sufficient to look for factors in Z[X]. Our next result is a
special case of the Eisenstein Irreducibility Test. The version here is slightly more general than
the more usual one which corresponds to taking s = 0.
1.38. Proposition (Eisenstein Test). Let f (X) ∈ Z[X] and s ∈ Z. Choose ai ∈ Z so that
f (X) = a0 + a1 (X − s) + · · · + ad−1 (X − s)d−1 + ad (X − s)d ,
where d = deg f (X). Suppose that p > 0 is a prime for which the following three conditions
hold:
• ak ≡ 0 (mod p) for k = 0, . . . , d − 1;
• a0 ≡ 0 (mod p2 );
• ad ≡ 0 (mod p).
Then f (X) is irreducible in Q[X] and hence also in Z[X].
1.39. Example. Let p
2 be a prime. Then the polynomial
Φp (X) = 1 + X + · · · + X p−1 ∈ Z[X]
is irreducible in Q[X] and hence also in Z[X].
10
Proof. Working in Z[X],
Φp (X)(X − 1) = (1 + X + · · · + X p−1 )(X − 1)
= Xp − 1
= (1 + (X − 1))p − 1
p
p
(X − 1)k
k
=
k=1
≡ (X − 1)p
(mod p),
since by (1.2a), p divides
p
k
=
p!
k! (p − k)!
when k = 1, . . . , p − 1. Hence
Φp (X) ≡ (X − 1)p−1
(mod p)
Also,
p
1
=p≡0
(mod p2 ),
giving
Φp (X) = (X − 1)p−1 + cp−2 (X − 1)p−2 + · · · + c1 (X − 1) + c0
(1.4)
with cr ≡ 0 (mod p) and c0 = p. So the Eisenstein Test can be applied here with s = 1 to show
that Φp (X) is irreducible in Z[X].
1.40. Example. As examples we have the irreducible polynomials
Φ2 (X) = 1 + X,
Φ3 (X) = 1 + X + X 2 ,
Φ5 (X) = 1 + X + X 2 + X 3 + X 4 ,
Φ7 (X) = 1 + X + X 2 + X 3 + X 4 + X 5 + X 6 ,
Φ11 (X) = 1 + X + X 2 + X 3 + X 4 + X 5 + X 6 + X 7 + X 8 + X 9 + X 10 .
n
These are examples of the cyclotomic polynomials Φn (X) ∈ Z[X] which are defined for all
1 by
(1.5a)
Xn − 1 =
Φd (X),
dn
where the product is taken over all the positive divisors of n. For example,
X 2 − 1 = (X − 1)(X + 1) = Φ1 (X)Φ2 (X),
X 3 − 1 = (X − 1)(X 2 + X + 1) = Φ1 (X)Φ3 (X),
X 4 − 1 = (X − 1)(X + 1)(X 2 + 1) = Φ1 (X)Φ2 (X)Φ4 (X),
X 5 − 1 = (X − 1)(X 4 + X 3 + X + 1) = Φ1 (X)Φ5 (X),
X 6 − 1 = (X − 1)(X + 1)(X 2 + X + 1)(X 2 − X + 1) = Φ1 (X)Φ2 (X)Φ3 (X)Φ6 (X),
X 12 − 1 = (X − 1)(X + 1)(X 2 + X + 1)(X 2 + 1)(X 2 − X + 1)(X 4 − X 2 + 1)
= Φ1 (X)Φ2 (X)Φ3 (X)Φ4 (X)Φ6 (X)Φ12 (X).
11
Cyclotomic polynomials can be computed recursively using Equation (1.5a). If we know Φk (X)
for k < n, then
Xn − 1
.
(1.5b)
Φn (X) =
Φd (X)
dn
d 0, then
ϕ(n) = ϕ(pr11 ) · · · ϕ(prss ).
Furthermore, if p is a prime and r > 0, then
(1.9)
ϕ(pr ) = (p − 1)pr−1 .
Notice that as a result, ϕ(n) is even when n > 2.
1.42. Remark. For those who know about the M¨
obius function µ (which takes values 0, ±1)
and M¨
obius inversion, the latter can be used to solve Equation (1.6) for ϕ, giving
n
(1.10)
ϕ(n) =
µ(d) .
d
dn
Similarly, the formulæ of (1.5) lead to
(1.11)
(X n/d − 1)µ(d) .
Φn (X) =
dn
So for example, if p, q are distinct primes, then using standard properties of µ,
Φpq (X) = (X pq − 1)µ(1) (X pq/p − 1)µ(p) (X pq/q − 1)µ(q) (X pq/pq − 1)µ(pq)
= (X pq − 1)(X q − 1)−1 (X p − 1)−1 (X − 1) =
(X pq − 1)(X − 1)
.
(X q − 1)(X p − 1)
Recall that an element ζ of a field K is a primitive nth root of unity if
min{k : 1
k and ζ k = 1} = n.
We think of ζn = e2πi/n as the standard complex primitive nth root of unity. Then every
complex nth root of unity has the form ζnk = e2πik/n for k = 0, 1, . . . , n − 1.
12
1.43. Theorem. For each n 1, the cyclotomic polynomial Φn (X) is irreducible in Q[X]
and hence in Z[X]. The complex roots of Φn (X) are the primitive nth roots of unity,
ζnk = e2πik/n
(0
n − 1, gcd(k, n) = 1).
k
and the number of these is deg Φn (X) = ϕ(n). Hence,
(X − ζnt ).
Φn (X) =
t=1,...,n−1
gcd(t,n)=1
Proof. We will give a reformulation and proof of this in Theorem 6.2.
1.44. Example. For n = 6 we have
2πi/6
ζ6 = e
πi/3
=e
√
3
1
i.
= +
2
2
Then ϕ(6) = 2 and
Φ6 (X) = X 2 − X + 1 = (X − ζ6 )(X − ζ65 ).
It is also worth recording a related general result on cyclic groups.
1.45. Proposition. Let n 1 and C = g be a cyclic group of order n and a generator g.
Then an element g r ∈ C is a generator if and only if gcd(r, n) = 1; the number of such elements
of C is ϕ(n).
This leads to a useful group theoretic result.
1.46. Lemma. Let G be a finite group satisfying the following condition:
• For each n 1, there are at most n solutions of xn = ι in G.
Then G is cyclic and in particular is abelian.
♠
♥ ♦
♣
Proof. Let θG (d) denote the number of elements in G of order d. By Lagrange’s
Theorem, θG (d) = 0 unless d divides G. Since
{g ∈ G : g = d},
G=
dG
we have
G =
θG (d).
dG
Recall the Euler ϕfunction satisfies Equation (1.6), hence
G =
ϕ(d).
dG
Combining these we obtain
(1.12)
θG (d) =
dG
ϕ(d).
dG
Let d be a divisor of G. By Proposition 1.45, for each element g ∈ G of order d, the cyclic
subgroup g
G has ϕ(d) generators, each of order d. As there are at most d such elements g
in G, this gives θG (d) ϕ(d). So
θG (d)
ϕ(d).
dG
dG
Now if θG (d) < ϕ(d) for some d, we would have a strict inequality in place of Equation (1.12).
Hence θG (d) = ϕ(d) for all d. In particular, there are ϕ(G) elements of order G, hence there
must be an element of order G, so G is cyclic.
13
The above results for polynomials over Q and Z have analogues over the field of fractions
k(T ) and polynomial ring k[T ], where k is a field.
A polynomial f (X) ∈ k[T ][X] is an element of k(T )[X]. If R = k[T ] or k(T ), we say that
f (X) has a proper factorisation over R if f (X) = g(X)h(X) for some g(X), h(X) ∈ R[X] with
deg g(X) > 0 and deg h(X) > 0.
1.47. Proposition (Gauss’s Lemma). Let f (X) ∈ k[T ][X]. Then f (X) has a proper factorisation over k[T ] if and only it has a proper factorisation over k(T ).
Here is another version of the Eisenstein Test; again we state a version which is slightly
more general than the usual one which corresponds to the case where s = 0.
1.48. Proposition (Eisenstein Test). Let f (X) ∈ k[T ][X] and s ∈ k[T ]. Choose ai ∈ k[T ]
so that
f (X) = a0 + a1 (X − s) + · · · + ad−1 (X − s)d−1 + ad (X − s)d ,
where d = deg f (X). Suppose that p(T ) ∈ k[T ] is an irreducible for which the following three
conditions hold:
• ak ≡ 0 (mod p(T )) for k = 0, . . . , d − 1;
• a0 ≡ 0 (mod p(T )2 );
• ad ≡ 0 (mod p(T )).
Then f (X) is irreducible in k(T )[X] and hence also in k[T ][X].
1.49. Example. Let k be a field. Then the polynomial X n − T is irreducible in k(T )[X].
1.4. Finding roots of complex polynomials of small degree
♠
♥ ♦
♣
In this section we work within the complex numbers and take k ⊆ C. In practice we
will usually have k = R or k = C.
For monic linear (degree 1) or quadratic (degree 2) polynomials, methods of finding roots are
very familiar. Let us consider the cases of cubic (degree 3) and quartic (degree 4) polynomials.
Cubic polynomials: Cardan’s method. The following 16th century method of finding
roots of cubics is due to Jerˆome Cardan who seems to have obtained some preliminary versions
from Niccol`a Tartaglia by somewhat disreputable means! For historical details see [2, 3].
A monic cubic
f (X) = X 3 + a2 X 2 + a1 X + a0 ∈ C[X]
can be transformed into one with no quadratic term by a change of variables X −→ X − a2 /3
giving
1
a1 a2 2a32
+
g(X) = f (X − a2 /3) = X 3 − a1 − a22 X − a0 +
3
3
27
∈ C[X].
Clearly finding the roots of f (X) is equivalent to finding those of g(X), so we may as well
assume that we want to find the complex roots of
f (X) = X 3 + pX + q ∈ C[X].
Suppose that x ∈ C is a root of f (X), i.e.,
x3 + px + q = 0.
(1.13)
If we introduce u ∈ C for which
x=u−
then
u−
p
3u
3
p
,
3u
+p u−
and so
u3 −
p
+q =0
3u
p3
+ q = 0,
27u3
14
hence
u6 + qu3 −
p3
= 0.
27
Solving for u3 we obtain
q q
u3 = − ±
2 2
where
equation
q2 +
q2 +
4p3
,
27
4p3
denotes one of the complex square roots of the discriminant of the quadratic
27
p3
= 0.
27
Now if we take u to be a cube root of one of the complex numbers
U 2 + qU −
4p3
q q
q2 +
− ±
2 2
27
we obtain the desired root of f (X) as x = u − p/3u. Notice that we have a choice of 2 values
for u3 and for each of these a choice of 3 values for u, differing by factors of the form ω r for
r = 0, 1, 2 where ω = e2πi/3 is a primitive cube root of 1. However, since
4p3
3
2+
4p
q
−q
−
−q − q 2 +
27
1
27 = −27
= 2
,
3
q − (q 2 + 4p3 /27)
4p3
4p
−q + q 2 +
27
it is easy to verify that there are in fact only 3 choices of the root x which we can write
symbolically as
(1.14)
x=
3
q 1
− +
2 2
q2 +
4p3
+
27
3
q 1
− +
2 2
q2 +
4p3
−
27
3
q 1
− −
2 2
q2 +
4p3
27
or more precisely as
(1.15)
x=
p
3
3
q 1
− +
2 2
.
q2 +
4p3
27
1.50. Example. Find the complex roots of the polynomial
f (X) = X 3 + 3X − 10 ∈ R[X].
Solution. Applying the method above, we reduce to the quadratic equation
U 2 − 10U − 1 = 0
√
√
√
whose roots are 5 ± 26 ∈ R. Notice that 5 + 26 > 0 and 5 − 26 < 0; we also have
√
−1
√ .
5 − 26 =
5 + 26
√
Now 5 + 26 has the complex cube roots
√
√
√
3
3
3
5 + 26,
5 + 26 ω,
5 + 26 ω 2 .
Here we have x = u − 1/u, so the 3 complex roots of f (X) are
3
5+
√
26 −
1
3
5+
√
15
ωr
26
(r = 0, 1, 2).
Notice that one of these is real, namely
3
5+
√
3
26 −
1
3
5+
√
=
26
3
√
2
26 − 1
.
√
5 + 26
5+
Quartic polynomials: Ferrari’s method. The following method of finding roots of
quartics was publicised by Cardan who attributed it to his student Lodovicio Ferrari.
A general monic quartic polynomial
f (X) = X 3 + a3 X 3 + a2 X 2 + a1 X + a0 ∈ C[X]
can be transformed into one with no cubic term by a change of variables X −→ X − a2 /3 giving
g(X) = f (X − a3 /4) =
3
Y 4 + a2 − a23 Y 2 +
8
1 3 1
a − a2 a3 + a1 Y −
8 3 2
1
3 4 1
a2 a23 −
a + a1 a3 + a0 .
16
256 3 4
Clearly finding the roots of f (X) is equivalent to finding those of g(X), so we may as well
assume that we want to find the complex roots of
f (X) = X 4 + pX 2 + qX + r ∈ C[X].
Suppose that x is a root and introduce numbers y, z such that z = x2 + y (we will fix the
values of these later). Then
z 2 = x4 + 2x2 y + y 2
= −px2 − qx − r + 2xy + y 2
= (2y − p)x2 − qx + y 2 − r.
Now choose y to make the last quadratic expression in x a square,
(2y − p)x2 − qx + (y 2 − r) = (Ax + B)2 .
(1.16)
This can be done by requiring the vanishing of the discriminant
q 2 − 4(2y − p)(y 2 − r) = 0.
(1.17)
Notice that if y = p/2 then we would require q = 0 and then
f (X) = X 4 + pX 2 + r = (X 2 )2 + p(X 2 ) + r = 0
can be solved by solving
Z 2 + pZ + r = 0.
Since Equation (1.17) is a cubic in y, we can use the method of solution of cubics to find a root
y = t say. Then for Equation (1.16) we have
(x2 + t) = (Ax + B)2 ,
whence
x2 = −t ± (Ax + B).
Thus taking the two square roots of the right hand side we obtain 4 values for x, which we write
symbolically as
x = ± −t ± (Ax + B).
1.51. Remark. In the case of cubic and quartic polynomials over C we can obtain all the
roots by repeatedly taking square or cube roots (or radicals). Consequently such polynomials are
said to be solvable by radicals. Later we will see that this is not true in general for polynomials
of degree at least 5; this is one of the great early successes of this theory.
16
1.5. Automorphisms of rings and fields
1.52. Definition. Let R be a ring and R0 ⊆ R a subring.
• An automorphism of R is a ring isomorphism α : R −→ R. The set of all such automorphisms is denoted Aut(R).
• An automorphism of R over R0 is a ring isomorphism α : R −→ R for which α(r) = r
whenever r ∈ R0 . The set of all automorphisms of R over R0 is denoted AutR0 (R).
1.53. Proposition. For a ring R with a subring R0 ⊆ R, Aut(R) and AutR0 (R) form
groups under composition of functions.
Proof. The composition α ◦ β of two automorphisms α, β : R −→ R is also an automorphism of R as is the inverse of α. The identity function id = idR : R −→ R is an automorphism.
Hence Aut(R) forms a group under composition. The argument for AutR0 (R) is similar.
1.54. Proposition. Let R be one of the core rings Z or Z/n with n > 1. Then
(i) The only automorphism of R is the identity, i.e., Aut(R) = {id}.
(ii) If S is a ring containing a core ring R and α ∈ Aut(S), then α restricts to the identity
on R, i.e., α(r) = r for all r ∈ R. Hence, Aut(S) = AutR (S).
Proof. (i) For such a core ring R, every element has the form k1 for some k ∈ Z. For an
automorphism α of R,
α(1) + · · · + α(1) if k > 0,
k
α(k1) = −(α(1) + · · · + α(1)) if k < 0,
−k
α(0)
if k = 0
1 + · · · + 1 if k > 0,
k
= −(1 + · · · + 1) if k < 0,
−k
0
if k = 0
=k1.
Thus α = id.
(ii) For α ∈ Aut(S), α(1) = 1 and a similar argument to that for (i) shows that α(r) = r for all
r ∈ R.
1.55. Proposition. Let D be an integral domain and α : D −→ D be an automorphism.
Then the induced homomorphism gives an automorphism α∗ : Fr(D) −→ Fr(D).
Proof. Given α, the induced homomorphism α∗ : Fr(D) −→ Fr(D) exists and we need
to show it has an inverse. The inverse automorphism α−1 : D −→ D also gives rise to an
induced homomorphism (α−1 )∗ : Fr(D) −→ Fr(D). Since α−1 ◦ α = id = α ◦ α−1 , we can apply
Corollary 1.20 to show that
(α−1 )∗ ◦ (α)∗ = id = (α)∗ ◦ (α−1 )∗ .
Hence (α)∗ is invertible with inverse (α−1 )∗ .
1.56. Corollary. There is a monomorphism of groups
( )∗ : Aut(D) −→ Aut(Fr(D));
α −→ α∗ .
1.57. Example. The field of fractions of the ring of integers Z is the field of rationals Q.
The homomorphism
( )∗ : Aut(Z) −→ Aut(Q); α −→ α∗
17
is an isomorphism and hence Aut(Q) = {id}.
Combining this example with Proposition 1.54(ii) we obtain another useful result.
1.58. Proposition. Let k be one of the prime fields Q or Fp with p > 0 prime. If R is
a ring containing k as a subring, then every automorphism of R restricts to the identity on k,
i.e., Aut(R) = Autk (R).
Recalling Definition 1.36, we have an example which shows that the monomorphism of
Corollary 1.56 need not be an epimorphism. Here we take D = Q[X] and Fr(Q[X]) = Q(X).
1.59. Example. The homomorphism
( )∗ : Aut(Q[X]) −→ Aut(Q(X));
α −→ α∗
is a monomorphism but it is not an epimorphism since there is an automorphism
γ : Q(X) −→ Q(X);
γ(f (X)) = f (1/X)
which sends X ∈ Q[X] ⊆ Q(X) to 1/X ∈
/ Q[X] and so does not restrict to an automorphism of
Q[X].
Let k be a field. The group of invertible 2 × 2 matrices over k is the 2 × 2 general linear
group over k,
a11 a12
GL2 (k) =
: aij ∈ k, a11 a22 − a12 a21 = 0
a21 a22
The scalar matrices form a normal subgroup
Scal2 (k) = {diag(t, t) : t ∈ k, t = 0} GL2 (k).
The quotient group is called the 2 × 2 projective general linear group over k,
PGL2 (k) = GL2 (k)/ Scal2 (k).
Notice that GL2 (k) has another interesting subgroup called the affine subgroup,
Aff 1 (k) =
a b
: a, b ∈ k, a = 0
0 1
GL2 (k).
1.60. Example. Let k be a field and X an indeterminate. Then Autk (k[X]) and hence
Autk (k(X)), contains a subgroup isomorphic to Aff 1 (k). In fact, Autk (k[X]) ∼
= Aff 1 (k).
Proof. We begin by showing that to each affine matrix
A=
a b
∈ Aff 1 (k)
0 1
there is an associated automorphism αA : k[X] −→ k[X].
For this we use the element aX + b ∈ k[X] together with the extension result of Theorem 1.22(i) to obtain a homomorphism αA : k[X] −→ k[X] with αA (X) = aX + b. Using the
inverse matrix
a−1 −a−1 b
A−1 =
0
1
we similarly obtain a homomorphism αA−1 : k[X] −→ k[X] for which
αA−1 (X) = a−1 X − a−1 b.
Using the same line of argument as in the proof of Proposition 1.55 (or doing a direct calculation)
we see that αA−1 is the inverse of αA an so αA ∈ Autk (k[X]). It is straightforward to check
that for A1 , A2 ∈ Aff 1 (k),
αA2 A1 = αA1 ◦ αA2 ,
(note the order!) hence there is a homomorphism of groups
Aff 1 (k) −→ Autk (k[X]);
18
A −→ αA−1 ,
which is easily seen to be a monomorphism. Composing with ( )∗ we see that there is a
monomorphism Aff 1 (k) −→ Autk (k(X)). In fact, this is also an epimorphism and we leave the
proof of this as an exercise.
1.61. Example. Let k be a field and X an indeterminate. Then
(i) Autk (k(X)) contains a subgroup isomorphic to PGL2 (k).
(ii) In fact, Autk (k(X)) ∼
= PGL2 (k).
♠
♥ ♦
♣
Proof. (i) We begin by showing that to each invertible matrix
A=
a11 a12
∈ GL2 (k)
a21 a22
there is an associated automorphism αA : k(X) −→ k(X).
We begin by choosing the element (a11 X + a12 )/(a21 X + a22 ) ∈ k(X) and then using Theorem 1.22(i) to obtain a homomorphism k[X] −→ k(X) that sends X to (a11 X+a12 )/(a21 X+a22 ).
By applying ( )∗ to this we obtain a homomorphism (known as a fractional linear transformation) αA : k(X) −→ k(X) for which
αA (X) =
a11 X + a12
.
a21 X + a22
Again we find that
αA2 A1 = αA1 ◦ αA2 .
−1
There is an associated homomorphism of groups GL2 (k) −→ Autk (k(X)) sending A to αA .
However, this is not an injection in general since for each scalar matrix diag(t, t),
αdiag(t,t) (X) =
tX
= X,
t
showing that αdiag(t,t) is the identity function.
In fact it is easy to see that Scal2 (k) GL2 (k) is the kernel of this homomorphism. Therefore
passing to the quotient PGL2 (k) = GL2 (k)/ Scal2 (k) we obtain a monomorphism PGL2 (k) −→
Autk (k(X)). There is one case where Scal2 (k) is the trivial group, namely k = F2 .
(ii) To show that every automorphism of k(X) is a fractional linear transformation is less
elementary. We give a sketch proof for the case of k = C; actually this argument can be modified
to work for any algebraically closed field, but an easy argument then shows the general case.
Let α ∈ AutC (C(X)). There is an associated rational (hence meromorphic) function f given
by z −→ f (z), where α(X) = f (X), defined on C with the poles of f deleted. If we write
f (X) =
p(X)
q(X)
where p(X), q(X) ∈ C[X] have no common factors of positive degree, then the order of f (X) is
ord f = max{deg p(X), deg q(X)}.
Now let c ∈ C. Then the number of solutions counted with algebraic multiplicity of the equation
f (z) = c turns out to be ord f . Also, if deg p(X)
deg q(X) then the number of poles of f
counted with algebraic multiplicity is also ord f . Finally, if deg p(X) > deg q(X) then we can
write
p0 (X)
f (X) = p1 (X) +
,
q(X)
where p0 (X), p1 (X) ∈ C[X] and deg p0 (X) < deg q(X). Then the number of poles of f counted
with algebraic multiplicity is
p0
deg p1 (X) + ord .
q
19
Now it is easy to see that since α is invertible so is the function f . But this can only happen
if the function f is injective which means that all of these numbers must be 1, hence ord f = 1.
Thus
aX + b
= constant
f (X) =
cX + d
and the matrix
a b
must be invertible.
c d
Clearly not every fractional linear transformation αA : k(X) −→ k(X) maps polynomials to
polynomials so ( )∗ : Autk (k[X]) −→ Autk (k(X)) is not an epimorphism.
Now we turn to a more familiar field R, the real numbers.
1.62. Proposition. The only automorphism of the field R is the identity function, hence
Aut(R) = {id}.
♠
♥ ♦
♣
Proof. First we note that Q ⊆ R is a subring and if α ∈ Aut(R) then α(q) = q for
q ∈ Q by Example 1.57.
We recall from Analysis that the rational numbers are dense in the real numbers in the
sense that each r ∈ R can be expressed as a limit r = limn→∞ qn , where qn ∈ Q. Then for a
continuous function f : R −→ R, its value at r depends on its values on Q since
f (r) = f ( lim qn ) = lim f (qn ).
n→∞
n→∞
We will show that an automorphism α ∈ Aut(R) is continuous.
First recall that for x, y ∈ R,
x 0. Then we can choose a rational number q such that 0 < q
Taking δ = q we find that for y ∈ R with y − x < δ (i.e., −δ < y − x < δ) we have
ε.
−δ = α(−δ) < α(y) − α(x) < α(δ) = δ,
hence
α(y) − α(x) < δ
ε.
This shows that α is continuous at x.
Thus every automorphism of R is continuous function which fixes all the rational numbers,
hence it must be the identity function.
1.63. Remark. If we try to determine Aut(C) the answer turns out to be much more
complicated. It is easy to see that complex conjugation ( ) : C −→ C is an automorphism of C
and fixes every real number, i.e., ( ) ∈ AutR (C); in fact, AutR (C) = {id, ( )}. However, it is
not true that every α ∈ Aut(C) fixes every real number! The automorphism group Aut(C) is
actually enormous but it is hard to find an explicit element other than id and ( ). Note that
given an automorphism α ∈ Aut(C), the composition α ◦ ( ) ◦ α−1 is also self inverse, so there
are many elements of order 2 in the group Aut(C).
20
Exercises on Chapter 1
1.1. Let R be a ring. Show that
{n ∈ Z : n > 0 and n1 = 0} = {n ∈ Z : n > 0 and nr = 0 for all r ∈ R}.
Deduce that if char R > 0 then these sets are nonempty and
char R = min{n ∈ Z : n > 0 and nr = 0 for all r ∈ R}.
1.2. Let R be an integral domain.
(a) Show that every subring S ⊆ R is also an integral domain. What is the relationship
between char S and char R ?
(b) If R is a field, give an example to show that a subring of R need not be a field.
1.3. For each of the following rings R, find the characteristic char R and the characteristic
subring of R. Determine which of these rings is an integral domain. In (b) and (c), A is an
arbitrary commutative ring.
(a) Any subring R ⊆ C.
(b) The polynomial ring R = A[X].
(c) The ring of n × n matrices over A,
a11 . . . a1n
..
.
.
..
.. : aij ∈ A .
R = Matn (A) = .
an1 . . . ann
1.4. If R is a commutative ring with unit containing the prime field Fp for some prime p > 0,
show that the function ϕ : R −→ R given by ϕ(t) = tp , defines a ring homomorphism. Give
examples to show that ϕ need not be surjective or injective.
1.5. Let R and S be rings with unity and Q S a prime ideal.
(a) If ϕ : R −→ S is a ring homomorphism, show that
ϕ−1 Q = {r ∈ R : ϕ(r) ∈ Q} ⊆ R
is a prime ideal of R.
(b) If R ⊆ S is a subring, show that Q ∩ R is a prime ideal of R.
(c) If the word ‘prime’ is replaced by ‘maximal’ throughout, are the results in parts (a)
and (b) still true? [Hint: look for a counterexample.]
(d) If R ⊆ S is a subring and P R is a maximal ideal, suppose that Q S is a prime ideal
for which P ⊆ Q. Show that Q ∩ R = P .
1.6. Let k be a field, R be a ring with unit and ϕ : k −→ R a ring homomorphism. Show that
ϕ is a monomorphism.
1.7. Consider the sets
Z(i) = {u + vi : u, v ∈ Z} ⊆ C,
Q(i) = {u + vi : u, v ∈ Q} ⊆ C.
(a) Show that Z(i) and Q(i) are subrings of C. Also show that Z(i) is an integral domain,
Q(i) is a field and Z(i) is a subring of Q(i).
(b) Show that the inclusion homomorphism inc : Z(i) −→ Q(i) extends to a monomorphism
inc∗ : Fr(Z(i)) −→ Q(i).
(c) Show that inc∗ is an isomorphism, so Fr(Z(i)) = Q(i).
1.8. Let R be a commutative ring.
(a) If a, b ∈ R, show that there is a unique ring homomorphism ψa,b : R[X] −→ R[X] for
which ψa,b (r) = r if r ∈ R and ψa,b (X) = aX + b. If c, d ∈ R, determine ψa,b ◦ ψc,d . If
a is a unit, show that ψa,b is an isomorphism and find its inverse.
(b) Now suppose that R = k is a field and a, b ∈ k with a = 0. Prove the following.
21
(i) If f (X) ∈ k[X], the deg ψa,b (f (X)) = deg f (X).
(ii) If p(X) ∈ k[X] is a prime then so is ψa,b (p(X)).
(iii) If p(X) ∈ k[X] is an irreducible then so is ψa,b (p(X)).
1.9. Let k be a field and k[[X]] be the set consisting of all power series
∞
ak X k = a0 + a1 X + · · · + ak X k + · · · ,
k=0
with ak ∈ k.
(a) Show that this can be made into an integral domain containing k[X] as a subring by
defining addition and multiplication in the obvious way.
k
(b) Show that ∞
k=0 ak X ∈ k[[X]] is a unit if and only if a0 = 0.
(c) Show that Fr(k[[X]]) consists of all finitetailed Laurent series
∞
ak X k = a X + a
+1 X
+1
+ · · · + ak X k + · · ·
k=
for some
∈ Z and ak ∈ k.
1.10. Taking k = Q, find the quotient and remainder when performing long division of f (X) =
6X 4 − 6X 3 + 3X 2 − 3X − 2 by d(X) = 2X 3 + X + 3.
1.11. Taking k = F3 , find the quotient and remainder when performing long division of
f (X) = 2X 3 + 2X 2 + X + 1 by d(X) = 2X 3 + 2X.
1.12. Let p > 0 be a prime. Suppose that f (X) = a0 + a1 X + · · · + an X n ∈ Z[X] with
p an and that f (X) ∈ Fp [X] denotes the polynomial obtained by reducing the coefficients of
f (X) modulo p. If f (X) is irreducible, show that f (X) is irreducible. Which of the following
polynomials in Z[X] is irreducible?
X 3 − X + 1, X 3 + 2X + 1, X 3 + X − 1, X 5 − X + 1, X 5 + X − 1, 5X 3 − 10X + X 2 − 2.
1.13. Find generators for each of the following ideals:
√
I2 = {f (X) ∈ Q[X] : f ( 2 i) = 0} Q[X],
√
I4 = {f (X) ∈ R[X] : f ( 2) = 0} R[X],
I1 = {f (X) ∈ Q[X] : f (i) = 0} Q[X],
√
I3 = {f (X) ∈ Q[X] : f ( 2) = 0} Q[X],
√
I5 = {f (X) ∈ R[X] : f ( 2 i) = 0} R[X],
I6 = {f (X) ∈ R[X] : f (ζ3 ) = 0} R[X].
1.14. Consider the inclusion inc : Q −→ C and its extension to ε√2 : Q[X] −→ C.
Determine the image ε√2 Q[X] ⊆ C. What is ε−√2 Q[X] ⊆ C? Find ker ε√2 Q[X] and
ker ε−√2 Q[X]; are these maximal ideals?
√
1.15. Let ω = (−1 + 3i)/2 ∈ C. Consider the inclusion inc : Q −→ C and its extension
to εω : Q[X] −→ C. Determine the image εω Q[X] ⊆ C. Determine ker εω Q[X] and decide
whether it is maximal. Find another evaluation homomorphism with the same kernel and image.
1.16. Consider the inclusion inc : Q −→ C and its extension to εα : Q[X] −→ C where α is
one of the 4 complex roots of the polynomial f (X) = X 4 − 2 ∈ Q[X]. Determine the image
εα Q[X] ⊆ C and the ideal ker εα Q[X]; is the latter ideal maximal? What happens if α is
replaced by one of the other roots of f (X)?
Repeat this problem starting with the inclusion of the real numbers into the complex numbers inc : R −→ C and εα : R[X] −→ C.
1.17. Use Cardan’s method to find the complex roots of the polynomial
f (X) = X 3 − 9X 2 + 21X − 5.
22
1.18. Consider the real numbers
α=
3
10 +
√
108 +
3
10 −
√
108,
3
β=
1+
2
3
7
+
3
3
1−
2
3
7
.
3
Find rational cubic polynomials f (X) and g(X) for which f (α) = 0 = g(β). Hence determine
these real numbers.
1.19. Prove the final part of Example 1.60 by showing that there is an isomorphism of groups
Aff 1 (k) ∼
= Autk (k[X]).
1.20. Let k be any field. Consider the 6 automorphisms αj : k(X) −→ k(X) (j = 1, . . . , 6)
defined by
α1 (f (X)) = f (X),
α2 (f (X)) = f (1 − X),
α3 (f (X)) = f (1/X),
α4 (f (X)) = f ((X − 1)/X),
α5 (f (X)) = f (1/(1 − X)),
α6 (f (X)) = f (X/(X − 1)).
Show that the set consisting of these elements is a subgroup Γk
the symmetric group S3 . When k = F2 , show that Γk ∼
= GL2 (k).
Autk (k(X)) isomorphic to
1.21. Determine the cyclotomic polynomial Φ20 (X).
1.22. Let p > 0 be a prime.
(a) Show that for k 1, the cyclotomic polynomial Φpk (X) satisfies
k−1
Φpk (X) = Φp (X p
)
and has as its complex roots the primitive pk th roots of 1.
(b) Show that Φpk (X) ∈ Q[X] is irreducible.
(c) Generalize part (a) to show that if n = pr11 · · · prkk is the prime power factorization of n
with the pi being distinct primes and ri > 0, then
r −1
r1 −1
···pkk
Φn (X) = Φp1 ···pk (X p1
1.23. For n
).
2, show that
X ϕ(n) Φn (X −1 ) = Φn (X).
1.24. Show that for n 1, ζn + ζn−1 = 2 cos(2π/n).
Find expressions for ζ5 +ζ5−1 and ζ52 +ζ5−2 in terms of cos(2π/5). Hence find a rational polynomial
which has cos(2π/5) as a root.
1.25. Let p > 0 be a prime and K be a field with char K = p.
(a) Show that if ζ ∈ K is a pth root of 1 then ζ = 1. Deduce that if m, n > 0 and p n,
then every npm th root of 1 in K is an nth root of 1.
(b) If a ∈ K, show that the polynomial X p − a ∈ K[X] has either no roots or exactly one
root in K.
23
CHAPTER 2
Fields and their extensions
2.1. Fields and subfields
2.1. Definition. Let K and L be fields and suppose that K ⊆ L is a subring. Then we
say that K is a subfield of L; L is also said to be an extension (field ) of K. We write K L or
L/Kto indicate this and K < L if K is a proper subfield of L, i.e., if K = L.
An important fact about an extension of fields L/K is that L is a Kvector space whose
addition is the addition in the field L while scalar multiplication is defined by
u · x = ux
(u ∈ K, x ∈ L).
2.2. Definition. We will call dimK L the degree or index of the extension L/K and use the
notation [L : K] = dimK L. An extension of fields L/K is finite (dimensional ) if [L : K] < ∞,
otherwise it is infinite (dimensional ).
2.3. Example. Show that the extension C/R is finite, while R/Q and C/Q are both infinite.
Solution. We have
C = {x + yi : x, y ∈ R},
so 1, i span C as a vector space over R. Since i ∈
/ R, these elements are also linearly independent
over R and therefore they form a basis, whence [C : R] = 2. The infiniteness of R/Q and C/Q are
consequences of the fact that any finite dimensional vector space over Q is countable, however
R and C are uncountable. A basis for the Qvector space R is known as a Hamel basis.
√
2.4. Example. Consider the extension Q( 2)/Q where
√
√
Q( 2) = {x + y 2 : x, y ∈ Q}.
√
Show that [Q( 2) : Q] = 2.
√
√
√ Solution. The elements
√ 1, 2 clearly span the Qvector space Q( 2). √Now recall that
2∈
/ Q. If the elements 1, 2 were linearly dependent we would have u + v 2 = 0 for some
u, v ∈ Q not both zero; in fact it is easy to see that we would then also have u, v both nonzero.
Thus we would have
√
u
2 = − ∈ Q,
v
√
√
which we know √
to be false. Hence 1, 2 are linearly independent and so form a basis for Q( 2)
over Q and [Q( 2) : Q] = 2.
If we have two extensions L/K and M/L then it is a straightforward to verify that K
and so we have another extension M/K.
M
2.5. Definition. Given two extensions L/K and M/L, we say that L/K is a subextension
of M/K and sometimes write L/K M/L.
2.6. Theorem. Let L/K be a subextension of M/K.
(i) If one or both of the dimensions [L : K] or [M : L] is infinite then so is [M : K].
(ii) If the dimensions [L : K] and [M : L] are both finite then so is [M : K] and
[M : K] = [M : L] [L : K].
25
Proof. (i) If [M : K] is finite, choose a basis m1 , . . . , mr of M over K. Now any element
u ∈ M can be expressed as
u = t1 m1 + · · · + tr mr ,
where t1 , . . . , tr ∈ K; but since K ⊆ L, this means that m1 , . . . , mr spans M over L and so
[M : L] < ∞. Also L is a Kvector subspace of the finite dimensional Kvector space M , hence
[L : K] < ∞.
(ii) Setting r = [L : K] and s = [M : L], choose a basis 1 , . . . , r of L over K and a basis
m1 , . . . , ms of M over L.
Now let v ∈ M . Then there are elements y1 , . . . , ys ∈ L for which
v = y1 m1 + · · · + ys ms .
But each yj can be expressed in the form
yj = x1j
1
+ · · · + xrj
r
for suitable xij ∈ K. Hence,
s
r
j=1
i=1
s
v=
xij
r
mj =
i
xij ( i mj ),
j=1 i=1
where each coefficient xij is in K. Thus the elements
Kvector space M .
Now suppose that for some tij ∈ K we have
s
i mj
(i = 1, . . . , r, j = 1, . . . , s) span the
r
tij ( i mj ) = 0.
j=1 i=1
On collecting terms we obtain
s
r
tij
j=1
where each coefficient
means that for each j,
r
i=1 tij i
i
mj = 0,
i=1
is in L. By the linear independence of the mj over L, this
r
tij
i
= 0.
i=1
By the linear independence of the i over K, each tij = 0.
Hence the i mj form a basis of M over K and so
[M : K] = rs = [M : L] [L : K].
We will often indicate subextensions in diagrammatic form where larger fields always go
above smaller ones and the information on the lines indicates dimensions
MF
[M :L]
A
L
[L:K]
!
6
&
1
[M :K]=[M :L] [L:K]
K
We often suppress ‘composite’ lines such as the dashed one. Such towers of extensions are our
main objects of study. We can build up sequences of extensions and form towers of arbitrary
26
length. Thus, if L1 /K, L2 /L1 , . . . , Lk /Lk−1 is a such a sequence of extensions, there is a
diagram
Lk
Lk−1
L1
K
2.2. Simple and finitely generated extensions
F . Given elements u1 , . . . , ur ∈ F we set
2.7. Definition. Let F be a field and K
K(u1 , . . . , ur ) =
L
K L F
u1 ,...,ur ∈L
which is the smallest subfield in F that contains K and the elements u1 , . . . , ur . The extension K(u1 , . . . , ur )/K is said to be generated by the elements u1 , . . . , ur ; we also say that
K(u1 , . . . , ur )/K is a finitely generated extension of K. An extension of the form K(u)/K is
called a simple extension of K with generator u.
We can extend this to the case of an infinite sequence u1 , . . . , ur , . . . in F and denote by
K(u1 , . . . , ur , . . .) F the smallest extension field of K containing all the elements ur .
It can be shown that
(2.1) K(u1 , . . . , ur ) =
f (u1 , . . . , ur )
∈ F : f (X1 , . . . , Xr ), g(X1 , . . . , Xr ) ∈ K[X1 , . . . , Xr ], g(u1 , . . . , ur ) = 0 .
g(u1 , . . . , ur )
Reordering the ui does not change K(u1 , . . . , un ).
2.8. Proposition. Let K(u)/K and K(u, v)/K(u) be simple extensions. Then
K(u, v) = K(u)(v) = K(v)(u).
More generally,
K(u1 , . . . , un ) = K(u1 , . . . , un−1 )(un )
and this is independent of the order of the sequence u1 , . . . , un .
2.9. Theorem. For a simple extension K(u)/K, exactly one of the following conditions
holds.
(i) The evaluation at u homomorphism εu : K[X] −→ K(u) is a monomorphism and on
passing to the fraction field gives an isomorphism (εu )∗ : K(X) −→ K(u). In this case,
K(u)/K is infinite and u is said to be transcendental over K.
(ii) The evaluation at u homomorphism εu : K[X] −→ K(u) has a nontrivial kernel
ker εu = (p(X)) where p(X) ∈ K[X] is an irreducible monic polynomial of positive degree and the quotient homomorphism εu : K[X]/(p(X)) −→ K(u) is an isomorphism.
In this case K(u)/K is finite with [K(u) : K] = deg p(X) and u is said to be algebraic
over K.
27
Proof. (i) If ker εu = (0), all that needs checking is that (εu )∗ is an epimorphism; but as
u is in the image of (εu )∗ this is obvious.
(ii) When ker εu = (0), Theorem 1.31(iv) implies that the image of εu is a subfield of K(u) and
since it contains u it must equal K(u). Hence εu is an isomorphism. Using Long Division, we
find that every element of K[X]/(p(X)) can be uniquely expressed as a coset of the form
f (X) + (p(X)),
where deg f (X) < deg p(X). Hence every element of K[X]/(p(X)) can be uniquely expressed
as a linear combination over K of the d cosets
1 + (p(X)), X + (p(X)), X 2 + (p(X)), . . . , X d−1 + (p(X)),
where d = deg p(X). Via the isomorphism εu under which εu (X k + (p(X))) = uk , we see that
the elements 1, u, . . . , ud−1 form a basis for K(u) over K.
√ √
√ √
2.10. Example. For the extension Q( 2, 3)/Q we have [Q( 2, 3) : Q] = 4.
√
Proof. By Example 2.4 we know that [Q( 2) : Q] = 2. We have the following tower of
extensions.
√ √
Q( 2, 3)
√ √
√
[Q( 2, 3):Q( 2)]
√
Q( 2)
√ √
√ √
√
[Q( 2, 3):Q]=2[Q( 2, 3):Q( 2)]
2
Q
√
√ √
: Q( 2)] =√2. √
We will show that [Q( 2, √3) √
√
√
Notice
that
if
u
∈
Q(
2,
3)
=
Q(
2)(
3)
then
u
=
a
+
b
3
for
some
a,
b
∈
Q(
√ √
√
√
√2),
√
so 1, 3 span Q( 2, 3) over Q( 2). But if these are linearly dependent then 3 ∈ Q( 2).
Writing
√
√
3=v+w 2
with v, w ∈ Q, we find that
√
v 2 + 2w2 + 2vw 2 = 3 ∈ Q,
√
and hence 2vw 2 ∈√Q. The possibilities v = 0 or w√= 0 are easily ruled out, while v, w √
=0
would implies that 2 ∈ Q which
is
false.
So
1,
3
are
linearly
independent
over
Q(
2)
√ √
√ √
√
and√therefore
form a basis of Q( 2, 3). This shows that [Q( 2, 3) : Q( 2)] = 2 and so
√
[Q( 2, 3) : Q] = 4.
√ √
2.11. Remark. There are some other subfields of Q( 2, 3) which are conveniently displayed in the following diagram.
√ √
Q( 2, 3)
r
rrr
√ r
Q( 2)
r
2 rrrr
2
√
Q( 3)
ww
ww
ww
ww
2
ww
w
2
Q
vvv
vvv2
vvv
v
rr
rrr
r
r
rr 2
rrr
√
Q( 6)
One idea in the verification of Example 2.10 can be extended to provide a useful general
result whose proof is left as an exercise.
28
2.12. Proposition. Let p1 , . . . , pn be a sequence of distinct primes pi > 0. Then
√
√
√
pn ∈
/ Q( p1 , . . . , pn−1 ).
√
√
√
√
√
√
Hence [Q( p1 , . . . , pn ) : Q( p1 , . . . , pn−1 )] = 2 and [Q( p1 , . . . , pn ) : Q] = 2n .
√
√
2.13. Example. For the extension Q( 2, i)/Q we have [Q( 2, i) : Q] = 4.
√
√
√
2) : Q] =√2. Also, i ∈
/ Q(
2) since√i is not real and Q( 2) R.
Proof. We know that [Q(
√
√
Since i2 + 1 = 0, we have Q( 2, i) = Q( 2)(i) and [Q( 2, i) : Q( 2)] = 2. Using the formula
√
√
√
√
[Q( 2, i) : Q] = [Q( 2, i) : Q( 2)] [Q( 2) : Q],
√
we obtain [Q( 2, i) : Q] = 4.
√
√
This example also has several other subfields, with only Q( 2) = Q( 2, i) ∩ R being a
subfield of R.
C
2
∞
R
∞
√
Q( 2, i)
t
tt
tt
t
t
tt
2
√
Q( 2)
uuu
uuu2
uuu
u
2
Q(i)
uuu
uuu
u
2 uuu
u
2
Q
√
Q( 2 i)
rr
rrr
r
r
r 2
rrr
2.14. Example. For n 1, let En = Q(21/n ) R, where 21/n ∈ R denotes the positive real
nth root of 2.
(i) Show that [En : Q] = n.
(ii) If m 1 with m  n, show that Em En and determine [En : Em ].
(iii) If m, n are coprime, show that Emn = Q(21/m , 21/n ).
Solution. (i) Consider the evaluation homomorphism ε21/n : Q[X] −→ En . Applying the
Eisenstein Test 1.38 using the prime 2 to the polynomial X n − 2 ∈ Z[X], we find that
ker ε21/n = (X n − 2) Q[X],
and the induced homomorphism ε21/n : Q[X]/(X n − 2) −→ En is an isomorphism. Hence
[En : Q] = n.
(ii) Since n/m is an integer,
21/m = (21/n )n/m ∈ En ,
so
Em = Q(21/m ) ⊆ En .
By Theorem 2.6 we have
n = [En : Q] = [En : Em ] [Em : Q] = m[En : Em ],
whence [En : Em ] = n/m.
(iii) By (ii) we have Em Emn and En Emn , hence Q(21/m , 21/n )
there are integers r, s for which rm + sn = 1 and so
rm + sn
r
s
1
=
= + .
mn
mn
n m
29
Emn . As gcd(m, n) = 1,
This shows that
21/mn = (21/n )r (21/m )s ∈ Q(21/m , 21/n ),
whence Emn
Q(21/m , 21/n ). Combining these inclusions we obtain Emn = Q(21/m , 21/n ).
Exercises on Chapter 2
√
√
2.1. Let p ∈ N be an prime. Show that the extension Q( p)/Q has [Q( p) : Q] = 2.
√ √
√
2.2. Let p, q > 0 be distinct primes. Show that [Q( p, q) : Q( p)] = 2.
2.3. Prove Proposition 2.12 by induction on n.
2.4. Let K a field with char K = 2 and suppose that L/K is an extension. If a, b ∈ K are
distinct, suppose that u, v ∈ L satisfy u2 = a and v 2 = b. Show that K(u, v) = K(u + v).
[Hint: first show that u±v = 0 and deduce that u−v ∈ K(u+v); then show that u, v ∈ K(u+v).]
2.5. Show that [Q(i) : Q] = 2.
√
√
2.6. Show that [Q( 3, i) : Q] = 4. Find the three subfields L Q( 3, i) with [L : Q] = 2 and
display their relationship in a diagram, indicating which ones are subfields of R.
2.7. Let ζ5 = e2πi/5 ∈ C.
(a) Explain why [Q(ζ5 ) : Q] = 4.
(b) Show that cos(2π/5), sin(2π/5) i ∈ Q(ζ5 ).
(c) Show that for t ∈ R,
cos 5t = 16 cos5 t − 20 cos3 t + 5 cos t.
(d) Show that the numbers cos(2kπ/5) with k = 0, 1, 2, 3, 4 are roots of the polynomial
f (X) = 16X 5 − 20X 3 + 5X − 1 = (X − 1)(4X 2 + 2X − 1)2
and deduce that [Q(cos(2π/5)) : Q] = 2.
(e) Display the relationship between the fields Q, Q(cos(2π/5)), and Q(ζ5 ) in a suitable
diagram.
2.8. This question is for those who like lots of calculation or using Maple. Let ζ7 = e2πi/7 ∈ C.
(a) Explain why [Q(ζ7 ) : Q] = 6.
(b) Show that cos(2π/7), sin(2π/7) i ∈ Q(ζ7 ).
(c) Show
cos 7t = 64 cos7 t − 112 cos5 t + 56 cos3 t − 7 cos t.
Show that the numbers cos(2kπ/7) with k = 0, 1, . . . , 6 are roots of the polynomial
f (X) = 64X 7 − 112X 5 + 56X 3 − 7X − 1 = (X − 1)(8X 3 + 4X 2 − 4X − 1)2
and deduce that [Q(cos(2π/7)) : Q] = 3.
(d) Show that sin(2π/7) i is a root of
g(X) = 64X 7 + 112X 5 + 56X 3 + 7X = X(64X 6 + 112X 4 + 56X 2 + 7)
and that 64X 6 + 112X 4 + 56X 2 + 7 ∈ Q[X] is irreducible. What is [Q(sin(2π/7) i) : Q]?
(e) Display the relationship between the fields Q, Q(cos(2π/7)), Q(sin(2π/7) i) and Q(ζ7 )
in a diagram.
(f) Is i ∈ Q(ζ7 )?
2.9. In this question we continue to consider the situation described in Example 2.14.
30
(a) Show that
{id}
if n is odd,
∼
{id, τn } = Z/2 if n is even,
where τn has composition order 2.
(b) Let E =
En R. Show that AutQ (E) = {id}.
AutQ (En ) =
n 1
(c) Display the 6 subfields of E12 in a diagram.
(d) Which of the subfields in part (c) contain the element 21/2 + 21/3 ?
31
CHAPTER 3
Algebraic extensions of fields
3.1. Algebraic extensions
Let L/K be an extension of fields. From Theorem 2.9(ii), recall the following notion.
3.1. Definition. An element t ∈ L is algebraic over K if there is a nonzero polynomial
p(X) ∈ K[X] for which p(t) = 0.
Notice in particular that for an element t ∈ K, the polynomial p(X) = X − t ∈ K[X]
satisfies p(t) = 0, so t is algebraic over K.
Theorem 2.9 allows us to characterize algebraic elements in other ways.
3.2. Proposition. Let t ∈ L. Then the following conditions are equivalent.
(i) t is algebraic over K.
(ii) The evaluation homomorphism εt : K[X] −→ L has nontrivial kernel.
(iii) The extension K(t)/K is finite dimensional.
3.3. Definition. If t ∈ L is algebraic over K then by Proposition 3.2,
ker εt = (minpolyK,t (X)) = (0),
where minpolyK,t (X) ∈ K[X] is an irreducible monic polynomial called the minimal polynomial
of t over K. The degree of minpolyK,t (X) is called the degree of t over K and is denoted
degK t.
3.4. Proposition. If t ∈ L is algebraic over K then
[K(t) : K] = deg minpolyK,t (X) = degK t.
Proof. This follows from Theorem 2.9(ii).
3.5. Remark. Suppose that t ∈ L is algebraic over K and that p(X) ∈ ker εt with
deg p(X) = deg minpolyK,t (X). Then minpolyK,t (X)  p(X) and so
p(X) = u minpolyK,t (X)
for some u ∈ K. In particular, when p(X) is monic,
p(X) = minpolyK,t (X).
We will often use this without further comment.
3.6. Example. Consider C/Q. The minimal polynomial of
√
2 ∈ C over Q is
2
minpolyQ,√2 (X) = X − 2.
√
Proof. Clearly X 2 − 2 ∈ ker ε√2 since ( 2)2 − 2 = 0. By Example 2.4,
√
deg minpolyQ,√2 (X) = [Q( 2) : Q] = 2,
hence
minpolyQ,√2 (X) = X 2 − 2.
3.7. Example. Consider C/Q. The minimal polynomial of i ∈ C over Q is X 2 + 1.
33
Proof. Clearly X 2 + 1 ∈ ker εi since i2 + 1 = 0. As [Q(i) : Q] = 2, we have
minpolyQ,i (X) = X 2 + 1.
3.8. Example. Consider C/Q. Find the minimal polynomial of the primitive 6th root of
unity, ζ6 ∈ C over Q.
Solution. Recall from Example 1.44 that ζ6 is a root of the irreducible cyclotomic polynomial
Φ6 (X) = X 2 − X + 1.
Then Φ6 (X) ∈ ker εζ6 so minpolyQ,ζ6 (X)  Φ6 (X). Since Φ6 (X) is irreducible and monic, we
must have
minpolyQ,ζ6 (X) = Φ6 (X)
and so degQ ζ6 = 2.
√
√
3.9. Example. Consider C/Q. Find the minimal polynomial of 2 + 3 over Q.
Solution. Notice that
√
√ √
√
√
√
√
√
( 3 − 2)( 3 + 2)
1
√
√
√ ∈ Q( 2 + 3).
3− 2=
=√
( 3 + 2)
2+ 3
So we have
√
√
√
√
√
√
1 √
( 2 + 3) − ( 3 − 2) ∈ Q( 2 + 3),
2=
2
√
√
√
√
√
√
1 √
3=
( 2 + 3) + ( 3 − 2) ∈ Q( 2 + 3),
2
√ √
√
√
√
√
√ √
hence Q( 2, 3) Q( 2 + 3). Since Q( 2 + 3) Q( 2, 3) we must have
√
√
√ √
Q( 2 + 3) = Q( 2, 3).
Referring to Example 2.10 we see that
√
√
degQ ( 2 + 3) = 4.
Let us find a nonzero polynomial in ker ε√2+√3 Q[X].
√
√
√
Referring to Example 2.10 or Proposition 2.12 we see that 2 + 3 ∈
/ Q( 2), hence
√
√
degQ(√2) ( 2 + 3) = 2.
√
One polynomial in ker ε√2+√3 Q( 2)[X] is
√
√
√
√
√
(X − ( 2 + 3))(X − ( 2 − 3)) = X 2 − 2 2X − 1.
Since this is monic and of degree 2,
√
minpolyQ(√2),√2+√3 (X) = X 2 − 2 2X − 1.
Similarly,
√
minpolyQ(√2),−√2+√3 (X) = X 2 + 2 2X − 1.
Consider
p(X) = minpolyQ(√2),√2+√3 (X) minpolyQ(√2),−√2+√3 (X)
√
√
= (X 2 − 2 2X − 1)(X 2 + 2 2X − 1)
= X 4 − 10X 2 + 1.
√
√
Then p( 2 + 3) = 0 so p(X) ∈ ker εt . Since deg p(X) = 4 and p(X) is monic, we have
minpolyQ,√2+√3 (X) = X 4 − 10X 2 + 1.
3.10. Definition. Let L/K be a finite extension. An element u ∈ L for which L = K(u)
is called a primitive element for the extension L/K.
34
Later we will see that when char K = 0 every finite extension L/K has a primitive element.
3.11. Lemma. Let L/K be a finite extension and u ∈ L. Then u is a primitive element for
L/K if and only if degK u = [L : K].
Proof. K(u) ⊆ L is a finite dimensional Kvector subspace. Then K(u) = L if and only
dimK K(u) = dimK L. Since degK u = dimK K(u) and [L : K] = dimK L the result follows.
Sometimes the minimal polynomial of an element in an extension is introduced in a different
but equivalent way.
3.12. Proposition. Let t ∈ L be algebraic over K. Then
I(t) = {f (X) ∈ K[X] : f (t) = 0} ⊆ K[X]
is an ideal which is principal and has an irreducible monic generator q(X) ∈ K[X]. In fact,
q(X) = minpolyK,t (X).
Proof. It is easy to see that I(t) K[X] and therefore I(t) = (q(X)) for some monic
generator q(X). To see that q(X) is irreducible, suppose that q(X) = q1 (X)q2 (X) with
deg qi (X) < deg q(X). Now as q1 (t)q2 (t) = 0, we must have q1 (t) = 0 or q2 (t) = 0, hence
q1 (X) ∈ I(t) or q2 (X) ∈ I(t). These possibilities give q(X)  q1 (X) or q(X)  q2 (X) and
so deg q(X)
deg q1 (X) or deg q(X)
deg q2 (X), contradicting the above assumption that
deg qi (X) < deg q(X).
The irreducible monic polynomial minpolyK,t (X) is in I(t) so q(X)  minpolyK,t (X) and
therefore q(X) = minpolyK,t (X).
The next Lemma will often be useful.
3.13. Lemma. Let L/K be an extension and suppose that u1 , . . . , un ∈ L are algebraic. Then
K(u1 , . . . , un )/K is a finite extension.
Proof. Use induction on n together with Proposition 2.8 and Theorem 2.6(ii).
We now come to an important notion for extensions.
3.14. Definition. The extension L/K is algebraic or L is algebraic over K if every element
t ∈ L is algebraic over K.
3.15. Proposition. Let L/K be a finite extension. Then L/K is algebraic.
Proof. Let t ∈ L. Since the Kvector space L is finite dimensional, when viewed as
elements of this vector space, the powers 1, t, . . . , tn , . . . must be linearly dependent over K.
Hence for suitable coefficients cj ∈ K not all zero and some m 1 we have
c0 + c1 t + · · · + cm tm = 0.
But this means that t is algebraic over K.
3.16. Proposition. Let M/L and L/K be algebraic extensions. Then the extension M/K
is algebraic.
Proof. Let u ∈ M . Then u is algebraic over L, so there is a polynomial
p(X) = p0 + p1 X + · · · + pm X m ∈ L[X]
of positive degree with p(u) = 0. By Lemma 3.13, the extension K(p0 , . . . , pm )/K is finite and
so is K(p0 , . . . , pm , u)/K(p0 , . . . , pm ). By Theorem 2.6(ii), K(p0 , . . . , pm , u)/K is finite, so by
Proposition 3.15, u is algebraic over K.
3.17. Definition. For an extension L/K, let
Lalg = {t ∈ L : t is algebraic over K} ⊆ L.
3.18. Proposition. For an extension L/K, Lalg is a subfield containing K and Lalg /K is
algebraic.
35
Proof. Clearly K ⊆ Lalg . We must show that Lalg L.
Let u, v ∈ Lalg . Then by Lemma 3.13, K(u, v)/K is a finite dimensional extension, hence
every element of K(u, v) is algebraic over K. In particular, u + v and uv are in K(u, v) and if
u = 0, u−1 is also in K(u, v). Therefore u + v, uv and u−1 are all algebraic over K.
3.19. Example. In the extension C/Q we can consider Calg C which is called the subfield
of algebraic numbers. Similarly, in the extension R/Q the subfield
Ralg = Calg ∩ R
C
consists of all the real algebraic numbers. Elements of C−Calg are called transcendental complex
numbers; examples are e and π. The sets Calg and Ralg are both countable, whereas C and R
are uncountable, so there are in fact many more transcendental numbers but it can be hard to
determine whether a given number is transcendental or not. A more usual notation for Calg
is Q since this is the algebraic closure of Q which will be discussed later. When dealing with
algebraic extensions of Q we will usually work with subfields of Q = Calg .
We end this section with a technical result.
3.20. Proposition. Let K(u)/K be a finite simple extension. Then there are only finitely
many subextensions F/K K(u)/K.
Proof. Consider the minimal polynomial minpolyK,u (X) ∈ K[X]. Now for any subextension F/K K(u)/K we can also consider
minpolyF,u (X) = c0 + c1 X + · · · + ck−1 X k−1 + X k ∈ F [X],
which divides minpolyK,u (X) in F [X]. The Unique Factorization Property 1.33 implies that
minpolyK,u (X) has only finitely many monic divisors in K(u)[X], so there are only a finite
number of possibilities for minpolyF,u (X). Now consider F0 = K(c0 , c1 , . . . , ck−1 ), the extension
field of K generated by the coefficients of minpolyF,u (X). Then F0 F and so minpolyF,u (X) ∈
F0 [X] is irreducible since it is irreducible in F [X]; hence minpolyF,u (X) = minpolyF0 ,u (X). We
have
[K(u) : F ] = deg minpolyF,u (X) = deg minpolyF0 ,u (X) = [K(u) : F0 ],
hence F = F0 .
This shows that there are only finitely many subextensions F/K K(u)/K, each of which
has the form K(a0 , a1 , . . . , a −1 ), where
a0 + a1 X + · · · + a
−1 X
−1
+ X ∈ K(u)[X]
is a factor of minpolyK,u (X) in K(u)[X].
3.2. Splitting fields and Kronecker’s Theorem
We can now answer a basic question. Let K be a field and p(X) ∈ K[X] be a polynomial
of positive degree.
3.21. Question. Is there an extension field L/K for which p(X) has a root in L?
A stronger version of this question is the following.
3.22. Question. Is there an extension field E/K for which p(X) factorizes into linear
factors in E[X]?
3.23. Definition. p(X) ∈ K[X] splits in E/K or over E if it factorizes into linear factors
in E[X].
Of course, if we have such a field E then the distinct roots u1 , . . . , uk of p(X) in E generate
a subfield K(u1 , . . . , uk ) E which is the smallest subfield of E that answers Question 3.22.
3.24. Definition. Such a minimal extension of K is called a splitting field of p(X) over K
and we will sometimes denote it by K(p(X)) or Kp .
36
We already know how to answer Question 3.21.
3.25. Theorem (Kronecker’s Theorem: first version). Let K be a field and p(X) ∈ K[X]
be a polynomial of positive degree. Then there is a finite extension L/K for which p(X) has a
root in L.
Proof. We begin by factorizing p(X) ∈ K[X] into irreducible monic factors qj (X) together
with a constant factor c:
p(X) = cq1 (X) · · · qr (X).
Now for any j we can form the quotient field K[x]/(qj (X)) which is a finite dimensional (simple)
extension of K and in which the coset X + (qj (X)) satisfies the equation
qj (X + (qj (X))) = 0 + (qj (X)).
Hence p(X) has a root in K[x]/(qj (X)).
Of course, this construction is only interesting if qj (X) to has degree bigger than 1 since a
linear polynomial already has a root in K.
To answer Question 3.22 we iterate this construction. Namely, having found one root u1 in
an extension L1 /K we discard the linear factor X − u1 and consider the polynomial
p(X)
∈ L1 [X].
X − u1
We can repeat the argument to form a finite extension of L1 (and hence of K) containing a
root of p1 (X) and so on. At each stage we either already have another root in L1 or we need
to enlarge the field to obtain one.
p1 (X) =
3.26. Theorem (Kronecker’s Theorem: second version). Let K be a field and p(X) ∈ K[X]
be a polynomial of positive degree. Then there is a finite extension E/K which is a splitting
field of p(X) over K.
In practise we often have extension fields ‘lying around in nature’ containing roots and we
can work inside of these. When working over Q (or any other subfield of C) we can always find
roots in C by the Fundamental Theorem of Algebra. We then refer to a subfield of C which is
a splitting field as the splitting subfield.
3.27. Example. Find a splitting field E/Q for p(X) = X 4 −4 over Q and determine [E : Q].
Solution. Notice that
p(X) = (X 2 − 2)(X 2 + 2),
√
√
√
√
so first we adjoin
the roots ± 2 of (X 2 − 2) to form Q( 2, − 2) = Q( 2) which gives an
√
extension Q( 2)/Q of degree 2.
√
√
Next consider the polynomial X 2 + 2 ∈ Q( 2)[X]. The √
complex roots of X 2 + 2 are ± 2i
and√these
real, so this polynomial is irreducible
in Q( 2)[X]. Hence we need to consider
√ are not √
√
√
Q( 2, 2i) = Q( 2, i) and the extension Q( 2, i)/Q( 2) which has degree 2.
C
∞
√
Q( 2, i)
adjoin roots of X 2 + 2
2
√
Q( 2)
adjoin roots of X 2 − 2
2
Q
√
√
Thus the splitting subfield of p(X) over Q in C is Q( 2, i) and [Q( 2, i) : Q] = 4.
37
Of course we could have started by first adjoining roots of X 2 + 2 and then adjoining roots
of X 2 − 2, thus giving the tower
C
∞
√
Q( 2, i)
adjoin roots of X 2 − 2 2
√
Q( 2i)
adjoin roots of X 2 + 2 2
Q
An important point is that if a splitting field exists inside of a given extension field F/K, it is
unique as a subfield of F .
3.28. Proposition. Let F/K be an extension field and p(X) ∈ K[X]. If E1 , E2
splitting subfields for p(X) over K then E1 = E2 .
F are
Proof. Let u1 , . . . , uk ∈ F be the distinct roots of p(X) in F . By definition, K(u1 , . . . , uk )
is the smallest subfield containing K and all the uj . But K(u1 , . . . , uk ) must be contained in
any splitting subfield, so E1 = K(u1 , . . . , uk ) = E2 .
Since we will frequently encounter quadratic polynomials we record a useful result on roots
of such polynomials. Recall that p(X) = aX 2 + bX + c ∈ K[X] is quadratic if a = 0 and its
discriminant is
∆ = b2 − 4ac ∈ K.
The proof of the next result is the standard one which works provided 2 has an inverse in K,
i.e., when char K = 2.
3.29. Proposition. Let K be a field of characteristic different from 2. Then the quadratic
polynomial p(X) = aX 2 + bX + c ∈ K[X] has
• no roots in K if ∆ is not a square in K;
• one root −b/(2a) = −(2a)−1 b if ∆ = 0;
• two distinct roots
−b − δ
−b + δ
= (2a)−1 (−b + δ),
= (2a)−1 (−b − δ),
2a
2a
if ∆ = δ 2 for some nonzero δ ∈ K.
In particular, the splitting field of p(X) over K is K if ∆ is a square in K and K(δ) otherwise,
where δ is one of the two square roots of ∆ in some extension of K such as the algebraic closure
K which we will introduce in Section 3.4.
3.30. Example. Find a splitting field E/Q for p(X) = X 3 −2 over Q and determine [E : Q].
Solution. By the Eisenstein Test 1.38, p(X) is irreducible
over Q. One root of p(X) is
√
√
3
2 ∈ R so we adjoin this to Q to form an extension Q( 3 2)/Q of degree 3. Now
√
√
√
3
3
3
p(X) = (X − 2)(X 2 + 2X + ( 2)2 )
√
√
and√ the second
factor has the nonreal complex roots 3 2 ζ3 , 3 2 ζ32 lying in the extension
√
√
Q( 3 2, ζ3√
)/Q( 3 2) of degree 2. So the splitting subfield of X 3 − 2 in C over Q is Q( 3 2, ζ3 )
with [Q( 3 2, ζ3 ) : Q] = 6.
√
√
An alternative strategy would have been to adjoin one of the other roots 3 2 ζ3 or 3 2 ζ32
first. We could also have begun by adjoining ζ3 to form the extension Q(ζ3 )/Q, but none of
38
√
the roots of p(X) lie in this field so the extension Q( 3 2, ζ3 )/Q(ζ3 ) of degree 3 is obtained by
adjoining one and hence all of the roots.
√
Figure 3.1 shows all the subfields of the extension Q( 3 2, ζ3 )/Q.
C
2
∞
R
√
Q( 3 2, ζ )
∞
VV3
iiiqqq
i
i
i
V
i
2 i
2 q
2 VV
iiii qqqq
i
i
VV
i
ii
qq
VV
iiii
√
√
√
3
3
3
3
Q( 2 ζ32 ) VVV
Q( 2 ζ3 )
Q( 2)
``
www
VV
``
www
VV
``
www
VV
``
www
3
`
3
www
3
Q(ζ3 )
www ```
r
www ``
r
r
2 r
www ``
rrr
www``
r
r
w rr
Q
√
Figure 3.1. The subfields of Q( 3 2, ζ3 )/Q
3.3. Monomorphisms between extensions
3.31. Definition. For extensions F/K and L/K, let MonoK (L, F ) denote the set of all
monomorphisms L −→ F which fix the elements of K.
3.32. Remark. We always have AutK (F ) ⊆ MonoK (F, F ) and MonoK (F, F ) is closed under
composition but is not always a group since elements are not necessarily invertible. If F/K is
finite, then we do have MonoK (F, F ) = AutK (F ) since every injective Klinear transformation
is surjective and so invertible.
We will also use the following notation.
3.33. Definition. Let F/K be an extension and p(X) ∈ K[X]. Set
Roots(p, F ) = {u ∈ F : p(u) = 0},
the set of roots of p(X) in F . This is always a finite set which may of course be empty (this
happens precisely when p(X) has no root in K).
Suppose that p(X) ∈ K[X] is an irreducible polynomial which we might as well assume is
monic, and let F/K be an extension. Then if t ∈ F is a root of p(X), the evaluation homomorphism εt : K[X] −→ F factors through the quotient monomorphism εt : K[X]/(p(X)) −→ F
whose image is K(t) F . Of course, there is one such monomorphism for each root of p(X) in
F . If we fix one such root t0 and identify K[X]/(p(X)) with K(t0 ) via εt0 , then each root of
p(X) in F gives rise to a monomorphism ϕt = εt ◦ ε−1
t0 : K(t0 ) −→ F for which ϕt (t0 ) = t.
ϕt =e
εt ◦e
ε−1
t
0
K(t0 ) o
e
εt0
∼
=
K[X]/(p(X))
39
e
εt
/* F
Notice that if ϕ : K[X]/(p(X)) −→ F is any homomorphism extending the identity function
on K, then the coset X + (p(X)) must be sent by ϕ to a root of p(X) in F , hence every such
homomorphism arises this way. This discussion is summarized in the following result.
3.34. Proposition. Let F/K be a field extension. Let p(X) ∈ K[X] be an irreducible
polynomial with t0 ∈ F be a root of p(X). Then there is a bijection
Roots(p, F ) ←→ MonoK (K(t0 ), F )
given by t ←→ ϕt , where ϕt : K(t0 ) −→ F has the effect ϕt (t0 ) = t.
√
3.35. Example. Show that MonoQ (Q( 2), C) has two elements.
√
2 − 2) where X 2 − 2 is irreducible over Q. Hence
Solution. We have Q( 2) ∼
= Q[X]/(X
√
√
2
the Qmonomorphisms we want send 2 to ± 2 which are the
√ complex roots of X − 2. In
fact both possibilities occur, giving monomorphisms id, α : Q( 2) −→ C, where
√
√
α(a + b 2) = a − b 2.
√
We can replace C by Q( 2) to obtain
√
√
√
√
MonoQ (Q( 2), C) = MonoQ (Q( 2), Q( 2)) = AutQ (Q( 2)).
We will see that this is not always true.
√
√
√
3.36. Example. Show that MonoQ (Q( 3 2), C) has 3 elements but MonoQ (Q( 3 2), Q( 3 2))
contains only the identity function.
√ √
3 − 2 and there are 3 complex roots 3 2, 3 2 ζ ,
Solution. Here minpolyQ, √
3 (X) = X
3
2
√
√
√
3
3
3
2
2 ζ3 . As
√ two of these roots are not real, MonoQ (Q( 2), Q( 2)) contains only the identity
since Q( 3 2) R.
√
√
√
3
3
3
Each of the above roots corresponds to one of the
subfields
Q(
2),
Q(
2
ζ
)
or
Q(
2 ζ32 )
3
√
3
of C and there are 3 monomorphisms α0 , α1 , α2 : Q( 2) −→ C given by
√
√
√
√
3
3
3
3
α0 (a + b 2 + c( 2)2 ) = a + b 2 + c( 2)2 ,
√
√
√
√
3
3
3
3
α1 (a + b 2 + c( 2)2 ) = a + b 2 ζ3 + c( 2)2 ζ32 ,
√
√
√
√
3
3
3
3
α2 (a + b 2 + c( 2)2 ) = a + b 2 ζ32 + c( 2)2 ζ3 .
These mappings have images
√
√
3
3
α0 Q( 2) = Q( 2),
√
√
3
3
α1 Q( 2) = Q( 2 ζ3 ),
√
√
3
3
α2 Q( 2) = Q( 2 ζ32 ).
3.37. Proposition. Let F/K and L/K be extensions.
(i) For p(X) ∈ K[X], each monomorphism α ∈ MonoK (L, F ) restricts to a function
αp : Roots(p, L) −→ Roots(p, F ) which is an injection.
(ii) If α ∈ MonoK (L, L), then αp : Roots(p, L) −→ Roots(p, L) is a bijection.
Proof. (i) For u ∈ Roots(p, L) we have
p(α(u)) = α(p(u)) = α(0) = 0,
so α maps Roots(p, L) into Roots(p, F ). Since α is an injection its restriction to Roots(p, L) ⊆ L
is also an injection.
(ii) From (i), αp : Roots(p, L) −→ Roots(p, L) is an injective function from a finite set to itself,
hence it is also surjective by the Pigeon Hole Principle. Thus αp : Roots(p, L) −→ Roots(p, L)
is a bijection.
Part (ii) says that any automorphism of L/K permutes the set of roots in L of a polynomial
p(X) ∈ K[X]. This gives us a strong hold on the possible automorphisms. In the case of finite,
or more generally algebraic, extensions it is the key to understanding the automorphism group
and this is a fundamental insight of Galois Theory.
40
√
3.38. Example. Determine MonoQ (Q( 3 2, ζ3 ), C).
√
Solution. We have already met the extension Q( 3 2, ζ3 )/Q in Example 3.30 and we will
make use of information from there. We build
up the list of monomorphisms √
in stages.
√
3
First consider monomorphisms that fix 2 and hence fix the subfield Q( 3 2). These form
the subset
√
√
3
3
MonoQ( √
3
2) (Q( 2, ζ3 ), C) ⊆ MonoQ (Q( 2, ζ3 ), C).
√
√
We know that Q( 3 2, ζ3 ) = Q( 3 2)(ζ3√
) and that ζ3 is a root of the irreducible cyclotomic
2 + X + 1 ∈ Q( 3 2)[X]. So there are two monomorphisms id, α fixing
polynomial
Φ
(X)
=
X
3
0
√
Q( 3 2), where α0 has the effect
√
√
3
2 −→ 3 2
α0 :
.
ζ3 −→ ζ32
√
√
2 to 3 2 ζ3 . This time we have 2 distinct ways
Next we consider monomorphisms that
send 3√
√
to extend to elements of MonoQ (Q( 3 2, ζ3 ), Q( 3 2, ζ3 )) since again we can send ζ3 to either ζ3
or ζ32 . The possibilities are
√
√
√
√
3
3
2 −→ 3 2 ζ3
2 −→ 3 2 ζ3
, α1 :
α1 :
.
ζ3 −→
ζ3
ζ3 −→
ζ32
√
√
Finally we consider monomorphisms that send 3 2 to 3 2 ζ32 . There are again two possibilities
√
√
√
√
3
3
2 −→ 3 2 ζ32
2 −→ 3 2 ζ32
α2 :
, α2 :
.
ζ3 −→
ζ3
ζ3 −→
ζ32
These are all 6 of the required monomorphisms. It is also the case here that
√
√
√
√
3
3
3
3
MonoQ (Q( 2, ζ3 ), C) = MonoQ (Q( 2, ζ3 ), Q( 2, ζ3 )) = AutQ (Q( 2, ζ3 )),
√
3
so these form a group. It is a nice exercise to show that AutQ (Q(
2, ζ3 )) ∼
= S√
3 , the symmetric
√
3
group on 3 objects. It is also worth remarking that  AutQ (Q( 2, ζ3 )) = [Q( 3 2, ζ3 ) : Q].
We end this section with another useful result.
3.39. Proposition. Let L/K be an extension and α ∈ MonoK (L, L). Then α restricts to
an automorphism αalg : Lalg −→ Lalg .
Proof. Suppose that u ∈ Lalg , say p(u) = 0 for some p(X) ∈ K[X] of positive degree.
Then
p(α(u)) = α(p(u)) = α(0) = 0,
alg
so α maps L ⊆ L into itself and therefore gives rise to a restriction αalg : Lalg −→ Lalg which
is also a monomorphism. We must show that αalg is a bijection by showing it is surjective.
Let v ∈ Lalg and suppose that q(v) = 0 for some q(X) ∈ K[X] of positive degree. Now
Roots(q, L) = ∅ since it contains v, and it is also finite. Then αq : Roots(q, L) −→ Roots(q, L)
is a bijection by Proposition 3.37(ii), hence v = αq (w) = α(w) for some w ∈ Roots(q, L) ⊆ Lalg .
This shows that v ∈ im α and so αalg is surjective.
3.4. Algebraic closures
An important property of the complex numbers is that C is algebraically closed.
3.40. Theorem (Fundamental Theorem of Algebra for C). Every nonconstant polynomial
p(X) ∈ C[X] has a root in C.
3.41. Corollary. Every nonconstant polynomial p(X) ∈ C[X] has a factorization
p(X) = c(X − u1 ) · · · (X − ud ),
where c, u1 , . . . , ud ∈ C and this is unique apart from the order of the roots uj .
It is natural to pose the following question.
41
3.42. Question. Let K be a field. Is there an algebraically closed field F containing K?
By taking F alg we might as well ask that such a field be algebraic over K.
3.43. Definition. Let K be a field. An extension F/K is called an algebraic closure of K
if F is algebraic over K and algebraically closed.
3.44. Theorem. Let K be a field.
(i) There is an algebraic closure of K.
(ii) Let F1 and F2 be algebraic closures of K. Then there is an isomorphism ϕ : F1 −→ F2
which fixes the elements of K.
F1
}}
}}
}
}
}~ }
Ke
ϕ
ee
ee
ee
e
/ F2
Hence algebraic closures are essentially unique.
♠
♥ ♦
♣
Proof. See [3] for a proof using Zorn’s Lemma (see Axiom 3.48) which is logically
equivalent to the Axiom of Choice.
Because of the uniqueness we usually fix some choice of algebraic closure of K and write K
or K alg cl , referring to it as the algebraic closure of K. We are already familiar with the example
C = C. There are some immediate consequences of Theorem 3.44. We will temporarily write
.
E1 = E2 to indicate that for extensions E1 /K and E2 /K there is an isomorphism E1 −→ E2
fixing the elements of K.
3.45. Proposition. Let K be a field.
.
(i) If L/K is an algebraic extension, then L = K.
.
(ii) If L/K is an extension, then so is L/K and (L)alg = K.
♠
♥ ♦
♣
Proof. (i) By Proposition 3.16, every element of L is algebraic over K. Since L is
algebraically closed it is an algebraic closure of K.
(ii) Every nonconstant polynomial in (L)alg [X] has a root in L; indeed, by Proposition 3.16,
all of its roots are in fact algebraic over K since (L)alg is algebraic over K. Hence these roots
lie in (L)alg , which shows that it is algebraically closed.
For example, we have Q = Calg and R = C.
There is a stronger result than Theorem 3.44(ii), the Monomorphism Extension Theorem,
which we will find useful. Again the proof uses Zorn’s Lemma which we state below. First we
need some definitions.
3.46. Definition. A partially ordered set (X, ) consists of a set X and a binary relation
such that whenever x, y, z ∈ X,
• x x;
• if x y and y z then x z;
• if x y and y x then x = y.
(X, ) is totally ordered if for every pair x, y ∈ X, at least one of x y or y x is true.
3.47. Definition. Let (X, ) be a partially ordered set and Y ⊆ X.
• y ∈ X is an upper bound for Y if for every y ∈ Y , y y.
• An element x ∈ X is a maximal element of X if
x
y
=⇒
y = x.
3.48. Axiom (Zorn’s Lemma). Let (X, ) be a partially ordered set in which every totally
ordered subset has an upper bound. Then X has a maximal element.
42
3.49. Theorem (Monomorphism Extension Theorem). Let M/K be an algebraic extension
and L/K
M/K. Suppose that ϕ0 : L −→ K is a monomorphism fixing the elements of K.
Then there is an extension of ϕ0 to a monomorphism ϕ : M −→ K.
8K
Ñ@
Ñ
ÑÑ
ÑÑ
Ñ
Ñ
M
ÑÑ ϕ0
Ñ
ÑÑ
ÑÑ
Ñ
Ñ
ϕ
L
♠
♥ ♦
♣
/K
=
K
Proof. We consider the set X consisting of all pairs (F, θ), where F/L
θ : F −→ K extends ϕ0 . We order X using the relation for which (F1 , θ1 )
F1 F2 and θ2 extends θ1 . Then (X, ) is a partially ordered set.
Suppose that Y ⊆ X is a totally ordered subset. Let
F =
M/L and
(F2 , θ2 ) whenever
F.
(F,θ)∈Y
Then F /L
M/L. Also there is a function θ : F −→ K defined by
θ(u) = θ(u)
whenever u ∈ F for (F, θ) ∈ Y . It is straightforward to check that if u ∈ F for (F , θ ) ∈ Y
then
θ (u) = θ(u),
so θ is welldefined. Then for every (F, θ) ∈ Y we have (F, θ)
(F , θ), so (F , θ) is an upper
bound for Y . By Zorn’s Lemma there must be a maximal element of X, (M0 , θ0 ).
Suppose that M0 = M , so there is an element u ∈ M for which u ∈
/ M0 . Since M is algebraic
over K it is also algebraic over M0 , hence u is algebraic over M0 . If
minpolyM0 ,u (X) = a0 + · · · + an−1 X n−1 + X n ,
then the polynomial
f (X) = θ0 (a0 ) + · · · + θ0 (an−1 )X n−1 + X n ∈ (θ0 M0 )[X]
is also irreducible and so it has a root v in K (which is also an algebraic closure of θ0 M0
K). The Homomorphism Extension Property 1.22 of the polynomial ring M0 [X] applied to
the monomorphism θ0 : M0 −→ K yields a homomorphism θ0 : M0 [X] −→ K extending θ0
and for which θ0 (u) = v. This factors through the quotient ring M0 [X]/(minpolyM0 ,u (X)) to
give a monomorphism θ0 : M0 (u) −→ K extending θ0 . But then (M0 , θ0 )
(M0 (u), θ0 ) and
(M0 , θ0 ) = (M0 (u), θ0 ), contradicting the maximality of (M0 , θ0 ). Hence M0 = M and so we
can take ϕ = θ0 .
3.50. Example. Let u ∈ K and suppose that p(X) = minpolyK,u (X) ∈ K[X]. Then for
any other root of p(X), v ∈ K say, there is a monomorphism ϕv : K(u) −→ K with ϕv (u) = v.
This extends to a monomorphism ϕ : K −→ K.
3.51. Definition. Let u, v ∈ K. Then v is conjugate to u over K or is a conjugate of u
over K if there is a monomorphism ϕ : K −→ K for which v = ϕ(u).
3.52. Lemma. If u, v ∈ K, then v is conjugate to u over K if and only if minpolyK,u (v) = 0.
43
Proof. Suppose that v = ϕ(u) for some ϕ ∈ MonoK (K, K). If
minpolyK,u (X) = a0 + a1 X + · · · + ad−1 X d−1 + X d ,
then
a0 + a1 u + · · · + ad−1 ud−1 + ud = 0
and so
a0 + a1 v + · · · + ad−1 v d−1 + v d = ϕ(a0 + a1 u + · · · + ad−1 ud−1 + ud ) = 0.
The converse follows from Example 3.50.
3.5. Multiplicity of roots and separability
Let K be a field. Suppose that f (X) ∈ K[X] and u ∈ K is a root of f (X), i.e., f (u) = 0.
Then we can factor f (X) as f (X) = (X − u)f1 (X) for some f1 (X) ∈ K[X].
3.53. Definition. If f1 (u) = 0 then u is a multiple or repeated root of f (X). If f1 (u) = 0
then u is a simple root of f (X).
We need to understand more clearly when an irreducible polynomial has a multiple root
since this turns out to be important in what follows. Consider the formal derivative on K[X],
i.e., the function ∂ : K[X] −→ K[X] given by
∂(f (X)) = f (X) = a1 + 2a2 X + · · · + dad X d−1 ,
where f (X) = a0 + a1 X + a2 X 2 + · · · + ad X d with aj ∈ K.
3.54. Proposition. The formal derivative ∂ : K[X] −→ K[X] has the following properties.
(i) ∂ is Klinear.
(ii) ∂ is a derivation, i.e., for f (X), g(X) ∈ K[X],
∂(f (X)g(X)) = ∂(f (X))g(X) + f (X)∂(g(X)).
(iii) If char K = 0, then ker ∂ = K and ∂ is surjective.
(iv) If char K = p > 0, then
ker ∂ = {h(X p ) : h(X) ∈ K[X]}
and im ∂ is spanned by the monomials X k with p (k + 1).
Proof. (i) This is routine.
(ii) By Klinearity, it suffices to verify this for the case where f (X) = X r and g(X) = X s with
r, s 0. But then
∂(X r+s ) = (r + s)X r+s−1 = rX r−1 X s + sX r X s−1 = ∂(X r )X s + X r ∂(X s ).
(iii) If f (X) = a0 + a1 X + a2 X 2 + · · · + ad X d then
∂(f (X)) = 0
⇐⇒
a1 = 2a2 = · · · = dad = 0.
So ∂(f (X)) = 0 if and only if f (X) = a0 ∈ K. It is also clear that every polynomial g(X) ∈
K[X] has the form g(X) = ∂(f (X) where f (X) is an antiderivative of g(X).
(iv) For a monomial X m , ∂(X m ) = mX m−1 and this is zero if and only if p  m. Using this we
see that
∂(a0 + a1 X + a2 X 2 + · · · + ad X d ) = 0
Also, im ∂ is spanned by the monomials
p (k + 1).
Xk
⇐⇒
am = 0 whenever p m.
for which ∂(X k+1 ) = 0, which are the ones with
We now apply the formal derivative to detect multiple roots.
3.55. Proposition. Let f (X) ∈ K[X] have a root u ∈ L for some extension L/K. Then u
is a multiple root of f (X) if and only if f (X) and f (X) have a common factor of positive
degree in K[X] which vanishes at u.
44
Proof. Working in L[X], let f (X) = (X − u)f1 (X). Then
f (X) = f1 (X) + (X − u)f1 (X),
so f (u) = f1 (u). Hence u is a multiple root if and only if f (X) and f (X) have a common
factor in L[X] (and hence in K[X] by Proposition 3.12) and which vanishes at u.
3.56. Corollary. If f (X) is irreducible in K[X] then a root u is a multiple root if and
only if f (X) = 0. In particular, this can only happen if char K > 0.
3.57. Corollary. If char K = 0 and f (X) is irreducible in K[X], then every root of f (X)
is simple.
3.58. Example. For n
1, show that each of the roots of f (X) = X n − 1 in C is simple.
Solution. We have f (X) = ∂(X n − 1) = nX n−1 , so for any root ζ of f (X),
f (ζ) = nζ n−1 = 0.
3.59. Example. Show that 2i is a multiple root of f (X) = X 4 + 8X 2 + 16.
Solution. We have f (X) = 4X 3 +16X. Using Long Division and the Euclidean Algorithm
we find that gcd(f (X), f (X)) = X 2 + 4, where 2i is also a root of X 2 + 4. Hence 2i is a multiple
root of f (X). In fact, X 4 + 8X 2 + 16 = (X 2 + 4)2 , so this is obvious.
3.60. Example. Let p > 0 be a prime and suppose that L/Fp is an extension. Show that
each of the roots of f (X) = X p − 1 in L is multiple.
Solution. We have f (X) = ∂(X p − 1) = pX p−1 = 0, so if ζ is any root of f (X) then
f (ζ) = 0. Later we will see that 1 is the only root of X p − 1.
3.61. Definition. An irreducible polynomial p(X) ∈ K[X] is separable over K if every
root of p(X) in an extension L/K is simple. By Corollary 3.56, this is equivalent to requiring
that p (X) = 0. If u ∈ L is a multiple root of p(X), then the multiplicity of u in p(X) is the
maximum m such that p(X) = (X − u)m q(X) for some q(X) ∈ L[X].
3.62. Proposition. Let K be a field and let K be an algebraic closure. If the irreducible
polynomial p(X) ∈ K[X] has distinct roots u1 , . . . , uk ∈ K, then the multiplicities of the uj are
equal. Hence in K[X],
p(X) = c(X − u1 )m · · · (X − uk )m ,
where c ∈ K and m 1.
Proof. Let u ∈ K be a root of p(X) and suppose that it has multiplicity m, so we can
write p(X) = (X − u)m p1 (X) where p1 (X) ∈ K(u)[X] and p1 (u) = 0.
Now let v ∈ K be any other root of p(X). By Proposition 3.34, there is a monomorphism
ϕv : K(u) −→ K for which ϕv (u) = v. When p(X) is viewed as an element of K(u)[X], the
coefficients of p(X) are fixed by ϕv . Then
ϕv ((X − u)m p1 (X)) = (X − u)m p1 (X),
and so
(X − v)m p1 (X) = (X − u)m p1 (X),
where p1 (X) ∈ K[X] is obtained applying ϕv to the coefficients of p1 (X). Now by Corollary 1.34,
(X − v)m must divide p1 (X) in K[X], and therefore the multiplicity of v must be at least m.
Interchanging the rˆoles of u and v we find that the multiplicities of u and v are in fact equal.
3.63. Corollary. Let K be a field and let K be an algebraic closure. If the irreducible
polynomial p(X) ∈ K[X] has distinct roots u1 , . . . , uk ∈ K which are all simple then in K[X],
p(X) = c(X − u1 ) · · · (X − uk ),
where c ∈ K and k = deg p(X).
45
3.64. Corollary. Let K be a field and let u ∈ K. Then the number of distinct conjugates
of u is
deg minpolyK,u (X)
,
m
where m is the multiplicity of u in minpolyK,u (X).
3.65. Definition. An algebraic element u ∈ L in an extension L/K is separable if its
minimal polynomial minpolyK,u (X) ∈ K[X] is separable.
3.66. Definition. An algebraic extension L/K is called separable if every element of L is
separable over K.
3.67. Example. An algebraic extension L/K of a field of characteristic 0 is separable by
Corollary 3.57.
3.68. Definition. Let L/K be a finite extension. The separable degree of L over K is
(L : K) =  MonoK (L, K).
3.69. Lemma. For a finite simple extension K(u)/K,
(K(u) : K) =  Roots(minpolyK,u , K).
If K(u)/K is separable, then [K(u) : K] = (K(u) : K).
Proof. This follows from Proposition 3.34 applied to the case L = K.
Any finite extension L/K can be built up from a succession of simple extensions
(3.1)
K(u1 )/K, K(u1 , u2 )/K(u1 ), · · · , L = K(u1 , . . . , uk )/K(u1 , . . . , uk−1 ).
So we can use the following to compute (L : K) = (K(u1 , . . . , uk ) : K).
3.70. Proposition. Let L/K and M/L be finite extensions. Then
(M : K) = (M : L)(L : K).
Proof. For α ∈ MonoK (M, K) let αL ∈ MonoK (L, K) be its restriction to L. By the
Monomorphism Extension Theorem 3.49, each element of MonoK (L, K) extends to a monomorphism M −→ K, so every element β ∈ MonoK (L, K) has the form β = αL for some α ∈
MonoK (M, K). Since (L : K) =  MonoK (L, K), we need to show that the number of such α
is always (M : L) =  MonoL (M, K).
So given β ∈ MonoK (L, K), choose any extension to a monomorphism β : K −→ K; by
Proposition 3.39, β is an automorphism. Of course, restricting to M K we obtain a monomorphism M −→ K. Now for any extension β : M −→ K of β we can form the composition
β −1 ◦ β : M −→ K; notice that if u ∈ L, then
β −1 ◦ β (u) = β −1 (β(u)) = u,
hence β −1 ◦ β ∈ MonoL (M, K). Conversely, each γ ∈ MonoL (M, K) gives rise to a monomorphism β ◦ γ : M −→ K which extends β. In effect, this shows that there is a bijection
extensions of β to monomorphism a M −→ K ←→ MonoL (M, K),
so (M : L) =  MonoL (M, K) agrees with the number of extensions of β to a monomorphism
M −→ K. Therefore we have the desired formula (M : K) = (M : L)(L : K).
3.71. Corollary. Let L/K be a finite extension. Then (L : K)  [L : K].
46
Proof. If L/K is a simple extension then by Propositions 3.62 and 3.34 we know that this
is true. The general result follows by building up L/K as a sequence of simple extensions as
in (3.1) and then using Theorem 2.6(ii) which gives
[L : K] = [K(u1 ) : K] [K(u1 , u2 ) : K(u1 )] · · · [K(u1 , . . . , uk ) : K(u1 , . . . , uk−1 )].
For each k, (K(u1 , . . . , uk ) : K(u1 , . . . , uk−1 )) divides [K(u1 , . . . , uk ) : K(u1 , . . . , uk−1 )], so the
desired result follows.
3.72. Proposition. Let L/K be a finite extension. Then L/K is separable if and only if
(L : K) = [L : K].
Proof. Suppose that L/K is separable. If K
E
L, then for any u ∈ L, u is algebraic over E, and in the polynomial ring E[X] we have minpolyE,u (X)  minpolyK,u (X). As
minpolyK,u (X) is separable, so is minpolyE,u (X), and therefore L/E is separable. Clearly E/K
is also separable. We have (L : K) = (L : E) (E : K) and [L : K] = [L : E] [E : K], so to
verify that (L : K) = [L : K] it suffices to show that (L : E) = [L : E] and (E : K) = [E : K].
Expressing L/K in terms of a sequence of simple extensions as in (3.1), we have
(L : K) = (K(u1 ) : K) · · · (L : K(u1 , . . . , uk−1 )),
[L : K] = [K(u1 ) : K] · · · [L : K(u1 , . . . , uk−1 )].
Now we can apply Lemma 3.69 to each of these intermediate separable simple extensions to
obtain (L : K) = [L : K].
For the converse, suppose that (L : K) = [L : K]. We must show that for each u ∈ L, u is
separable. For the extensions K(u)/K and L/K(u) we have (L : K) = (L : K(u)) (K(u) : K)
and [L : K] = [L : K(u)] [K(u) : K]. By Corollary 3.71, there are some positive integers r, s for
which [L : K(u)] = r(L : K(u)) and [K(u) : K] = s(K(u) : K). Hence
(L : K(u))(K(u) : K) = rs(L : K(u))(K(u) : K),
which can only happen if r = s = 1. Thus (K(u) : K) = [K(u) : K] and so u is separable.
3.73. Proposition. Let L/K and M/L be finite extensions. Then M/K is separable if and
only if L/K and M/L are separable.
Proof. If M/K is separable then [M : K] = (M : K) and so by Proposition 3.70,
[M : L][L : K] = (M : L)(L : K).
This can only happen if [M : L] = (M : L) and [L : K] = (L : K), since (M : L) [M : L] and
(L : K) [L : K]. By Proposition 3.72 this implies that L/K and M/L are separable.
Conversely, if L/K and M/L are separable then [M : L] = (M : L) and [L : K] = (L : K),
hence
[M : K] = [M : L][L : K] = (M : L)(L : K) = (M : K).
Therefore M/K is separable.
3.6. The Primitive Element Theorem
3.74. Definition. For a finite simple extension L/K, an element u ∈ L is called a primitive
element for the extension if L = K(u).
3.75. Theorem (Primitive Element Theorem). Let L/K be a finite separable extension.
Then L has a primitive element.
Proof. The case where K is a finite field will be dealt with in Proposition 5.16. So we will
assume that K is infinite.
Since L is built up from a sequence of simple extensions it suffices to consider the case
L = K(u, v). Let p(X), q(X) ∈ K[X] be the minimal polynomials of u and v over K. Suppose
that the distinct roots of p(X) in K are u = u1 , . . . , ur , while the distinct roots of q(X) are
v = v1 , . . . , vs . By the separability assumption, r = deg p(X) and s = deg q(X).
47
Since K is infinite, we can choose an element t ∈ K for which
u − ui
t=
vj − v
whenever j = 1. Then taking w = u + tv ∈ L, we find that w = ui + tvj whenever j = 1. Define
the polynomial (of degree r)
h(X) = p(w − tX) ∈ K(w)[X] ⊆ L[X].
Then h(v) = p(u) = 0, but h(vj ) = p(ui ) = 0 for any j = 1 by construction of t, so none of the
other vj is a zero of h(X).
Now since the polynomials h(X), q(X) ∈ K(w)[X] have exactly one common root in K,
namely v, by separability their greatest common divisor in K(w)[X] is a linear polynomial which
must be X − v, hence v ∈ K(w) and so u = w − tv ∈ K(w). This shows that K(u, v) K(w)
and therefore K(w) = K(u, v).
3.76. Corollary. Let L/K be a finite separable extension of a field of characteristic 0.
Then L has a primitive element.
Proof. Since Q
K, K is infinite and by Example 3.67 L/K is separable.
To find a primitive element we can always use the method suggested by the proof of Theorem 3.75, however a ‘try it and see’ approach will often be sufficient.
√
3.77. Example. Find a primitive element for the extension Q( 3, i)/Q.
√
√
√
3
+
i.
Then
working
over
the
subfield
Q(
3)
Q(
3, i) we find
Solution.
Consider
√
that i ∈
/ Q( 3) R and
√
√
√
√
(X − ( 3 + i))(X − ( 3 − i)) = X 2 − 2 3X + 4 ∈ Q( 3)[X],
hence
√
X 2 − 2 3X + 4 = minpolyQ(√3),√3+i (X).
Now taking
√
√
(X 2 − 2 3X + 4)(X 2 + 2 3X + 4) = X 4 − 4X 2 + 16 ∈ Q[X],
we see that minpolyQ,√3+i (X)  (X 4 − 4X 2 + 16) in Q[X]. Notice that
√
√
√
√
( 3 − i)
( 3 − i)
1 √
−1
√
( 3 + i) = √
=
= ( 3 − i) ∈ Q( 3 + i),
3+1
4
( 3 + i)( 3 − i)
√
√
since ( 3 + i)−1 ∈ Q( 3 + i). Hence
√
√
√
1 √
1 √
3 = (( 3 + i) + ( 3 − i)), i = (( 3 + i) − ( 3 − i)),
2
2
√
√
√
√
√
are both in Q( 3 + i), showing that Q( 3, i) Q( 3 + i) and so Q( 3, i) = Q( 3 + i). Thus
we must have deg minpolyQ,√3+i (X) = 4, and so minpolyQ,√3+i (X) = X 4 − 4X 2 + 16.
There is a general phenomenon illustrated by Example 3.77.
3.78. Proposition. Let u ∈ K be separable over K. Then
minpolyK,u (X) = (X − α1 (u)) · · · (X − αd (u)),
where α1 , . . . , αd are the elements of MonoK (K(u), K). In particular, the polynomial
(X − α1 (u)) · · · (X − αd (u)) ∈ K[X]
is in K[X] and is irreducible therein.
Proof. Since K(u) is separable then by Lemma 3.52,
d = deg minpolyK,u (X) = [K(u) : K] = (K(u) : K).
48
In Example 3.77 we have
√
√
√
√
[Q( 3, i) : Q] = [Q( 3, i) : Q( 3)][Q( 3) : Q] = 2 · 2 = 4.
√
√
There are four monomorphisms αk : Q( 3, i) −→ Q( 3, i) given by
√
√
√
√
√
√
3 −→
3
3 −→ − 3
3 −→ − 3
α1 = id, α2 =
, α3 =
, α4 =
.
i −→ −i
i −→
i
i −→ −i
Then
√
√
α2 ( 3 + i) = ( 3 − i),
so
3 − i)(X −
√
√
√
α4 ( 3 + i) = (− 3 − i),
√
3 − i)(X + 3 + i) = X 4 − 4X 2 + 16 ∈ Q[X].
√
√
√
Hence this polynomial is irreducible. So we have [Q( 3 + i) : Q] = 4 and Q( 3 + i) = Q( 3, i).
(X −
√
√
√
α3 ( 3 + i) = (− 3 + i),
3 + i)(X +
√
3.7. Normal extensions and splitting fields
Let K be an algebraic closure for the field K and let E/K K/K be a finite extension. If
ϕ ∈ MonoK (E, K), then by Remark 3.32, ϕE = E if and only if ϕE E.
3.79. Definition. E/K is normal if ϕE = E for every ϕ ∈ MonoK (E, K).
3.80. Remark. If E/K is a normal extension then whenever an irreducible polynomial
p(X) ∈ K[X] has a root in E, it splits in E since by Lemma 3.52 each pair of roots of p(X) is
conjugate over K and one can be mapped to the other by a monomorphism K −→ K which
must map E into itself.
3.81. Proposition. A finite extension E/K is normal if and only if it is a splitting field
over K for some polynomial f (X) ∈ K[X].
Proof. Suppose that E/K is normal. Then there is a sequence of extensions
K
K(u1 )
···
K(u1 , u2 )
K(u1 , . . . , un ) = E
Construct a polynomial by taking
f (X) = minpolyK,u1 (X) minpolyK,u2 (X) · · · minpolyK,un (X).
Then by Remark 3.80, f (X) splits in E. Also, E is generated by some of the roots of f (X).
Hence E is a splitting field for f (X) over K.
Now suppose that E is a splitting field for g(X) ∈ K[X], so that E = K(v1 , . . . , vk ), where
v1 , . . . , vk are the distinct roots of g(X) in E. Now any monomorphism θ ∈ MonoK (E, K) must
map these roots to θ(v1 ), . . . , θ(vk ) which are also roots of g(X) and therefore lie in E (see
Proposition 3.34). Since θ permutes the roots vj , we have
θE = θK(v1 , . . . , vk ) = K(θ(v1 ), . . . , θ(vk )) = K(v1 , . . . , vk ) = E.
This result makes it easy to recognize a normal extension since it is sufficient to describe it
as a splitting field for some polynomial over K. In Chapter 4 we will see that separable normal
extensions play a central rˆole in Galois Theory, indeed these are known as Galois extensions.
49
Exercises on Chapter 3
3.1. Prove Proposition 3.2.
3.2. Finding splitting subfields E
polynomials.
p1 (X) = X 4 −X 2 +1,
C over Q and determine [E : Q] for each of the following
p2 (X) = X 6 −2,
p3 (X) = X 4 +2,
p4 (X) = X 4 +5X 3 +10X 2 +10X+5.
[Hint: for p4 (X), consider p4 (Y − 1) ∈ Q[Y ].]
√
3.3. Prove that AutQ (Q( 3 2, ζ3 )) ∼
= S3 , the symmetric group on 3 elements, as claimed in the
solution of Example 3.38. [Hint: work out the effect of each automorphism on the three roots of
the polynomial X 3 − 2.]
3.4. Let k be a field of characteristic char k = p > 0 and k(T ) be the field of rational functions
in T over k. Show that the polynomial g(X) = X p − T ∈ k(T )[X] is irreducible and has a
multiple root in k(T ). How does g(X) factor in k(T )[X]?
√ √
√
√
3.5.√ Find primitive elements for the extensions Q( 5, 10)/Q, Q( 2, i)/Q, Q( 3, i)/Q,
Q( 4 3, i)/Q, in each case finding it minimal polynomial over Q. [Hint: look for elements of
high degree over Q, or use the method of proof of Theorem 3.75.]
3.6. Prove the following converse of Proposition 3.20:
Let L/K be a finite extension. If there are only finitely many subextensions F/K L/K,
then L/K is simple, i.e., L = K(w) for some w ∈ L.
[Hint: First deal with the case where L = K(u, v), then use induction on n to prove the general
case L = K(u1 , . . . , un ).]
3.7. Let K be a field. Show that every quadratic (i.e., of degree 2) extension E/K is normal.
Is such an extension always separable?
3.8. Let f (X) ∈ Q[X] be an irreducible polynomial of odd degree greater than 1 and having
only one real root u ∈ R. Show that Q(u)/Q is not a normal extension.
50
CHAPTER 4
Galois extensions and the Galois Correspondence
In this Chapter we will study the structure of Galois extensions and their associated Galois
groups, in particular we will explain how these are related through the Galois Correspondence.
Throughout the chapter, let K be a field.
4.1. Galois extensions
4.1. Definition. A finite extension E/K is a (finite) Galois extension if it is both normal
and separable.
From Section 3.5 we know that for such a Galois extension E/K, [E : K] = (E : K) and also
every monomorphism ϕ ∈ MonoK (E, K) maps E into itself, hence restricts to an automorphism
of E which will be denoted ϕE .
>K
~~
~
~~
~~ =
∼
E ϕ / E
ϕ
E
/K
K
Also, by the Monomorphism Extension Theorem 3.49, every automorphism α ∈ AutK (E) extends to a monomorphism E −→ K fixing elements of K. So there is a bijection
=
MonoK (E, K) ←→ AutK (E)
and we have
(4.1)
 AutK (E) = (E : K) = [E : K].
4.2. Definition. For a finite Galois extension E/K, the group
Gal(E/K) = AutK (E)
is called the Galois group of the extension or the Galois group of E over K. The elements of
Gal(E/K) are called (Galois) automorphisms of E/K.
Notice that Equation (4.1) implies
(4.2)
 Gal(E/K) = (E : K) = [E : K].
We can also reformulate the notion of conjugacy introduced in Definition 3.51.
4.3. Definition. Let E/K a finite Galois extension and u, v ∈ E. Then v is conjugate to
u if there is a ϕ ∈ Gal(E/K) for which v = ϕ(u); we also say that v is a conjugate of u.
It is easy to see that for u, v ∈ K, there is a finite Galois extension E/K in which v is a
conjugate of u if and only v is a conjugate of u over K in the old sense. Here is a slightly
different way to understand this. First notice that every element ϕ ∈ AutK (K, K) restricts to a
monomorphism E −→ K whose image is contained in E, hence gives rise to an automorphism
ϕE : E −→ E. Similarly, if F/K is any finite normal extension with E F , every automorphism
F : E −→ E. The proof of the next result is left as
θ : F −→ F restricts to an automorphism θE
an exercise.
51
4.4. Proposition. If E/K is a finite Galois extension, then the function
AutK (K, K) −→ AutK (E, E);
ϕ −→ ϕE
is a surjective group homomorphism. If F/K K/K is any finite normal extension with E
then there is a surjective group homomorphism
AutK (F, F ) −→ AutK (E, E);
F
F
θ −→ θE
.
Furthermore, for ϕ ∈ AutK (K, K) we have
(ϕF )FE = ϕE .
4.2. Working with Galois groups
Let E/K be a finite Galois extension. Then we know that E is a splitting field for some
polynomial over K since E/K is normal. We also know that E is a simple extension of K
since E/K is separable. Hence E is a splitting field for the minimal polynomial of any primitive
element for E/K; this minimal polynomial has degree [E : K]. It is often convenient to use these
facts to interpret elements of the Galois group as permutations of the roots of some polynomial
which splits over E.
√ √
4.5. Example. Describe the Galois group Gal(Q( 2, 3)/Q) as a subgroup of the group
of permutations of the roots of (X 2 − 2)(X 2 − 3) ∈ Q[X].
Solution. We have
√ √
√ √
√
√
[Q( 2, 3) : Q] = [Q( 2, 3) : Q( 2)] [Q( 2) : Q] = 4,
and the following nontrivial elements of the Galois group together with the element identity
α1 = id:
√
√
√
√
√
√
2
−→
−
2
2
−→
2
2
−→
−
√
√
√
√
√
√2
− 2 −→
− 2 −→ − 2
− 2 −→
2
√
√ , α3 = √
√ , α4 = √
√2 .
α2 =
√3 −→
√3
√3 −→ −√3
√3 −→ −√3
− 3 −→ − 3
− 3 −→
3
− 3 −→
3
√ √
√
√
Writing the roots in the list 2, − 2, 3, − 3 and numbering them from 1 to 4, these automorphisms correspond to the following permutations in S4 expressed in cycle notation:
α2 ←→ (1 2),
α3 ←→ (3 4),
α4 ←→ (1 2)(3 4).
4.6.√Example.
Using a primitive element u for the extension, describe the Galois group
√
Gal(Q( 2, 3)/Q) as a subgroup of the group of permutations of the roots of minpolyQ,u (X) ∈
Q[X].
√ √
√
√
√
√
2,
3)
=
Q(
2
+
3)
and
the
conjugates
of
u
=
2
+
3 are
Solution.
We
have
Q(
√
√
± 2 ± 3. Listing these as
√
√ √
√
√
√
√
√
2 + 3, 2 − 3, − 2 + 3, − 2 − 3,
and after numbering them accordingly, we find the correspondences
α2 ←→ (1 3)(2 4),
α3 ←→ (1 2)(3 4),
α4 ←→ (1 4)(2 3).
Next we summarize the properties of Galois groups that can be deduced from what we
have established so far. Recall that for an extension F/K and a polynomial f (X) ∈ K[X],
Roots(f, F ) denotes the set of roots of f (X) in F .
4.7. Recollection. Recall that an action of a group G on a set X is transitive if for every
pair of elements x, y ∈ X, there is an element g ∈ G such that y = gx (so there is only one
orbit); the action is faithful or effective if for every nonidentity element h ∈ G, there is a
element z ∈ X such that hz = z.
52
4.8. Theorem. Let E/K be a finite Galois extension. Suppose that E is the splitting field
of a separable irreducible polynomial f (X) ∈ K[X] of degree n. Then the following are true.
(i) Gal(E/K) acts transitively and faithfully on Roots(f, E).
(ii) Gal(E/K) can be identified with a subgroup of the group of permutations of Roots(f, E).
If we order the roots u1 , . . . , un then Gal(E/K) can be identified with a subgroup of Sn .
(iii)  Gal(E/K) divides n! and is divisible by n.
As we have seen in Examples 4.5 and 4.6, in practise it is often easier to use a not necessarily
irreducible polynomial to determine and work with a Galois group.
4.9. Example. The Galois extension Q(ζ8 )/Q has degree [Q(ζ8 ) : Q] = 4 and it has the
following automorphisms apart from the identity:
α : ζ8 −→ ζ83 ,
β : ζ8 −→ ζ85 ,
γ : ζ8 −→ ζ87 .
If we list the roots of the minimal polynomial
minpolyQ,ζ (X) = Φ8 (X) = X 4 + 1
in the order ζ8 , ζ83 , ζ85 , ζ87 , we find that these automorphisms correspond to the following permutations in S4 :
α ←→ (1 2)(3 4), β ←→ (1 3)(2 4), γ ←→ (1 4)(2 3).
So the Galois group Gal(Q(ζ8 )/Q) corresponds to
{id, (1 2)(3 4), (1 3)(2 4), (1 4)(2 3)}
S4 .
Noticing that
1
1
ζ8 = √ + √ i,
2
2
√
√
√
we√easily find that 2, i ∈ Q(ζ8 );√hence Q( 2, i)
Q(ζ8 ). Since [Q( 2, i) : Q] = 4, we have
Q( 2, i) = Q(ζ8 ). Notice that Q( 2, i) is the
field of f (X) = (X 2 − 2)(X 2 + 1) over
√ splitting
√
Q. Now list the roots of f (X) in the order 2, − 2, i, −i, and observe that
√
√
√
√
√2 −→ −√2
√2 −→ −√2
− 2 −→
− 2 −→
2
2
←→ (1 2)(3 4),
←→ (1 2),
α:
β
:
i −→ −i
i −→
i
−i −→
i
−i −→ −i
√
√
√2 −→
√2
− 2 −→ − 2
γ:
i −→ −i ←→ (3 4).
−i −→
i
√
In this description, the Galois group Gal(Q(ζ8 )/Q) = Gal(Q( 2, i)/Q) corresponds to the
subgroup
{id, (1 2), (3 4), (1 2)(3 4)} S4 .
While it can be hard to determine Galois groups in general, special arguments can sometimes
be exploited.
4.10. Example. Suppose that f (X) = X 3 + aX 2 + bX + c ∈ Q[X] is an irreducible cubic
and that f (X) has only one real root. Then Gal(Q(f (X))/Q) ∼
= S3 .
Proof. Let u1 ∈ R be the real root of f (X) and let u2 , u3 be the remaining complex
roots. Then Q(f (X)) = Q(u1 , u2 , u3 ) and in fact [Q(f (X)) : Q] = 6 since [Q(f (X)) : Q]  6
and u2 ∈
/ Q(u1 )
R. Hence Gal(Q(f (X))/Q) is isomorphic to a subgroup of S3 and so
Gal(Q(f (X))/Q) ∼
= S3 since the orders agree. We also have Q(f (X)) ∩ R = Q(u1 ).
The Galois group Gal(Q(f (X))/Q) contains an element of order 3 which corresponds to a
3cycle when viewed as a permutation of the roots u1 , u2 , u3 ; we can assume that this is (1 2 3).
53
It also contains an element of order 2 obtained by restricting complex conjugation to Q(f (X));
this fixes u1 and interchanges u2 , u3 , so it corresponds to the transposition (2 3).
4.11. Remark. Such examples occur when the cubic polynomial f (X) has local maximum
and minimum at real values c+ and c− with f (c+ ), f (c− ) > 0 or f (c+ ), f (c− ) < 0. This happens
for example with f (X) = X 3 − 3X + 3 which has local extrema at ±1 and f (1) = 1, f (−1) = 5.
Given a Galois extension E/K, we will next study subextensions L/K
E/K and subgroups Γ Gal(E/K), focusing on the relationship between objects of these types.
4.3. Subgroups of Galois groups and their fixed fields
Let E/K a Galois extension and suppose that Γ
Gal(E/K). Consider the subset of
elements of E fixed by Γ,
E Γ = {u ∈ E : ∀γ ∈ Γ, γ(u) = u}.
4.12. Lemma. E Γ
E is a subfield of E containing K.
Proof. For u, v ∈ E Γ and γ ∈ Γ,
γ(u + v) = γ(u) + γ(v) = u + v,
γ(uv) = γ(u)γ(v) = uv.
Also, if u = 0,
γ(u−1 ) = γ(u)−1 = u−1 .
Finally, if t ∈ K then γ(t) = t, so K
4.13. Definition. E Γ
EΓ.
E is the fixed subfield of Γ.
By Proposition 3.73, the extensions E/E Γ and E Γ /K are separable. E/E Γ is also normal,
so this is a Galois extension; we will identify its Galois group. Notice that
[E : E Γ ] = (E : E Γ ) =  Gal(E/E Γ ).
Now each element of Gal(E/E Γ ) is also an element of Gal(E/K) and Gal(E/E Γ ) Gal(E/K).
Notice that by definition Γ
Gal(E/E Γ ), so Lagrange’s Theorem implies that Γ divides
Γ
 Gal(E/E ). In fact we have
4.14. Proposition. For Γ
Gal(E/K), we have Gal(E/E Γ ) = Γ and the equations
[E : E Γ ] =  Gal(E/E Γ ) = Γ,
[E Γ : K] =
 Gal(E/K)
.
Γ
Proof. We know that E/E Γ is separable, so by the Primitive Element Theorem 3.75 it is
simple, say E = E Γ (u). Now let the distinct elements of Γ be γ1 = id, γ2 , . . . , γh , where h = Γ.
Consider the polynomial of degree h
f (X) = (X − u)(X − γ2 (u)) · · · (X − γh (u)) ∈ E[X].
Notice that f (X) is unchanged by applying any γk to its coefficients since the roots γj (u) are
permuted by γk . Hence, f (X) ∈ E Γ [X]. This shows that
[E : E Γ ] = [E Γ (u) : E Γ ]
Since Γ
h = Γ.
Gal(E/E Γ ), we also have
h = Γ
 Gal(E/E Γ ) = [E : E Γ ].
Combining these two inequalities we obtain
[E : E Γ ] =  Gal(E/E Γ ) = Γ = h
and therefore Γ = Gal(E/E Γ ).
54
4.4. Subfields of Galois extensions and relative Galois groups
Let E/K a Galois extension and suppose that L/K E/K (i.e., K L E). Then E/L
is also a Galois extension whose Galois group Gal(E/L) is sometimes called the relative Galois
group of the pair of extensions E/K and L/K. The following is immediate.
4.15. Lemma. The relative Galois group of the pair of extensions L/K E/K is a subgroup
of Gal(E/K), i.e., Gal(E/L) Gal(E/K), and its order is  Gal(E/L) = [E : L].
4.16. Proposition. Let L/K
E/K. Then L = E Gal(E/L) .
Proof. Clearly L
E Gal(E/L) . Suppose that u ∈ E − L. By Theorem 4.8(i), there is
an automorphism θ ∈ Gal(E/L) such that θ(u) = u, hence u ∈
/ E Gal(E/L) . This shows that
E Gal(E/L) L and therefore E Gal(E/L) = L.
We need to understand when Gal(E/L) Gal(E/K) is actually a normal subgroup. The
next result explains the connection between the two uses of the word normal which both ultimately derive from their use in Galois theory.
4.17. Proposition. Let E/K be a finite Galois extension and L/K E/K.
(i) The relative Galois group Gal(E/L) of the pair of extensions L/K E/K is a normal
subgroup of Gal(E/K) if and only if L/K is a normal extension.
(ii) If L/K is normal and hence a Galois extension, then there is a group isomorphism
∼
=
Gal(E/K)/ Gal(E/L) −
→ Gal(L/K);
α Gal(E/L) −→ αL .
Proof. (i) Suppose that Gal(E/L) Gal(E/K), i.e., for all α ∈ Gal(E/L) and β ∈
Gal(E/K), we have βαβ −1 ∈ Gal(E/L). Now if u ∈ L, then for any γ ∈ Gal(E/K) and
α ∈ Gal(E/L), γ(u) ∈ E satisfies
αγ(u) = γ(γ −1 αγ(u)) = γ(u),
since γ −1 αγ ∈ Gal(E/L); hence γ(u) ∈ E Gal(E/L) = K. By the Monomorphism Extension
Theorem 3.49, every monomorphism L −→ K fixing K extends to a monomorphism E −→ K
which must have image E, so the above argument shows that L/K is normal.
Conversely, if L/K is normal, then for every ϕ ∈ Gal(E/K) and v ∈ L, ϕ(v) ∈ L, so for
every θ ∈ Gal(E/L), θ(ϕ(v)) = ϕ(v) and therefore
ϕ−1 θϕ(v) = v.
This shows that ϕ−1 θϕ ∈ Gal(E/L). Hence for every ϕ ∈ Gal(E/K),
ϕ Gal(E/L)ϕ−1 = Gal(E/L),
which shows that Gal(E/L) Gal(E/K).
(ii) If α ∈ Gal(E/K), then αL = L since L/K is normal. Hence we can restrict α to an
automorphism of L,
αL : L −→ L; αL (u) = α(u).
Then αL is the identity function on L if and only if α ∈ Gal(E/L). It is easy to see that the
function
Gal(E/K) −→ Gal(L/K); α −→ αL
is a group homomorphism whose kernel is Gal(E/L). Thus we obtain an injective homomorphism
Gal(E/K)/ Gal(E/L) −→ Gal(L/K)
for which
[E : K]
= [L : K] =  Gal(L/K).
 Gal(E/K)/ Gal(E/L) =
[E : L]
Hence this homomorphism is an isomorphism.
55
4.5. The Galois Correspondence and the Main Theorem of Galois Theory
We are now almost ready to state our central result which describes the Galois Correspondence associated with a finite Galois extension. We will use the following notation. For a finite
Galois extension E/K, let
S(E/K) = the set of all subgroups of Gal(E/K);
F(E/K) = the set of all subextensions L/K of E/K.
Each of these sets is ordered by inclusion. Since every subgroup of a finite group is a finite
subset of a finite set, S(E/K) is also a finite set. Define two functions by
ΦE/K : F(E/K) −→ S(E/K);
ΦE/K (L) = Gal(E/L),
ΘE/K : S(E/K) −→ F(E/K);
ΘE/K (Γ) = E Γ .
4.18. Theorem (Main Theorem of Galois Theory). Let E/K be a finite Galois extension.
Then the functions ΦE/K and ΘE/K are mutually inverse bijections which are orderreversing.
F(E/K) o
ΦE/K
/
S(E/K)
ΘE/K
Under this correspondence, normal subextensions of E/K correspond to normal subgroups of
Gal(E/K) and vice versa.
Proof. We know from Proposition 4.16 that for an extension L/K in F(E/K),
ΘE/K (ΦE/K (L)) = ΘE/K (Gal(E/L)) = E Gal(E/L) = L.
Also, by Proposition 4.14 for H ∈ S(E/K) we have
ΦE/K (ΘE/K (Γ)) = ΦE/K (E Γ ) = Gal(E/E Γ ) = Γ.
This shows that ΦE/K and ΘE/K are mutually inverse and so are inverse bijections.
Let L1 /K, L2 /K ∈ F(E/K) satisfy L1 /K
L2 /K. Then Gal(E/L2 ) Gal(E/L1 ) since
L1 ⊆ L2 and so if α ∈ Gal(E/L2 ) then α fixes every element of L1 . Hence ΦE/K (L2 )
ΦE/K (L1 ) and so ΦE/K reverses order.
Similarly, if Γ1 , Γ2 ∈ S(E/K) and Γ1 Γ2 , then E Γ2 E Γ1 since if w ∈ E Γ2 then it is fixed
by every element of Γ1 (as Γ1 is a subset of Γ2 ). Hence ΘE/K reverses order.
There is an immediate consequence of the Main Theorem 4.18 which is closely related to
Proposition 3.20.
4.19. Corollary. Let E/K be a finite Galois extension. Then there are only finitely many
subextensions L/K E/K.
Proof. Since the set S(E/K) is finite, so is F(E/K).
When dealing with a finite Galois extension E/K, we indicate the subextensions in a diagram
with a line going upwards indicating an inclusion. We can also do this with the subgroups of
the Galois group Gal(E/K) with labels indicating the index of the subgroups. In effect, the
Galois Correspondence inverts these diagrams.
4.20. Example. Figure 4.1 shows the Galois Correspondence for the extension of Example 3.30.
√
As noted at the end of Example 3.38, the Galois group here is Gal(Q( 3 2, ζ3 )/Q) ∼
= S3 . It
3 − 2 for
is useful to make this isomorphism explicit. First
take
the
3
roots
of
the
polynomial
X
√
√
√
3
3
3
which E is the splitting field over Q; these are 2, 2 ζ3 , 2 ζ32 which we number in the order
they are listed. Then the monomorphisms id, α0 , α1 , α1 , α2 , α2 extend to automorphisms of E,
each of which permutes these 3 roots in the following ways given by cycle notation:
α0 = (2 3),
α1 = (1 2 3),
α1 = (1 2),
56
α2 = (1 3 2),
α2 = (1 3).
√
3
E = Q( 2, ζ3 )
H
ii
eeeekekkk
e
e
e
ii
e
e
eee
kk
e
e
k
ii
e
k
e
2
e
k
e
e
k
e
ii
e
k
2
2
eeee
kk
e
k
e
ii
e
e
e
e
√
√
√
e
e
ii3
ii
Q( 3 2 ζ32 )
Q( 3 2)
Q( 3 2 ζ3 )
ii
r
N
N
I
rr
ii
rr
ii
rr
ii
r
r
rr 3
3
r
rr
3
Q(ζ3 )
rr
Q
kkk
k
k
rrr
2kkk
rr
k
kk
rr
kkk
r
kkkk
QS
ΦE/K
Gal(E/Q)
2
kkk v
kkk vvvv
k
k
k
v
k
k
v
k
k
v
k
v
k
3 vv
3 kkkk
v
k
Gal(E/Q(ζ
3
k
3 ))
v
k
vv
kkk
y
k
v
k
k
yy
vv
kkk
yy
vv
kkk
v
y
k
k
v
y
√
√ kk
√v
y
3 yy
Gal(E/Q( 3 2 ζ32 )) yyy
Gal(E/Q( 3 2 ζ3 ))
Gal(E/Q( 3 2))
yy
yy
y
2
y
2
2
yyy
{id}
√
Figure 4.1. The Galois Correspondence for E = Q( 3 2, ζ3 )/Q
We find that
√
3
Gal(E/Q(ζ3 )) = {id, α1 , α2 } ∼
Gal(E/Q( 2)) = {id, α0 } ∼
={id, (1 2 3), (1 3 2)},
={id, (2 3)},
√
√
3
3
Gal(E/Q( 2 ζ3 )) = {id, α2 } ∼
Gal(E/Q( 2 ζ32 )) = {id, α1 } ∼
={id, (1 3)},
={id, (1 2)}.
Notice that {id, (1 2 3), (1 3 2)} S3 and so Q(ζ3 )/Q is a normal extension. Of course Q(ζ3 )
is the splitting field of X 3 − 1 over Q.
4.6. Galois extensions inside the complex numbers and complex conjugation
When working with Galois extensions contained in the complex numbers it is often useful
to make use of complex conjugation as an element of a Galois group. Let E/Q be a finite Galois
extension with E/Q C/Q. Setting ER = R ∩ E, we have Q ER E.
4.21. Proposition. Complex conjugation ( ) : C −→ C restricts to an automorphism of E
over Q, ( )E/Q : E −→ E. Furthermore,
(i) ( )E/Q agrees with the identity function if and only if ER = E.
(ii) If ER = E, then
( )E/Q = {id, ( )E/Q } ∼
= Z/2,
hence, ER = E
( )E/Q
and [E : ER ] = 2.
Proof. Let u ∈ E. As E/Q is normal, minpolyQ,u (X) ∈ Q[X] splits over E, so all of its
complex roots lie in E. But ( ) permutes the roots of this minimal polynomial. Therefore ( )
maps E into itself.
57
(i) For z ∈ C, z = z if and only if z ∈ R.
(ii) Here  ( )E/Q  = 2, and
E
( )E/Q
= {u ∈ E : u = u} = ER .
C
2
∞
R
nE
nnn
n
n
nn
nnn 2
n
n
n
ER e
ee
ee
ee
e
∞
Q
We will usually write ( ) rather than ( )E/Q when no confusion seems likely to result.
4.22. Example. Consider the cyclotomic extension Q(ζ8 )/Q where
1
1
ζ8 = eπi/4 = √ + √ i.
2
2
From Example 4.9 we know that
√
Q(ζ8 ) = Q( 2, i), [Q(ζ8 ) : Q] = 4,
and we easily see that
√
Q(ζ8 )R = Q( 2).
4.7. Galois groups of even and odd permutations
We have seen that for a monic separable polynomial f (X) ∈ K[X] of degree n, the Galois
group of its splitting field E over K can naturally be thought of as a subgroup of the symmetric
group Sn , where we view the latter as permuting the roots of f (X). It is reasonable to ask when
Gal(E/K) An rather than just Gal(E/K) Sn .
We first recall an interpretation of the sign of a permutation σ ∈ Sn , sgn σ = ±1. For each
pair i, j with 1
i 0 be a prime. Let K be a field with char K = p. Suppose that 0 = a ∈ K and
f (X) = X p − a ∈ K[X]. Let L/K where L is a splitting field for f (X) over K.
(a) Show that f (X) has p distinct roots in L. If u ∈ L is one such root, describe the
remaining roots and show that L contains p distinct pth roots of 1.
(b) Suppose that K contains p distinct pth roots of 1. Show that either f (X) is irreducible
over K or it factors into p distinct linear factors over K.
(c) Suppose that the only pth root of 1 in K is 1. Show that either f (X) is irreducible
over K or it has a root in K.
4.8. Let K be a field of characteristic char K = p where p > 0 is a prime. Suppose that
0 = a ∈ K and f (X) = X p − a ∈ K[X]. Show that if f (X) has no root in K then it is
irreducible over K.
64
CHAPTER 5
Galois extensions for fields of positive characteristic
In this chapter we will investigate extensions of fields of positive characteristic, especially
finite fields. A thorough account of finite fields and their applications can be found in [5].
Throughout this chapter we will assume that K is a field of characteristic char K = p (where
p > 0 is a prime) containing the prime subfield Fp .
5.1. Finite fields
If K is a finite field, then K is an Fp vector space. Our first goal is to count the elements
of K. Here is a more general result.
5.1. Lemma. Let F be a finite field with q elements and let V be an F vector space. Then
dimF V < ∞ if and only if V is finite in which case V  = q dimF V .
Proof. If d = dimF V < ∞, then for a basis v1 , . . . , vd we can express each element v ∈ V
uniquely in the form v = t1 v1 + · · · + td vd , where t1 , . . . , td ∈ F . Clearly there are exactly q d
such expressions, so V  = q d .
Conversely, if V is finite then any basis has finitely many elements and so dimF V < ∞.
5.2. Corollary. Let F be a finite field and E/F an extension. Then E is finite if and
only if E/F is finite and then E = F [E:F ] .
5.3. Corollary. Let K be a finite field. Then K/Fp is finite and K = p[K:Fp ] .
Our next task is to show that for each power pd there is a finite field with pd elements. We
start with the algebraic closure Fp of Fp and consider the polynomial
d
Θpd (X) = X p − X ∈ Fp [X].
Notice that Θpd (X) = −1, hence by Proposition 3.55 every root of Θpd (X) in Fp is simple. Therefore by Corollary 1.35 Θpd (X) must have exactly pd distinct roots in Fp , say
0, u1 , . . . , upd −1 . Then in Fp [X] we have
d
X p − X = X(X − u1 ) · · · (X − upd −1 ),
and each root is separable over Fp . Let
Fpd = {u ∈ Fp : Θpd (u) = 0} ⊆ Fp ,
d −1
Notice that u ∈ F0pd if and only if up
F0pd = {u ∈ Fpd : u = 0}.
= 1.
5.4. Proposition. For each d
1, Fpd is a finite subfield of Fp with pd elements and
F0pd = F×
. Furthermore, the extension Fpd /Fp is a separable splitting field.
pd
Proof. If u, v ∈ Fpd then by the Idiot’s Binomial Theorem 1.10,
d
d
d
d
d
(u + v)p − (u + v) = (up + v p ) − (u + v) = (up − u) + (v p − v) = 0,
d
d
d
(uv)p − uv = up v p − uv = uv − uv = 0.
d
d
Furthermore, if u = 0 then up −1 = 1 and so u has multiplicative inverse up −2 . Hence Fpd Fp .
Notice that Fp
Fpd , so Fpd /Fp is a finite extension. In any field the nonzero elements are
always invertible, hence F0pd = F×
.
pd
65
5.5. Definition. The finite subfield Fpd
Fp is called the Galois field of order pd .
The notation GF(pd ) is often used in place of Fpd . Of course, Fp1 = GF(p1 ) = GF(p) = Fp
and [Fpd : Fp ] = d.
5.6. Proposition. Let d 1.
d
d
Fp is the splitting subfield for each of the polynomials X p − X and X p −1 − 1
(i) Fpd
over Fp .
(ii) Fpd Fp is the unique subfield with pd elements.
(iii) If F is any field with pd elements then there is an monomorphism F −→ Fp with image
Fpd , hence F ∼
= Fpd .
Proof. (i) As Fpd consists of exactly the roots of Θpd (X) in Fp , it is the splitting subfield.
d
The nonzero elements of Fpd are the roots of X p −1 − 1, so Fpd is also the splitting subfield for
this polynomial.
(ii) Let F
Fp have pd elements. Notice that the nonzero elements of F form a group F ×
under multiplication. This group is abelian and has pd − 1 elements, so by Lagrange’s Theorem,
d
d
each element u ∈ F × has order dividing pd − 1, therefore up −1 = 1 and so up = u. But this
Fpd ; equality follows since these
means every element of F is a root of Θpd (X) and so F
subfields both have pd elements.
(iii) Apply the Monomorphism Extension Theorem 3.49 for K = L = Fp and M = F . By (ii),
the image of the resulting monomorphism must be Fpd , therefore F ∼
= Fpd .
It is worth noting the following consequence of this result and the construction of Fpd .
5.7. Corollary. Let K be a finite field of characteristic p. Then K/Fp is a finite Galois
extension.
5.8. Example. Consider the polynomial X 4 − X ∈ F2 [X]. By inspection, in the ring F2 [X]
we find that
X 4 − X = X 4 + X = X(X 3 + 1) = X(X + 1)(X 2 + X + 1).
Now X 2 + X + 1 has no root in F2 so it must be irreducible in F2 [X]. Its splitting field is a
quadratic extension F2 (w)/F2 where w is one of the roots of X 2 + X + 1, the other being w + 1
since the sum of the roots is the coefficient of X. This tells us that every element of F4 = F2 (w)
can be uniquely expressed in the form a + bw with a, b ∈ F2 . To calculate products we use the
fact that w2 = w + 1, so for a, b, c, d ∈ F2 we have
(a + bw)(c + dw) = ac + (ad + bc)w + bdw2 = (ac + bd) + (ad + bc + bd)w.
5.9. Example. Consider the polynomial X 9 − X ∈ F3 [X]. Let us find an irreducible
polynomial of degree 2 in F3 [X]. Notice that X 2 + 1 has no root in F3 , hence X 2 + 1 ∈ F3 [X] is
irreducible; so if u ∈ F3 is a root of X 2 + 1 then F3 (u)/F3 has degree 2 and F3 (u) = F9 . Every
element of F9 can be uniquely expressed in the form a + bu with a, b ∈ F3 . Multiplication is
carried out using the relation u2 = −1 = 2.
By inspection, in the ring F3 [X] we find that
X 9 − X = X(X 8 − 1) = (X 3 − X)(X 2 + 1)(X 2 + X − 1)(X 2 − X − 1).
So X 2 + X − 1 and X 2 − X − 1 are also quadratic irreducibles in F3 [X]. We can find their roots
in F9 using the quadratic formula since in F3 we have 2−1 = (−1)−1 = −1. The discriminant of
X 2 + X − 1 is
1 − 4(−1) = 5 = 2 = u2 ,
so its roots are (−1)(−1 ± u) = 1 ± u. Similarly, the discriminant of X 2 − X − 1 is
1 − 4(−1) = 5 = 2 = u2
and its roots are (−1)(1 ± u) = −1 ± u. Then we have
F9 = F3 (u) = F3 (1 ± u) = F3 (−1 ± u).
66
There are two issues we can now clarify.
5.10. Proposition. Let Fpm and Fpn be two Galois fields of characteristic p. Then Fpm
Fpn if and only if m  n.
Proof. If Fpm
Fpn , then by Corollary 5.2,
pn = (pm )[Fpn :Fpm ] = pm[Fpn :Fpm ] ,
so m  n.
If m  n, write n = km with k
n
mk
up = up
m
1. Then for u ∈ Fpm we have up = u, so
m
m(k−1)
= (up )p
Hence u ∈ Fpn and therefore Fpm
m(k−1)
= up
m
= · · · = up = u.
Fpn .
This means that we can think of the Galois fields Fpn as ordered by divisibility of n. The diagram of subfields for Fp24 can be seen in Figure 5.1 which shows extensions with no intermediate
subextensions.
Fp24
{{
{{
{
{
{{
Fp8
Fp4
Fp2
Fp12
{{
{{
{
{
{{
{{
{{
{
{
{{
Fp6
gg
gg
gg
gg
gg
gg
gg
gg
{{
{{
{
{
{{
Fp3
Fp
Figure 5.1. The subfields of Fp24
5.11. Theorem. The algebraic closure of Fp is the union of all the Galois fields of characteristic p,
Fp =
Fpn .
n 1
Furthermore, each element u ∈ Fp is separable over Fp .
Proof. Let u ∈ Fp . Then u is algebraic over Fp and the extension Fp (u)/Fp is finite. Hence
by Corollary 5.2, Fp (u) Fp is a finite subfield. Proposition 5.10 now implies that Fp (u) = Fpn
for some n. The separability statement follows from Corollary 5.7.
We will require a useful fact about Galois fields.
5.12. Proposition. The group of units F×
in Fpd is cyclic.
pd
This is a special case of a more general result about arbitrary fields.
5.13. Proposition. Let K be a field. Then every finite subgroup U
K × is cyclic.
Proof. Use Corollary 1.35 and Lemma 1.46.
5.14. Definition. w ∈ F×
is called a primitive root if it is a primitive (pd − 1)th root of
pd
unity, i.e., its order in the group F×
is (pd − 1), hence w = F×
.
pd
pd
67
5.15. Remark. Unfortunately the word primitive has two confusingly similar uses in the
context of finite fields. Indeed, some authors use the term primitive element for what we have
called a primitive root, but that conflicts with our usage, although as we will in the next result,
every primitive root is indeed a primitive element in our sense!
5.16. Proposition. The extension of Galois fields Fpnd /Fpd is simple, i.e., Fpnd = Fpd (u)
for some u ∈ Fpnd .
Proof. By Proposition 5.12, Fpnd has a primitive root w say. Then every element of
Fpnd can be expressed as a polynomial in w, so Fpnd
Fpd (w)
Fpnd . This implies that
Fpnd = Fpd (w).
5.17. Remark. This completes the proof of the Primitive Element Theorem 3.75 which we
had previously only established for infinite fields.
5.18. Example. In Example 5.8 we find that F4 = F2 (w) has the two primitive roots w
and w + 1.
5.19. Example. In Example 5.9 we have F9 = F3 (u) and F×
9 is cyclic of order 8. Since
ϕ(8) = 4, there are four primitive roots and these are the roots of the polynomials X 2 + X − 1
and X 2 − X − 1 which we found to be ±1 ± u.
We record a fact that is very important in Number Theory.
5.20. Proposition. Let p > 0 be an odd prime.
(i) If p ≡ 1 (mod 4), the polynomial X 2 + 1 ∈ Fp [X] has two roots in Fp .
(ii) If p ≡ 3 (mod 4) the polynomial X 2 +1 ∈ Fp [X] is irreducible, so Fp2 ∼
= Fp [X]/(X 2 +1).
×
Proof. (i) We have 4  (p − 1) = F×
p , so if u ∈ Fp is a generator of this cyclic group, the
×
×
order of uFp /4 is 4, hence this is a root of X 2 + 1 (the other root is −uFp /4 ).
×
(ii) If v ∈ Fp is a root of X 2 + 1 then v has order 4 in F×
p . But then 4  (p − 1) = Fp , which is
impossible since p − 1 ≡ 2 (mod 4).
Here is a generalization of Proposition 5.20.
5.21. Proposition. Fpd contains a primitive nth root of unity if and only if pd ≡ 1 (mod n)
and p n.
5.2. Galois groups of finite fields and Frobenius mappings
Consider an extension of Galois fields Fpnd /Fpd . By Proposition 5.6(i), Corollary 5.7 and
Proposition 3.73, this extension is Galois and
 Gal(Fpnd /Fpd ) = [Fpnd : Fpd ] = n.
We next introduce an important element of the Galois group Gal(Fpnd /Fpd ).
5.22. Definition. The (relative) Frobenius mapping for the extension Fpnd /Fpd is the funcd
tion Fd : Fpnd −→ Fpnd given by Fd (t) = tp .
5.23. Proposition. The relative Frobenius mapping Fd : Fpnd −→ Fpnd is an automorphism of Fpnd that fixes the elements of Fpd , so Fd ∈ Gal(Fpnd /Fpd ). The order of Fd is n, so
Gal(Fpnd /Fpd ) = Fd , the cyclic group generated by Fd .
Proof. For u, v ∈ Fpnd , we have the identities
d
d
d
Fd (u + v) = (u + v)p = up + v p ,
so Fd is a ring homomorphism. Also, for u ∈ Fpd we have
d
Fd (u) = up = u,
68
d
d
d
Fd (uv) = (uv)p = up v p ,
so Fd fixes the elements of Fpd . To see that Fd is an automorphism, notice that the composition
power Fnd = Fd ◦ · · · ◦ Fd (with n factors) satisfies
nd
Fnd (t) = tp
=t
n−1
for all t ∈ Fpnd , hence Fnd = id. Then Fd is invertible with inverse F−1
. This also shows
d = Fd
that the order of Fd in the group AutFpd (Fpnd ) is at most n. Suppose the order is k with k n;
kd
then every element u ∈ Fpnd satisfies the equation Fkd (u) = u which expands to up
u ∈ Fpkd . But this can only be true if k = n.
Frobenius mappings exist on the algebraic closure Fp . For d
= u, hence
1, consider the function
pd
Fd : Fp −→ Fp ;
Fd (t) = t .
5.24. Proposition. Let d 1.
(i) Fd : Fp −→ Fp is an automorphism of Fp which fixes the elements of Fpd . In fact for
u ∈ Fp , Fd (u) = u if and only if u ∈ Fpd .
(ii) The restriction of Fd to the Galois subfield Fpdn agrees with the relative Frobenius
mapping Fd : Fpnd −→ Fpnd .
(ii) If k 1, then Fkd = Fkd . Hence in the automorphism group AutFpd (Fp ), Fd has infinite
order, so AutFpd (Fp ) is infinite.
Proof. This is left as an exercise.
The Frobenius mapping F = F1 is often called the absolute Frobenius mapping since it exists
as an element of each of the groups AutFp (Fp ) and AutFp (Fpn ) = Gal(Fpn /Fp ) for every n 1.
In Gal(Fpnd /Fpd ) = Fd , for each k with k  n there is the cyclic subgroup Fkd of order
 Fkd  = n/k.
Fk
5.25. Proposition. For k  n, the fixed subfield of Fkd in Fpnd is Fpndd = Fpdk .
Fpnd
n/k
Fk
Fpndd = Fpdk
k
Fpd
dk
Proof. For u ∈ Fpnd we have Fkd (u) = up , hence Fkd (u) = u if and only if u ∈ Fpdk .
Figure 5.2 shows the subgroup diagram corresponding to the lattice of subfields of Fp24
shown in Figure 5.1.
5.3. The trace and norm mappings
For an extension of Galois fields Fpnd /Fpd , consider the function TFpnd /Fpd : Fpnd −→ Fpnd
defined by
d
2d
TFpnd /Fpd (u) = u + up + up
(n−1)d
+ · · · + up
= u + Fd (u) + F2d (u) + · · · + F(n−1)d (u).
Notice that
d
2d
+ up
d
2d
+ up
Fd (TFpnd /Fpd (u)) = up + up
= up + up
3d
+ · · · + up
nd
3d
+ · · · + up
69
(n−1)d
+ u = TFpnd /Fpd (u).
Gal(Fp24 /Fp ) = F ∼
= Z/24
lll
lll
l
l
l
lll
lll
F3
F2
ll
l
l
lll
lll
l
l
ll
lll
4
6
F
F
F8
F12
F24 = {id}
Figure 5.2. The subgroups of the Galois groups of Fp24 /Fp
So by Proposition 5.24(i), TFpnd /Fpd (u) ∈ Fpd . If we modify TFpnd /Fpd to have codomain Fpd ,
we obtain the relative trace
TrFpnd /Fpd : Fpnd −→ Fpd ;
d
2d
TrFpnd /Fpd (u) = u + up + up
(n−1)d
+ · · · + up
.
5.26. Proposition. The relative trace TrFpnd /Fpd is a surjective Fpd linear mapping and
whose kernel is an Fpd vector subspace of dimension n − 1.
d
Proof. Clearly TrFpnd /Fpd is additive. For t ∈ Fpd we have tp = t, so Fpd linearity follows
from the formula
d
2d
tu + (tu)p + (tu)p
(n−1)d
+ · · · + (tu)p
d
2d
= tu + tup + tup
(n−1)d
+ · · · + tup
.
To see that TrFpnd /Fpd is surjective, notice that TrFpnd /Fpd (u) = 0 if and only if u is a root of
the polynomial
d
2d
(n−1)d
X + Xp + Xp + · · · + Xp
∈ Fpd [X]
which has degree p(n−1)d and so has at most p(n−1)d < pnd roots in Fpnd . This means that
ker TrFpnd /Fpd cannot be the whole of Fpnd . TrFpnd /Fpd is surjective since its codomain has
dimension 1.
There is a multiplicative version of this construction. Consider the function
NFpnd /Fpd : F×
−→ F×
;
pnd
pd
d
2d
(n−1)d
NFpnd /Fpd (u) = uup up · · · up
= u Fd (u) F2d (u) · · · F(n−1)d (u).
Then we have
d
2d
3d
nd
d
2d
3d
(n−1)d
Fd (NFpnd /Fpd (u)) = up up up · · · up
= up up up · · · up
pd
= uu u
p2d
u
p3d
···u
u
p(n−1)d
= NFpnd /Fpd (u).
So by Proposition 5.24(i), NFpnd /Fpd (u) ∈ Fpd . By redefining the codomain we obtain the relative
norm
2d
(n−1)d
d
.
NormFpnd /Fpd : F×
−→ F×
; NormFpnd /Fpd (u) = uup up · · · up
pnd
pd
5.27. Proposition. The relative norm NormFpnd /Fpd is a surjective group homomorphism.
70
Proof. Multiplicativity is obvious. The kernel of NormFpnd /Fpd consists of the roots in Fpnd
of the polynomial
d
(n−1)d
X 1+p +···+p
− 1 ∈ Fpd [X],
so
pnd − 1
.
 ker NormFpnd /Fpd  1 + pd + · · · + p(n−1)d = d
p −1
Hence
pnd − 1
 im NormFpnd /Fpd  =
pd − 1.
 ker NormFpnd /Fpd 
Since im NormFpnd /Fpd
F×
, we also have
pd
 im NormFpnd /Fpd 
pd − 1,
therefore
im NormFpnd /Fpd = F×
.
pd
Exercises on Chapter 5
5.1. Show that Proposition 5.13 also applies to an integral domain in place of a field.
5.2. What happens to Theorem 5.20 if we try to take p = 2.
5.3. Let f (X) ∈ Fpd [X] be an irreducible polynomial with deg f (X) = n. Find the splitting field
of f (X). Deduce that for any other irreducible polynomial g(X) ∈ Fpd [X] with deg g(X) = n,
the splitting fields of f (X) and g(X) over Fpd agree.
5.4. Find the smallest Galois fields containing all the roots of the following polynomials, in
each case find a primitive root of this Galois field:
(a) X 8 − 1 ∈ F41 [X];
(b) X 8 − 1 ∈ F5 [X];
be a primitive root. If
5.5. Let w ∈ F×
pd
(c) X 8 − 1 ∈ F11 [X];
(d) X 8 − 1 ∈ F2 [X].
. Deduce that degFp w = d
< d, show that w ∈
/ F×
p
and d  ϕ(pd − 1).
5.6. Let p > 0 be a prime. Suppose that d 1, and K/Fpd is an extension. For a ∈ K, let
d
ga (X) = X p − X − a ∈ K[X].
(a) If the polynomial ga (X) is irreducible over K, show that the splitting field E of ga (X)
over K is separable and Gal(E/K) ∼
= Fpd . [Hint: show that if u ∈ E is a root of ga (X)
in an extension E/K, then so is u + t for every t ∈ Fp .]
(b) If d = 1, show that ga (X) is irreducible over K if and only if it has no root in K.
(c) If K is a finite field and d > 1, explain why ga (X) can never be irreducible over K.
5.7. Let p be an odd prime, d 1 and write q = pd .
(a) Consider {±1} = {1, −1} as a group under multiplication. Show that there is a unique
group homomorphism λq : F×
q −→ {±1} which is characterized by the requirement
that for every u ∈ F×
,
λ
(u)
= 1 if and only if u = v 2 for some v ∈ F×
q
q
q . Is λq always
surjective?
(b) Consider the set of all squares in Fq ,
Σq = {u2 ∈ Fq : u ∈ Fq } ⊆ Fq .
Show that the number of elements of Σq is Σq  = (q + 1)/2. Deduce that if t ∈ Fq then
the set
t − Σq = {t − u2 ∈ Fq : u ∈ Fq }
has t − Σq  = (q + 1)/2 elements.
71
(c) If t ∈ Fq , show that
Σq ∩ (t − Σq ) 1.
Deduce that every element of Fq is either a square or can be written as the sum of two
squares.
(d) Deduce that the equation x2 + y 2 + z 2 = 0 has at least one nontrivial solution in Fq .
(e) What can you say about the case p = 2?
72
CHAPTER 6
A Galois Miscellany
In this chapter we will explore some miscellaneous topics in Galois Theory. Historically,
Galois Theory has always been an important tool in Number Theory and Algebra, stimulating
the development of subjects such as Group Theory, Ring Theory and such diverse areas as Differential Equations, Complex Analysis and Algebraic Geometry. Many of the ideas introduced
in this chapter are of great importance in these and other mathematical areas.
6.1. A proof of the Fundamental Theorem of Algebra
We will prove the Fundamental Theorem of Algebra for the complex numbers C. This proof
is essentially due to Gauss but he did not use the historically more recent Sylow theory. It is
interesting to compare the proof below with others which use the topology of the plane and circle
or Complex Analysis; our proof only uses the connectivity of the real line (via the Intermediate
Value Theorem) together with explicit calculations in C involving square roots.
6.1. Theorem (The Fundamental Theorem of Algebra). The field of complex numbers C
is algebraically closed and R = C.
Proof. We know that [C : R] = 2, so C/R is algebraic. Let p(X) ∈ C[X] be irreducible.
Then any root u of p(X) in the algebraic closure C is algebraic over R, so in C[X] we have
p(X)  minpolyR,u (X). The splitting field of p(X) over C is contained in the splitting field E of
minpolyR,u (X)(X 2 + 1) over R. Since C E, we have 2  [E : R] and so 2   Gal(E/R).
Now consider a 2Sylow subgroup P
Gal(E/R) and recall that  Gal(E/R)/P  is odd.
For the fixed subfield of P , we have
[E P : R] =
 Gal(E/R)
,
P 
which shows that E P /R has odd degree. The Primitive Element Theorem 3.75 allows us to
write E P = R(v) for some v whose minimal polynomial over R must also have odd degree.
But by the Intermediate Value Theorem, every real polynomial of odd degree has a real root,
so irreducibility implies that v has degree 1 over R and therefore E P = R. This shows that
Gal(E/R) = P , hence Gal(E/R) is a 2group.
As C/R is a Galois extension, we can consider the normal subgroup Gal(E/C) Gal(E/R) for
which  Gal(E/R) = 2  Gal(E/C). We must show that  Gal(E/C) = 1, so suppose not. From
the theory of 2groups, there is a normal subgroup N Gal(E/C) of index 2, so we can consider
the Galois extension E N /C of degree 2. But from known properties of C (see Proposition 3.29),
every quadratic aX 2 + bX + c ∈ C[X] has complex roots (because we can find square roots
of every complex number). So we cannot have an irreducible quadratic polynomial in C[X].
Therefore  Gal(E/C) = 1 and E = C.
6.2. Cyclotomic extensions
We begin by discussing the situation for cyclotomic extensions over Q using material discussed in Section 1.3. Let ζn = e2πi/n , the standard primitive nth root of 1 in C. In Theorem 1.43, it was claimed that the irreducible polynomial over Q which has ζn as a root was the
73
nth cyclotomic polynomial
(X − ζnt ).
Φn (X) =
t=1,...,n−1
gcd(t,n)=1
6.2. Theorem. Let n 2. Then
• Q(ζn ) = Q[X]/(Φn (X));
• [Q(ζn ) : Q] = ϕ(n);
• Gal(Q(ζn )/Q) ∼
= (Z/n)× , where the element tn ∈ (Z/n)× acts on Q(ζn ) by tn · ζn = ζnt .
♠
♥ ♦
♣
Proof. Since the complex roots of Φn (X) are the powers ζnt with t = 1, . . . , n − 1
and gcd(t, n) = 1, Q(ζn ) is the splitting field of Φn (X) over Q and indeed Q(ζn ) = Q(ζnt )
whenever t has the above properties and so ζnt is a primitive nth root of unity. The main step
in the proof is to show that Φn (X) ∈ Z[X] is irreducible. To do this we will show that every
power ζnt as above is actually a Galois conjugate of ζn over Q, therefore
Φn (X) = minpolyQ,ζn (X) = minpolyQ,ζnt (X)
and hence Φn (X) is irreducible.
Consider
Z(ζn ) = {a0 + a1 ζn + · · · + ar ζnr : r 0, aj ∈ Z} ⊆ Q(ζn ).
Then Z(ζn ) is a subring of Q(ζn ) and so is an integral domain. Its group of units contains the
cyclic subgroup ζn of order n.
Let p > 0 be a prime which does not divide n. Let P Z(ζn ) be a maximal ideal which
contains p; then the quotient ring Z(ζn )/P is a field of characteristic p. In fact, it is a finite
field, say Fpd for some d. Let π : Z(ζn ) −→ Fpd be the quotient homomorphism.
Inside the group of units of Z(ζn ) is the subgroup of powers of ζn , ζn
Z(ζn )× ; this
is a cyclic subgroup of order n. We claim that when restricted to ζn , π gives an injective
. To see this, suppose that π (ζnr ) = 1 for some r =
group homomorphism, π : ζn −→ F×
pd
1, 2, . . . , n − 1; then ζnr − 1 ∈ P . By elementary Group Theory we can assume that r  n and so
p r. On factoring we have
(ζn − 1)(ζnr−1 + · · · + ζn + 1) ≡ (ζn − 1)r
(mod P ),
so ζn − 1 ∈ P or r ∈ P since maximal ideals are prime. But Z ∩ P = (p) and so r ∈
/ P , hence
ζn − 1 ∈ P . Recalling that
ζnn−1 + · · · + ζn + 1 = 0,
we see that n ∈ P and hence p  n, thus contradicting our original assumption on n. So π is
injective.
Writing u = π (u), we can consider the effect of the absolute Frobenius map F : Fpd −→ Fpd
t
on ζ n = ζnt ,
t
t
F(ζ n ) = (ζ n )p = ζntp .
t
This shows that in the Galois extension Fpd /Fp , ζ n is conjugate to ζntp ; by iterating this we find
t
k
that ζ n is conjugate to every power of the form ζntp .
Now let t = 1, . . . , n − 1 and gcd(t, n) = 1. Suppose there is a factorization
Φn (X) = f (X) minpolyQ,ζn (X)
for some monic polynomial f (X) ∈ Z[X] and f (ζnt ) = 0. Consider the prime power factorization
t = pr11 · · · prmm , where the pj are primes with 2
p1 < · · · < pm and rj
1 with. Since
gcd(t, n) = 1 we also have pj n s.
Now consider a maximal ideal P1 Z[ζn ] containing p1 . Reducing modulo P1 and working
r1
p
in the resulting extension Fpd1 /Fp1 , we find that ζ n is conjugate to ζ n1 . By separability and
1
the fact that the reduction map π1 : Z[ζn ] −→ Fpd1 is injective on the powers of ζn , we find that
1
74
r1
r1
p
r1
p
p
r1
p
f (ζn1 ) = 0 and so f (ζn1 ) = 0 in Z[ζn ]. This shows that minpolyQ,ζn (ζn1 ) = 0 and so ζn1 is
conjugate to ζn .
r1
p
Repeating this argument starting with ζn1 and using the prime p2 we find that
r1 r2
p2
p
minpolyQ,ζn (ζn1
r1 r2
p2
p
and so ζn1
)=0
is conjugate to ζn . Continuing in this fashion, for each j = 1, . . . , m we have
r
r1 r2
p2 ···pj j
p
minpolyQ,ζn (ζn1
)=0
r
r1
···p j
p
and so ζn1 j is conjugate to ζn . When j = m, this shows that minpolyQ,ζn (ζnt ) = 0. Hence
ζnt is conjugate to ζn in the extension Q(ζn )/Q.
6.3. Theorem. For n > 2, consider the cyclotomic extension Q(ζn )/Q where ζn = e2πi/n .
Then Q(ζn )R = Q(ζn ). Furthermore,
Q(ζn )R = Q(ζn )
( )
= Q(ζn + ζ n ) = Q(cos(2π/n)),
and
[Q(cos(2π/n)) : Q] =
ϕ(n)
.
2
Proof. Recall that
×
Gal(Q(ζn )/Q) ∼
= Z/n ,
where the residue class of r acts by sending ζn to ζnr . Complex conjugation corresponds to the
residue class of −1 ≡ n − 1 (mod n). Making use of the identities
eθi = cos θ + sin θ i,
1
cos θ = (eθi + e−θi ),
2
we obtain
1
1
cos(2π/n) = (ζn + ζ n ) = (ζn + ζn−1 ).
2
2
Complex conjugation fixes each of the real numbers cos(2πk/n) for k = 1, 2, . . . , n − 1. The
residue class of r acts by sending cos(2π/n) to cos(2πr/n); it is elementary to show that
cos(2πr/n) = cos(2π/n) unless r ≡ 1 (mod n). Hence
( ) = {id, ( )} = Gal(Q(cos(2π/n))/Q).
Thus we have
Q(ζn )
( )
= Q(cos(2π/n)),
and so [Q(cos(2π/n)) : Q] = ϕ(n)/2. Notice that ζn is a root of the polynomial
X 2 − 2 cos(2π/n)X + 1 ∈ Q(cos(2π/n))[X],
so we also have
(6.1)
minpolyQ(cos(2π/n)),ζn (X) = X 2 − 2 cos(2π/n)X + 1.
6.4. Example. We have
[Q(ζ24 ) : Q] = ϕ(24) = 8
and
Gal(Q(ζ24 )/Q) ∼
= Z/2 × Z/2 × Z/2.
75
Proof. By Theorem 1.43 we have [Q(ζ24 ) : Q] = 8. Also,
√
√
3
3
1
1
6
3
8
ζ24 = i, ζ24 = +
i, ζ24 = − +
i,
2
2
2
2
√ √
and all of these numbers are in Q(ζ24 ), hence Q( 2, 3, i) Q(ζ24 ). It is easy to check that
√ √
[Q( 2, 3, i) : Q] = 8,
which implies that
√ √
Q(ζ24 ) = Q( 2, 3, i).
Using this we find that
Gal(Q(ζ24 )/Q) ∼
= Z/2 × Z/2 × Z/2.
We also have cos(2π/24) = cos(π/12) ∈ Q(ζ24 ). Since
√
3
cos(2π/12) = cos(π/6) =
,
2
we have
√
3
2
2 cos (π/12) − 1 =
2
and so
3
4 cos4 (π/12) − 4 cos2 (π/12) + 1 = ,
4
giving
16 cos4 (π/12) − 16 cos2 (π/12) + 1 = 0.
Then
16X 4 − 16X 2 + 1 = 16 minpolyQ,cos(π/12) (X).
Note that case (i) of Kaplansky’s Theorem 4.28 applies to the polynomial minpolyQ,cos(π/12) (X).
For this example, Gal(Q(ζ24 )/Q) has 23 − 1 = 7 subgroups of each of the orders 2 and 4; it
is an interesting exercise to find them all together with their fixed subfields.
6.5. Remark. The minimal polynomial for cos(π/12) can also be found as follows. We have
Φ24 (ζ24 ) = 0, hence since
Φ24 (X) = X 8 − X 4 + 1,
we obtain
8
4
ζ24
− ζ24
+ 1 = 0.
−4
Then after multiplying by ζ24 we have
−4
4
= 0,
ζ24
− 1 + ζ24
giving
−4
4
) − 1 = 0.
(ζ24
+ ζ24
Now
−1 4
−4
−2
4
2
(ζ24 + ζ24
) = (ζ24
+ ζ24
) + 4(ζ24
+ ζ24
) + 6,
hence
−1 4
−2
−4
4
2
ζ24
+ ζ24
= (ζ24 + ζ24
) − 4(ζ24
+ ζ24
) − 6.
Similarly,
−1 2
−2
2
(ζ24 + ζ24
) = ζ24
+ ζ24
+ 2,
so
−2
−1 2
2
ζ24
+ ζ24
= (ζ24 + ζ24
) − 2.
Combining these we have
−1 4
−1 2
(ζ24 + ζ24
) − 4(ζ24 + ζ24
) + 1 = 0,
and so
16 cos4 (π/12) − 16 cos2 (π/12) + 1 = 0.
This method will work for any n where ϕ(n) is even, i.e., when n > 2.
76
6.6. Remark. The polynomial that expresses cos nθ as a polynomial in cos θ is the nth Chebsyhev polynomial of the first kind Tn (X) ∈ Z[X]. Here are the first few of these
polynomials:
T2 (X) = 2X 2 − 1,
T3 (X) = 4X 3 − 3X,
T4 (X) = 8X 4 − 8X 2 + 1,
T5 (X) = 16X 5 − 20X 3 + 5X,
T6 (X) = 32X 6 − 48X 4 + 18X 2 − 1,
T7 (X) = 64X 7 − 112X 5 + 56X 3 − 7X.
These form a system of orthogonal polynomials which can be computed in Maple using the
command orthopoly[T](n,X).
Now let K be a field with characteristic char K n. The polynomial Φn (X) has integer
coefficients, so we can view it as an element of K[X] since either Q
K or Fp
K and we
can reduce the coefficients modulo p. In either case it can happen that Φn (X) factors in K[X].
However, we can still describe the splitting field of X n − 1 over K and its Galois group.
6.7. Theorem. If char K n, then the splitting field of X n −1 over K is K(ζ), where ζ ∈ K
is a primitive nth root of unity. The Galois group Gal(K(ζ)/K) is isomorphic to a subgroup
of (Z/n)× , hence it is abelian with order dividing ϕ(n).
Proof. Working in K, we know that Φn (ζ) = 0, hence the roots of minpolyK,ζ (X) ∈ K[X]
are primitive roots of 1. So X n − 1 splits over K(ζ) and each element α ∈ Gal(K(ζ)/K) has
the action α(ζ) = ζ rα , where gcd(rα , n) = 1. Hence Gal(K(ζ)/K) is isomorphic to a subgroup
of Gal(Q(ζn )/Q) ∼
= (Z/n)× which implies that it is abelian and its order divides ϕ(n).
6.8. Remark. When p = char K > 0, this Galois group only depends on the largest subfield
of K which is algebraic over Fp . For example, if K = Fpd (T ) then the value of d is the crucial
factor. The precise outcome can be determined with the aid of Proposition 5.21.
6.9. Example. We have the following splitting fields and Galois groups.
(i) The splitting field of X 4 − 1 over F3 (T ) is F9 (T ) and
Gal(F9 (T )/F3 (T )) ∼
= Z/2.
= (Z/4)× ∼
(ii) By Proposition 5.20, X 4 − 1 splits over F5 (T ) and the Galois group Gal(F5 (T )/F5 (T ))
is trivial.
Proof. (i) By Proposition 5.20, X 4 − 1 is separable over F3 (T ) and has irreducible factors
(X − 1), (X + 1) and (X 2 + 1). The splitting field of (X 2 + 1) over F3 is F9 = F3 (ζ), where
ζ 2 + 1 = 0, so (X 2 + 1) splits over F9 (T ). Also,
Gal(F9 /F3 ) ∼
= (Z/4)× ∼
= Z/2,
with generator σ satisfying σ(ζ) = ζ −1 = −ζ. This generator clearly extends to an automorphism of F9 (T ) which fixes T .
(ii) By Proposition 5.20, X 4 − 1 splits over F5 .
6.3. Artin’s Theorem on linear independence of characters
Let G be a group and K a field.
6.10. Definition. A group homomorphism χ : G −→ K × is called a character of G with
values in K.
6.11. Example. Given any ring homomorphism ϕ : R −→ K we obtain a character of R×
in K by restricting ϕ to a map χϕ : R× −→ K × .
6.12. Example. Given an automorphism α : K −→ K, χα : K × −→ K × is a character of
K × in K.
6.13. Example. Let E/K be a Galois extension and σ ∈ Gal(E/K). Then χσ : E × −→ E ×
is a character.
77
6.14. Definition. Let χ1 , . . . , χn be characters of a group G in a field K. Then χ1 , . . . , χn
are linearly independent if for t1 , . . . , tn ∈ K,
t1 χ1 + · · · + tn χn = 0
=⇒
t1 = · · · = tn = 0.
If χ1 , . . . , χn are not linearly independent then they are linearly dependent.
In this definition, the functional equation means that for all g ∈ G,
t1 χ1 (g) + · · · + tn χn (g) = 0.
6.15. Theorem (Artin’s Theorem). Let χ1 , . . . , χn be distinct characters of a group G in a
field K. Then χ1 , . . . , χn are linearly independent.
Proof. We proceed by induction on n. For n = 1 the result is easily verified. For the
inductive assumption, suppose that it holds for any n k.
Let χ1 , . . . , χk+1 be a set of k + 1 distinct characters for which there are t1 , . . . , tk+1 ∈ K
not all zero and such that
t1 χ1 + · · · + tk+1 χk+1 = 0.
(6.2)
If one of the ti is zero, say tr = 0, then χ1 , . . . , χr−1 , χr+1 , . . . , χk+1 is linearly dependent,
contradicting the inductive assumption. Hence all of the ti must be nonzero. As χ1 = χ2 ,
there must be an element g0 ∈ G for which χ1 (g0 ) = χ2 (g0 ). So for all g ∈ G, Equation (6.2)
applied to g0 g yields
t1 χ1 (g0 g) + · · · + tk+1 χk+1 (g0 g) = 0,
and therefore since χj (g0 g) = χj (g0 )χj (g), we see that
t1 χ1 (g0 )χ1 + · · · + tk+1 χk+1 (g0 )χk+1 = 0.
Multiplying Equation (6.2) by χ1 (g0 ) and subtracting gives
t2 (χ2 (g0 ) − χ1 (g0 ))χ2 + t3 (χ3 (g0 ) − χ1 (g0 ))χ3 + · · · + tk+1 χk+1 = 0,
in which the coefficient t2 (χ2 (g0 )−χ1 (g0 )) is not zero. Hence χ2 , . . . , χk+1 is linearly dependent,
again contradicting the inductive assumption. So χ1 , . . . , χk+1 is linearly independent, which
demonstrates the inductive step.
6.16. Corollary. Suppose that α1 , . . . , αn are distinct automorphisms of the field K. Let
t1 , . . . , tn ∈ K be a sequence of elements, not all of which are 0. Then there is a z ∈ K for
which
t1 α1 (z) + · · · + tn αn (z) = 0.
Hence the Klinear transformation t1 α1 + · · · + tn αn : K −→ K is nontrivial.
6.17. Corollary. Let E/K be a finite Galois extension of degree n and let α1 , . . . , αn
be the distinct elements of Gal(E/K). Then the function α1 + · · · + αn : E −→ E is a nontrivial Klinear transformation whose image is contained in K. Hence the associated Klinear
transformation
TrE/K : E −→ K; TrE/K (x) = α1 (x) + · · · + αn (x)
is surjective.
The function TrE/K : E −→ K is called the trace mapping of E/K.
Proof. First note that for x ∈ E and γ ∈ Gal(E/K),
γ(α1 (x) + · · · + αn (x)) = γα1 (x) + · · · + γαn (x) = α1 (x) + · · · + αn (x),
since the list γα1 , . . . , γαn is the same as α1 , . . . , αn apart from its order. Hence,
α1 (x) + · · · + αn (x) ∈ E Gal(E/K) = K.
The rest of the statement follows directly from Corollary 6.17.
78
Suppose that E/K is a finite Galois extension with cyclic Galois group Gal(E/K) = σ of
order n. For each u ∈ E × , the element uσ(u) · · · σ n−1 (u) ∈ E satisfies
σ(uσ(u) · · · σ n−1 (u)) = σ(u) · · · σ n−1 (u)σ n (u) = σ(u) · · · σ n−1 (u)u,
hence in uσ(u) · · · σ n−1 (u) ∈ E
σ
×
= K. Now using this we define a group homomorphism
NE/K : E −→ K × ;
NE/K (u) = uσ(u) · · · σ n−1 (u).
NE/K is called the norm mapping for E/K and generalizes the norm mapping for finite fields
of Section 5.3.
There is another homomorphism
δE/K : E × −→ E × ;
δE/K (u) = uσ(u)−1 .
Notice that for u ∈ E × ,
NE/K (δE/K (u)) = (uσ(u)−1 )(σ(u)σ 2 (u)−1 · · · σ n−1 (u)σ n (u)−1 ) = 1,
since σ n (u) = u. So im δE/K
Proposition 5.27.
ker NE/K . Our next result is an important generalization of
6.18. Theorem (Hilbert’s Theorem 90). Let E/K be a finite Galois extension with cyclic
Galois group Gal(E/K) = σ of order n. Then im δE/K = ker NE/K . Explicitly, if u ∈ E × and
u σ(u) · · · σ n−1 (u) = 1, then there is a v ∈ E × such that u = vσ(v)−1 .
Proof. Let u ∈ ker NE/K .
The characters σ k : E × −→ E × with k = 0, 1, . . . , n−1 are distinct and linearly independent
by Artin’s Theorem 6.15. Consider the function
id +uσ + uσ(u)σ 2 + · · · + uσ(u) · · · σ n−2 (u)σ n−1 : E × −→ E.
This cannot be identically zero, so for some w ∈ E, the element
v = w + uσ(w) + uσ(u)σ 2 (w) + · · · + uσ(u) · · · σ n−2 (u)σ n−1 (w)
is nonzero. Notice that
uσ(v) = uσ(w) + uσ(u)σ 2 (w) + uσ(u)σ 2 (u)σ 3 (w) + · · · + uσ(u)σ 2 (u) · · · σ n−1 (u)σ n (w) = v,
since
Thus we have u = vσ(v)−1
uσ(u)σ 2 (u) · · · σ n−1 (u)σ n (w) = w.
as required.
6.4. Simple radical extensions
In this section we will investigate splitting fields of polynomials of the form X n − a, where
char K n. We call these simple radical extensions and later in Definition 6.33 we introduce a
more general notion of radical extension.
6.19. Proposition. Let f (X) = X n − a ∈ K[X] be irreducible and separable over K. Then
the splitting field of f (X) over K has the form K(u, ζ), where u is a root of f (X) and ζ is a
primitive nth root of 1.
6.20. Corollary. If K contains a primitive nth root of 1, ζ, then the splitting field of
f (X) = X n − a over K has the form K(u), where u is a root of f (X). The Galois group
Gal(K(u)/K) is cyclic of order n with a generator σ for which σ(u) = ζu.
In the more general situation of Proposition 6.19,
{id} Gal(K(ζ, u)/K(ζ)) Gal(K(ζ, u)/K),
where Gal(K(ζ, u)/K(ζ)) is cyclic and
Gal(K(ζ)/K)) ∼
= Gal(K(ζ, u)/K))/ Gal(K(ζ, u)/K(ζ))
79
is abelian. The Galois Correspondence identifies the following towers of subfields and subgroups.
K(ζ, u)
Gal(K(ζ, u)/K)
K(ζ)
Gal(K(ζ, u)/K(ζ))
a
K
7
8
f
}
'
{id}
6.21. Definition. Let K be a field with char K n and which contains a primitive nth
root of 1, ζ say. Then L/K is a simple nKummer extension if L = K(u) where un = a
for some a ∈ K. L/K is an (iterated ) nKummer extension if L = K(u1 , . . . , uk ) where
un1 = a1 , . . . , unk = ak for some elements a1 , . . . , ak ∈ K.
Note that in this definition we do not require the polynomials X n − aj ∈ K[X] to be
irreducible.
6.22. Proposition. Let K(u)/K be a simple nKummer extension. Then K(u)/K is a
Galois extension and Gal(K(u)/K) is cyclic with order dividing n.
Proof. Suppose that un = a ∈ K. Then in K[X] we have
X n − a = (X − u)(X − ζu) · · · (X − ζ n−1 u).
Clearly the roots of X n − a are distinct and so K(u)/K is separable over K; in fact, K(u) is a
splitting field of X n − a over K. This means that K(u)/K is Galois.
For each α ∈ Gal(K(u)/K) we have α(u) = ζ rα u for some rα = 0, 1 . . . , n − 1. Notice that
for β ∈ Gal(K(u)/K),
βα(u) = β(ζ rα u) = ζ rα β(u) = ζ rα ζ rβ u = ζ rα +rβ u,
and so rβα = rα + rβ . Hence the function
ρ : Gal(K(u)/K) −→ ζ ;
ρ(α) = ζ rα ,
is a group homomorphism. As ζ is cyclic of order n, Lagrange’s Theorem implies that the
image of ρ has order dividing n. Since every element of Gal(K(u)/K) is determined by its effect
on u, ρ is injective, hence  Gal(K(u)/K) divides n. In fact, Gal(K(u)/K) is cyclic since every
subgroup of a cyclic group is cyclic.
√
6.23. Example. Let n
1 and q ∈ Q. Then Q(ζn , n q)/Q(ζn ) is a simple nKummer
extension.
√
√
6.24. Example. Q(i, 2)/Q(i) is a simple 4Kummer extension with Gal(Q(i, 2)/Q(i))
cyclic of order 2.
√
Proof. We have ( 2)4 − 4 = 0, but
X 4 − 4 = (X 2 − 2)(X 2 + 2),
and
X 2 − 2 = minpolyQ(i),√2 (X).
√
The corresponding group homomorphism ρ : Gal(Q(i)( 2)/Q(i)) −→ i has image
im ρ = {1, −1}
Here is a converse to Proposition 6.22.
80
i .
6.25. Proposition. Suppose that char K n and there is an element ζ ∈ K which is a
primitive nth root of unity. If E/K is a finite Galois extension with cyclic Galois group of
order n, then there is an element a ∈ E such that E = K(a) and a is a root of a polynomial of
the form X n − b with b ∈ K. Hence E/K is a simple nKummer extension.
Proof. We have
NE/K (ζ −1 ) = ζ −n = 1,
so by Hilbert’s Theorem 6.18, there is an element a ∈ E for which ζ −1 = aσ(a)−1 . Then
σ(a) = ζa and the elements σ k (a) = ζ k a for k = 0, 1, . . . , n − 1 are distinct, so they must be
the n conjugates of a. Also note that
X n − an = (X − a)(X − ζa) · · · (X − ζ n−1 a) = (X − a)(X − σ(a)) · · · (X − σ n−1 (a)),
hence an ∈ K since it is fixed by σ. Since K(a)
E, this shows that
n = [K(a) : K]
[E : K] = n
and therefore
[K(a) : K] = [E : K] = n,
whence K(a) = E.
6.5. Solvability and radical extensions
We begin by recalling some ideas about groups, see [3, 4] for further details.
6.26. Definition. A group G is solvable, soluble or soluable if there is a chain of subgroups
(called a subnormal series)
{1} = G
G
−1
···
G1
G0 = G
in which Gk+1 Gk and each composition factor Gk /Gk+1 is abelian; we usually write
{1} = G
G
−1
· · · G1 G0 = G.
If each composition factor is a cyclic group of prime order the subnormal series is called a
composition series. A group which is not solvable is called insolvable.
6.27. Remark. It is a standard result that we can always refine (i.e., add extra terms) a
subnormal series of a solvable group to obtain a composition series. The primes appearing as
well as the number of times each occurs are all determined by G, only their order varying for
different composition series.
6.28. Example. Let G be a finite abelian group. Then G is solvable.
6.29. Example. Let G be a finite pgroup, where p is a prime. Then G is solvable.
In fact, for a finite pgroup G, there is always a normal subgroup of a pgroup with index
p, so in this case we can assume each quotient Gk /Gk+1 is cyclic of order p.
6.30. Proposition. Let G be a group.
(i) If G is solvable then every subgroup H G and every quotient group G/N is solvable.
(ii) If N G and G/N are solvable then so is G.
In the opposite direction we can sometimes see that a group is insolvable. Recall that a
group is simple if it has no nontrivial proper normal subgroups.
6.31. Proposition. Let G be a finite group. Then G is insolvable if any of the following
conditions holds:
(i) G contains a subgroup which is a nonabelian simple group (or has a quotient group
which is a nonabelian simple group).
(ii) G has a quotient group which is a nonabelian simple group.
(iii) G has a composition series in which one of the terms is a nonabelian simple group.
81
6.32. Example. For n
5, the alternating and symmetric groups An and Sn are insolvable.
Proof. This follows from the fact that if n
quotient group Sn /An ∼
= Z/2.
5, An is a simple group and An
Sn with
Now we explain how this relates to fields and their extensions. Let K be a field and L/K a
finite extension. For simplicity, we assume also that char K = 0.
6.33. Definition. L/K is a radical extension of K if it has the form L = K(a1 , a2 , . . . , an )
with
adkk ∈ K(a1 , a2 , . . . , ak−1 )
for some dk 1. Thus every element of L is expressible in terms of iterated roots of elements
of K.
6.34. Definition. If L is the splitting field of a polynomial f (X) ∈ K[X], then f (X) is
solvable by radicals over K if L is contained in a radical extension of K.
6.35. Definition. L/K is solvable if L
of K.
L where L /K is a finite radical Galois extension
6.36. Theorem. Let E/K be a finite Galois extension. Then E/K is solvable if and only
if the group Gal(E/K) is solvable.
♠
♥ ♦
♣
Proof. Suppose that E
E where E /K is a radical Galois extension. Then
Gal(E/K) is a quotient group of Gal(E /K), so it is solvable by Proposition 6.30.
Now suppose that Gal(E/K) is solvable and let n =  Gal(E/K). Let E be the splitting
field of X n − 1 over E, so E contains a primitive nth root of unity ζ and therefore it contains
a primitive dth root of unity for every divisor d of n. Now Gal(E /E) Gal(E /K) and
by Theorem 6.7, Gal(E /E) is abelian. Also, Gal(E /K)/ Gal(E /E) ∼
= Gal(E/K) which is
solvable, so Gal(E /K) is solvable by Proposition 6.30. We will now show that E /K is a
radical extension.
Clearly K(ζ)/K is radical. Then Gal(E /K(ζ)) Gal(E /K) is solvable. Let
{1} = G
G
−1
· · · G1 G0 = Gal(E /K(ζ))
be a composition series. The extension (E )G1 /K(ζ) is radical by Proposition 6.25. Similarly,
each extension (E )Gk+1 /(E )Gk is radical. Hence E /K(ζ) is radical, as is E /K.
√
6.37. Example. The Galois group of the extension Q(ζ3 , 3 2)/Q is solvable.
√
Proof. We have already studied this extension in Example 3.30 and 4.20. Clearly Q(ζ3 , 3 2)
is a radical extension of Q and
√
√
3
3
Q(ζ3 , 2) = Q(ζ3 )( 2).
√
We know that Gal(Q(ζ3 , 3 2)/Q) ∼
each element of the Galois group with
= S3 , where we identify
√
3
3
a permutation of the three roots of X − 2 in Q(ζ3 , 2) which we list in the order
√
√
√
3
3
3
2, 2 ζ3 , 2 ζ32 .
We have the following towers of subfields and subgroups related under the Galois Correspondence.
√
5 S3
Q(ζ3 , 3 2)
e
3
2
Q(ζ3 ) = Q(ζ3 ,
√
3
i
2)A3
A3 = Gal(Q(ζ3 ,
5
2
Q
3
y
)
82
{id}
√
3
2)/Q(ζ3 ))
Here Q(ζ3 )/Q is itself a Galois extension and A3 S3 . Notice that A3 ∼
= Z/2,
= Z/3 and S3 /A3 ∼
so we have the following composition series for S3 :
{id} A3 S3 .
It is also interesting to reverse the question and ask whether there are extensions which are
not solvable. This was a famous problem pursued for several hundred years. To find examples,
we first recall that the smallest nonabelian simple group is A5 which has order 60. We should
therefore expect to look for a polynomial of degree at least 5 to find a Galois group for a splitting
field to be simple or occur as a composition factor of such a Galois group. Here is an explicit
example over Q.
6.38. Example. The splitting field of the polynomial f (X) = X 5 − 35X 4 + 7 ∈ Q[X] is not
solvable.
Proof. Let E
C be the splitting field of f (X) over Q. Using the Eisenstein Test 1.38
with p = 7, we find that f (X) is irreducible over Q. By Theorem 4.8(iii), 5 divides the order of
Gal(E/Q), so by Cauchy’s Lemma this group contains an element of order 5.
Now observe that
f (X) = 5X 4 − 140X 3 = 5X 3 (X − 28),
f (X) = 20X 4 − 420X 2 = 20X 2 (X − 21).
There are two turning points, namely a maximum at x = 0 and a minimum at x = 28. Then
f (0) = 7 > 0 > f (28) = −4302585,
hence there are three real roots of f (X) and two nonreal complex ones. Then complex conjugation restricts to an element of order 2 in Gal(E/Q) which interchanges the nonreal roots and
fixes the others. If we list the roots of f (X) as u1 , u2 , u3 , u4 , u5 with u1 , u2 being the nonreal
roots, then the transposition (1 2) ∈ S5 corresponds to this element. Furthermore, the only
elements of S5 of order 5 are 5cycles; by taking an appropriate power we can assume that there
is a 5cycle of the form (1 2 3 4 5) corresponding to an element of Gal(E/Q) which we can view
as a subgroup of S5 . The next lemma shows that Gal(E/Q) ∼
= S5 .
6.39. Lemma. Let n
1. Suppose that H
(1 2 · · · n). Then H = Sn .
Sn and H contains the elements (1 2) and
The proof is left as an exercise. This completes the verification of Example 6.38.
It is worth remarking that the most extreme version of this occurs when we ask for a Galois
group which is simple. There has been a great deal of research activity on this question in the
past few decades, but apparently not all simple groups are known to occur as Galois groups of
extensions of Q or other finite subextensions of C/Q. Here is an example whose Galois group
is A5 ; this is verified using Proposition 4.26.
6.40. Example. The Galois group of f (X) = X 5 + 20X + 16 over Q is Gal(Q(f (X))/Q) ∼
=
A5 , hence it is not solvable.
6.6. Symmetric functions
Let k be a field. Consider the polynomial ring on n indeterminates k[X1 , . . . , Xn ] and its
field of fractions K = k(X1 , . . . , Xn ). Each permutation σ ∈ Sn acts on k[X1 , . . . , Xn ] by
σ · f (X1 , . . . , Xn ) = f σ (X1 , . . . , Xn ) = f (Xσ(1) , . . . , Xσ(n) ).
Viewed as a function σ· : k[X1 , . . . , Xn ] −→ k[X1 , . . . , Xn ] is a ring isomorphism; this extends to
a ring isomorphism σ· : k(X1 , . . . , Xn ) −→ k(X1 , . . . , Xn ). Varying σ we obtain actions of the
group Sn on k[X1 , . . . , Xn ] and k(X1 , . . . , Xn ) by ring isomorphisms fixing k and in the latter
case it is by field automorphisms fixing k.
83
6.41. Definition. The field of symmetric functions on n indeterminates is
Symn (k) = k(X1 , . . . , Xn )Sn
k(X1 , . . . , Xn ).
So if f (X1 , . . . , Xn ) ∈ k(X1 , . . . , Xn ), then
f (X1 , . . . , Xn ) ∈ Symn (k)
⇐⇒
∀σ ∈ Sn f (X1 , . . . , Xn ) = f (Xσ(1) , . . . , Xσ(n) ).
6.42. Theorem. The extension k(X1 , . . . , Xn )/ Symn (k) is a finite Galois extension for
which Gal(k(X1 , . . . , Xn )/ Symn (k)) ∼
= Sn .
Proof. There are elements of k[X1 , . . . , Xn ] ⊆ k(X1 , . . . , Xn ) called elementary symmetric
functions,
Xi1 Xi2 · · · Xik ,
ek =
i1 0 be a prime and G a group of order G = pn for some n 1. Show by induction
on n that there is a normal subgroup N G with N  = pn−1 . [Hint: what do you know about
the centre of G? Use this information to produce a quotient group of smaller order than G.]
6.2. Let K be a field for which char K = 2 and n 1 be odd. If K contains a primitive nth
root of unity, show that then K contains a primitive 2nth root of unity.
6.3. Find all values of n
in the following fields:
1 for which ϕ(n)  4. Using this, determine which roots of unity lie
√
√
√
Q(i), Q( 2 i), Q( 3 i), Q( 5 i).
6.4. (a) Describe the elements of (Z/24)× explicitly and verify that this group is isomorphic
to Z/2 × Z/2 × Z/2. Describe the effect of each element on Q(ζ24 ) and Q(cos(π/12)) under the
action described in Theorem 6.2.
(b) Determine the group (Z/20)× and describe the effect of each of its elements on Q(ζ20 ) and
Q(cos(π/10)) under the action described in Theorem 6.2.
6.5. Let n
1.
84
(a) What can you say about sin(2π/n) and Gal(Q(sin(2π/n))/Q))?
(b) Determine sin(π/12) and Gal(Q(sin(π/12))/Q)).
6.6. In this question, work in the cyclotomic field Q(ζ5 ) where ζ5 = e2πi/5 .
(a) Describe the Galois group Gal(Q(ζ5 )/Q) and its action on Q(ζ5 ).
(b) Determine the minimal polynomial of cos(2π/5) over Q. Hence show that
√
−1 + 5
.
cos(2π/5) =
4
For which other angles θ is cos θ a root of this minimal polynomial? What is the value
of sin(2π/5) ?
(c) Find the tower of subfields of Q(ζ5 ) and express them as fixed fields of subgroups of
Gal(Q(ζ5 )/Q).
6.7. In this question, let p be an odd prime and let ζp = e2πi/p ∈ Q(ζp )
(a) Consider the product
C.
(p−1)/2
(ζpr − ζp−r ) ∈ Q(ζp ).
ξ=
r=1
Show that
p−1
2
(p−1)/2
(1 − ζpr ).
ξ = (−1)
r=1
(b) Deduce that
ξ2 =
p
−p
if p ≡ 1
if p ≡ 3
(mod 4),
(mod 4).
(c) Conclude that
√
if p ≡ 1 (mod 4),
± p
√
± p i if p ≡ 3 (mod 4).
√
√
and also p ∈ Q(ζp ) if p ≡ 1 (mod 4) and p i ∈ Q(ζp ) if p ≡ 3 (mod 4).
ξ=
6.8. Prove Lemma 6.39. [Hint: show that every 2cycle of the form (i i + 1) is in H by
considering elements of the form (1 2 · · · n)r (1 2)(1 2 · · · n)n−r .]
6.9. This question is about an additive version of Hilbert’s Theorem 90, see Theorem 6.18.
Let E/K be a Galois extension with cyclic Galois group Gal(E/K) = σ of order n.
(a) Show that the function
T : E −→ E;
T (u) = u + σ(u) + σ 2 (u) + · · · + σ n−1 (u),
takes values in K and use this to define a Klinear mapping TrE/K : E −→ K.
(b) If v ∈ E has TrE/K (v) = 0, show that there is a w ∈ E such that v = w − σ(w).
[Hint: Show that there is an element t ∈ E for which TrE/K t = 0, then consider
w=
1
(TrE/K t)
vσ(t) + (v + σ(v))σ 2 (t) + · · · + (v + σ(v)σ 2 (t) + · · · + σ n−2 (v))σ n−1 (t)
and adapt the proof of Hilbert’s Theorem 90 in Theorem 6.18, using TrE/K in place of NE/K .]
6.10. (a) For n
1 and 1
k
n, the kth power sum sk ∈ k[X1 , . . . , Xn ]Sn is defined by
Xik .
sk =
1 i n
Prove the formula
sk = e1 sk−1 − e2 sk−2 + · · · + (−1)k−1 ek−1 s1 + (−1)k kek .
85
(b) For n
1 and 1
k
n, the total symmetric function is defined by
Xj1 Xj2 · · · Xjk ,
hk =
j1 j2 ··· jk
i.e., the sum of all the monomials in the Xi of degree k.
(i) For large values of n, express h1 , h2 , h3 in terms of the elementary symmetric functions
e1 , e2 , e3 .
(ii) Show that the power sum functions sk of the previous question satisfy
sk = −(h1 sk−1 + h2 sk−2 + · · · + hk−1 s1 ) + khk .
86
Bibliography
[1] E. Artin, Galois Theory, Dover Publications (1998); ISBN 0 486 62342 4.
[2] JP. Escofier, Galois theory, SpringerVerlag, New York (2001); ISBN 0387987657. [Highly recommended,
especially for its historical notes]
[3] J. B. Fraleigh, A First Course in Abstract Algebra, Addison Wesley (1999); ISBN 0 201 33596 4. [Highly
recommended ]
[4] S. Lang, Algebra, Addison Wesley (1993); ISBN 0 201 55540 9.
[5] R. Lidl & H. Niederreiter, Finite Fields, Cambridge University Press (1997); ISBN 0 521 39231 4.
[6] J. Rotman, Galois Theory, SpringerVerlag (1998); ISBN 0 387 98541 7.
[7] I. Stewart, Galois Theory, Chapman and Hall (1989); ISBN 0 412 345501. [Very highly recommended.]
87
Solutions
Chapter 1
1.1. Clearly {n ∈ Z : n > 0 and nr = 0 for all r ∈ R} ⊆ {n ∈ Z : n > 0 and n1 = 0}. If
0 < n ∈ Z and n1 = 0, then for every r ∈ R,
nr = r + · · · + r = (1 + · · · + 1)r = (n1)r = 0r = 0,
n
n
so {n ∈ Z : n > 0 and n1 = 0} ⊆ {n ∈ Z : n > 0 and nr = 0 for all r ∈ R}. Hence these sets
are in fact equal. When char R = p > 0 they must both be nonempty. Now by definition of
characteristic,
char R = min{n ∈ Z : n > 0 and n1 = 0} = min{n ∈ Z : n > 0 and nr = 0 for all r ∈ R}.
1.2. (a) Let u, v ∈ S and suppose that uv = 0; then u = 0 or v = 0 since u, v ∈ R and R is
an integral domain. Consider the unit homomorphisms η : Z −→ R and η : Z −→ S. Then for
n ∈ Z, η (n) = η(n), so ker η = ker η and therefore char S = char R.
(b) Q is a field and Z ⊆ Q is a subring which is not a field.
1.3. (a) For any subring R ⊆ C, R is an integral domain with characteristic subring Z and
char R = 0.
(b) The characteristic subring of A[X] is the same as that of A and char A[X] = char A. A[X]
is an integral domain if and only if A is an integral domain.
(c) If we identify A with the subring of scalar matrices in Matn (A), then the characteristic
subring of Matn (A) is the same as that of A and char Matn (A) = char A. If n > 1 then
Matn (A) is not commutative, in any case it always has zerodivisors since any singular matrix
is a zerodivisor.
1.4. The main thing to check is that ϕ(u + v) = ϕ(u) + ϕ(v) which is a consequence of the
Idiot’s Binomial Theorem. For R = Fp [X], ϕ is not surjective, while for R = Fp [X]/(X 2 ), ϕ is
not injective.
1.5. (a) Recall from the Isomorphism Theorems of basic Ring Theory that ϕ−1 Q R; we need
to show it is a prime ideal. Suppose that u, v ∈ R with uv ∈ ϕ−1 Q; then ϕ(u)ϕ(v) = ϕ(uv) ∈ Q
and so ϕ(u) ∈ Q or ϕ(v) ∈ Q, hence u ∈ ϕ−1 Q or v ∈ ϕ−1 Q.
(b) Consider the inclusion function inc : R −→ S; then inc−1 Q = Q ∩ R, so this result follows
from (a).
(c) Consider Z ⊆ Q; then the zeroideal (0)Q Q has (0)Q ∩ Z = (0)Z Z but this is not maximal
in Z since for any prime p > 0, (p)Z Z is a (maximal) ideal that properly contains (0)Z .
(d) We have P ⊆ Q ∩ R R with P R maximal; so P ⊆ Q ∩ R. In fact Q only needs to be a
proper ideal of S for this argument to work.
1.6. The only proper ideal of k is the zero ideal (0), so ker ϕ = (0).
1.7. (a) Addition and multiplication are given by the obvious formulæ,
(u1 +v1 i)+(u2 +v2 i) = (u1 +u2 )+(v1 +v2 )i,
(u1 +v1 i)(u2 +v2 i) = (u1 u2 −v1 v2 )+(u1 v2 +u2 v1 )i,
with Z[i] and Q[i] both closed under these operations and containing 1 = 1 + 0i as a unity, so
they are subrings of the field C; by Qu. 1.1, they are both integral domains. To see that Q[i] is
a field, notice that if u + vi = 0 with u, v ∈ Q,
(u − vi)(u + vi) = (u + vi)(u − vi) = u2 + v 2 = 0,
so
u
v
+ 2
i ∈ Q(i)
2
+v
u + v2
is the inverse of u + vi. Hence every nonzero element of Q[i] has an inverse, therefore Q[i] is a
field.
1
(b) & (c) The crucial point is that every element of Q[i] can be written as (u + vi) with
n
n, u, v ∈ Z and n = 0. Then
u2
inc∗
(u + vi)
n
(u + vi)
n + 0i
= inc∗
1
(u + vi),
n
=
so the latter element is in the image of inc∗ which must therefore be a surjection.
1.8. (a) Existence and uniqueness of such an ψa,b follow from the Homomorphism Extension
Property 1.22 and its effect on f (X) = ni=0 ri X i ∈ R[X] where ri ∈ R is
n
ri (aX + b)i .
ψa,b (f (X)) = f (aX + b) =
i=0
We have
ψa,b ◦ ψc,d (X) = ψa,b (cX + d) = c(aX + b) + d = caX + (cb + d) = ψca,cb+d (X).
By the uniqueness part of the Homomorphism Extension Property, we have ψa,b ◦ψc,d = ψca,cb+d .
If a is a unit then ψa−1 ,−ba−1 : R[X] −→ R[X] has the property that ψa−1 ,−ba−1 (aX +b) = X and
ψa,b (a−1 X − ba−1 ) = X, so by the uniqueness part of the Homomorphism Extension Property,
ψa,b ◦ ψa−1 ,−ba−1 = id = ψa−1 ,−ba−1 ◦ ψa,b .
−1
.
Therefore these are inverse isomorphisms, ψa−1 ,−ba−1 = ψa,b
n
i
(b) (i) If f (X) = i=0 ci X ∈ k[X] with ci ∈ k and cn = 0, then deg f (X) = n. Now
n
ci (aX + b)i
ψa,b (f (X)) =
i=0
= cn an X n + terms of lower degrees in X.
Since cn an = 0, this shows that deg ψa,b (f (X)) = deg f (X).
(ii) Suppose that ψa,b (p(X))  g(X)h(X) for g(X), h(X) ∈ k[X]. Choose k(X) ∈ k[X] so that
g(X)h(X) = k(X)ψa,b (p(X)). Since ψa,b is an isomorphism, we have
−1
−1
−1
ψa,b
(g(X))ψa,b
(h(X)) = ψa,b
(k(X))p(X)
−1
−1
and as p(X) is prime, p(X)  ψa,b
(g(X)) or p(X)  ψa,b
(h(X)). Hence ψa,b (p(X))  g(X) or
ψa,b (p(X))  h(X) and so ψa,b (p(X)) is prime.
(iii) This follows from (ii) and Proposition 1.30.
1.9. (a) Addition and multiplication are given by the usual formulæ,
∞
∞
k
(
ak X ) + (
k=0
∞
k
bk X ) =
k=0
∞
k
(ak + bk )X ,
(
k=0
∞
k
ak X )(
k=0
∞
k
bk X ) =
k=0
k
a bk− )X k .
(
k=0
=0
Clearly k[X] ⊆ k[[X]] is a subring. Given two nonzero elements a, b ∈ k[[X]] we may write
∞
∞
k
a=
ak X ,
k=k0
b=
bX
0
90
with ak0 = 0 = b 0 . Then the lowest degree term in ab is ak0 b 0 X k0 + 0 with ak0 b 0 = 0. Hence
ab = 0. So k[[X]] is an integral domain.
∞
k
(b) Let a =
k=0 ak X ∈ k[[X]]. Then a has an inverse in k[[X]] only if there is a b =
∞
k=0 b X ∈ k[[X]] with ab = 1, in particular this forces a0 = 0 since otherwise the lowest term
in X in ab would be of degree greater than 0. Conversely, if a0 = 0, then we can inductively
solve the system of equations
n
a bn− = a0 bn + a1 bn−1 + · · · + an bn = 0
a0 b0 = 1,
(n
1),
=0
to ensure that ab = 1.
(c) We can define make the set k((X)) of all such finite tailed Laurent series into a ring with
addition and multiplication defined by
∞
∞
∞
ak X k ) + (
(
k=k1
bk X k ) =
k=min{k1 ,k2 }
k=k2
∞
∞
∞
ak X k )(
(
k=k0
(ak + bk )X k ,
b X )=
=
k
(
a bk−j )X k .
k=min{k0 , 0 } j=0
0
Clearly k[[X]] ⊆ k((X)) is a subring. Notice that every element
k0 < 0 can be written as
(
ar+k0 X r )X k0 .
∞
k=k0
ak X k ∈ k((X)) with
r=0
The inclusion inc : k[[X]] −→ k((X)) extends to the monomorphism inc∗ : Fr(k[[X]]) −→ k((X))
for which
∞
∞
r
r=0 ar+k0 X
=
(
ar+k0 X r )X k0 ,
inc∗
X −k0
r=0
so inc∗ is surjective.
1.10. Here f (X) = (3X − 3)d(X) + (−9X + 7).
1.11. Here f (X) = −X 3 − X 2 + X + 1 and d(X) = −X 3 − X with
f (X) = d(X) + (−X − X 2 + 1) = d(X) + (2X 2 + 2X + 1).
1.12. The reduction modulo p function
ρ : Z[X] −→ Fp [X];
ρ(f (X)) = f (X),
is a ring homomorphism. If f (X) = g(X)h(X) with g(X), h(X) ∈ Z[X], deg g(X) < deg f (X)
and deg h(X) < deg f (X), then
f (X) = ρ(g(X)h(X)) = ρ(g(X))ρ(h(X)) = g(X)h(X),
where deg g(X) < deg f (X) = deg f (X) and deg h(X) < deg f (X) = deg f (X). But this is
impossible since f (X) is irreducible. So f (X) must be irreducible.
X 3 − X + 1 reduces modulo 3 to an irreducible since it has no roots modulo 3. So X 3 − X + 1
is irreducible.
X 3 + 2X + 1 ≡ X 3 − X + 1 (mod 3) so this polynomial reduces modulo 3 to an irreducible and
so is irreducible.
X 3 + X − 1 reduces modulo 2 to an irreducible since it has no roots modulo 2. So X 3 + X − 1
is irreducible.
X 5 − X + 1 is irreducible modulo 3 and 5 so is itself irreducible.
X 5 + X − 1 = (X 3 + X 2 − 1)(X 2 − X + 1) and 5X 3 − 10X + X 2 − 2 = (5X + 1)(X 2 − 2) so
neither of these is irreducible.
√
1.13. I1 = (X 2 +1), I2 = (X 2 +2), I3 = (X 2 −2), I4 = (X − 2), I5 = (X 2 +2), I6 = X 2 +X +1.
91
1.14. The image is
The image of ε−√2
√
√
ε√2 Q[X] = Q[ 2] = {a + b 2 : a, b ∈ Q}.
√
is ε−√2 Q[X] = Q[ 2] = ε√2 Q[X]. We have
ker ε√2 = ker ε−√2 = (X 2 − 2) Q[X]
which is a maximal ideal.
√
1.15. Notice that ω = (−1 + 3i)/2 = ζ3 is a primitive 3rd root of unity and is a root of the
irreducible polynomial X 2 + X + 1 ∈ Q[X]. Then
εω Q[X] = Q[ω] = {a + bω : a, b ∈ Q},
ker εω = (X 2 + X + 1) Q[X],
where (X 2 + X + 1) Q[X] is a maximal ideal. The other complex root of X 2 + X + 1 is ω 2 ,
so the evaluation homomorphism εω2 has εω2 Q[X] = εω Q[X] and ker εω2 = ker εω .
1.16. We have
εα Q[X] = Q[α] = {a + bα + cα2 + dα3 : a, b, c, d ∈ Q},
ker εα = (X 4 − 2) Q[X],
and the latter ideal is maximal. The other complex roots of X 4 − 2 are −α, α i, −α i (notice
that two of these are real while the other two are not). Then
ker ε−α = ker εα i = ker ε−α i = (X 4 − 2) Q[X]
but although ε−α Q[X] = Q[α], we have
εα i Q[X] = ε−α i Q[X] = Q[α i] = {a + bα i + cα2 + dα3 i : a, b, c, d ∈ Q} = Q[α],
so εα i Q[X] = εα Q[X] since one of these is a subset of R but the other is not.
If we replace Q by R, then X 4 − 2 = (X 2 − 2)(X 2 + 2) in R[X] and we have to consider
which factor α is a root of. If α2 − 2 = 0 then
εα R[X] = ε−α R[X] = R[α] = {a + bα : a, b ∈ R} ⊆ R,
ker εα = ker ε−α = (X 2 − 2) R[X].
If α2 + 2 = 0 then
εα R[X] = ε−α R[X] = R[α] = {a + bα : a, b ∈ R} ⊆ R,
ker εα = ker ε−α = (X 2 + 2) R[X].
1.17. First change variable to obtain
g(X) = f (X + 3) = X 3 − 6X + 4.
Using Cardan’s method we have to solve the quadratic equation
U 2 + 4U + 8 = 0,
which has roots
√
−2 ± 2i = ( 2)3 e3πi/4 .
Thus we can take
√
2
u = 2e
ω = √ (1 + i)ω r = (1 + i)ω r (r = 0, 1, 3).
2
√
√
√
√
For the roots of g(X) we obtain 2, 3 − 1, − 3 − 1, while for f (X) we have 5, 3 + 2, − 3 + 2.
√
πi/4 r
1.18. Work backwards with Cardan’s method. For α, take
q
27q 2 + 4p3
− = 10,
= 108,
2
108
so q = −20 and p = 6. Thus α is a real root of f (X) = X 3 + 6X − 20. Notice that 2 is a real
root of this polynomial and
f (X) = (X − 2)(X 2 + 2X + 10),
92
where X 2 + 2X + 10 has no real roots. Therefore α = 2.
For β, take
q
27q 2 + 4p3
28
− = 1,
= ,
2
108
27
so q = −2 and p = 1. Thus β is a real root of g(X) = X 3 + X − 2 for which 1 is also a root and
g(X) = (X − 1)(X 2 + X + 2),
where X 2 + X + 2 has no real roots. Therefore β = 1.
1.19. To see that the homomorphism
Aff 1 (k) −→ Autk (k[X]);
A −→ αA−1 ,
described in the Proof of Example 1.60 is surjective, suppose that ϕ ∈ Autk (k[X]) is any
automorphism. Let
ϕ(X) = a0 + a1 X + · · · + an X n
with ai ∈ k and an = 0. If n = 0 then ϕk[X] = k ⊆ k[X] so ϕ would not be surjective, hence
we must have n 1. Suppose that show that n > 1. Then
ϕk[X] = {c + 0 + c1 ϕ(X) + · · · + ck ϕ(X)k : c0 , c1 , . . . , ck ∈ k} = k[X].
But if k > 0 and ck = 0 then deg(c + 0 + c1 ϕ(X) + · · · + ck ϕ(X)k ) = kn > 1, so X ∈
/ ϕk[X],
which gives a contradiction. So we must have n = 1. Therefore ϕ(X) = a0 +a1 X and so ϕ = αA
for some A ∈ Aff 1 (k).
1.20. Calculation.
1.21. We have
deg Φ20 (X) = ϕ(20) = ϕ(4)ϕ(5) = 2 × 4 = 8
and
X 20 − 1 = (X 10 − 1)(X 10 + 1) = (X 10 − 1)(X 2 + 1)(X 8 − X 6 + X 4 − X 2 + 1).
Since the roots of X 10 − 1 are the 10th roots of unity, we find that
Φ20 (X)  (X 2 + 1)(X 8 − X 6 + X 4 − X 2 + 1);
since cyclotomic polynomials are irreducible, we must have Φ20 (X) = X 8 − X 6 + X 4 − X 2 + 1.
1.22. (a) We have
k
k−1
X p − 1 = (X p
k−1
)p − 1 = (X p
k−1
− 1)Φp (X p
),
so by (1.5),
k−1
Φpj (X) = Φp (X p
)
0 j k
Φpj (X),
0 j k−1
k−1
and therefore Φpk (X) = Φp (X p ). The complex roots of Φp (X) are the primitive pth roots
of 1, so the roots of Φpk (X) are their pk−1 st roots which are the primitive pk th roots of 1.
(b) Using the formula of Equation 1.4, we have
k−1
Φpk (X) = Φp (X p
k−1
) = (X p
k−1
− 1)p−1 + cp−2 (X p
k−1
− 1)p−2 + · · · + c1 (X p
− 1) + c0 ,
where cr ≡ 0 (mod p) and c0 = p. The Idiot’s Binomial Theorem gives
k−1
Xp
k−1
− 1 ≡ (X − 1)p
(mod p)
so
k−1
Φpk (X) = (X − 1)(p−1)p
k−1
+ cp−2 (X − 1)(p−2)p
where cr ≡ 0 (mod p). In fact,
c0 = Φpk (1) = Φp (1) = c0 = p,
93
k−1
+ · · · + c1 (X − 1)p
+ c0 ,
so the Eisenstein Test can be applied to show that Φpk (X) is irreducible over Q.
(c) First notice that
deg Φn (X) = ϕ(n) = (p1 − 1) · · · (pk − 1)pr11 −1 · · · prkk −1 ,
and
r −1
r1 −1
···pkk
deg Φp1 ···pk (X p1
) = ϕ(p1 · · · pk )pr11 −1 · · · prkk −1 = (p1 − 1) · · · (pk − 1)pr11 −1 · · · prkk −1 ,
r −1
r1 −1
···pkk
so deg Φn (X) = deg Φp1 ···pk (X p1
). Also, each root ξ of Φn (X),
r −1
r1 −1
···pkk
(ξ p1
r −1
r1 −1
···pkk
and no smaller power of (ξ p1
)p1 ···pk = ξ n = 1,
r −1
r1 −1
···pkk
) has this property, hence (ξ p1
Φp1 ···pk (X). This shows that Φn (X)  Φp1 ···pk (X
the same degree they are equal.
r −1
p11
) is a root of
r −1
···pkk
). As these are monic polynomials of
(X − ζnt ), so
1.23. By Theorem 1.43, Φn (X) =
t=1,...,n−1
gcd(t,n)=1
(X −1 − ζnt )
Φn (X −1 ) =
t=1,...,n−1
gcd(t,n)=1
= X −ϕ(n)
(1 − Xζnt )
t=1,...,n−1
gcd(t,n)=1
= X −ϕ(n)
(1 − Xζnn−t )
t=1,...,n−1
gcd(t,n)=1
= X −ϕ(n)
(1 − Xζn−t )
t=1,...,n−1
gcd(t,n)=1
= X −ϕ(n)
(ζnt − X)
t=1,...,n−1
gcd(t,n)=1
= (−1)ϕ(n) X −ϕ(n)
(X − ζnt )
t=1,...,n−1
gcd(t,n)=1
= (−1)ϕ(n) X −ϕ(n) Φn (X).
Since 2  ϕ(n) when n > 2 and the result is immediate when n = 2, we see that desired equation
always holds.
1.24. We have
ζn + ζn−1 = e2πi/n + e−2πi/n
= (cos(2πi/n) + sin(2πi/n) i) + (cos(2πi/n) − sin(2πi/n) i) = 2 cos(2πi/n).
Now we have
ζ5 + ζ5−1 = 2 cos(2π/5),
ζ52 + ζ5−2 = (ζ5 + ζ5−1 )2 − 2 = 4 cos2 (2π/5) − 2.
We also have Φ5 (X) = X 4 + X 3 + X 2 + X + 1, so
ζ54 + ζ53 + ζ52 + ζ5 + 1 = 0.
Rearranging and using the formulæ ζ54 = ζ5−1 , ζ53 = ζ5−2 , we have
(ζ52 + ζ5−2 ) + (ζ5 + ζ5−1 ) + 1 = 0,
94
hence
4 cos2 (2π/5) + 2 cos(2π/5) − 1 = 0.
Thus a suitable polynomial is 4X 2 + 2X − 1 ∈ Q[X].
1.25. (a) In K[X], by the Idiot’s Binomial Theorem 1.10,
X p − 1 = X p + (−1)p = (X + (−1))p = (X − 1)p .
By the Unique Factorization Property 1.33, the only root of this polynomial in K must be 1.
Similarly,
m
m
X np − 1 = (X n − 1)p
and the only roots of this must be nth roots of 1.
(b) If u ∈ K is a root of this polynomial then up = a. As in (a) we have
X p − a = X p − up = (X − u)p ,
so u is the only root in K.
Chapter 2
2.1. This is similar to Example 2.4.
√ √
√
√ √
√
√
√
2.2. It is obvious that [Q( p, q) : Q( p)] 2; if [Q( p, q) : Q( p)] = 1 then q ∈ Q( p),
√
√
say q = a + b p for some a, b ∈ Q. Then
√
√
q = (a + b p)2 = (a2 + b2 p) + 2ab p,
giving the simultaneous pair of equations
a2 + b2 p = q,
2ab = 0.
√
If b = 0 then q ∈ Q which contradicts the result of Qu. 2.1. If a = 0 then
b = b1 /b2 with b1 , b2 ∈ Z and gcd(b1 , b2 ) = 1, we obtain
√
√
q = b p. Writing
b22 q = b21 p
and so p  b2 and q  b1 . Writing b1 = b1 q and b2 = b2 q for suitable b1 , b2 ∈ Z, we obtain
(b2 )2 p2 q = (b1 )2 q 2 p,
hence
(b2 )2 p = (b1 )2 q.
From this we obtain p  b1 and q  b2 ; but then p  b1 as well as p  b2 , contradicting the fact
√
√
that gcd(b1 , b2 ) = 1. So q ∈
/ Q( p).
2.3. Arrange the induction carefully.
2.4. Notice that if v = ±u then b = v 2 = u2 = a which is impossible; so v = ±u. Then
u−v =
(u − v)(u + v)
u2 − v 2
a−b
=
=
∈ K(u + v).
u+v
u+v
u+v
Hence
u=
So K(u, v)
1
1
((u + v) + (u − v)) ∈ K(u + v), v = ((u + v) − (u − v)) ∈ K(u + v).
2
2
K(u + v) K(u, v) and therefore K(u + v) = K(u, v).
2.5. Since 1, i span the Qvector space Q(i), we have [Q(i) : Q] 2. But also if x, y ∈ R, then
x + yi = 0 ⇐⇒ x = y = 0, so 1, i is a basis for Q(i) over Q. Hence [Q(i) : Q] = 2.
√
√
√
√
2.6. First notice that [Q(
3) : Q] =
2 (with Qbasis
1, 3) √
and Q( 3) R. Also, i ∈
/ Q( 3)
√
√
√
and since i2 + 1 = 0, Q( 3, i) = Q( 3)(i) has [Q( 3, i) : Q( 3)] = 2. By Theorem 2.6(ii),
√
√
√
√
[Q( 3, i) : Q] = [Q( 3, i) : Q( 3)] [Q( 3) : Q] = 2 × 2 = 4.
95
√
The following
three subfields of Q(√ 3, i) are distinct and are extensions of Q having degree 2:
√
L1 = Q( 3), L2 = Q(i), L3 = Q( 3 i). Then [Lr ∩ Ls : Q] > 1 ⇐⇒ Lr ∩ Ls = Lr = Ls , so
Lr ∩ Ls = Q whenever r = s. The only real subfield amongst these is L1 .
C
2
∞
R
√
Q( 3, i)
∞
t
tt
tt
t
tt
√ t
2
Q( 3)
uuu
uuu2
uuu
u
2
Q(i)
uuu
uuu
u
2 uuu
u
2
Q
√
Q( 3 i)
r
rrr
r
r
rr 2
rrr
2.7. (a) Since 5 is a prime,
[Q(ζ5 ) : Q] = [Q[X]/(Φ5 (X)) : Q] = ϕ(5) = 5 − 1 = 4.
(b) We have ζ5 = cos(2π/5) + sin(2π/5) i ∈ Q(ζ5 ). But also ζ5−1 ∈ Q(ζ5 ) and ζ5−1 = cos(2π/5) −
sin(2π/5) i ∈ Q(ζ5 ). Hence we have
cos(2π/5) =
1
ζ5 + ζ5−1 ∈ Q(ζ5 ),
2
sin(2π/5) i =
1
ζ5 − ζ5−1 ∈ Q(ζ5 ).
2
(c) This can be found by repeated use of the double angle formula
cos(A + B) = cos A cos B − sin A sin B.
The polynomial Tn (X) expressing cos nθ in terms of cos θ is called the nth Chebyshev polynomial, see Remark 6.6.
(d) For k = 0, 1, 2, 3, 4, cos(5(2kπ/5)) = cos(2kπ) = 1, so T5 (cos 2kπ/5) − 1 = 0. So each of
the numbers cos(2kπ/5) is a root of the polynomial T5 (X) − 1 = (X − 1)(4X 2 + 2X − 1)2 . For
k = 1, 2, 3, 4, cos(2kπ/5) is a root of 4X 2 + 2X − 1, therefore
Q(cos(2π/5)) ∼
= Q[X]/(4X 2 + 2X − 1),
(e)
[Q(cos(2kπ/5)) : Q] = 2.
Q(ζ5 )
2
Q(cos(2π/5))
2
Q
2.8. This is similar to the previous question.
2.9. (a) If α ∈ AutQ (En ) then α(21/n )n = α(2) = 2, so α(21/n ) ∈ En is also a real nth root of 1. If n is odd, the only possibility is α(21/n ) = 21/n , so α = id. If n is even, the
possibilities are α(21/n ) = ±21/n . We can realize this automorphism starting with the evaluation
homomorphism ε21/n : Q[X] −→ En and precomposing with the isomorphism ψ : Q[X] −→ Q[X]
for which ψ(X) = −X to form ε21/n = ε21/n ◦ ψ. On passing to the quotient homomorphism of
96
ε21/n we obtain an automorphism τn of En under which τn (21/n ) = −21/n .
(b) Since E R, an automorphism α ∈ AutQ (E) has the effect
21/n
±21/n
α(21/n ) =
if n is odd,
if n is even.
If for some n we have α(21/n ) = −21/n then
−21/n = α(21/n ) = α(21/2n )2 > 0
since α(21/2n ) ∈ R. This contradiction shows that α(21/n ) = 21/n for every n, so α = id.
(c) Assuming there are only 6 such subfields, they form the following tower.
E4
2
{{
{{
E12
{
3 {{{
{{
{{
{
{
{{ 3
2
E6 g
gg
gg2
gg
g
E2 g
gg
gg
g
2 gg
{{
{{
{
{
{{ 3
E3
Q
(d) This element is a root of the polynomial
(X − (21/2 + 21/3 ))(X − (−21/2 + 21/3 )) = X 2 − 2(21/3 )X + 22/3 − 2 ∈ E3 [X],
so it is certainly an element of E6 which is the only degree 2 extension of E3 . If 21/2 + 21/3 ∈ E3
then 21/2 ∈ E3 , which would imply 2 = [E2 : Q]  [E3 : Q] = 3 which is false, so 21/2 + 21/3 ∈
/ E3 ;
a similar argument shows that 21/2 + 21/3 ∈
/ E2 . Writing ω = e2πi/3 , 21/2 + 21/3 is a root of
(X − (21/2 + 21/3 ))(X − (21/2 + 21/3 ω))(X − (21/2 + 21/3 ω 2 ))
= X 3 − 3(21/2 )X 2 + 6X − (2 + 2(21/2 )) ∈ E2 [X],
so it cannot lie in E4 since 21/2 + 21/3 ∈
/ E2 and 3 [E4 : E2 ] = 2. So 21/2 + 21/3 is in E6 and
E12 and none of the others.
Chapter 3
3.1. Clearly, t is algebraic over K if and only if ker εt = (0), i.e., (i) ⇐⇒ (ii). By Theorem 2.9,
(ii) ⇐⇒ (iii). Hence these three conditions are indeed equivalent.
3.2. The diagrams at the bottom indicate useful subfields of the splitting fields occurring in
each of these examples.
√
1 ± 3i
4
2
2
±πi/3
, so
p1 (X) = X − X + 1: The polynomial X − X + 1 has the complex roots e
=
2
the four roots of p1 (X) are the complex square roots of these numbers, i.e., ±e±πi/6 . Explicitly
these are
√
√
√
√
3 1
3 1
3 1
3 1
+ i, −
− i,
− i, −
+ i.
2 √ 2
2
2
2
2
2
2
The splitting field is E = Q( 3, i) and [E : Q] = 4.
√
√
p2 (X) = X 6 − 2: The roots are the six complex 6th roots of 2, i.e., 6 2e2kπi/6 = 6 2ekπi/3 for
k = 0, 1, 2, 3, 4, 5. Explicitly, these are
√
√
√
√
√
√
√
√
√
√
√
√
6
6
6
6
6
6
6
6
√
√
2
2 3
2
2 3
2
2 3
2
2 3
6
6
+
i, −
+
i, − 2, −
−
i,
−
i.
2,
2
2
2
2
2
2
2
2
√
√
√
√
The splitting field is E = Q( 6 2, 3i) = Q( 6 2)( 3i) which has degree [E : Q] = 12.
97
√
p3 (X) = X 4 + 2: The roots are the four 4th roots of −2, i.e., 4 2e(2k+1)πi/4 for k = 0, 1, 2, 3.
Explicitly these are
1
1
1
1
1
1
1
1
√
+√
i, − √
+√
i, − √
−√
i, √
−√
i.
4
4
4
4
4
4
4
4
2
2
2
2
2
2
2
2
√
The splitting field is E = Q( 4 2, i) and [E : Q] = 8.
p4 (X) = X 4 + 5X 3 + 10X 2 + 10X + 5: Notice that
p4 (Y − 1) = Y 4 + Y 3 + Y 2 + Y + 1 = Φ5 (Y ),
so the splitting field of p4 (X) over Q is the same as that of Φ5 (Y ) over Q and this is the
cyclotomic field Q(ζ5 ) where ζ5 = cos(2π/5) + sin(2π/5)i with [Q(ζ5 ) : Q] = 4; in fact we have
Q(ζ5 ) = Q(cos(2π/5), sin(2π/5)i).
√ √
Q( 6 2, 3, i)
2
√
Q( 3, i)
√ √
Q( 6 2, 3)
2
√
Q( 4 2, i)
√
Q( 3)
2
√
Q( 4 2)
Q
Q(cos(2π/5))
4
2
Q
2
2
6
√
Q( 3)
Q(cos(2π/5), sin(2π/5)i)
Q
2
Q
√
√
√
3.3. List the three
roots of X 3 −2 as u1 = 3 2, u2 = 3 2ζ3 , u3 = 3 2ζ32 . Then each automorphism
√
α ∈ AutQ (Q( 3 2, ζ3 )) permutes these roots, so can be identified with the unique permutation
σα ∈ S3 for which
α(ui ) = uσα (i) (i = 1, 2, 3).
We find that (using cycle notation for permutations)
σid = id,
σα0 = (2 3),
σα1 = (1 2 3),
σα1 = (1 2), σα2 = (1 3 2),
√
These are the six elements of S3 , therefore AutQ (Q( 3 2, ζ3 )) ∼
= S3 .
σα2 = (1 3).
3.4. Irreducibility is a consequence of the polynomial version of the Eisenstein Test 1.48.
Suppose that t ∈ k(T ) is a root of g(X); then using the Idiot’s Binomial Theorem we have
(X − t)p = X p − tp = X p − T,
so t is in fact a root of multiplicity p, hence it is the only root of g(X) in k(T ). This also gives
the factorization of g(X) into linear factors over k(T ).
√ √
√ √
√
√
3.5. Q( 5, 10)/Q: Here [Q( 5, 10) : Q] = 4 and the√element
5
+
10 has degree 4 with
√
4 − 30X 2 + 25 which has roots ± 5 ± 10.
minimal
polynomial
X
√
√
√
the element 2 + i has degree 4 with minimal
Q( 2, i)/Q: Here [Q( 2, i) : Q] = 4 and √
polynomial
X 4 − 2X 2 +
√
√9 which has roots ± 2 ± i.
√
Q( 3, i)/Q: Here [Q( 3, i) : Q] = 4 and the
element 3 + i has degree 4 with minimal
√
polynomial
X 4 − 4X 2 +
16 which has roots ± 3 ± i.
√
√
√
4
4
4
Q( 3, i)/Q: Here [Q( 3, i) : Q] = 4 and the element
3 + i has
√
√ degree 8 with minimal
polynomial X 8 + 4X 6 + 40X 2 + 4 which has roots ± 4 3 ± i and ± 4 3i ± i.
3.6. The induction is straightforward. Here is the argument that K(u, v)/K is simple. We
assume that K is infinite since otherwise the result will be proved in Proposition 5.16.
Consider the subfields K(u + tv) K(u, v) with t ∈ K. Then there are only finitely many
of these, so there must be s, t ∈ K such that s = t and K(u + sv) = K(u + tv). Then
(s − t)v = (u + sv) − (u + tv) ∈ K(u + tv),
98
hence v ∈ K(u + tv). This implies that
u = (u + tv) − tv ∈ K(u + tv),
hence K(u, v)
K(u + tv)
K(u, v) and so K(u, v) = K(u + tv).
3.7. If E/K is a quadratic extension then for any u ∈ E − K we have 1 < [K(u) : K]
2,
so [K(u) : K] = 2 = [E : K] and therefore K(u) = E. Then minpolyK,u (X) must factor into
linear factors over E, so both its roots in K lie in E. This shows that E is normal over K.
The example F2 (Z)/F2 (Z 2 ) is not separable since X 2 − Z 2 ∈ F2 (Z 2 )[X] is irreducible but
not separable (see Qu. 3.4). If char K = 2 then all quadratic polynomials over K are separable.
3.8. Let E
C be a splitting subfield for f (X) over Q. Then if v ∈ C is a nonreal root of
f (X) we have v ∈
/ Q(u), so f (X) does not split over Q(u) even though it has a root in this field.
This means that there is a monomorphism ϕ ∈ MonoQ (Q(u), C) = MonoQ (Q(u), Q) for which
ϕ(u) = v, hence ϕQ(u) = Q(u) and so Q(u)/Q is not normal.
Chapter 4
4.1. By Proposition 3.81 we know that splitting fields are always normal, so it is only necessary
to show that the splitting field E of p(X) over K is separable over K. Since E is obtained by
repeatedly adjoining roots of p(X), the result follows from Proposition 3.73 together with the
fact that if L/K E/K is separable and v ∈ E is a root of p(X), then L(v)/K is separable.
4.2. (a) Suppose that f (X) = c3 X 3 + c2 X 2 + c1 X + c4 with c3 = 0. Then
f (uX + v) =
c3 u3 X 3 + (3c3 vu2 + c2 u2 )X 2 + (3c3 uv 2 + c1 u + 2c2 uv)X + (c3 v 3 + c4 + c1 v + c2 v 2 ),
so if we take u to be any cube root of c3 and u = −c2 /3c3 then f (uX + v) has the desired form.
Notice that v ∈ K(u) and then f (uX + v) ∈ K(u), so provided that we can find a cube root of
1/c3 in K, we have f (uX + v) ∈ K.
(b) Viewing Gal(E/K) as a subgroup of S3 , by Theorem 4.8 we know that 3 divides  Gal(E/K);
but the only subgroups of S3 with this property are S3 and A3 .
(c) This is a tedious calculation! See Section 4.7 for the rest of this question.
4.3. If a/b is a rational root of f (X), we may assume that gcd(a, b) = 1. Now a3 − 3ab2 + b3 =
0, which easily implies that a, b = ±1; but 1 is certainly not a root. Hence there are no
rational roots and so no proper rational factors. By the formula following Proposition 4.25, the
discriminant of f (X) is
∆ = −27 − 4(−3)3 = 81 = 92 .
If the distinct roots of f (X) in C are u, v, w, the splitting subfield K(v, w) = Q(u, v, w)
C
satisfies 3  [Q(u, v, w) : Q] and [Q(u, v, w) : Q]  3! = 6. The Galois group Gal(Q(u, v, w)/Q) is a
subgroup of S3 (viewed as the permutation group of {u, v, w}). Since the discriminant is a square
in Q, Proposition 4.26 implies that Gal(Q(u, v, w)/Q) A3 ∼
= Z/3. So  Gal(Q(u, v, w))/Q) = 3
and Gal(Q(u, v, w)) is cyclic of order 3 whose generator is a 3cycle which cyclically permutes
u, v, w.
4.4. (a) This should be a familiar result.
(b) The centre of D8 is α2 which has order 2, and there are three normal subgroups of order
4, namely
α = {ι, α, α2 , α3 },
α2 , β = {ι, α2 , β, βα2 },
α2 , βα = {ι, α2 , βα, βα3 }.
Notice that there are also four nonnormal subgroups of order 2,
β = {ι, β},
βα = {ι, βα},
βα2 = {ι, βα2 },
99
βα3 = {ι, βα3 }.
4.5. This is an example of case (iii) of Kaplansky’s Theorem and
√ we use the notation of the
2 = −12, so we can take δ = 2 3i. The roots of X 2 + 3 are
proof.
The
discriminant
here
is
δ
√
± 3i, so we may assume
√ √
√ √
√
√
243
243
4
4
−1
u = 3 ζ8 =
(1 + i), v = 3 ζ8 =
(1 − i),
2
2
√
√
where as usual ζ8 = e2πi/8 = (1 + i)/ 2. Hence we have uv = 3 and uvδ = 6i. This gives the
diagram of subfields of E
√
√
√
E = Q( 4 3 ζ8 , 4 3 ζ8−1 ) = Q( 4 3, ζ8 )
√
Q( 3, i)
jjjj
jjjj
j
j
j
j
j
jjj
j
j
j
√
√
Q( 3i)
Q( 3)
j Q(i)
jjjj
j
j
j
jjj
2
jjjj 2
2
j
j
j
jjjj
Q
√
√
√
√
Then α is the restriction of complex conjugation to E, while β( 3i) = 3i and β( 3) = − 3,
hence also β(i) = −i. Using the choices of the proof, we have
√
√
√
√
√
4
4
4
4
4
β( 3 ζ8 ) = − 3 ζ8 , β( 3 ζ8−1 ) = β(− 3 ζ8 i) = − 3 ζ8 i.
√
√
√
√
The effects of σ and γ on the four roots 4 3 ζ8 , 4 3 ζ8−1 , − 4 3 ζ8 , − 4 3 ζ8−1 of f (X) are given in
permutation notation by σ = (1 4 3 2) and α = (1 2)(3 4), and these generate a dihedral
subgroup of S4 . Using the previous question (but beware that the notation there is inconsistent
with that of the present situation!) we have the normal subgroups
σ2 ,
and these have fixed fields
√
2
E σ = Q( 3, i),
E
σ
σ ,
= Q(i),
σ2, α ,
E
σ 2 ,α
each of which is a normal extension of Q.
σ 2 , ασ ,
√
= Q( 3),
E
σ 2 ,ασ
√
= Q( 3 i),
√
3 − 10)/Q: This is similar to Example 4.20, with splitting field Q( 3 10, ζ ) and
4.6. Q(X
3
√
3
∼
10,
ζ
)/Q)
Gal(Q(
S
.
=
3
3
√
√
√ √
√ √
√
Q( 2)(X 3 − 10)/Q( 2): The splitting field is Q( 2, 3 10, ζ3 ), [Q( 2, 3 10) : Q( 2)] = 3 and
√ √
√ √
3
3
Q( 2, 10) Q( 2, 10, ζ3 ).
√
√ √
3
Since
√ ζ3 is 3not real, [Q(
√ 2, 10, ζ3 ) :√Q( 2)] = 6. The Galois group is isomorphic to S3 .
Q( √3 i)(X − 10)/Q(
3 i): Here Q( 3 i) = Q(ζ3 ), with
[Q(ζ3 ) : Q] = 2. The splitting field is
√
√
3
3
3
Q( 10, ζ3 ) and
[Q( √
10, ζ3 ) : Q(ζ3 )] = 3, hence Gal(Q( 10, ζ3 )/Q(ζ3 )) ∼
= Z/3 with generator σ
√
3
3
for√which σ( 10) = 10 ζ√
3.
Q( 23 i)(X 3 − X − 1)/Q( 23 i): First note that X 3 − X − 1 ∈ Z[X] must be irreducible since
its reduction modulo 2, X 3 + X + 1 ∈ F2 [X], has no root in F2 and hence has no linear factor
(see Qu. 1.10). To proceed further we can use the ideas of Qu. 4.2 above (see also
√ Section 4.7).
3 − X − 1 is ∆ = −23 and so δ =
The discriminant
of
the
polynomial
X
23 i. Then if
√
3 − X − 1) is the splitting field of X 3 − X − 1 over Q, Gal(E/Q) ∼ S and
E = Q( 23
i)(X
= 3
√
Gal(E/Q( 23 i)) ∼
= A3 .
√
√
K(X 3 − X − 1)/K for K = Q,
√ Q( 5), Q( 5 i), Q(i): Continuing the preceding discussion,
notice that [E ∩ R : Q] = 3, so 5 ∈
/ E, hence
√
√
Q( 5)(X 3 − X − 1) = Q(X 3 − X − 1)( 5)
100
and
√
√
[Q( 5)(X 3 − X − 1) : Q( 5)] = [Q(X 3 − X − 1) : Q] = 6,
√
√
√
/ E and i ∈
/ E, hence
hence Gal(Q( 5)(X 3 − X − 1)/Q( 5)) ∼
= S3 . Similarly, 5 i ∈
√
√
Gal(Q( 5 i)(X 3 − X − 1)/Q( 5 i)) ∼
= Gal(Q(i)(X 3 − X − 1)/Q(i)).
= S3 ∼
4.7. (a) Since char K = 0, f (X) = pX p−1 = 0, so if u ∈ L is any root of f (X) then
f (u) = pup−1 = 0. By Proposition 3.55, there are no multiple roots, hence p distinct roots. If
u, v ∈ L are distinct roots, then (vu−1 )p = 1, so v = uξ for ξ ∈ K a pth root of 1 with ξ = 1.
(b) If there is a root u ∈
/ K, the Galois group Gal(L/K) acts in the following way. By Theorem 4.8, there must be an element γ ∈ Gal(L/K) with γ(u) = u. We can write γ(u) = uξγ
where ξγ = 1 is a pth root of 1. Since γ(ξγ ) = ξγ , for r 1 we have γ r (u) = uξγr , which can
only equal u if p  r. So u must have at least p conjugates which are all roots of f (X). Since
deg f (X)p , every root of f (X) is conjugate to u, so f (X) must be irreducible over K.
(c) Suppose that f (X) = g(X)h(X) with g(X) ∈ K[X] monic irreducible and 0 < d =
deg g(X) < p. Let L/K with L a splitting field for f (X) over K and let w ∈ L be a root
of g(X). Arguing as in (a), we know that each root of g(X) has the form wξ where ξ is some
pth root of 1; moreover, L must contain p distinct pth roots of 1. Now the constant coefficient
of g(X) is g(0) = (−1)d ξ0 wd ∈ K where ξ0 is a pth root of 1. So
g(0)p = (−1)dp ξ0p (wp )d = (−1)dp ad ,
from which it follows that ad is a pth power in K. As gcd(p, d) = 1, there are integers r, s such
that rp + sd = 1, so we have
a = (ar )p (ad )s = a pth power in K.
Hence if f (X) is not irreducible in K[X] it has a root in K.
4.8. If u ∈ L is a root of f (X) in an extension L/K then by the Idiot’s Binomial Theorem 1.10
X p − a = X p + (−u)p = (X − u)p ,
so u is the only such root in L and f (X) splits over L. If (X − u)d ∈ K[X] for some d with
1 < d < p then ud ∈ K. Since gcd(d, p) = 1, there are integers r, s such that rd + sp = 1. Hence
(ud )s (up )r = u, where the left hand side is in K. This shows that u ∈ K. Hence either f (X)
has a root in K or it must be irreducible over K.
Chapter 5
5.1. By Theorem 1.17, an integral domain D always admits a monomorphism into a field
j : D −→ F (e.g., F can be taken to be the field of fractions of D), so any subgroup U
D×
×
becomes isomorphic to a subgroup jU
F , and if U is finite so is jU . Therefore jU and U
are cyclic.
5.2. The only root of X 2 + 1 in F2 is the multiple root 1.
5.3. The field Fpd [X]/(f (X)) is an extension of Fpd which has degree n, hence it is a finite
field with pdn elements, hence Proposition 5.6 implies that it is isomorphic to Fpdn . Since the
extension Fpdn /Fpd is normal, Fpdn is a splitting field for f (X) over Fpd .
5.4. (a) Here 41 is prime. Since 8  (41 − 1), there is a primitive 8th root of unity in F41 . 6 is
a primitive root for F41 and 65 ≡ 27 (mod 4)1 has order 8.
(b) Here 5 is prime 4  (5 − 1), so there is a primitive 4th root of unity in F×
5 , but no primitive
8th root of unity. In fact, 2 and 3 have order 4, so these are primitive roots for F5 . Notice that
in F5 [X],
X 8 − 1 = (X 4 − 1)(X 4 + 1) = (X 4 − 1)(X 2 − 2)(X 2 − 3),
where the polynomials X 2 − 2 and X 2 − 3 are irreducible. Therefore F25 is the splitting field for
X 8 − 1 over F5 and we have F25 ∼
= F5 (u) = F5 (v), where u2 = 2 and v 2 = 3, so ±u and ±v are
101
primitive 8th roots of unity. To find an element of order 24 in F×
25 , we first find one of order 3.
Consider the polynomial X 2 + X + 1 ∈ F5 [X]; in F5 , this has roots which have order 3. These
roots are given by (−1 ± w)/2, where w2 = (1 − 4) = −3 = 2, hence they are
(−1 ± u)
= −3 ± 3u.
2
Now the elements ±(2 ± 2u)u = ±(±4 + 2u) = ±4 ± 2u all have order 8 × 3 = 24.
(c) Here 11 is prime and 8  (121 − 1) = 120, so F121 is the splitting field of X 8 − 1 over F11 . The
polynomial X 2 + 1 is irreducible over F11 so F121 = F11 (z) where z 2 = −1. Since 120 = 8 × 3 × 5,
it is sufficient to find elements of order 8, 3 and 5 whose product will have order 120.
Suppose that a + bz ∈ F121 with a, b ∈ F11 . If this element has order 8, then (a + bz)2 = ±z.
So let us solve
(a2 − b2 ) + 2abz = z.
Then 2ab = 1 and b2 = a2 , hence b = ±a. Now we have 2a2 = ±1 and so a2 = ±1/2 = ±6.
Now 6 is not a square in F11 but
72 ≡ −6 ≡ 42
(mod 1)1,
so we have a = 4, b = ±4 and a = 7, b = ±7. Therefore the elements of order 8 in F×
121 are
4 ± 4z and 7 ± 7z.
By the same approach as in (b), the elements of order 3 in F121 are (−1 ± 5z)/2 = 5 ± 8z.
2 is a primitive root for F11 so 4 = 22 has order 5.
Combining these we obtain the following primitive roots for F121 : 7 ± z, 10 ± 4z.
(d) In F2 [X] we have X 8 − 1 = (X − 1)8 , whose only root in F2 is 1. So the splitting field is F2 .
d
5.5. Notice that Fp (w) is a splitting field of the separable polynomial X p −1 − 1 over Fp , so
then Fp (w)
Fp . Since Fp (w) = Fpd we have d
; we also have degFp w = d.
if w ∈ F×
p
The number of conjugates of w is d, hence each primitive root of Fpd has d conjugates and the
∼ Z/(pd − 1), i.e.,
total number of these is the number of generators of the cyclic group F×
pd =
ϕ(pd − 1). Hence d  ϕ(pd − 1). This can also be interpreted in terms of the evident action of
Gal(Fpd /Fp ) ∼
= Z/d on the set of all primitive roots of Fpd ; each orbit has exactly d elements,
so the number of orbits is ϕ(pd − 1)/d which is an integer.
5.6. (a) First note that ga (X) = −1, so ga (X) is separable, hence E/K is separable. If u ∈ E
is a root of ga (X), then for t ∈ Fpd ,
d
d
d
d
ga (u + t) = (u + t)p − (u + t) − a = (up − u − a) + (tp − t) = (up − u − a) = 0,
hence u + t is also a root of ga (X). This means that E = K(u) since all the other roots of ga (X)
lie in K(u). As ga (X) is irreducible over K, [E : K] = pd =  Gal(E/K) and so the following
pd automorphisms are the elements of Gal(E/K):
σt : E −→ E;
σt (u) = u + t
(t ∈ Fpd ).
It is easy to check that for s, t ∈ Fpd , σs ◦σt = σs+t . Hence there is an isomorphism Gal(E/K) ∼
=
Fpd with σt corresponding to t ∈ Fpd .
(b) If ga (X) is irreducible over K then it cannot have a root in K since p > 1.
Conversely, suppose that ga (X) has no root in K. Then if u ∈ E is any root of ga (X) in
a splitting field over K, the other roots are the p elements u + t ∈ E (t ∈ Fp ). If u + t0 = u
is a conjugate of u with 0 = t0 ∈ Fp , there must be an element τt0 ∈ Gal(E/K) for which
τt0 (u) = u + t0 . Then τt0 must be isomorphic to a nontrivial subgroup of Fp , but this must
be Fp since this group is simple. Hence, u must have p conjugates and so ga (X) is irreducible
over K.
(c) If K is a finite field and d > 1 then if ga (X) were irreducible over K, then by (a), E would
be finite and Gal(E/K) ∼
= Fpd . But Fpd is not cyclic, yet we know from Proposition 5.23 that
Gal(Fpd /Fp ) ∼
Z/d
is
cyclic.
=
102
×
d
5.7. (a) By Proposition 5.12, F×
q is a cyclic group. If p = 2 then F2d  = 2 − 1, which is odd,
so every element of F×
is a square; we may therefore take λ2d (u) = 1 for all u ∈ F×
. So now
2d
2d
d − 1, which is even. The set of squares in F× is the
suppose that p is odd. Then F×

=
p
pd
pd
normal subgroup
(F×
)2 = {u2 : u ∈ F×
} F×
pd
pd
pd
and it is easily seen that its quotient group has order 2, hence
F×d /(F×d )2 ∼
= {±1}.
p
p
We may use this group isomorphism to define λq . Clearly we have
ker λq = (F×
)2 .
pd
λq is surjective if and only if p is odd.
u
Remark: when d = 1, λp (u) =
, the Legendre symbol of u from Number Theory.
p
(b) If u ∈ Σq , then either u = 0 or u = 0 and u = (±v)2 for some v ∈ F×
q . Thus we have
Σq  = 1 +
(q − 1)
(q + 1)
=
.
2
2
Then
t − Σq  = Σq  =
(q + 1)
.
2
(c) Since Σq ∪ (t − Σq ) ⊆ Fq , we have
q
Σq ∪ (t − Σq ) = Σq  + t − Σq  − Σq ∩ (t − Σq ).
This implies that
q
(q + 1) − Σq ∩ (t − Σq )
and so
Σq ∩ (t − Σq ) 1.
Thus for every t ∈ Fq , there are u, v ∈ Fq (possibly 0) for which u2 = t − v 2 , whence t = u2 + v 2 .
(d) By (c), we may write −1 = a2 + b2 for some a, b ∈ Fq , i.e.,
12 + a2 + b2 = 0.
Chapter 6
∼ Z/p, which is abelian. Suppose that the result holds whenever
6.1. Now when n = 1, G =
k
G = p with k < n. Now if G = pn , recall that by Cauchy’s Lemma, the centre Z of G
is nontrivial. Hence G/Z has order G/Z = pk with k < n. By the inductive hypothesis,
there is a normal subgroup M G/Z with M  = pk−1 . By one of the Isomorphism Theorems,
there is a normal subgroup N G containing Z and satisfying N/Z = M ⊆ G/Z. Clearly
N  = Z M  = pn−1 . This establishes the inductive step and hence the desired result.
6.2. In this situation, for any nonzero t ∈ K, −t = t (since otherwise 2t = 0 and so t = 0).
If ζ ∈ K is a primitive nth root of unity, then (−ζ)n = (−1)n ζ n = −1, while (−ζ)2n =
(−1)2n ζ 2n = 1. Hence −ζ ∈ K is a primitive 2nth root of unity.
6.3. Write n = 2k pr11 · · · prss , where each pj is an odd prime, p1 < p2 < · · · < ps , rj
k 0. Then
1 and
ϕ(n) = ϕ(2k )ϕ(pr11 ) · · · ϕ(prss ) = ϕ(2k )(p1 − 1)pr11 −1 · · · (ps − 1)prss −1 .
If s > 0 then ϕ(n)  4 happens precisely when r1 = · · · = rs = 1 and one of the following
possibilities occurs:
• p1 = 5, s = 1 and k = 0 (hence n = 5);
103
• p1 = 3, s = 1 and k = 0, 1, 2 (hence n = 3, 6, 12);
• s = 0 and k = 0, 1, 3 (hence n = 1, 2, 4, 8).
Q(i): Here degree [Q(i) : Q] = 2 and clearly the four 4th roots of unity ±1, ±i lie in this field.
As ϕ(5) = 4, it√has no 5th roots
√ of unity except 1. If it contained a 3rd√root of unity then it
would contain 3 and so Q( 3, i) Q(i) which is impossible since [Q( 3, i) : Q] = 4. From
this we see that the only roots of unity in Q(i) are ±1, ±i.
√
√
1
2
Q( 2 i): This field contains the 4 primitive 8th roots of unity ± ±
i as well as the 4th
2
2
roots.
√
√
1
3
Q( 3 i): This contains the six 6th roots of unity ±1, ± ±
i.
2
2
√
Q( 5 i): This field contains only the square roots of unity ±1.
6.4. (a) We have ϕ(24) = ϕ(8)ϕ(3) = 4 × 2 = 8. The elements of Z/24 which are invertible
are the residue classes modulo 24 of the numbers 1, 5, 7, 11, 13, 17, 19, 23. For each of these
numbers r, the residue class modulo 24, r, satisfies r2 = 1, hence these all have order 2 except
1 which has order 1. Since (Z/24)× is abelian, it is isomorphic to Z/2 × Z/2 × Z/2. The effect
r . Notice that 23 acts like complex conjugation.
of these elements on Q(ζ24 ) is given by r · ζ24
The effect on Q(cos(π/12)) is given by
r · cos(π/12) = cos(πr/12),
so in particular,
−r · cos(π/12) = cos(−πr/12) = cos(πr/12) = r · cos(π/12).
(b) This is similar to (a). We have ϕ(20) = ϕ(4)ϕ(5) = 2 × 4 = 8 and the elements of (Z/20)×
are the residue classes modulo 20 of the numbers 1, 3, 7, 9, 11, 13, 17, 19. This time there are
elements of order 4, for instance 7 and 13. Then we have (Z/20)× ∼
= Z/2 × Z/4.
6.5. For any n 1, let ζn = e2πi/n = cos(2π/n) + sin(2π/n) i. Notice that if n is odd, then
Q(ζn ) = Q(−ζn ) where −ζn is a primitive 2nth root of unity, so we might as well assume that
n is even from now on. We also have
ζn − ζn−1 = 2 sin(2π/n) i ∈ Q(ζn ).
(a) If 4 n then writing n = 2k with k odd, we have
[Q(ζn ) : Q] = ϕ(2k) = ϕ(2)ϕ(k) = ϕ(k),
while
[Q(ζ2n ) : Q] = ϕ(4k) = ϕ(4)ϕ(k) = 2ϕ(k).
k . So
Hence, Q(ζn ) cannot contain ζ2n and by another simple argument it cannot contain i = ζ2n
k ,
we see that sin(2π/n) ∈
/ Q(ζn ) in this situation. Notice that since i = ζ2n
sin(2π/n) =
2 − ζ −2
ζ2n
2n
∈ Q(ζ2n ),
2i
and by Theorem 6.3,
sin(2π/n) ∈ Q(ζ2n ) ∩ R = Q(cos(π/n)).
Also, we have
[Q(cos(π/n)) : Q] = 2[Q(cos(2π/n)) : Q],
hence
Q(cos(π/n)) = Q(cos(2π/n))(sin(2π/n))
and
[Q(cos(2π/n))(sin(2π/n)) : Q(cos(2π/n))] = 2,
with
minpolyQ(cos(2π/n)),sin(2π/n) (X) = X 2 + cos2 (2π/n) − 1.
104
If 4  n, we can write n = 4 . Then i = ζn , so i ∈ Q(ζn ), whence
sin(π/2 ) = sin(2π/n) =
ζn − ζn−1
∈ Q(ζn ).
i
Clearly
sin(π/2 ) ∈ Q(ζn ) ∩ R = Q(cos(2π/n)).
Consider the automorphism σ ∈ Gal(Q(ζn )/Q)) for which σ(ζn ) = ζn2
see that σ has order 2. Then
+1
= −ζn ; it is easy to
σ(cos(2π/n)) = σ(cos(π/2 )) = − cos(π/2 ),
σ(cos(π/ )) = cos(π/ ),
σ(sin(2π/n)) = σ(sin(π/2 )) =
From this we find that when
−ζn + ζn−1
=
2(−ζn )
sin(π/2 )
− sin(π/2 )
if
if
is odd,
is even.
is odd,
Q(cos(2π/n)) = Q(cos(π/2 )) = Q(cos(π/ ))(sin(π/2 )) = Q(sin(π/2 )),
since cos(π/ ) = 1 − 2 sin2 (π/2 ) ∈ Q(sin(π/2 )). Thus we have [Q(sin(π/2 )) : Q] = 2ϕ( ) and
Gal(Q(sin(π/2 ))/Q) = Gal(Q(cos(π/2 ))/Q) = (Z/4 )× /{1, −1}.
Similarly, if
is even,
[Q(cos(π/ ))(sin(π/2 )) : Q(cos(π/ ))] = 2
and we must have
Q(cos(2π/n)) = Q(cos(π/2 )) = Q(sin(π/2 ))
with
Gal(Q(sin(π/2 ))/Q) = Gal(Q(cos(π/2 ))/Q) = (Z/4 )× /{1, −1}
(b) We have
√
2− 3
1 − cos(π/6)
=
,
sin (π/12) =
2
4
2
and so
sin(π/12) =
Then
2−
2
√
3
√
=
6−
4
√
2
.
√
√
√ √
Q(sin(π/12)) = Q( 6 − 2) = Q( 2, 3).
and
Gal(Q(sin(π/12))/Q)) ∼
= (Z/4 )× /{1, −1} ∼
= Z/2 × Z/2.
Here the effect of the coset of the residue class of r ∈ (Z/4 )× is given by
r · sin(π/12) =
r − ζr
ζ24
24
= sin(rπ/12) i1−r .
ir
Explicitly we have
√
√
6− 2
1 · sin(π/12) = −1 · sin(π/12) = sin(π/12) =
,
√4 √
6+ 2
5 · sin(π/12) = −5 · sin(π/12) = sin(5π/12) =
,
4√
√
− 6− 2
7 · sin(π/12) = −7 · sin(π/12) = − sin(7π/12) =
,
4
√
√
− 6+ 2
11 · sin(π/12) = −11 · sin(π/12) = − sin(11π/12) =
.
4
105
In terms of the generators
1·
7·
√
√
2=
2=
√
√
2,
2,
√
2 and
1·
√
√
3=
3 these act by
√
√
2 = − 2,
√
√
11 · 2 = − 2,
5·
3,
√
√
5 · 3 = − 3,
√
5·
11 ·
√
√
3=
√
3,
√
3 = − 3.
6.6. (a) We have
 Gal(Q(ζ5 )/Q) = [Q(ζ5 ) : Q] = deg Φ5 (X) = ϕ(5) = 4,
and (Z/5)× is cyclic generated by the residue class 2. The action is given by
2 · ζ5 = ζ52 ,
2
3
2 · ζ5 = ζ54 ,
2 · ζ5 = ζ53 ,
4
2 · ζ5 = ζ5 .
(b) We have ζ5 + ζ5−1 = 2 cos(2π/5) and Φ5 (ζ5 ) = 0, so since ζ53 = ζ5−2 and ζ54 = ζ5−1 ,
(ζ52 + ζ5−2 ) + (ζ5 + ζ5−1 ) + 1 = 0
and therefore
(ζ5 + ζ5−1 )2 + (ζ5 + ζ5−1 ) − 1 = 0.
Hence
4 cos2 (2π/5) + 2 cos(2π/5) − 1 = 0.
The quadratic polynomial 4X 2 + 2X − 1 ∈ Z[X] has discriminant 20 which is not a square in
Q, so this is this polynomial is irreducible over Q, therefore
1
1
minpolyQ,cos(2π/5) (X) = X 2 + X − .
2
4
√
√
−1 ± 5
−1 + 5
The roots of this are
. As cos(2π/5) > 0 we must have cos(2π/5) =
. We
4 √
4
−1 − 5
also have cos(4π/5) =
. As sin(2π/5) > 0,
4
√
√
1+5−2 5
5+ 5
2
2
sin (2π/5) = 1 − cos (2π/5) = 1 −
=
,
16
8
√
5+ 5
hence sin(2π/5) =
.
8
∼
(c) Gal(Q(ζ5 ) = Z/4 and has 3 subgroups {1}
of subfields.
{1, 4}
Gal(Q(ζ5 ), giving the following tower
Q(ζ5 )
2
Q(ζ5 )
4
√
= Q(cos(2π/5)) = Q( 5)
2
Q
106
6.7. (a) We have
(p−1)/2
2
(p−1)/2
(ζpr
ξ =
−
ζp−r )2
(ζpr − ζp−r )(ζp−r − ζpr )
(p−1)/2
= (−1)
r=1
r=1
(p−1)/2
(1 − ζp−2r )(1 − ζp2r )
= (−1)(p−1)/2
r=1
p−1
= (−1)(p−1)/2
(1 − ζp2r )
r=1
(p−1)
(1 − ζps )
= (−1)(p−1)/2
s=1
since each congruence 2x ≡ t (mod p) has exactly one solution modulo p for each t.
(b) Since
−1 if p ≡ 1 (mod 4),
(−1)(p−1)/2 =
1 if p ≡ 3 (mod 4),
and
p−1
(1 − ζps ) = Φp (1) = p,
s=1
the result follows.
(c) Taking square roots we find that
√
if p ≡ 1
± p
√
± p i if p ≡ 3
(mod 4),
(mod 4).
√
√
As ξ ∈ Q(ζp ), we see that p ∈ Q(ζp ) if p ≡ 1 (mod 4) and p i ∈ Q(ζp ) if p ≡ 3 (mod 4).
ξ=
6.8. Recall the wellknown formula
σ(i1 · · · ir )σ −1 = (σ(i1 ) · · · σ(ir )).
Then for 1
r
n − 2 we have
(1 2 · · · n)r (1 2)(1 2 · · · n)n−r = (1 2 · · · n)r (1 2)((1 2 · · · n)r )−1 = (r + 1 r + 2),
while
(1 2 · · · n)n−1 (1 2)((1 2 · · · n)n−1 )−1 = (1 2 · · · n)−1 (1 2)((1 2 · · · n)−1 )−1 = (n 1) = (1 n).
This means that every such 2cycle (r + 1 r + 2) is in H. Also recall that every permutation
ρ ∈ Sn is a product of 2cycles, so it suffices to show that every 2cycle (a b) ∈ Sn is a product
of 2cycles of the form (r + 1 r + 2). Assuming that a < b, we also have
(a b) = (b − 1 b) · · · (a + 2 a + 3)(a + 1 a + 2)(a a + 1)(a + 1 a + 2)(a + 2 a + 3) · · · (b − 1 b),
and this is in H. Hence H = Sn .
6.9. (a) For each u ∈ E,
σ(T (u)) = σ(u + σ(u) + σ 2 (u) + · · · + σ n−1 (u))
= σ(u) + σ 2 (u) + · · · + σ n (u)
= σ(u) + σ 2 (u) + · · · + σ n−1 (u) + u = T (u),
so T (u) is fixed by σ and all its powers, hence by Gal(E/K). Therefore T (u) is in E Gal(E/K) = K.
It is straightforward to verify that the resulting function TrE/K : E −→ K is Klinear.
(b) Let v ∈ E and suppose that TrE/K (v) = 0. By Artin’s Theorem 6.15, the linear combination
107
of characters id +σ + · · · + σ n−1 must be linearly independent, so there is an element t ∈ E for
which
TrE/K t = t + σ(t) + · · · + σ n−1 (t) = 0.
Then
u = vσ(t) + (v + σ(v))σ 2 (t) + · · · + (v + σ(v)σ 2 (t) + · · · + σ n−2 (v))σ n−1 (t)
satisfies
u − σ(u) = v σ(t) + σ 2 (t) + · · · + σ n−1 (t) − σ(v) + · · · + σ n−1 (v) t
= v t + σ(t) + σ 2 (t) + · · · + σ n−1 (t) − v + σ(v) + · · · + σ n−1 (v) t
= (TrE/K t)v − (TrE/K v)t = (TrE/K t)v.
So we obtain
v=
1
TrE/K t
u−σ
1
TrE/K t
u .
6.10. (a) This can be proved by induction on n. Write
e[m]
r =
Xi1 · · · Xir ,
s[m]
r =
Xir .
1 i m
i1 m. The desired result is that for all n
[n] [n]
[n] [n]
[n]
[n]
1 and k
1,
[n]
sk = e1 sk−1 − e2 sk−2 + · · · + (−1)k−1 ek−1 s1 + (−1)k kek .
[1]
[1]
When n = 1 we have sr = X1r and e1 = X1 from which the result follows. Now suppose that
[n+1]
[n]
k , while
the result is true for some n 1. Then sk
= sk + Xn+1
[n+1] [n+1]
sk−1
[n]
(e1
e1
[n+1] [n+1]
[n+1] [n+1]
[n+1]
sk−2 + · · · + (−1)k−1 ek−1 s1
+ (−1)k kek
=
[n]
[n]
[n]
[n]
k−1
k−2
Xn+1 )(sk−1 + Xn+1
) − (e2 + e1 Xn+1 )(sk−2 + Xn+1
) + ···
[n]
[n]
[n]
[n]
[n]
+ (−1)k−1 (ek−1 + ek−2 Xn+1 )(s1 + Xn+1 ) + (−1)k k(ek + ek−1 Xn+1 )
[n]
[n] k−1
[n] k−2
[n]
= sk + (e1 Xn+1
− e2 Xn+1
+ · · · + (−1)k−1 ek−1 Xn+1 )
[n]
[n] [n]
[n] [n]
[n]
+ (sk−1 − e1 sk−2 + · · · + (−1)k−1 ek−2 s1 + (−1)k kek−1 )Xn+1
[n] k−1
[n]
k
2
+ (Xn+1
− e1 Xn+1
+ · · · + (−1)k−1 ek−2 Xn+1
)
[n]
[n+1]
k
= sk + Xn+1
= sk
,
− e2
+
which demonstrates the inductive step.
(b)(i) We have h1 = e1 , h2 = e21 − e2 and h3 = e3 − 2e1 e2 + e31 .
(ii) This can be done by induction on n in a similar way to part (a).
108