( )E/Q

hence ER = E and [E : ER ] = 2.

Proof. Let u ∈ E. As E/Q is normal, minpolyQ,u (X) ∈ Q[X] splits over E, hence all of

its complex roots lie in E. But ( ) permutes the roots of this polynomial. Hence ( ) maps E

into itself.

(i) For z ∈ C, z = z if and only if z ∈ R.

(ii) Here | ( )E/Q | = 2, and

( )E/Q

= {u ∈ E : u = u} = ER .

E C

2

R ∞

nn E

∞

nnn

nn

nnn2

n

nnn

ER e

ee

ee

ee

e

Q

We will usually write ( ) rather than ( )E/Q when no confusion seems likely to result.

Example 4.21. Consider the cyclotomic extension Q(ζ8 )/Q where

1 1

ζ8 = eπi/4 = √ + √ i.

2 2

From Example 4.8 we know that

√

Q(ζ8 ) = Q( 2, i), [Q(ζ8 ) : Q] = 4,

and we easily see that √

Q(ζ8 )R = Q( 2).

54 4. GALOIS EXTENSIONS AND THE GALOIS CORRESPONDENCE

4.7. Kaplansky™s Theorem

In this section we give a detailed account of the Galois theory of irreducible rational poly-

nomials f (X) = X 4 + aX 2 + b ∈ Q[X]. The following result describes the Galois groups that

occur and the proof introduces some useful computational techniques.

Theorem 4.22 (Kaplansky™s Theorem). Let f (X) = X 4 + aX 2 + b ∈ Q[X] be irreducible.

(i) If b is a square in Q then Gal(Q(f (X))/Q) ∼ Z/2 — Z/2.

=

2 ’ 4b) is a square in Q then Gal(Q(f (X))/Q) ∼ Z/4.

(ii) If b(a =

2 ’ 4b) is a square in Q then Gal(Q(f (X))/Q) ∼ D .

(iii) If neither b nor b(a =8

™

Proof. Let g(X) = X 2 +aX +b ∈ Q[X]. Notice that g(X) must be irreducible since

™™¦

™

otherwise f (X) would factorize, hence (a2 ’ 4b) is not a square in Q. Setting d = (a2 ’ 4b) ∈ Q

and δ to be a square root of d, we see that the roots of g(X) are (’a ± δ)/2, where δ ∈ Q. Then

/

the roots of f (X) are ±u, ±v, where

(’a + δ) (’a ’ δ)

u2 = , v2 = ,

2 2

so the splitting ¬eld of f (X) over Q is E = Q(u, v) which contains the quadratic extension

Q(δ)/Q. Since deg f (X) = 4, we must also have 4 | [E : Q]. In fact, since E is obtained by at

most 3 successive quadratic extensions we also have [E : Q] | 8.

(i) We have

a2 ’ d 4b

(uv)2 = u2 v 2 = = = b,

4 4

hence uv is a square root of b which is in Q. Setting c = uv ∈ Q, we ¬nd that v = c/u ∈ Q(u).

This shows that E = Q(u) and we have the following Galois tower.

E = Q(u)

2

Q(δ)

2

Q

In particular [E : Q] = 4 = | Gal(E/Q)|. Notice that for the Galois extension Q(δ)/Q there

must be a normal subgroup N Gal(E/Q) with

Q(δ) = E N , Gal(Q(δ)/Q) = Gal(E/Q)/N.

Hence there is an element σ ∈ Gal(E/Q) for which σ(δ) = ’δ. This element must also have

the e¬ects σ(u) = ±v and σ(v) = ±u. Given u we might as well choose v so that σ(u) = v.

There is also an element „ ∈ N for which „ (u) = ’u and we also have „ (v) = ’v. Notice that

if σ(v) = ’u then easy calculation shows that

„ σ(v) = σ„ (v) = u, „ σ(δ) = σ„ (δ) = ’δ,

hence we might as assume that σ(v) = u since if necessary we can replace our original choice

by „ σ.

We now have

c c

σ(u) = , „ (u) = ’u, „ σ(u) = σ„ (u) = ’ .

u u

These satisfy

σ 2 = „ 2 = (σ„ )2 = id = the identity, σ„ = „ σ.

This shows that

Gal(Q(f (X))/Q) = Gal(E/Q) = {id, σ, „, σ„ } ∼ Z/2 — Z/2 = the Klein 4-group.

=

4.7. KAPLANSKY™S THEOREM 55

(ii) If bd is a square in Q, then

(uvδ)2 = u2 v 2 d = bd,

which is a square in Q, so we can write uvδ = c ∈ Q or equivalently v = c/(uδ) ∈ Q(u) since

Q(δ) Q(u). This shows that E = Q(u, v) = Q(u) and again we have a Galois tower

E = Q(u)

2

Q(δ)

2

Q

with [E : Q] = 4 = | Gal(E/Q)|.

Since Q(δ)/Q is Galois there is an element σ ∈ Gal(E/Q with σ(δ) = ’δ and this has the

e¬ect σ(u) = ±v; given u we might as well choose v so that σ(u) = v. Notice that

c c

σ(v) = = ’ = ’u,

σ(uδ) vδ

so σ 2 (u) = ’u. This shows that

Gal(Q(f (X))/Q) = Gal(E/Q) = {id, σ, σ 2 σ 3 } ∼ Z/4 = a cyclic group of order 4.

=

(iii) Suppose that d, b and bd are not squares in Q. By an easy calculation we ¬nd that

(uv)2 = b, so uv ∈ E is a square root of b in E. Suppose that uv ∈ Q(δ); then uv = p + qδ for

some p, q ∈ Q. By squaring we obtain

b = (p2 + q 2 d) + 2pqδ,

and so pq = 0. We cannot have q = 0 since this would imply that b was a square in Q; if

p = 0 then b = q 2 d and so bd = (qd)2 , implying that bd was a square in Q. Thus we have

Q(uv) © Q(δ) = Q. A similar discussion shows that

Q(uvδ) © Q(δ) = Q = Q(uvδ) © Q(uv).

So we have a Galois tower which includes the following sub¬elds.

E = Q(u, v)

Q(uv, δ)

xxx

rr xxx

rrr xxx

r

rrr xx

r

Q(δ) Q(uv) Q(uvδ)

vvv p

ppp

vvv

pp

vvv 2

ppp 2

vvv

2

ppp

Q

Choose

± ∈ Gal(E/Q(uv)) Gal(E/Q)

so that ±(δ) = ’δ. By renaming ’v to v if necessary, we may assume that v = ±(u) and so

u = ±(v). Notice that ±2 = id.

Choose

β ∈ Gal(E/Q(δ)) Gal(E/Q)

with β(uv) = ’uv. We must have either β(u) = ’u or β(v) = ’v, so by interchanging ±δ if

necessary we can assume that β(u) = ’u and β(v) = v. Notice that β 2 = id.

Choose

γ ∈ Gal(E/Q(δ, uv)) Gal(E/Q)

56 4. GALOIS EXTENSIONS AND THE GALOIS CORRESPONDENCE

so that γ(u) = ’u. Then we must have γ(v) = ’v since γ(uv) = uv. Notice that γ 2 = id.

Setting σ = ±β we ¬nd σ(u) = ’v and σ(v) = u. Then σ 2 = γ and σ has order 4. Also,

±σ± = βσβ = σ ’1 .

The eight elements

id, σ, γ, σ ’1 , ±, ±σ, ±γ, ±σ ’1

form a group isomorphic to the dihedral group of order 8, D8 . Therefore we have

Gal(Q(f (X))/Q) = Gal(E/Q) ∼ D8 , =

and [E : Q] = 8. The corresponding Galois tower is

E = Q(u, v)

2

Q(uv, δ)

xxx

r

2 rrrr xxx2

xxx

r 2

rrr xx

r

Q(δ) Q(uv) Q(uvδ)

vvv

pp

vvv

ppp

vvv pp

2

ppp 2

vvv

2

pp

Q

Example 4.23. We have the following Galois groups:

Gal(Q(X 4 + 1)/Q) ∼ Z/2 — Z/2; Gal(Q(X 4 + 4X 2 + 2)/Q) ∼ Z/4;

= =

Gal(Q(X 4 + 2X 2 + 2)/Q) ∼ D8 .

=

Exercises on Chapter 4

4.1. If f (X) ∈ K[X] is a separable polynomial, prove that the splitting ¬eld of f (X) over K is

a ¬nite Galois extension of K.

4.2. Let K be a ¬eld for which char K = 2, 3 and suppose that f (X) ∈ K[x] is a cubic

polynomial.

(a) Show that there u, v ∈ K with u = 0 such that f (uX + v) = X 3 + aX + b for some

a, b ∈ K. If f (X) is monic, deduce that a, b ∈ K; under what conditions is this always

true?

(b) If g(X) = X 3 + aX + b ∈ K[x] is irreducible and E = K(g(X)) is its splitting ¬eld

over K, explain why Gal(E/K) is isomorphic to one of the groups S3 or A3 .

(c) Continuing with the notation and assumptions of (b), suppose that w1 , w2 , w3 are the

distinct roots of g(X) in E and let

∆ = (w1 ’ w2 )2 (w2 ’ w3 )2 (w1 ’ w3 )2 ∈ E.

Show that

∆ = ’4b3 ’ 27a2 ,

and hence ∆ ∈ K. If δ = (w1 ’ w2 )(w3 ’ w3 )(w1 ’ w3 ), show that

A3 if δ ∈ K,

Gal(E/K) ∼

=

S3 if δ ∈ K.

/

[Hint: Consider K(δ) E and the e¬ect of even and odd permutations in Gal(E/K)

S3 on the element δ.]

4.3. This is a revision exercise on ¬nite groups of small order.

57

(a) Show that every non-abelian ¬nite group has order at least 6.

(b) Let D8 be the dihedral group with the eight elements

ι, ±, ±2 , ±3 , β, β±, β±2 , β±3

satisfying

±4 = ι, β 2 = ι, β±β = ±’1 = ±3 .

Find all the normal subgroups of D8 .

4.4. Use Kaplansky™s Theorem 4.22 to ¬nd the Galois group of the splitting ¬eld E of the

polynomial X 4 + 3 ∈ Q[X] over Q. Determine all the subextensions F E for which F/Q is

Galois.

4.5. Find the Galois groups for each of the following extensions:

√ √ √ √

Q(X 3 ’ 10)/Q; Q( 2)(X 3 ’ 10)/Q( 2); Q( 3 i)(X 3 ’ 10)/Q( 3√ i);

√ √ √

3 ’ X ’ 1)/Q( 23 i); 3 ’ X ’ 1)/K for K = Q, Q( 5), Q( 5 i), Q(i).

Q( 23 i)(X K(X

4.6. Let p > 0 be a prime. Let K be a ¬eld 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 ¬eld 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 p-th roots of 1.

(b) Suppose that K contains p distinct p-th 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 p-th root of 1 in K is 1. Show that either f (X)is irreducible

over K or it has a root in K.

4.7. Let K be a ¬eld 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.

CHAPTER 5

Galois extensions for ¬elds of positive characteristic

In this chapter we will investigate extensions of ¬elds of positive characteristic, especially

¬nite ¬elds. A thorough account of ¬nite ¬elds and their applications can be found in [5]. We

will assume that p > 0 is a prime and K is a ¬eld of characteristic char K = p ; we will also