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=
( )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

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