2.3 Galois Extensions, Galois Groups and the Fundamental Theorem of Galois Theory

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The following is frequently useful in a variety of contexts.

Lemma 2.3.1 [Dedekind Independence Lemma]

(i) Let E,K be fields, and let {_1, _2, · · · , _r} be distinct monomorphisms

E ! K. If y1, y2, · · · , yr 2 K are not all zero, then the map

E −! K, _ 7!

X

yi_i(_)

is not the zero map.

(ii) Let E be a field, and let G be a group of automorphisms of E. Set

K = invG(E) and let x1, x2, · · · , xr 2 E be K-linearly independent. If

y1, y2, · · · , yr 2 E are not all zero, then the map

G −! E, _ 7!

X

yi_(xi)

is not the zero map.

If F _ K is a field extension, we set Gal(E/F) = {automorphisms _ :

K ! K| _|F = 1F}. We call Gal(E/F) the Galois group of K over F. Note

that if F _ K _ E, then Gal(E/K) is a subgroup of Gal(E/F). For the

next couple of results we assume a fixed extension E _ F, with Galois group

G = Gal(E/F).

Proposition 2.3.2 Assume that E _ E1 _ E2 _ F, and set H1 = Gal(E/E1), H2 =

Gal(E/E2). If [E1 : E2] < 1, then

[H2 : H1] _ [E1 : E2].

Proposition 2.3.3 Let E _ F, and set G = Gal(E/F). Assume that 1 _

H1 _ H2 _ G, and let E1 = invH1(E), E2 = invH2(E). If [H2 : H1] < 1,

then

[E1 : E2] _ [H2 : H1].

Corollary 2.3.3.1 Let E _ F be a field extension, let G = Gal(E/F), and

let F0 = invG(E).

(i) If [E : F0] < 1, then |G| < 1 and [E : F0] = |G|.

2.3. GALOIS EXTENSIONS AND GALOIS GROUPS 51

(ii) If |G| < 1, then [E : F0] < 1 and [E : F0] = |G|.

Next set

E/F = {subfields K| E _ K _ F},

G = {subgroups H _ G}.

We have the maps

Gal(E/•) : E/F −! G,

inv•(E) : G −! E/F.

Note that Propositions 14 and 15 say that Gal(E/•) and inv•(E) are

“contractions” relative to [· , ·].

We now define two concepts of “closure.”

(i) If K 2 E/F, set

clE(K) = invGal(E/K)(E),

the closure of K in E. If K = clE(K), say that K is closed in E.

(ii) If H _ G, set

clG(H) = Gal(E/invH(E)),

the closure of H in G. If H = clG(H), say that H is closed in G.

Corollary 2.3.3.2

(i) Let E _ E1 _ E2 _ F, and assume that [E1 : E2] < 1 and that E2 is

closed in E. Then E1 is closed in E.

(ii) Let {e} _ H1 _ H2 _ G and assume that [H2 : H1] < 1 and that H1

is closed in G. Then H2 is closed in G.

Theorem 2.3.4 Let E _ F be an algebraic extension with F closed in E.

Then every element of E/F is closed in E.

The field extension E _ F is called a Galois extension (we sometimes

say that E is Galois over F) if F is closed in E. Let E _ F be a field extension

with Galois group G, and let K 2 E/F. We say that K is stable if _K = K

for each _ 2 G. We denote by c

G the closed subgroups of G.

Theorem 2.3.5 (Fundamental Theorem of Galois Theory) Let E _

F be an algebraic Galois extension.

52 CHAPTER 2. FIELD AND GALOIS THEORY

(i) The mappings

Gal(E/•) : E/F ! c

G, inv• : c

G ! E/F

are inverse isomorphisms.

(ii) Let K 2 E/F correspond to the closed subgroup H _ G under the above

correspondence. Then K is stable in E if and only if H / G. In this

case, K is Galois over F and

Gal(K/F) _= G/H.

The next result, the so-called “Theorem on Natural Irrationalities,” is

frequently useful in computations.

Theorem 2.3.6 Assume that we have an extension of fields F _ E _ K,

where E is a Galois extension of F. Assume that also F _ L _ K, and that

K is the composite EL. then K is a Galois extension of L and Gal(K/L) _=

Gal(E/E \ L).

Exercises 2.3

1. Let F _ K be a finite Galois extension. Either prove the following

statements, or give counterexample(s).

(a) Any automorphism of F extends to an automorphism of K.

(b) Any automrophism of K restricts to an automomrphism of F.

2. Recall the “Galois correspondence:”

􀀀 = Gal(E/•) : E/F −! G,

_ = inv•(E) : G −! E/F.

Prove that 􀀀 _ _ _ 􀀀 = 􀀀, and that _ _ 􀀀 _ _ = _. Thus images under

either map are always closed.

3. Let _ = 4

p

2 2 R, and set K = Q(_). Compute the closure of Q in K.

2.3. GALOIS EXTENSIONS AND GALOIS GROUPS 53

4. As in Exercise 3 of Section 2.2, let f(x) = x3 + x2 − 2x − 1 2 Q[x],

and let _ 2 C be a root of f(x). Compute the closure of Q in Q(_).

5. If E _ F is a finite Galois extension, prove that every subgroup of

G = Gal(E/F) is closed.

6. Let E _ K _ F with E _ F algebraic. If E is Galois over K and K is

Galois over F, must it be true that E is Galois over F?

7. Let F _ E be an extension of fields, with E an algebraically closed

field. Assume that F _ K1,K2 _ E are subfields, both algebraic and

Galois over F. If K1 and K2 are F-isomorphic, then they are equal.

Show that the result need not be true if K1 and K2 are not Galois over

F.

8. Let F _ K _ E be an extension of fields with both E and K Galois over

F. Let _ 2 E, with minimal polynomial m_(x) 2 F[x]. If m_(x) =

f1(x)f2(x) · · · fr(x) is the prime factorization of m_(x) in K[x], prove

that fi(x) 6= fj(x) when i 6= j, and that deg fi(x) = deg fj(x) for all

i, j.

9. Let p1, p2, . . . pk be distinct prime numbers, and let E =

Q(p

p1,

p

p2, . . . ,

p

pk). Show that E is a Galois extension of Q, whose

Galois group is an elementary abelian group of order 2k. (Hint: For

each non-empty subset M _ {1, Q 2, . . . , k}, form the integer qM =

i2M pi. Thus the field KM = Q(p

qM) is a subfield of E; prove

that if M1 6= M2 then KM1 6= KM2 . Since there are 2k − 1 nonempty

subsets of {1, 2, . . . , k}, one can apply Exercise 22 of Section 1.7.)

10. Retain the notation and assumptions of the above exercise. Prove that

Q(p

p1 +p

p2 + . . . +p

pk) = Q(p

p1,

p

p2, . . . ,

p

pk).

11. (The Galois group of a simple transcendental extension.) Let F be

a field and let x be indeterminate over F. Set E = F(x), a simple

transcendental extension of F.

(i) Let _ 2 E; thus _ = f(x)/g(x), where f(x), g(x) 2 F[x], and

where f(x) and g(x) have no common factors. Write

f(x) =

Xn

i=0

aixi, g(x) =

Xn

i=0

bixi

54 CHAPTER 2. FIELD AND GALOIS THEORY

where an 6= 0, or bn 6= 0. Therefore, n = max {deg f(x), deg g(x)}.

Note that

(an − _bn)xn + (an−1 − _bn−1)xn−1 + · · · + (a0 − _b0) = 0.

If we set

F(X) =

Xn

i=0

(ai − _bi)Xi 2 F(_)[X],

then x is a root of F(X). Show that F(X) is irreducible in

F(_)[X]. (Hint: By Gauss’ Lemma, F(X) is irreducible in F(_)[X]

if and only if F(X) is irreducible in F[_][X] = F[_,X]. However,

F(X) = F(_,X) = f(X) − _g(X)

which is linear in _. Therefore, the only factors of F(_,X) are

common factors of f(X) and g(X); there are no nontrivial common

factors.)

(ii) From part (i), we see that [F(x) : F(_)] = n. This implies that

any automorphism of F(x) must carry x to f(x)/g(x) where one

of f(x) or g(x) is linear, the other has degree less than or equal to

1, and where f(x) and g(x) have no common non-trivial factors:

x 7!

a + bx

c + dx

, ad − bc 6= 0.

Conversely, any such choice of a, b, c, d determines an automorphism

of F(x). Therefore, we get a surjective homomorphism

GL2(F) −! Gal(F(x)/F).

Note that the kernel of the above homomorphism is clearly Z(GL2(F)),

the set of scalar matrices in GL2(F). In other words,

Gal(F(x)/F) _= PGL2(F).

12. Let F = F2, the field of 2 elements, and let x be indeterminate over

F. From the above exercise, we know that Gal(F(x)/F) _= PGL2(2) _=

GL2(2) _= S3, a group of 6 elements. For each subgroup H _ G =

Gal(F(x)/F), compute invH(F(x)). From this, compute the closure of

F in F(x). (Hint: This takes a bit of work. For example, if _ 2 G is the

involution given by x 7! 1/x, then one sees that invH(F(x)) = F(x +

1/x), where H = h_i. In turns out that invG(F(x)) = F( (x3+x+1)(x3+x2+1)

x2(x2+1) ).)

2.4. SEPARABILITY AND THE GALOIS CRITERION 55

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