Excerpted from Beachy/Blair, Abstract Algebra, 2nd Ed. © 1996

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§ 8.3 The Fundamental Theorem of Galois Theory

 
Proposition 8.3.1. Let F be a field, and let G be a subgroup of Aut(F). Then

{ a in F | (a) = a   for all   in G }

is a subfield of F.

 
Definition 8.3.2. Let F be a field, and let G be a subgroup of Aut (F). Then

{ a in F | (a) = a   for all   in G }

is called the G-fixed subfield of F, or the G-invariant subfield of F, and is denoted by FG.

 
Proposition 8.3.3. If F is the splitting field over K of a separable polynomial and G = Gal(F/K), then FG = K.

 
Lemma 8.3.4. [Artin] Let G be a finite group of automorphisms of the field F, and let K = FG. Then

[F:K] | G |.

 
Definition 8.3.5. Let F be an algebraic extension of the field K. Then F is said to be a normal extension of K if every irreducible polynomial in K[x] that contains a root in F is a product of linear factors in F[x].

 
Theorem 8.3.6. The following conditions are equivalent for an extension field F of K:

(1) F is the splitting field over K of a separable polynomial;

(2) K = FG for some finite group G of automorphisms of F;

(3) F is a finite, normal, separable extension of K.

 
Corollary 8.3.7. If F is an extension field of K such that K = FG for some finite group G of automorphisms of F, then G = Gal(F/K).

 
Example 8.3.1. The Galois group of GF(pn) over GF(p) is cyclic of order n, generated by the automorphism defined by (x) = xp, for all x in GF(pn). This automorphism is usually known as the Frobenius automorphism of GF(pn).

 
Theorem 8.3.8. [Fundamental Theorem of Galois Theory] Let F be the splitting field of a separable polynomial over the field K, and let G = Gal(F/K).

(a) There is a one-to-one order-reversing correspondence between subgroups of G and subfields of F that contain K:

(i) If H is a subgroup of G, then the corresponding subfield is FH, and

H = Gal(F/FH).

(ii) If E is a subfield of F that contains K, then the corresponding subgroup of G is H = Gal(F/E), and

E = FH.

(b) For any subgroup H of G, we have

[F:FH] = | H | and [FH:K] = [G:H].

(c) Under the above correspondence, the subgroup H is normal if and only if the subfield E = FH is a normal extension of K. In this case,

Gal(E/K) Gal(F/K) / Gal(F/E).

 
In the statement of the fundamental theorem we could have simply said that normal subgroups correspond to normal extensions. In the proof we noted that if E is a normal extension of K, then (E) E for all in Gal(F/K). In the context of the fundamental theorem, we say that two intermediate subfields E1 and E2 are conjugate if there exists in Gal(F/K) such that ( E1 ) = E2. The next result shows that the subfields conjugate to an intermediate subfield E correspond to the subgroups conjugate to Gal(F/E). Thus E is a normal extension if and only if it is conjugate only to itself.

 
Proposition 8.3.9. Let F be the splitting field of a separable polynomial over the field K, and let E be a subfield such that K E F, with H = Gal(F/E). If is in Gal(F/K), then

Gal(F/(E)) = H -1.

 
Theorem 8.3.10. [Fundamental Theorem of Algebra] Any polynomial in C[x] has a root in C.



§ 8.3 Solved problems

 
6. Prove that if F is a field extension of K and K = FG for a finite group G of automorphisms of F, then there are only finitely many subfields between F and K.     Solution

7. Let F be the splitting field over K of a separable polynomial. Prove that if Gal (F/K) is cyclic, then for each divisor d of [F:K] there is exactly one field E with K E F and [E:K] = d.     Solution

8. Let F be a finite, normal extension of Q for which | Gal (F/Q) | = 8 and each element of Gal (F/Q) has order 2. Find the number of subfields of F that have degree 4 over Q.     Solution

9. Let F be a finite, normal, separable extension of the field K. Suppose that the Galois group Gal (F/K) is isomorphic to D7. Find the number of distinct subfields between F and K. How many of these are normal extensions of K?     Solution

10. Show that F = Q(, i) is normal over Q; find its Galois group over Q, and find all intermediate fields between Q and F.     Solution

 


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