Page 282 - DMTH403_ABSTRACT_ALGEBRA

Unit 30: Invariant Subfield

The following conditions are equivalent for an extension field F of K: Notes
(1) F is the splitting field over K of a separable polynomial;

(2) K = F for some finite group G of automorphisms of F;
G
(3) F is a finite, normal, separable extension of K.
If F is an extension field of K such that K = F for some finite group G of automorphisms of F,
G
then G = Gal(F/K).

Example: The Galois group of GF(p ) over GF(p) is cyclic of order n, generated by the
n
automorphism  defined by  (x) = x , for all x in GF(p ). This automorphism is usually known as
n
p
the Frobenius automorphism of GF(p ).
n
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 F , and
H
H = Gal(F/F ).
H
(ii) If E is a subfield of F that contains K, then the corresponding subgroup of G is
H = Gal(F/E), and
E = F .
H
(b) For any subgroup H of G, we have
[F : F ] = | H| and [F : K] = [G : H].
H
H
(c) Under the above correspondence, the subgroup H is normal if and only if the subfield
E = F is a normal extension of K. In this case,
H
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 E and E are conjugate if there exists in Gal(F/K) such that (E ) = E .
2
1
1
2
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.
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
[Fundamental Theorem of Algebra]: Any polynomial in C[x] has a root in C.

Example: Prove that if F is a field extension of K and K = F for a finite group G of
G
automorphisms of F, then there are only finitely many subfields between F and K.
Solution: The given condition is equivalent to the condition that F is the splitting field over K of
a separable polynomial. Since we must have G = Gal (F/K), the fundamental theorem of Galois
theory implies that the subfields between F and K are in one-to-one correspondence with the
subgroups of F. Because G is a finite group, it has only finitely many subgroups.

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