Galois Theory: A Study of Cyclotomic Field ... - Scripps College

Proof of the Main Theorem of Galois Theory. 13
Since K/F is a Galois extension and F ⊂ L ⊂ K, the extension K/L is
a Galois extension by Corollary 7. So, the degree of the extension [K : L] is
equal to the order of H, denoted [K : L] = |H|. Using Lemma 14, we know
that the order of H equals the degree of [K : K H ], or |H| = [K : K H ]. Since
[K : L] = [K : K H ], and L ⊂ K H , we have L = K H .
Now, suppose we start with H being a subgroup of G and we let L be
the fixed field K H . Then H is a subgroup of G(K/L) since any element in
H fixes L. That is, H ⊂ G(K/L). Now, the order of H is equal to the degree
of the field extension from K H to K. But L = K H , so |H| = [K : K H ]. We
know [K : L] is equal to |G(K/L)| since K/L is Galois. So we can conclude
that [K : K H ] = [K : L] = |G(K/L)|. This gives us H = G(K/L).
Now, we know that the two maps given by
L ↦→ G(K/L) and H ↦→ K H
are inverses. The Main Theorem of Galois Theory, as you may recall, stated
that if K was a Galois extension of a field F , and G = G(K/F ) was its Galois
group, then there is a bijective map H ↦→ K H from the set of subgroups
of G to the set of intermediate fields F ⊂ L ⊂ K. Further, its inverse function
given by L ↦→ G(K/L). Further still, if H = G(K/L), then the degree
of H is equal to [K : L], and [L : F ] = [G : H]. Since K/F was chosen to be
a Galois extension of L, the degree of G is [K : F ]. Since |H| = [K : L] and
[K : F ]/[K : L] = [L : K], with |G| being equal to [K : F ] and |H| equal
to [K : L], we know [L : F ] = [G : H], and thus we have proved the Main
Theorem of Galois Theory.
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