Math, asked by mintukumarjen, 1 month ago

State and prove fundamenta theorem of Galois theory​

Answers

Answered by MissMridu
2

(The Fundamental Theorem of Galois Theory).

Let  L/K be a finite Galois extension. Then there is an inclusion reversing

bijection between the subgroups of the Galois group Gal(L/K) and intermediary subfields L/M/K. Given a subgroup H, let M = L

H and

given an intermediary field L/M/K, let H = Gal(L/M).

Proof. This will be an easy consequence of all that has gone before.

Suppose that we are given a subgroup H of G. Let M = L

H and

then set K = Gal(L/M). We want to show that K = H. As we have

already proved that

H ⊂ K,

and |G| = [L : K] is finite, it suffices to prove that the cardinality of

K and is at most the cardinality of H. But But

|H| = [L : M],

and there are at most [L : M] automorphisms of L/M, so that

|K| ≤ [L : M] = |H|.

Thus H = K, and the composition one way is the identity.

Now suppose that we start with L/M/K. Let H = Gal(L/M) and

let N = L

H. We already know that

M ⊂ N,

and so by the Tower Law it suffices to prove that

[L : N] ≥ [L : M].

As L/K is Galois, then so is L/M. But then

[L : M] = |H|.

As H is a set of automorphisms of L/N, we have

[L : N] ≥ |H| = [L : M].

Thus M = N and the composition the other way is the identity. Thus

we have a correspondence. We have already seen that this correspondence is inclusion reversing.

The rest of the course will be adressed to deriving consequences of

the Fundamental Theorem. We start by observing

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