Fixed field of an automorphism group

Let $𝐹 : 𝐾$ be a field extension and $𝐺 \leq A u t (𝐹 : 𝐾)$ be a group of field extension automorphisms. The fixed field of $𝐺$ is the intermediate field #m/def/field

𝐹 𝐺 : = {𝛼 \in 𝐹 : (\forall 𝑔 \in 𝐺) [𝑔 𝛼 = 𝛼]}

The induced correspondence from intermediate fields $𝐹 : 𝐸 : 𝐾$ to subgroups of $𝐺 \leq A u t (𝐹 : 𝐾)$ is called the Galois correspondence, which is an example of a Galois connection in that it is order-reversing:

𝐹 𝐺 \leq 𝐹 𝐻 ⟺ 𝐻 \leq 𝐺 .

Moreover, for any $𝐺 \leq A u t (𝐹 : 𝐾)$ and $𝐹 : 𝐸 : 𝐾$ ,

𝐸 \leq 𝐹 A u t (𝐹 : 𝐸), 𝐺 \leq A u t (𝐹 : 𝐹 𝐺);

Further still, if $𝐺 1, 𝐺 2 \leq A u t (𝐹 : 𝐾)$ and $𝐹 : 𝐸 1, 𝐸 2 : 𝐾$ ,

A u t (𝐹 : 𝐸 1 𝐸 2) = A u t (𝐹 : 𝐸 1) \cap A u t (𝐹 : 𝐸 2), 𝐹 ⟨ 𝐺 1, 𝐺 2 ⟩ = 𝐹 𝐺 1 \cap 𝐹 𝐺 2;

where $𝐸 1 𝐸 2$ and $⟨ 𝐺 1, 𝐺 2 ⟩$ are the generated fields and groups respectively.

Proof

The only of these which is not immediate are the last ones. Suppose $𝜎 \in A u t (𝐹 : 𝐸 1) \cap A u t (𝐹 : 𝐸 2)$ . Then $𝜎$ must also fix anything which can be written as a product of elements in $𝐸 1$ ands $𝐸 2$ , hence $𝜎 \in A u t (𝐹 : 𝐸 1 𝐸 2)$ , and therefore $A u t (𝐹 : 𝐸 1) \cap A u t (𝐹 : 𝐸 2) \leq A u t (𝐹 : 𝐸 1 𝐸 2)$ . Since $𝐸 1, 𝐸 2 \leq 𝐸 1 𝐸 2$ , the other inclusion is immediate.

Similarly, suppose $𝛼 \in 𝐹 𝐺 1 \cap 𝐹 𝐺 2$ . Then $𝛼$ must be fixed by any product of elements from $𝐺 1$ and $𝐺 2$ , hence $𝛼 \in 𝐹 ⟨ 𝐺 1, 𝐺 2 ⟩$ . Since $𝐺 1, 𝐺 2 \leq ⟨ 𝐺 1, 𝐺 2 ⟩$ , the other inclusion is immediate.

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