This is secretly a post to experiment with Luca Trevisan‘s amazing LaTeX-to-Wordpress tool, and maybe do a token amount of revision in the process. Enjoy, and please point out if I could have achieved anything more efficiently. Note that though this proof is logically self-contained, it requires the reader to already be pretty familiar with the subject, so is probably quite unfriendly to others (sorry).

We roughly follow Teruyoshi Yoshida‘s `minimal notes’ to prove the fundamental theorem as efficiently as possible using basic linear algebra. I have probably weakened the generality of his results a bit but hopefully made the presentation slightly friendlier along the way. We pass via two important propositions (the tower laws and Dedekind’s lemma) to a proof that whenever is Galois, and a subextension, is Galois. We then construct the final important algebraic ingredient, which gives us an all-important upper bound on the degree of an extension over a *fixed field*.

If , finite, then is finite and

Also, for any ,

*Proof:* If a basis for , and a basis for then is a basis for , proving the first part. The second is a simple check that the sets on both sides contain each other, by restricting a map to and then viewing as an appropriate embedding of (the other direction being trivial).

If are distinct elements of , they are linearly independent considered as elements of . In particular, since they are all linear maps of -vector spaces , we have

*Proof:* Suppose not, so there exists a dependence relation between the . Let be such a relation with minimal. Since they are distinct morphisms, we can pick with . Applying the dependence relation twice and subtracting, for any ,

So we have found a dependence relation with smaller, a condtradiction.

If is Galois (in the sense that ) then for any intermediate field , is Galois.

*Proof:* Using the tower law for homomorphisms,

So since there are at most -homomorphisms , and at most -homomorphisms in each factor, the tower law for fields implies we have equality everywhere, and in particular that (taking as the inclusion map) .

We have still not yet talked about fixed fields. This will be an important part of the fundamental theorem of Galois theory, which is about subfields fixed by subgroups of the automorphism group of an extension. Indeed, one way to define a Galois extension is as one with the property that is *precisely* the set of elements of fixed by . Let us see why this is true.

If a finite field extension, , then

*Proof:*It is clear that from proposition 1.2. To do the other direction, let us use a similar algebraic method to the proof of proposition 1.2. Suppose , so let , and be a basis for , and consider the system of linear equations (in variables taking values in )

There are fewer equations than unknowns, so there must exist nontrivial solutions to the system. Let us take such a solution with minimal

Dividing through by and noting that since the are a basis, some must lie outside , take any not fixing . Since is a group, premultiplying by permutes its elements. Therefore, taking of each of the above equations and subtracting them from another corresponding equation will cancel the term in the th position and fail to cancel a term in the th position, giving a nontrivial solution with smaller .

We have now assembled enough tools to prove the fundamental theorem of Galois theory.

Let be Galois. The mappings and are mutually inverse.

*Proof:* Obviously and , and we shall establish both are equalities by dimension counting. Since is Galois by theorem 1.3, proposition 1.4 gives . Since is Galois by theorem 1.3, proposition 1.4 gives .

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