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 {F/K} is Galois, and {F/L} a subextension, {F/L} 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 {F/L}, {L/K} finite, then {F/K} is finite and

\displaystyle [F:L] [L:K] = [F:K].

Also, for any {E/K},

\displaystyle |\text{Hom}_K(F, E)| = \coprod_{\sigma \in \text{Hom}_K(L,E)} |\text{Hom}_L(F, E_\sigma)|.

Proof: If {e_1,...,e_n} a basis for {L/K}, and {f_1,...,f_m} a basis for {F/K} then {e_1f_1, e_1f_2, ....,e_nf_m} is a basis for {M/K}, proving the first part. The second is a simple check that the sets on both sides contain each other, by restricting a map to {L} and then viewing as an appropriate {L} embedding of {F} (the other direction being trivial). \Box

If {\sigma_1,...,\sigma_n} are distinct elements of {\text{Hom}_K(F,E)}, they are linearly independent considered as elements of {E^F}. In particular, since they are all linear maps of {K}-vector spaces {F \rightarrow E}, we have

\displaystyle |\text{Hom}_K(F,E)| \leq [F:K].

Proof: Suppose not, so there exists a dependence relation between the {\sigma_i}. Let {\sum_{i=1}^k \lambda_i \sigma_i = 0} be such a relation with {k} minimal. Since they are distinct morphisms, we can pick {x \in F} with {\sigma_k(x) \not= \sigma_1(x)}. Applying the dependence relation twice and subtracting, for any {y \in F},

\displaystyle 0 = \sum_{i=1}^k \lambda_i (\sigma_i(xy) - \sigma_k(x)\sigma_i(y)) = (\sum_{i=1}^{k-1} \lambda_i (\sigma_i(x) - \sigma_k(x))\sigma_i)y.

So we have found a dependence relation with {k} smaller, a condtradiction. \Box

If {F/K} is Galois (in the sense that {|\text{Aut}_K(F)| = [F:K]}) then for any intermediate field {F/L/K}, {F/L} is Galois.

Proof: Using the tower law for homomorphisms,

\displaystyle [F:K] = |\text{Aut}_K(F)| = \coprod_{\sigma \in \text{Hom}_K(L,F)} |\text{Hom}_L(F, F_\sigma)|.

So since there are at most {[L:K]} {K}-homomorphisms {L \rightarrow F}, and at most {[F:L]} {L}-homomorphisms in each factor, the tower law for fields implies we have equality everywhere, and in particular that (taking {\sigma: L \rightarrow F} as the inclusion map) {|\text{Hom}_L(F, F)| = [F:L]}. \Box

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 {F/K} is as one with the property that {K} is precisely the set of elements of {F} fixed by {\text{Aut}_K(F)}. Let us see why this is true.

If {F/K} a finite field extension, {H \leq \text{Aut}_K(F)}, then

\displaystyle [F:F^H] = |H|.

Proof:It is clear that {[F:F^H] \geq |H|} from proposition 1.2. To do the other direction, let us use a similar algebraic method to the proof of proposition 1.2. Suppose {n = [F:F^H] > |H| = m}, so let {\{\sigma_1,...,\sigma_m \} = H}, and {x_1,...,x_n} be a basis for {F/F^H}, and consider the system of linear equations (in variables {c_1,...,c_n} taking values in {F})

\displaystyle c_1 \sigma_i(x_1) + c_2 \sigma_i(x_2) + ... + c_n \sigma_i(x_n) = 0\ \ \ \ \ (i=1,2,...,m).

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

\displaystyle c_1 \sigma_i(x_1) + c_2 \sigma_i(x_2) + ... + c_k \sigma_i(x_k) = 0\ \ \ \ \ \forall i.

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

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

Let {F/K} be Galois. The mappings {L \mapsto \text{Aut}_L(F)} and {H \mapsto F^H} are mutually inverse.

Proof: Obviously {F^{\text{Aut}_L(F)} \supset L} and {\text{Aut}_{F^H}(F) \supset H}, and we shall establish both are equalities by dimension counting. Since {F/L} is Galois by theorem 1.3, proposition 1.4 gives {[F: F^{\text{Aut}_L(F)}] = |Aut_L(F)| = [F:L]}. Since {F/F^H} is Galois by theorem 1.3, proposition 1.4 gives {|Aut_{F^H}(F)| = [F:F^H] = |H|}. \Box

Advertisements