# Separable Algebras 3

Fix a field ${k}$ and let ${k_s}$ be a separable closure. Let ${G=Gal(k_s/k)}$. Today we prove the strongest structure theorem so far: The category of separable ${k}$-algebras is anti-equivalent to the category of finite ${G}$-sets (where the action is continuous). Recall that one way to phrase the ${G}$ action being continuous on ${X}$ is to say that ${X}$ is a union of sets on which the action factors through some finite quotient ${Gal(L/k)}$.

To show the theorem let’s construct our functor ${F: Sep_k \rightarrow G}$-Set (I made this notation up just now, so it isn’t standard). Define ${F(A)=Hom_k (A, k_s)}$. Recall from last time that our separable ${k}$-algebra must have the form ${\prod L_i}$ where ${L_i/k}$ are finite separable field extensions. Thus a map ${A\rightarrow k_s}$ kills all factors except one which lands inside a finite separable extension of ${k}$.

The ${G}$-action on ${F(A)}$ is the one given by acting on ${k_s}$. More specifically, given ${\sigma\in G}$ and ${f\in F(A)}$, we need to define a new ${k}$-algebra map ${\sigma\cdot f}$, but we do this by mapping ${x\mapsto \sigma(f(x))}$. If you try this trick in other situations, be careful. It works here because any element ${\sigma\in G}$ fixes ${k}$ and hence preserves the ${k}$-algebra structure. Suppose ${Im(f)\subset E}$, then the action factors through ${Gal(E/k)}$ and hence the action is continuous.

Now we get the rest of ${F}$ being a contravariant functor for free because we defined it to be ${Hom_k(-, k_s)}$, so any ${\phi: A\rightarrow B}$ gives us ${F(B)\rightarrow F(A)}$ by composing ${(B\stackrel{f}{\rightarrow} k_s)\mapsto (A\stackrel{\phi}{\rightarrow} B\stackrel{f}{\rightarrow} k_s)}$. Of course a morphism in ${G-set}$ must respect the ${G}$-action, but this is true by construction.

We must check that we have a bijection on Hom sets. Suppose we have a ${G}$-homomorphism ${F(B)\rightarrow F(A)}$, i.e. ${Hom_k(B, k_s)\rightarrow Hom_k(A, k_s)}$. We’ll show that ${Hom(F(B),k_s)\simeq B\otimes k_s}$ and ${Hom(F(A),k_s)\simeq A\otimes k_s}$. Thus applying ${Hom(-,k_s)}$ to both sides gives us a map ${A\otimes k_s\rightarrow B\otimes k_s}$. Keeping track of the action we can take the invariants to get ${(A\otimes k_s)^G=A\rightarrow (B\otimes k_s)^G=B}$. Thus from knowledge of ${F(B)\rightarrow F(A)}$ we can completely recover our map ${A\rightarrow B}$ which shows the functor is fully faithful.

The above argument requires us to keep careful track of the action to know it works. Let’s check the isomorphism ${G: A\otimes k_s\simeq Hom(F(A), k_s)}$. The map is given by ${a\otimes \lambda\mapsto f}$ where the map ${f(x)=x(a)\cdot \lambda}$ (evaluation on the first factor followed by multiplication). The action on the left is ${\sigma\cdot(a\otimes\lambda)=a\otimes \sigma(\lambda)}$ and the action on the right is conjugation ${\sigma\cdot f=f^\sigma}$. Let’s check equivariance of ${G}$. Consider ${G(\sigma\cdot(a\otimes\lambda))=x\mapsto x(a)\cdot \sigma(\lambda)}$
${=\sigma(\sigma^{-1}(x(a)\cdot\lambda)}$
${=f^\sigma (x)=\sigma\cdot G(a\otimes \lambda)}$. Thus the isomorphism preserves the ${G}$-action and we see the previous paragraph goes through.

Lastly we need to know the functor is essentially surjective. Let ${X}$ be an arbitrary ${G}$-set. Since ${X}$ is a disjoint union of its orbits and if ${X=F(A)\coprod F(B)}$, then ${X=F(A\times B)}$, we may assume without loss of generality assume the action of ${G}$ is transitive. We know that ${G}$ factors through ${Gal(L/k)}$ for some finite extension ${L}$. Let ${x\in X}$ so that the orbit of ${x}$ is all of ${X}$. The stabilizer of ${x}$ is some subgroup ${H}$ of ${G}$ and so we can define the fixed field ${A=L^H}$. Now we’re done, because ${F(A)=Hom_k(A, k_s)=Hom_k(A, L)}$ and the ${G}$-action is transitive by Galois theory. Thus the map ${X\rightarrow F(A)}$ determined by ${x\mapsto (A\hookrightarrow L)}$ is a ${G}$-isomorphism.

Our functor ${F}$ is fully faithful and essentially surjective and hence is an anti-equivalence of categories.