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.

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