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)}
{=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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