Let $K$ be a field and $G=Gal(K\_s/K)$ is its absolute Galois group. Then, by Galois theory, the category of finite separable algebras over $K$ (denoted by $Sep(K)$) and the category of finite continuous $G$-sets with discrete topology are (denoted by $G-Set\_f$) anti-equivalent. 


However, it is well known that the category $G-Set\_f$ is a (elementary) topos, and so is $Sep(K)^{op}$. Especially, it is cartesian closed. I think the product of this category (the coproduct of $Sep(K)$) is just a tensor product $A\otimes\_K B$. So, the exponentials $C^B$ have to satisfy $Hom\_{Sep(K)^{op}}(A\otimes \_K B, C)\simeq Hom\_{Sep(K)^{op}}(A,C^B)$ (equivalently
$Hom\_{Sep(K)}(C,A\otimes \_K B)\simeq Hom\_{Sep(K)}(C^B,A)$) , but I don't know such objects. 


So my question is the followings.


* What is the exponential object of $Sep(K)^{op}$?
* What is the subobject classifier of $Sep(K)^{op}$?


More generally, let $X$ be a connected scheme and $\pi(X,x)$ be its fundamental group for some fixed geometric point $x$. Then, the category of finite etale algebras over $X$ (denoted by $FEt(X)$) and $\pi(X,x)-Set\_{f}$ are anti-equivalent. In this case, what is its exponentials?



Exponentials in the opposite category of finite separable algebras
This has a very simple answer via Galois descent, and after giving it I will provide you with a broader algebro-geometric context (namely: Hom-schemes). Let $K'/K$ be a Galois extension such that $B\_{K'} := K' \otimes\_K B$ is a split $K'$-algebra (i.e., product of copies of $K'$); in other words, $K'$ is a common finite Galois splitting field of the finitely many factor fields of $B$. In concrete terms one would say $B' = {K'}^n$ where $n = \dim\_K(B)$, but it is more natural to regard the index set as not being $\{1,\dots, n\}$ but rather as being the set $I$ of primitive idempotents of $B'$. The point is that $G := {\rm{Gal}}(K'/K)$ naturally acts on $B'$ and hence on $I$, but doesn't "naturally" act on $\{1,\dots,n\}$. 


For $C' := K' \otimes\_K C$ it is clear that the finite etale $K'$-algebra ${C'}^{B'}$ is exactly (**EDIT**: the rest of this paragraph is fixes a mix-up in the original answer) the tensor power ${C'}^{\otimes I}$ indexed by $I$. Using this split-case conclusion over $K'$, the Galois group $G$ naturally acts on ${C'}^{\otimes I}$ by combining its natural action on $C'$ and its natural action on $I$ (via its action on $B'$). In other words, if we let $j\_i:C' \rightarrow {C'}^{\otimes I}$ be the inclusion into the $i$th tensor factor (for $i \in I$) then $G$ acts on the $K'$-algebra ${C'}^{\otimes I}$ via $(g.f)(j\_i(c')) = j\_{g^{-1}(i)}(g^{-1}(c'))$. 


Then $C^B$ is exactly the $K$-subalgebra of $G$-invariants in ${C'}^{\otimes I}$; that's the answer (and I think you won't find anything more "explicit" than that). This is the answer precisely because (by Galois theory!)
for any finite $K$-algebras $R$ and $S$, 
$${\rm{Hom}}\_K(R,S) = {\rm{Hom}}\_{K'}(R',S')^G$$
where $R' = K'\otimes\_K R$, $S' = K' \otimes\_K S$, and $G$ acts in the natural way on $K'$-algebra homomorphism (i.e., $(g.f)(r') = g(f(g^{-1}(r'))$). 


There is no "easy" way to describe this subalgebra of $G$-invariants as an operation directly on $B$ and $C$, and that is why it is a more subtle construction than tensor products, etc. which make sense for general algebras. 




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OK, now for the geometric interpretation.
Let $X$ be a scheme, and $Y \rightarrow X$ and $Z \rightarrow X$ two finite etale $X$-schemes. You seek an "exponential" $Z^Y$ as such; i.e., a finite etale $X$-scheme satisfying ${\rm{Hom}}\_X(W \times\_X Y, Z) = {\rm{Hom}}\_X(W, Z^Y)$ for finite etale $X$-schemes $W$. 


Rather generally, for any $X$-schemes $U$ and $V$ one can consider the Hom-functor $\mathbf{Hom}(U,V)$ whose value on an $X$-scheme $T$ is ${\rm{Hom}}\_T(U\_T,V\_T)$. If this functor is representable, we call such a representing object the "Hom-scheme"; informally, it classifies families of morphisms from $U$ to $V$ over varying $X$-schemes. 


Since the set of $X$-morphisms $f:W \times\_X Y \rightarrow Z$ is in bijective correspondence with the set of $W$-morphisms $W \times\_X Y \rightarrow W \times\_X Z$ (by assigning to $f$ the map given functorially by $(w,y) \mapsto (w, f(w,y))$), we see that $Z^Y$ is exactly a representing object for the Hom-functor $\mathbf{Hom}(Y,Z)$ within the category of finite etale $X$-schemes.


In vast generality (work of Grothendieck in the projective flat case, and Artin in the proper flat case via algebraic spaces) one knows that Hom-schemes do exist (sometimes as algebraic spaces). Of course, one doesn't need such deep results to understand the finite etale case, but it gives some useful context. In other words, you're really asking to represent $\mathbf{Hom}(Y,Z)$ as a finite etale $X$-scheme for $Y$ and $Z$ finite etale over $X$ (with the universal property in the more limited scope of finite etale $X$-schemes). 


Since the functor in question is visibly a Zariski sheaf, we may work Zariski-locally on the base (as we may glue local constructions for representing a globally given functor that is a Zariski sheaf), so we may assume $Y \rightarrow X$ has fiber degree that is some constant $n \ge 0$. By etale descent for morphisms and effectivity of etale descent for affine objects over the base, the functor in question is even an etale sheaf and it suffices to make a representing object (as finite etale scheme over the base) over an etale cover of $X$ (and then use descent theory to get it over the original base $X$). Since we arranged for the fibral degree of $Y$ over $X$ to be constant, there is always an etale cover $X' \rightarrow X$ such that the pullback $Y' = Y \times\_X X'$ is "split": a disjoint union of $n$ copies of $X'$. Then $\mathbf{Hom}(Y',Z') = {Z'}^n$; i.e., $Z^n$ does the job when $Y$ is split with constant degree $n$ over $X$, and the general case is an "etale-twisted form" of that.


If $X$ is connected with geometric point $x$ and associated fundamental group $\pi\_1$, then $\mathbf{Hom}(Y,Z)$ is the finite etale $X$-scheme corresponding to the $\pi\_1$-set ${\rm{Hom}}(Y\_x,Z\_x)$ equipped with the evident $\pi\_1$-action (as for $G$-action on the set of set maps Hom($E$, $E'$) for two $G$-sets $E$ and $E'$ with any group $G$; i.e. $(g.f)(e) = g(f(g^{-1}e))$).


