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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?

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  • $\begingroup$ A natural guess is $C^B=B^*\otimes C$ with $B^*$ the dual space of $B$ with some algebra structure; since $B$ is separable, there are special isomorphisms $B\to B^*$, and one can transport the algebra structure to $B^*$ (and probably $\mathrm{op}$ it?) $\endgroup$ Nov 19, 2013 at 7:02
  • $\begingroup$ This guess is unfortunately wrong. The answer has to make more effective use of the fact that $B$ is an etale $K$-algebra. See my answer below. $\endgroup$
    – Marguax
    Nov 19, 2013 at 16:29
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    $\begingroup$ Dear Mariano: To see explicitly where your suggestion goes awry, note that it would provide an algebra map from $C$ into $C^B$, hence from $C \otimes_K B$ into $C$, and so from $B$ into $C$ (chase $1 \otimes b$). But there is usually no such map. In effect, what is "missing" is to adequately combine the Galois actions on the sets of primitive idempotents of $B \otimes K_s$ and $C \otimes K_s$ (which is what underlies the construction in my answer). $\endgroup$
    – Marguax
    Nov 19, 2013 at 20:24
  • $\begingroup$ @Fujita: In geometric terms, the special case of the "subobject classifier" $2^Y$ for a given finite etale $X$-scheme $Y$ is a representing object for the functor ${\mathbf{Of}}(Y)$ assigning to any $X$-scheme $T$ the set of open and closed subschemes of $Y \times_X T$ (since monic maps in the category of finite etale $X$-schemes are open and closed immersions). See Lemma 18.5.3 in EGA IV$_4$ for a nice discussion of representing this latter functor more generally for $Y \rightarrow X$ any finite locally free morphism (though it is not a "subobject classifier" when $Y$ isn't etale over $X$). $\endgroup$
    – Marguax
    Nov 19, 2013 at 23:20

3 Answers 3

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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.


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. $(g.f)(e) = g(f(g^{-1}e))$).

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  • $\begingroup$ Thank you very much! Although I did know a little about algebraic spaces, I didn't know the notion of Hom-Schemes at all and these very interesting connection. Do you know some textbooks which include these interesting matters? $\endgroup$ Nov 19, 2013 at 21:59
  • $\begingroup$ However, is your answer for the case $B=K^n$ right? I suppose in this case, $C^B$ is seemed to be just a n-times tensor $C \otimes C \otimes \dots \otimes C$. $\endgroup$ Nov 19, 2013 at 22:36
  • $\begingroup$ @Fujita: sorry, I was getting the algebra and geometry mixed up; you are right that it is an $n$-fold tensor product, or more naturally a tensor power indexed by $I$. I will fix this. $\endgroup$
    – Marguax
    Nov 19, 2013 at 22:56
  • $\begingroup$ @Fujita: It is curious that you know a bit about algebraic spaces but not Hom-schemes, as Grothendieck's work on Hilbert schemes, Hom-schemes, Picard schemes etc. predates Artin's work on algebraic spaces and provides much of the motivation for it. I highly recommend the chapter on Hilbert schemes in the book "FGA Explained" (which should discuss Hom-schemes as an application). But for finite flat schemes over a locally noetherian base, Hom-schemes can be built by bare hands as affine over the base: just chase structure constants for algebra structure on vector bundles (good exercise). Enjoy. $\endgroup$
    – Marguax
    Nov 19, 2013 at 23:05
  • $\begingroup$ Oh, I didn't know its motivations. Thanks a lot! $\endgroup$ Nov 19, 2013 at 23:31
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Mariano Suárez-Alvarez' "natural guess" is approximately what is known as the Girard decomposition of implication in linear logic, which he invented.

In proof theory, the idea of linear logic is that it is about resources and instances of a hypothesis are "used up" in a proof. So if $A$ needs to be used twice to prove $B$ then we have $A\otimes A\multimap B$ but not $A\multimap B$. The connection between linear and intuitionistic logic is made by introducing an exponential, written $! A$ and called "of course $A$", where $! A$ is as many $\otimes$-copies of $A$ as are needed.

There are lots of semantic models of linear logic based on combinatorial or order-theoretic (or topological) ideas. By the analogy between lattices and topology on the one hand and rings and algebraic geometry on the other, there ought to be models based on commutative algebra. This is what the question asks, but I have never managed to make it work and I haven't heard that anyone else has done so.

I am therefore just putting the ideas from linear logic and category theory on the table, so that a commutative algebraist might try to turn them into a successful solution of the problem.

The models that are known are qualitatively like the situation in the question. There is a "spatial" view that is dual to the "algebraic" view. Where in proof theory we have intuitionistic and linear proofs (allowing multiple or single use of hypotheses), the maps in the algebraic view that are dual to continuous maps are homomorphisms for algebras that have multiplication as well as addition. Behind these are linear maps, respecting only the addition.

The usual binary connectives $\land$ and $\lor$ of intuitionistic logic split into additive and multiplicative forms in linear logic. Those for $\land$ are written $\times$ and $\otimes$, exactly corresponding to the (categorical) product and tensor product in linear algebra.

However, the familiar case of finite dimensional vector spaces over a field is of only limited help in understanding this structure. This is because the (categorical) product and coproduct (the additive forms of $\land$ and $\lor$) are the same. This is also the case for the multiplicative connectives.

Jean-Yves Girard writes the dual of $\otimes$ as an upside-down ampersand, but I will use $@$ here. If the category of linear maps is self dual, we have $A@B \equiv (A^{op}\otimes B^{op})^{op}$. For finite dimensional vector spaces over a field, $@$ and $\otimes$ are then the same thing, but in other models they need not be.

The $!$ operator links the linear and algebraic categories. but once again we need to go via the opposite categories.

The Girard decomposition is then $C^B\equiv (B\Rightarrow C) \equiv (!B)^{op} @ C$.

Here seems a good place to plug the book Proofs and Types that was written by Girard and translated by me and Yves Lafont. It is a first introduction to logic and proof theory, but was written while he was first developing linear logic and contains some material about it, though the subject has moved on a long way since then.

Now to return to the question, writing ${\bf A}(A,B)$ instead of $\mathrm{Hom}_{\mathrm{Sep}(K)^{op}}(A,B)$.

First let $C=K[x]$, the polynomial algebra over $K$ in one variable. Then we require $$ \mathbf{A}(C^B,A) \equiv \mathbf{A}(C,A\otimes B) \equiv {|A\otimes B|}, $$ the underlying set of the tensor product.

If I were doing topology here, I would let $A$ be (the lattice of open subspaces of) the one-point space, so $C^B$ would be the set $|A\otimes B|$ equipped with some topology. I will leave it to the commutative algebraist and algebraic geometers to fill in this step.

To generalise from $C=K[x]$ to an arbitrary algebra, we express it using generators $G$ and relations $R$ and therefore as a coequaliser. This makes the hom-set an equaliser.

However, I need to point out something that makes me suspect that this strategy is going to fail: assuming that the base field and therefore the algebras have infinite underlying sets, the algebra $C^B$ is going to be a polynomial ring with infinitely many variables, or a more general algebra with infinitely many generators.

It's not going to work with finite dimensional algebras over the field.

The thing that saves the day in the order-theoretic models is Scott continuity (preservation of infinitary joins). There is no analogue of this in commutative algebra.

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  • $\begingroup$ Thank you for your interesting observation and short summary of the strategy. However, $K[X]$ is not finite separable over $K$. Doesn't it matter? $\endgroup$ Nov 19, 2013 at 13:48
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    $\begingroup$ You have asked a difficult question whose solution (if there is one) will require collaboration between people from different disciplines (commutative algebra and models of linear logic), which makes it a good question for MathOverflow. In the first instance, both sides will need to offer and accept ideas without worrying about rigour. Can you fill in some of the obvious gaps in my argument? $\endgroup$ Nov 19, 2013 at 14:07
  • $\begingroup$ @PaulTaylor: If you make your answer CW then Fujita can edit directly into it. The CW option is sadly under-utilized, but it seems like a good way to get people from these different fields to be able to cobble together a meaningful answer in one place. $\endgroup$ Nov 19, 2013 at 14:18
  • $\begingroup$ Paul cannot make this CW himself (only mods can do that now, alas), but if he flags for CW, a mod can make it so. But it's only if he wants (CW answers do not result in gains or losses of rep). $\endgroup$
    – Todd Trimble
    Nov 19, 2013 at 18:16
  • $\begingroup$ As you see I have given a template for a community wiki answer below, although presumably we need to work from Marguax's answer now. $\endgroup$ Nov 19, 2013 at 18:21
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This is a scratchpad for experts in commutative algebra and models of linear logic to put together a possible solution of this problem.

It is probably best to take the ground field $K$ to be $\mathbb C$.

I suspect that Galois theory and insisting on fields are not going to help.

Also, this is not an undergraduate problem.

By linear algebra is meant the category $\mathbf V$ of vector spaces and linear maps over $K$, but it is likely that these will have to be topological.

In this category,

  • the initial object $\bf 0$ is the zero vector space,
  • the terminal object $\bf 1$ is also the zero vector space,
  • the additive sum $A+B$ is the coproduct, which is the direct sum of vector spaces,
  • the additive product $A\times B$ is the product, which is also the direct sum of vector spaces,
  • the multiplicative unit $I$ is the ground field $K$,
  • the multiplicative product $A\otimes B$ is the tensor product over $K$, and
  • the linear hom $A\multimap B$ is the hom-set ${\bf A}(A,B)$ together with the vector space structure inherited from $B$.

There is likely to be some debate over the following:

  • there is a dualising object $\bot$ that may also be the ground field $K$
  • the dual $A^\bot$ is $A\multimap\bot$, but this need not be involutive ($(A^\bot)^\bot=A$),
  • the par (using Girard's name) $A @ B$ must satisfy $(A @ B)^\bot=(A^\bot)\otimes(B^\bot)$ and may also be the tensor product, but need not be.

By algebras is meant the category $\mathbf A$ of commutative unital rings and homomorphisms over this field, but it is likely that these will have to be topological.

Often in models of linear logic the functor $? A$ (why not in Girard) is the free monoid on $A$, ie the free algebra generated by $A$ qua vector space. The functor $! A$ satisfies $!(A^\bot)=(? A)^\bot$.

However, there is a great deal of freedom in the choice of these operations. There can be models with the same underlying linear algebra but different $!$ operations.

We do need Seely's equations: $!(A\times B)=(!A\times !B)$ and $!\mathbf{1}=I$.

As a first step we also need to identify $\Sigma=!I$ and $R=!\Sigma$.

Very likely, $\Sigma=K[x]$, the polynomial ring in one variable.

$R$, considered as a variety, has as points the elements of the ground field. (Please would some commutative algebraist replace this with a more precise description.)

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    $\begingroup$ The case of ground field $\mathbf{C}$, or more generally when $B$ is a product of copies of the ground field, is simple: then the exponential is just $C^n$ where $n = \dim_K(B)$. I don't see how bringing in mathematical logic helps with this stuff. $\endgroup$
    – Marguax
    Nov 19, 2013 at 16:31
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    $\begingroup$ Dear Paul: I'm not using any commutative algebra either (and my view of mathematics goes way way beyond commutative algebra). This problem is purely one of Galois theory after one strips away all of the language; i.e., for the core problem about finite separable $K$-algebras, the mention of toposes and cartesian closed categories seems like a red herring to me. I was being honest I that don't see how those notions (or commutative algebra!) help to solve the problem. Look at the first 3 paragraphs of my solution: the main content is Galois theory in the guise of the displayed expression. $\endgroup$
    – Marguax
    Nov 19, 2013 at 16:56
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    $\begingroup$ Dear Paul: The key facts I am using from Galois theory are Galois descent for maps of finite-dimensional vector spaces (a jazzed-up version of the Galois correspondence) and that for any finite separable extension field $F/K$ and a finite Galois extension $K'/K$ into which $F$ embeds over $K$, the $K'$-algebra $K' \otimes_K F$ is a product of copies of $K'$ (indexed by its own primitive idempotents), as we see by applying the primitive element theorem to $F$. $\endgroup$
    – Marguax
    Nov 19, 2013 at 17:00
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    $\begingroup$ "This problem is purely one of Galois theory" if you're purely an algebraist. On the other hand, there are also people who are primarily interested in models of linear logic and have been scratching their heads for a long time trying to find one based on commutative rings. Also, if you haven't heard of linear logic before, please do not dismiss it as "just mathematical logic" because it has a lot more to do with mathematics and is less obviously logic than you think. $\endgroup$ Nov 19, 2013 at 17:08
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    $\begingroup$ I'm afraid I have to agree with the algebraists here: taking the base field to be $\mathbb{C}$ makes the question asked uninteresting, because the category of finite étale algebras over $\mathbb{C}$ is then equivalent to $\mathbf{FinSet}^\mathrm{op}$. (Note: we are not considering any category of modules here!) But that is not to say linear logic is uninteresting. $\endgroup$
    – Zhen Lin
    Nov 19, 2013 at 18:24

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