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This is probably a trivial question, but I don't see the answer, and I haven't found it on Wikipedia, nLab, nor MathOverflow.

Let $\text{ComAlg}$ denote the category whose objects are commutative algebras over a fixed field $\mathbb K$ and whose morphisms are homomorphisms of algebras, and let $\text{ComAlg}^{\rm op}$ denote its opposite category. Given commutative algebras $A,B$, let $\operatorname{hom}(A,B)$ denote the set of algebra homomorphisms $A\to B$, so that $\operatorname{hom}$ is the usual functor $\text{ComAlg}^{\rm op} \times \text{ComAlg} \to \text{Set}$. The short version of my question:

Is $\text{ComAlg}^{\rm op}$ Cartesian closed?

The long version of my question (if I've gotten all the signs right):

Is there a functor $[,] : \text{ComAlg} \times \text{ComAlg}^{\rm op} \to \text{ComAlg}$ such that there is an adjunction (natural in $A,B,C$, i.e. an isomorphism of functors $\text{ComAlg}^{\rm op} \times \text{ComAlg} \times \text{ComAlg} \to \text{Set}$) of the form: $$ \operatorname{hom}([A,B],C) \cong \operatorname{hom}(A,B\otimes C) ?$$

Recall: $\otimes$ is the coproduct in $\text{ComAlg}$, hence the product in $\text{ComAlg}^{\rm op}$.

Motivation: $\text{ComAlg}^{\rm op}$ is complete and cocomplete, and so many constructions that make sense in $\text{Set}$ and $\text{Top}$ transfer verbatim to the algebraic setting. I would like to know how many.

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If you restrict to the category of $k$-algebras of finite dimension over $k$, then the answer is yes, since Hom scheme constructions work well for flat proper maps. –  S. Carnahan Jul 12 '10 at 5:44
    
Scott, you're right that the theory of Hom-schemes underlies the issues here, though of course upon restricting to the finite-dimensional case everything can be seen "by hand" without recourse to Hom-schemes. (That is, it feels like killing a fly with a sledgehammer in this setting to haul out general theorems on Hom-schemes for flat proper maps.) And of course, to be correct, one really shouldn't say "Hom-scheme" if relaxing projective to proper, as then the Hom-functor is an algebraic space and generally not represented by a scheme. –  BCnrd Jul 12 '10 at 6:29
    
@BCnrd: Well, I haven't yet sat down to work out the counterexample. But you should leave your comment as an answer so that I can accept it. –  Theo Johnson-Freyd Jul 12 '10 at 21:07
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It might be of interest to know that the category of cocommutative coalgebras is cartesian closed (cf. fact that finite-dimensional cocommutative coalgebras is the category opposite to f.d. commutative algebras). –  Todd Trimble Sep 8 '10 at 11:13
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And while I'm at it (this goes back to the motivation given at the end of the question), the category of cocommutative coalgebras is also cocomplete, cocomplete, extensive, and locally finitely presentable. So a lot of arguments that apply to $Set$ also apply there. One thing missing though is that quotients are not preserved by pulling back along morphisms. –  Todd Trimble Sep 8 '10 at 12:00

2 Answers 2

up vote 7 down vote accepted

The existence of such an adjunction implies that $B \otimes -$ preserves limits, which doesn't seem very likely.

Here is a counterexample, though probably not the simplest one. Set $B = k[y]$ and consider the inverse limit of $k[x]/(x^{n+1})$. If we take the tensor products first, then we get $k[y][[x]]$ while if we take the limit first we obtain $k[[x]][y]$. These are distinct, since the first contains for example $(1-yx)^{-1} = \sum_{k \geq 0} y^k x^k$ and the second does not.

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More simple, $k[y] \otimes - \cong -[y]$ doesn't preserve products (column-finite matrices vs. matrices). –  Martin Brandenburg Jan 5 '13 at 0:39

Set $A = B = k[x]$ and figure out for yourself what that is a counterexample. (Hint: rigorously prove that there's no "universal polynomial" over $k$-algebras.)

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