Is the category of affine schemes (over a fixed field) Cartesian closed? - MathOverflow

Is the category of affine schemes (over a fixed field) Cartesian closed?

2010-07-12T04:36:30Z

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.

Answer by BCnrd for Is the category of affine schemes (over a fixed field) Cartesian closed?

2010-07-12T22:59:13Z

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.)

Answer by Josh Shadlen for Is the category of affine schemes (over a fixed field) Cartesian closed?

2010-07-13T00:21:32Z

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.