Is the category of affine schemes (over a fixed field) Cartesian closed? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-25T09:16:14Z http://mathoverflow.net/feeds/question/31502 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/31502/is-the-category-of-affine-schemes-over-a-fixed-field-cartesian-closed Is the category of affine schemes (over a fixed field) Cartesian closed? Theo Johnson-Freyd 2010-07-12T04:36:30Z 2010-07-13T00:21:32Z <p>This is probably a trivial question, but I don't see the answer, and I haven't found it on <a href="http://en.wikipedia.org/wiki/Cartesian_closed_category" rel="nofollow">Wikipedia</a>, <a href="http://ncatlab.org/nlab/show/cartesian+closed+category" rel="nofollow">nLab</a>, nor <a href="http://mathoverflow.net/questions/19004/is-the-category-commutative-monoids-cartesian-closed" rel="nofollow">MathOverflow</a>.</p> <p>Let <code>$\text{ComAlg}$</code> denote the category whose objects are commutative algebras over a fixed field <code>$\mathbb K$</code> and whose morphisms are homomorphisms of algebras, and let <code>$\text{ComAlg}^{\rm op}$</code> denote its opposite category. Given commutative algebras <code>$A,B$</code>, let <code>$\operatorname{hom}(A,B)$</code> denote the set of algebra homomorphisms <code>$A\to B$</code>, so that <code>$\operatorname{hom}$</code> is the usual functor <code>$\text{ComAlg}^{\rm op} \times \text{ComAlg} \to \text{Set}$</code>. The short version of my question:</p> <blockquote> <p>Is <code>$\text{ComAlg}^{\rm op}$</code> Cartesian closed?</p> </blockquote> <p>The long version of my question (if I've gotten all the signs right):</p> <blockquote> <p>Is there a functor <code>$[,] : \text{ComAlg} \times \text{ComAlg}^{\rm op} \to \text{ComAlg}$</code> such that there is an adjunction (natural in <code>$A,B,C$</code>, i.e. an isomorphism of functors <code>$\text{ComAlg}^{\rm op} \times \text{ComAlg} \times \text{ComAlg} \to \text{Set}$</code>) of the form: <code>$$\operatorname{hom}([A,B],C) \cong \operatorname{hom}(A,B\otimes C) ?$$</code></p> </blockquote> <p>Recall: <code>$\otimes$</code> is the coproduct in <code>$\text{ComAlg}$</code>, hence the product in <code>$\text{ComAlg}^{\rm op}$</code>.</p> <p>Motivation: <code>$\text{ComAlg}^{\rm op}$</code> is complete and cocomplete, and so many constructions that make sense in <code>$\text{Set}$</code> and <code>$\text{Top}$</code> transfer verbatim to the algebraic setting. I would like to know how many.</p> http://mathoverflow.net/questions/31502/is-the-category-of-affine-schemes-over-a-fixed-field-cartesian-closed/31628#31628 Answer by BCnrd for Is the category of affine schemes (over a fixed field) Cartesian closed? BCnrd 2010-07-12T22:59:13Z 2010-07-12T22:59:13Z <p>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.)</p> http://mathoverflow.net/questions/31502/is-the-category-of-affine-schemes-over-a-fixed-field-cartesian-closed/31634#31634 Answer by Josh Shadlen for Is the category of affine schemes (over a fixed field) Cartesian closed? Josh Shadlen 2010-07-13T00:21:32Z 2010-07-13T00:21:32Z <p>The existence of such an adjunction implies that $B \otimes -$ preserves limits, which doesn't seem very likely.</p> <p>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.</p>