Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Let $R$ be a ring and $X,Y$ two $R$-schemes, which you may assume to be noetherian or anything reasonable you like. Is it possible to "construct" $\text{Qcoh}(X \times_R Y)$ out of $\text{Qcoh}(X)$ and $\text{Qcoh}(Y)$ in the $2$-category of all cocomplete $R$-linear tensor categories?

Perhaps it is the $2$-coproduct? So the question is if for every cocomplete $R$-linear tensor category $C$ the canonical functor

$\text{Hom}(\text{Qcoh}(X \times_R Y),C) \to \text{Hom}(\text{Qcoh}(X),C) \times \text{Hom}(\text{Qcoh}(Y),C)$

$F \mapsto (F \circ (p_X)^*, F \circ (p_Y)^*)$

is an equivalence of categories. This is satisfied if $X,Y$ are affine, but I think also when $X,Y$ are projective over $R$ (EDIT: Yes, now I've proved this in detail, should I write it up?). Actually for my purposes it would be enough to prove that the functor is conservative, i.e. reflects isomorphisms.

Here was a similar question on MO, but it adresses (as with the answer by David Ben-Zvi) only the derived setting, but I want to work with the usual category of quasi-coherent modules.

share|improve this question
Typo in sentence two, you mean qcoh(y). –  B. Bischof Feb 17 '11 at 15:36
Meanwhile I can also prove that for $X/S$ quasi-projective and $Y/S$ qcqs, and $S$ qcqs, then $\mathsf{Qcoh}(X \times_S Y)$ is the $2$-pushout of $\mathsf{Qcoh}(X)$ and $\mathsf{Qcoh}(Y)$ over $\mathsf{Qcoh}(S)$. For the case of general qcqs $X,Y$ over a field, I have made some progress. It's almost proven ... –  Martin Brandenburg Oct 3 '13 at 9:28
add comment

2 Answers

I have recently proved a result that gives a partial answer to this question, see here (I need to make some assumptions on $X$ and $Y$). Let $X$ and $Y$ be noetherian schemes with the resolution property (that is, every coherent sheaf is a quotient of a locally free sheaf of finite rank). This includes all separated regular noetherian schemes and all projective noetherian schemes. I don't know a quasi-compact semi-separated scheme which does not have the resolution property.

If $X \times Y$ is also noetherian, then the category of coherent sheaves on $X \times Y$ is given by the Deligne tensor product of the categories $\operatorname{Coh}(X)$ and $\operatorname{Coh}(Y)$ of coherent sheaves on the two factors: There is an equivalence

$\operatorname{Coh}(X\times Y) \simeq \operatorname{Coh}(X) \boxtimes \operatorname{Coh}(Y)$

of symmetric monoidal $R$-linear categories.

There is an analogous result without the noetherian assumptions, but you have to consider the categories of finitely presentable quasi-coherent sheaves instead.

If $X$ and $Y$ have the strong resolution property (which means that the locally free sheaves of finite rank form a generator of the respective categories of quasi-coherent sheaves), then there is an equivalence

$\operatorname{QCoh}_{fp}(X\times Y) \simeq \operatorname{QCoh}_{fp}(X) \boxtimes \operatorname{QCoh}_{fp}(Y)$

of symmetric monoidal $R$-linear categories, where now $\boxtimes$ denotes Kelly's tensor product of finitely cocomplete $R$-linear categories. To prove this, I also show that this tensor product gives a bicategorical coproduct in the 2-category of "right exact symmetric monoidal categories," that is, symmetric monoidal finitely cocomplete $R$-linear categories for which tensoring with a fixed object is right exact. By passing to categories of ind-objects you find that the category of all quasi-coherent sheaves on the product is equivalent to the desired bicategorical coproduct.

share|improve this answer
Thanks a lot, this should - finally - answer my question! I have a look at your new paper (the last one about the characterization of categories of quasi-coherent sheaves has already been quite exciting, although I would like to get rid of the resolution properties). –  Martin Brandenburg Nov 16 '12 at 10:59
add comment

Since there's no response to this question so far I can break out my broken record and point out such theorems (sheaves on the product are tensor product of categories of sheaves on the factors) are true, and in fact extremely easy, in the derived setting (modulo the appropriate $\infty$-categorical technology). First for any quasicompact quasiseparated scheme we know the derived category is compactly generated (thanks to Thomason-Trobaugh --in fact by a single object, see Bondal-van den Bergh). Thus the (enhanced) derived category is just dg modules over the ext algebra of this generator (or over the small category of generators). Now it's easy to see that the external product of generators is a generator for the product. Hence sheaves on the product are modules for the tensor product of the dg algebras, ie sheaves on the product are the tensor product of categories of sheaves on each factor.

(This result is due to Toen in his great Inventiones paper on Morita theory. Nowadays it's much easier, thanks to Lurie foundations, literally this easy argument can be found in my paper with Francis and Nadler, modulo quoting DAG a lot..)

I would be curious to see the answer to your original question though --- my gut feeling is it's not nearly as easy if true.. but maybe an argument for schemes with ample sequences of line bundles can be made to imitate this (ie if you know you have generators with enough exactness you might have a shot..)

share|improve this answer
Thanks. I must admit that I have not enough background to understand your answer. But I actually doubt that we can just use "generators" here and related questions. The problem with these is that a represention with generators is always non-unique, and then it's a hard step to show functoriality. –  Martin Brandenburg Feb 21 '11 at 8:33
You can use generators to show a category is the tensor product of others, since modules over a tensor product of algebras are given by the categorical tensor product of module categories. So if eg you were in an Artinian situation and you had a projective generators, you'd describe your categories as modules for their endomorphisms and thus describe the tensor product category. The problem is that you don't tend to have projective generators! –  David Ben-Zvi Feb 21 '11 at 16:13
add comment

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.