3

3

Let $X$ be a ringed space. Recall that an $\mathcal{O}_X$-module $M$ is called coherent if it is of finite presentation and for every open $U \subseteq X$ every $\mathcal{O}_U$-submodule of finite type of $M|_U$ is of finite presentation. Coherent modules constitute an abelian category (in contrast to modules of finite presentation or just of finite type). See the Stacks project, modules, section 12. In general $\mathcal{O}_X$ might be not coherent.

Question. If $M,N$ are coherent $O_X$-modules, is it true that $M \otimes_{\mathcal{O}_X} N$ is coherent?

I guess that this will be false in this generality. So let us restrict to schemes, w.l.o.g. affine schemes. Here a module $M$ over a ring $A$ is coherent if it is of finite presentation and every submodule of finite type is of finite presentation.

flag

1 Answer

3

In the analytic category this is indeed true: if $\mathscr{F}$, $\mathscr{G}$ are coherent analytic sheaves on a complex space $X$, then $\mathscr{F} \otimes_{\mathscr{O}_X} \mathscr{G}$ is also coherent.

For a reference, look at [Grauert-Remmert, Coherent Analytic Sheaves], Proposition at the bottom of page 240.

It seems to me that their proof works also in the algebraic category, maybe you should check.

link|flag
3 
Thanks! The proof is very easy and actually shows something more general: If $M$ is of finite presentation and $N$ is coherent, then $M \otimes N$ is coherent. Proof: It is clear that $M \otimes N$ is of finite presentation. Now choose a local presentation of $M$ and thereby represent $M \otimes N$ as a cokernel of a morphism of the form $N^r \to N^s$. Since $N^r$ and $N^s$ are coherent, this cokernel is also coherent. – Martin Brandenburg Jan 19 2012 at 16:57

Your Answer

Get an OpenID
or

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