What is the tensor product of a sheaf and a module?

Question

The following object was studied in section III.12 of Hartshorne's Algebraic Geometry book. Let $A$ be a noetherian ring, $Y=\mathrm{Spec}A$, and $M$ be an $A$-module. Let $f:X\to Y$ be a morphism, and $\mathcal{F}$ be a quasicoherent sheaf on $X$. Then Hartshorne writes $\mathcal{F}\otimes_A M$ without defining what it is (technically, he already used it in Coroally III.9.4).

Here is what I have guessed. For each affine $U\subset X$, $A_U:=\mathcal{O}_X(U)$ is an $A$-algebra. So we have a presheaf by assigning $U$ with the module $\mathcal{F}(U)\otimes_A M$. Then we will call the associated sheaf of this presheaf $\mathcal{F}\otimes_A M$.

But what puzzles me is that in Proposition 12.2, it was claimed that if $C^*(\mathfrak{U}, \mathcal{F})$ is the Čech complex of $\mathcal{F}$, then for any $i_0, \ldots, i_p$, we have

$$\Gamma(U_{i_0,\ldots, i_p}, \mathcal{F}\otimes_A M)=\Gamma(U_{i_0,\ldots, i_p}, \mathcal{F})\otimes_A M.$$ In other words, it seems to me that the book claims that the presheaf that I have defined is in fact a sheaf, which is not obvious to me at all. Could anyone explain what I am missing here?

Edit: In proposition 12.2, it was assumed that $\mathcal{F}$ is flat over $Y$.

Answer 1

I cannot write comments yet so sorry for only slightly elaborating on points that have already been made. It is certain that Hartshorne forgot to assume that the cover $\mathfrak{U}$ in the proof of Proposition 12.2 consists of affine(!) open subschemes. Without this assumption there will not even be any connection between Cech and sheaf cohomology. He also needs to assume that the cover is a finite(!) cover for the tensor product to commute with the products occuring in the Cech complex.

The most natural description of $\mathcal{F}\otimes_A M$ is indeed $\mathcal{F}\otimes_{\mathcal{O}_X} f^* \tilde{M}$. If $U\subset X$ is an affine open subscheme, we have $(\mathcal{F}\otimes_A M)(U)=\mathcal{F}(U)\otimes_A M$. To see this note that $(\mathcal{F}\otimes_{\mathcal{O}_X} f^* \tilde{M})_{|U}=\mathcal{F}_{|U}\otimes_{\mathcal{O}_U} (f_{|U})^*\tilde{M}$. But if $U=\mathrm{Spec}\, B$ then by the description of the pullback of quasi-coherent sheaves between affine schemes we simply have $(f_{|U})^*\tilde{M}=\widetilde{B\otimes_A M}$ and hence $(\mathcal{F}_{|U}\otimes_{\mathcal{O}_U} (f_{|U})^*\tilde{M})(U)=\mathcal{F}(U)\otimes_B (B\otimes_A M)=\mathcal{F}(U)\otimes_A M$ as claimed.

Hence if the cover is a finite cover consisting of open affine subschemes and if the scheme is separated (so that all the intersections occuring in the Cech complex are affine) his claim about the Cech complex of $\mathcal{F}\otimes_A M$ is true.

Answer 2

If you want to interpret $\mathcal F \otimes_A M$ as a sheaf, then your description is correct. Another way to think about it is that $M$ defines a sheaf $\tilde{M}$ on Spec $A$ in the usual way. We can pull this back to $X$ to get an $\mathcal O_X$-module, and then form the tensor product of sheaves $\mathcal F \otimes_{\mathcal O_X} f^*\tilde{M}$. This gives the same answer as your definition. To check this, note that, with your definition, $\tilde{M} = \mathcal O_{\mathrm{Spec} A}\otimes_A M$.

It could be that Hartshorne means $\mathcal F \otimes_A M$ to be the presheaf you described, without sheafifying. (I can't say for sure, since I don't have the book with me.) This would explain his Cech computation.

Here is another possible explanation: if you choose the $U_i$ to be affine, and if $X$ is separated, so that the intersections of the $U_i$ are again affine, then Hartshorne's formula will still be valid even if you pass to the associated presheaf. The reason is that over an affine open set we have $\mathcal F = \widetilde{N}$ for some module $N$, and then Hartshorne's formula would just read $N\otimes M = N\otimes M$, where the left-hand side is the global sections of $\widetilde{N\otimes M}$, and the right-hand side is the global sections of $\widetilde{N}$ tensored with $M$ (both of course equaling $N\otimes M$).