I just started reading Hartshorne's "Deformation theory" book and occasionally I get confused by the way he treats various objects (and it seems like it is common in the literature overall, e.g. in the literature on the moduli spaces of sheaves).
A typical example would be as follows (theorem 2.7 in the book):
Given a scheme $X$ over $k$ with a coherent sheaf $\mathcal{F}$ we call a (first-order) deformation of $\mathcal{F}$ over $D=Spec \ k[t]/t^2$ the data of a coherent sheaf $\mathcal{F}'$ over $X'=X\times D$, flat over D, together with a homomorphism $\mathcal{F}' \to \mathcal{F}$ which induces an isomorphism $\mathcal{F}' \otimes k \to \mathcal{F}$.
Then the statement is that such deformations are in one-to-one correspondence with $\text{Ext}_{X}^{1} (\mathcal{F},\mathcal{F}).$
The very first line of the proof says the following: flatness of $\mathcal{F}'$ over $D$ is equivalent to the exactness of $$0\to \mathcal{F} \xrightarrow{t} \mathcal{F}' \to \mathcal{F} \to 0 $$obtained by tensoring $\mathcal{F}'$ with $0 \to k \xrightarrow{t} k[t]/t^2 \to k \to 0$.
Now the question is: what does it mean to tensor a sheaf of $\mathcal{O}_{X'}$-modules with a $k[t]/t^2$ module (not even a sheaf)?
There are various ways I tried to make sense of it in a more sheaf-theoretic way, namely:
- Change the ringed space structure on $X'$ by replacing $\mathcal{O}_{X'}$ with constant sheaf of rings $k[t]/t^2$, interpret $\mathcal{F}'$ as a sheaf of $k[t]/t^2$ modules and treat $k[t]/t^2,k$ as constant sheaves of $k[t]/t^2$ modules on $X'$. Then interpret $\mathcal{F}' \otimes k[t]/t^2, \mathcal{F}' \otimes k$ in the usual way as a tensor product of sheaves of $k[t]/t^2$ modules. Then probably run this machine back and check that the resulting sheaves are actually sheaves of $\mathcal{O}_{X'}$ modules. Seems like a very round-about way of doing things, and seems like it is becoming more and more clumsy if we are taking higher order deformations, or even an arbitrary Artin ring to deform over.
- More naively, interpret say $(\mathcal{F}' \otimes k) \ (U) = \mathcal{F}' (U) \otimes k$, check that it is a sheaf and check that it has whatever structure you are interested in (sheaf of $\mathcal{O}_{X'}, \mathcal{O}_{X}, k[t]/t^2$-modules, etc.) Seems like a nice shortcut way, but it also seems like certain functoriality properties are not so clear anymore, so that the existence of certain maps, behavior under exact sequences, etc. should now be separately checked locally on an open set.
After all I think that the above two interpretations and actually all other conceivable interpretations give the same answer. But is anyone aware of a concise and precise way of interpreting operations like tensoring a sheaf of $\mathcal{O}_{X}$ modules with a field, a vector space, an $R$-module, etc? A pointer to a resource where this is explicated in detail would be much welcomed.
We have a morphism $f: X'\rightarrow \operatorname{Spec} D$ such that the fibre over the closed point $\operatorname{Spec} k \hookrightarrow \operatorname{Spec} D$ is isomorphic to $X$.
Note that the exact sequence $0\rightarrow k \rightarrow k[t]/t^2 \rightarrow k \rightarrow 0$ of $k[t]/t^2$-modules is the same as an exact sequence $0\rightarrow \tilde{k} \rightarrow \widetilde{k[t]/t^2} \rightarrow \tilde{k} \rightarrow 0$ of quasi-coherent modules on $\operatorname{Spec} k[t]/t^2$. Then what is meant by tensoring this with $\mathcal{F}'$ is to apply $f^*$ to this exact sequence and then tensor by $\mathcal{F}'$.
You can think of $\mathcal{F}' \otimes k$ as restricting the coherent sheaf $\mathcal{F}'$ to the closed subscheme $X \hookrightarrow X'$.