I'm going through Vistoli's sections of FGA Explained to begin to understand stacks. It is well-known and proven in the text that the fibered category $QCoh$ of quasi-coherent sheaves is a stack in the fpqc topology. In particular then, given two quasicoherent sheaves $\xi$ and $\eta$ on a scheme $S$, the functor $Hom_U(\xi,\eta):Sch$/$U\rightarrow(Set)$ sending each object $X\rightarrow U$ to the set $Hom_{O_X}(\xi|_X,\eta|_X)$ should be a sheaf in the fpqc topology (this is proposition 4.7 for those who have the text).

So taking $U=SpecA$ affine, $X=U \rightarrow U$ the identity and $V=SpecB \rightarrow U$ a faithfully flat morphism (hence an fpqc covering of $U$), we should have an exact sequence

$0 \rightarrow Hom_{O_U}(\xi,\eta)\rightarrow Hom_{O_V}(\xi|_V,\eta|_V) \rightarrow Hom_{O_{V \times V}}(\xi|_{V \times _U V},\eta|_{V \times _U V})\rightarrow0$

where the second nonzero map is given by the difference between the pullbacks along the two projections.

Ok, so with all this fancy FGA stuff established, my question is actually pretty goofy and basic. Since everything above is affine, we can replace all the schemes by rings and all the sheaves by modules to obtain:

$0 \rightarrow Hom_{A}(M,N)\rightarrow Hom_{B}(M\otimes_A B,N\otimes_A B)$ $\rightarrow Hom_{B \otimes_A B}(M\otimes_A B\otimes_A B,N\otimes_A B\otimes_A B)\rightarrow0$

My question is: what exactly is the second nonzero map in simple terms? For example, if $\phi \in Hom_{B}(M\otimes_A B,N\otimes_A B)$ and $\psi$ is its image in $Hom_{B \otimes_A B}(M\otimes_A B\otimes_A B,N\otimes_A B\otimes_A B)$, can you give me a formula for $\psi(m\otimes b \otimes b')$ in terms of $\phi$ and the letters given. Because the way I'm interpreting the second map, it's always zero, which I'm pretty sure is wrong.

EDIT: Vistoli's section of FGA Explained is available here: http://arxiv.org/abs/math/0412512.

  • 1
    $\begingroup$ There is a more recent version of the document at Vistoli's own site. $\endgroup$ – Zhen Lin Jan 19 '14 at 9:32

In fact, you should be able to think of several homomorphisms $\mathrm{Hom}_B (B \otimes_A M, B \otimes_A N) \to \mathrm{Hom}_{B \otimes_A B} (B \otimes_A B \otimes_A M, B \otimes_A B \otimes_A N)$. The one we need here is induced by the two $A$-algebra homomorphisms $B \to B \otimes_A B$ given by $b \mapsto b \otimes 1$ and $b \mapsto 1 \otimes b$. Recall that, for any $A$-algebra homomorphism $g : B \to C$ whatsoever, there is a homomorphism \begin{align} \mathrm{Hom}_B (M', N') & \to \mathrm{Hom}_C (C \otimes_B M', C \otimes_B N') \\ \phi & \mapsto \mathrm{id}_C \otimes_B \phi \end{align} induced by the functor $C \otimes_B (-)$, where $C$ is a $B$-module via $g$. Now, we can define a $(B \otimes_A B)$-module isomorphism \begin{align} (B \otimes_A B) \otimes_B (B \otimes_A M) & \mapsto B \otimes_A B \otimes_A M \\ (b \otimes b') \otimes (1 \otimes m) & \mapsto b \otimes b' \otimes m \end{align} but note that the meaning of the above formula depends on the $B$-module structure we put on $B \otimes_A B$. So we get two homomorphisms $$\mathrm{Hom}_B (B \otimes_A M, B \otimes_A N) \to \mathrm{Hom}_{B \otimes_A B} (B \otimes_A B \otimes_A M, B \otimes_A B \otimes_A N)$$ induced by the two coprojections $B \to B \otimes_A B$. The differential in the cochain complex you seek is the difference of these two homomorphisms; you can check that, for an element of the form $\mathrm{id}_B \otimes_A \theta$ in $\mathrm{Hom}_B (B \otimes_A M, B \otimes_A N)$, both homomorphisms yield $\mathrm{id}_B \otimes_A \mathrm{id}_B \otimes_A \theta$.

  • $\begingroup$ Ok, thanks. I believe I have the correct pullbacks. If we suppose $\phi(1 \otimes m) = \sum{b_i \otimes n_i}$, then the one pullback of $\phi$ is $b \otimes b' \otimes m \mapsto \sum{b_i b \otimes b' \otimes n_i}$ while the other is $b \otimes b' \otimes m \mapsto \sum{b \otimes b_i b' \otimes n_i}$, so clearly the pullbacks are equal when $\phi(1 \otimes m)=1\otimes n$. On the other hand, how does one show that this is the only way to have equality between the pullbacks, which is what we need for exactness? $\endgroup$ – Cass Jan 19 '14 at 22:46
  • $\begingroup$ Actually, nevermind, since I believe this question is tantamount to the hard part of proving that $QCoh$ is a stack, which takes up several pages in FGA Explained. In any case, do you agree that the pullbacks I've written above are correct? $\endgroup$ – Cass Jan 19 '14 at 22:51
  • $\begingroup$ Yes, your formulae are correct. Checking that this is the only thing in the kernel of the differential is indeed the hard part of verifying faithfully flat descent. $\endgroup$ – Zhen Lin Jan 20 '14 at 0:52

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