2
$\begingroup$

If $\mathcal{M,N}$ are the associated sheaf of $A$ modules $M$ and $N$ on $X=Spec A,$ then what is $\mathcal{Hom_{O_X}(M,N)}$?Is that the associated sheaf of $Hom(M,N)\ ?$

$\endgroup$
6
  • 1
    $\begingroup$ One is a sheaf, the other is a group. How could you ever have an example where they are the same? $\endgroup$
    – rghthndsd
    Sep 26, 2013 at 13:30
  • $\begingroup$ I'm sorry its such a basic question.. You are right $\endgroup$
    – user40534
    Sep 26, 2013 at 13:31
  • $\begingroup$ If $\mathcal{M,N}$ are the associated sheaf of A modules M,N on X=Spec A, then what is $\mathcal{Hom_{O_X}(M,N)}$? $\endgroup$
    – user40534
    Sep 26, 2013 at 13:37
  • 1
    $\begingroup$ Module $M$ should be coherent. $\endgroup$ Sep 26, 2013 at 13:41
  • 3
    $\begingroup$ Assume $M$ is finitely presented and prove that for any multiplicative set $$S^{-1}(Hom(M,N)) \cong Hom_{S^{-1}(A)}(S^{-1}(M), S^{-1}(N))$$. Then play the usual game with sheaves on a good base... Or just look it up in EGA or Hartshorne chapter 3 section on local Ext or any beginning book ... $\endgroup$ Sep 26, 2013 at 15:39

1 Answer 1

8
$\begingroup$

Just to give a few more details on Daniel's comments:

In general, $\mathcal{H}om_{\mathcal{O}_X}(\mathcal{M},\mathcal{N})$ is not the associated sheaf to $Hom_A(M,N)$. Simple example: Take $A = \mathbb{Z}$, $M = \mathbb{Z}[\frac12]$ and $N = \mathbb{Z}$. Then $Hom_A(M,N) = 0$, but the Hom-sheaf is non-zero evaluated at the non-vanishing locus of $2$ (also known as $Spec \mathbb{Z}[\frac12]$).

Now assume that $M$ is a finitely presented module. Since taking associated module is left adjoint to global sections, we have a canonical map $$\widetilde{Hom_A(M,N)} \to \mathcal{H}om_{\mathcal{O}_X}(\mathcal{M},\mathcal{N}).$$ It is enough to show that this is an isomorphism on the standard opens $D(f)$ for $f \in A$. If we evaluate the source on $D(f)$, we get $Hom_A(M,N)[\frac 1f]$. If we evaluate the target on $D(f)$, we get $Hom_{A[\frac 1f]}(M[\frac1f], N[\frac1f]) \cong Hom_A(M, N[\frac1f])$. The localization $N[\frac1f]$ is isomorphic to the direct limit of

$$ N \xrightarrow{f\cdot} N \xrightarrow{f\cdot}\cdots. $$

Taking Hom out of a finitely presented module commutes with direct limits. Thus, $$Hom_A(M, N[\frac1f]) \cong Hom_A(M, N)[\frac1f]$$ as was to be shown.

$\endgroup$
0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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