To be precise, I want to know whether the following statement is true or false:

Let $A$ be a ring (it can be reduced), $f:\rm{Spec}(A/I) \to \rm{Spec}(A)$ is a closed immersion, $\tilde{M}$ is a quasi-coherent sheaf of $\mathcal{O}_{\rm{Spec}(A)}$-module, is it true $f^{-1}(\tilde{M})$ is also a quasi-coherent sheaf of $\mathcal{O}_{\rm{Spec}(A/I)}$-module?

  • 3
    $\begingroup$ No, it is even not a module over $O_{\mathrm{Spec}(A/I)}$ in general. Actually let $F=\tilde{M}$. The stalk $f^{-1}(F)_x=F_{f(x)}$ and the RHS is not a $A/I$-module (not killed by $I$) in general. $\endgroup$ – user18119 Jan 10 '12 at 20:43
  • $\begingroup$ Yes, I see. Thank you! $\endgroup$ – Li Zhan Jan 11 '12 at 17:58

I think the answer depends on what exactly you mean by $f^{-1}$.

One has the usual inverse image sheaf $f^{-1} \mathcal F$.

However, in the case of $\mathcal O$-modules (where $(X,\mathcal O)$ is a scheme), one usually considers a variant of $f^{-1}$. Namely:

Let $f : X \to Y$ be any morphism of schemes and let $\mathcal F$ be a $\mathcal O_Y$-module on $Y$. Then one defines the pullback $f^{*}\mathcal F := f^{-1}\mathcal F \otimes_{f^{-1}\mathcal O_Y} \mathcal O_X$, which is a $\mathcal O_X$-module on $X$.

Using this notion of pullback, one can show that quasi-coherent sheaves on $Y$ pull back to quasi-coherent sheaves on $X$ and also for example that rank $n$ vector bundles pull back to rank $n$ vector bundles.


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