# When is $f^{-1}(\tilde{M})$ a quasi-coherent sheaf?

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?

• 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. – user18119 Jan 10 '12 at 20:43
• Yes, I see. Thank you! – 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.