Inverse image of the sheaf associated to a module

Question

In Hartshorne, Algebraic geometry it's written, that for every scheme morphism $f: Spec B \to Spec A$ and $A$-module $M$ $f^*(\tilde M) = \tilde {(M \otimes_A B)}$. And that it immediately follows from the definition. But I don't know how to prove it in simple way. Could you help me?

Answer 1

Don't use the book by Hartshorne when you want to learn the foundations of algebraic geometry. There are many, many books which explain them far better (Liu, Görtz-Wedhorn, Bosch, ...). And in fact, in this case, this isomorphism does not immediately follow from the usual definition of $f^*$ via $f^{-1}$. Therefore, the claim by Hartshorne is misleading. One rather has to use the adjunction between scalar restriction and scalar extension:

The inverse image functor $f^*$ should be seen as (defined to be) the left adjoint of the direct image functor $f_*$. Therefore, it is enough to show that $f_* : \mathsf{Qcoh}(\mathrm{Spec}(B)) \to \mathsf{Qcoh}(\mathrm{Spec}(A))$ corresponds, under the equivalence of categories $ \mathsf{Qcoh}(\mathrm{Spec}(A)) \cong \mathsf{Mod}(A), F \mapsto \Gamma(F)$, to the scalar restriction functor $\mathsf{Mod}(B) \to \mathsf{Mod}(A)$. But this is clear, since we have more generally $\Gamma(f_*(F))=\Gamma(F)$ by definition of $f_*$.

One can, instead of adjointness, start with $f^* \tilde{M} = f^{-1} \tilde{M} \otimes_{f^{-1} \mathcal{O}_{\mathrm{Spec}(A)}} \mathcal{O}_{\mathrm{Spec}(B)}$, and use the complicated ad hoc definitions of the tensor product of sheaves and of $f^{-1}$ (remember that they are associated sheaves of certain presheaves). But this involves, in detail, many computations. In my opinion this approach is unnecessarily complicated.