Inverse image of the sheaf associated to a module 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?
 A: As Martin indicates in his answer, it is easiest to prove that $f_*$ corresponds to the forgetful functor from $B$-modules to $A$-modules, and then use adjointness to understand $f^*$.
Again as Martin notes, once we know that $f_*$ takes the quasi-coherent sheaf $\widetilde{N}$ to a quasi-coherent sheaf, we can determine which sheaf we get by taking global sections.
However, it is not quite automatic that $f_*$ takes a quasi-coherent sheaf to a quasi-coherent sheaf (e.g. this is not true in complete generality for arbitrary maps between arbitrary schemes).  
In the affine case, one can get around this issue as follows:


*

*To determine $f_*\widetilde{N}$, it is enough to understand its behaviour on the basis of open sets of the form Spec $A_a$ ($a \in A$).  

*The preimage of Spec $A_a$ is Spec $B_a$.

*So, by definition of $f_*$, the sections of $f_*\widetilde N$ on Spec $A_a$ are $N_a$, with the evident restriction maps if $a \mid a$'.  (Note that $N_a$ has the same meaning whether we regard $a$ as an element of $A$ and regard $N$ as an $A$-module, or regard $a$ as an element of $B$, via the morphism $A \to B$, and regard $N$ as a $B$-module.)

*When we regard $N$ as an $A$-module, then the associated sheaf again attaches $N_a$ to Spec $A_a$, with the evident restriction maps.  

*Thus indeed $f_* \widetilde N$ is naturally isomorphic to the sheaf attached to $N$ regarded as an $A$-module.  
One reason for spelling this argument out is to point out (a) that $f_*$ can be computed directly in terms of its sections (there is no sheafification process, unlike in the construction of $f^*$), and so is sometimes more accessible; and
(b) that computing $f_*$ requires understanding preimages of open subsets, and one has the important fact in the affine setting that the preimage of Spec $A_a$ is simply Spec $B_a$.  This fact has an obvious geometric meaning (say in the case of varieties) which is worth making clear for yourself.
A: 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. 
