Inverse image of the sheaf associated to a module (Hartshorne, Proposition II.5.2)

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.

Answer 2

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.

Answer 3

Q: "In Hartshorne, Algebraic geometry it's written, that for every scheme morphism $f: Spec (B) \to Spec (A)$ and $A$-module $M$ $f^*(\tilde M) = \widetilde {(M \otimes_A B)}$. 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: There is a "more direct" and "elementary" approach to this question. Let $\phi: A \rightarrow B$ be a map of commutative unital rings and let $E$ be an $A$-module. Let $f:=X:=Spec(B) \rightarrow S:=Spec(A)$ be the corresponding map of schemes and let $F:=\widetilde{E}$. Let $f^!(F)$ be the following presehaf on $X$:

$$f^!(F)(U):=\mathcal{O}_X(U) \otimes_{f^{-1}(\mathcal{O}_S)(U)} f^{-1}(F)(U).$$

It follows $f^*(F)$ is the sheaf associated to $f^!(F)$. There is a canonical map

$$B\otimes_A E \rightarrow \Gamma(X, f^!(F))$$

and a canonical map

$$B\otimes_A E \rightarrow \Gamma(X, f^*(F)).$$

This gives a map of sheaves of modules on $X$:

$$ i:\widetilde{B\otimes_A E} \rightarrow f^*(F).$$

When you localize the map $i$ you get for each prime ideal $\mathfrak{p} \subseteq B$ with $\mathfrak{q}:=A\cap\mathfrak{p}$ isomorphisms

$$(\widetilde{B\otimes_A E})_{\mathfrak{p}} \cong B_{\mathfrak{p}}\otimes_{A_{\mathfrak{q}}} E_{\mathfrak{q}} \cong \mathcal{O}_{X, \mathfrak{p}} \otimes_{\mathcal{O}_{S, \mathfrak{q}}} F_{\mathfrak{q}}\cong f^*(F)_{\mathfrak{p}} ,$$

Hence the localized map $i_{\mathfrak{p}}$ is an isomorphism for all points. This implies $i$ is an isomorphism.

The claim is not "immediate" but it follows from the theory developed in Hartshorne. One exercise (Ex.II.5.3) is needed: There is for any $\mathcal{O}_X$-module ($X:=Spec(B)$) $E$ and any $B$-module $M$ a natural isomorphism

$$ Hom_B(M ,\Gamma(X, E)) \cong Hom_{\mathcal{O}_X}(\widetilde{M}, E).$$

Note: Exercise II.5.3 comes after Proposition II.5.2.