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?
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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.
Martin Brandenburg
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1This is fine, but I think that at some point the bridge must be crossed in order to relate this abstract definition of $f^*$ and the more concrete definition of Hartshorne. In other words, I don't know if one can completely avoid the computation... – Bruno Joyal Feb 20 '13 at 19:47
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2You mean that $F \mapsto f^{-1} F \otimes_{f^{-1} \mathcal{O}} \mathcal{O}$ is left adjoint to $f_$? This follows from the universal property of the tensor product and the universal property of $f^{-1}$ as a left adjoint to $f_$ on the category of abelian sheaves. No computation is needed. These formalities hold in great generality ( -> ringed topoi), where in fact no element computations are possible. – Martin Brandenburg Feb 20 '13 at 19:55
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2Forgive me if I do not take your word for it. I can see the idea you have in mind, and I agree that it works if one carries it out formally. Would you mind stating precisely the context in which you are appealing to the universal property of $\otimes$? Because I know some care must be exercised in this regard (for instance, I don't think $\otimes$ and $Hom$ are adjoint in full generality in categories of sheaves, eg. Exercise 5.1c in Hartshorne, where one of the sheaves is assumed to be locally free of finite rank). Thanks :) – Bruno Joyal Feb 20 '13 at 20:14
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3Again, don't use Hartshorne ;). Of course $\otimes$ and $\underline{\hom}$ are adjoint in full generality. And $\otimes$ classifies, as usual, bilinear maps (in this case bilinear sheaf morphisms). – Martin Brandenburg Feb 21 '13 at 10:42
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So are you saying you can prove 5.1c without the finiteness assumption? My impression is that the proof heavily depends on it, but I may be wrong. – Bruno Joyal Feb 21 '13 at 19:23
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Dear Martin, I have thought about this a little when I had the time last night, and I must admit that you are right. The finiteness hypothesis is not required. Thank you for this, all the best, – Bruno Joyal Feb 27 '13 at 17:28
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1Dear Martin, One thing that has to be checked (at least in the way you wrote your answer above) is that $f_*$ actually takes quasi-coherent sheaves to quasi-coherent sheaves. Since this is not true in complete generality, some kind of argument in the affine case has to be given. Of course, it's not hard; e.g. one can take not just global sections, but global sections on each distinguished open affine Spec $A_a$ in the target. Regards, – Matt E Oct 19 '13 at 12:47
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Yes, you are right. Thank you for this clarifying comment and your answer. – Martin Brandenburg Oct 19 '13 at 13:26
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Out of curiosity: Can this argument be modified to show that $f^*(\widetilde M) \simeq \widetilde{M\otimes_S T}|_U$, where $f: U \rightarrow \mathrm{Proj},S$, $U \subseteq \mathrm{Proj},T$ is a morphism induced by a homogeneous homomorphism of graded algebras $S\rightarrow T$ (both generated in degree $1$)? The problem there is that $M \mapsto \widetilde M$ is not an equivlence of cats anymore. – Pavel Čoupek Aug 09 '17 at 15:48
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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.
Matt E
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Thank you for this addition. Somehow I am used to the fact that if $f : X \to Y$ is a qcqs morphism (here it is even affine), then $f_*$ preserves quasi-coherence (EGA I (new), 6.7.1). – Martin Brandenburg Oct 19 '13 at 13:24
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