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In the following we have $f:X \rightarrow Y$ morphism of schemes, $\mathcal{F}$ a quasi-coherent sheaf on $X$ and I am referring to proposition 5.8 page 115 in Hartshorne.

To prove that the pushforward of a quasi-coherent sheaf is quasi-coherent, there are the following assumptions: either $X$ is noetherian, or $f$ is quasi-compact and separated.

Both these assumptions lead to some finiteness of affine covers on $X$ where we can write $\mathcal{F}$ as $\widetilde{M}$.

My question is: Why do we need these assumptions? In the proof, after reducing to $Y$ affine, the argument leads to the exact sequence

\begin{equation} 0 \rightarrow f_* \mathcal{F} \rightarrow \bigoplus_i f_* \left(\mathcal{F}|_{U_i}\right) \rightarrow \bigoplus_{i,j,k} f_* \left(\mathcal{F}|_{U_{i,j,k}}\right), \end{equation}

where the $U_i$'s are affines in $X$ and the $U_{i,j,k}$'s are affines covering the intersection $U_i \cap U_j$. The assumption (at least it's what I understood) are needed to make this family of indexes finite. But wouldn't the proof work even in the infinite case?

Of course I have to be mistaken somewhere, but I cannot see where and why!

Stefano
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    Since Georges Elencwajg's answer below doesn't mention it : the mistake in your reasoning is that the exact sequence involves _products_, which are sums only in the finite case. – Segipp Feb 24 '18 at 16:24

4 Answers4

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1) There is no reason to believe that $ f_* \mathcal{F}$ is coherent whenever $X$ is affine : Hartshorne certainly does not claim that, contrary to your typo [now corrected !].

2) The reason that some hypotheses must be made on $f$ is that the result may be false without them!
There exists an example due to Altman-Hoobler-Kleiman, page 36 where indeed for some morphism $f:X \rightarrow Y$ and some quasi-coherent sheaf $\mathcal F$ on $X$ the image sheaf $ f_* \mathcal{F}$ is not quasi-coherent.
Such examples are non trivial: for example Dieudonné-Grothendieck claim to give one (in the new edition of EGA, page 314) but their description is incorrect.

Edit: A simple counterexample
Here is an example of a non quasi-coherent image of a quasi-coherent sheaf, simpler than the one in Altman-Hoobler-Kleiman :

a) Let $X_i\; (i\in \mathbb N)$ and $S$ be copies of $\operatorname {Spec}(\mathbb Z)$ and let $X=\coprod X_i$ be the disjoint union of the $X_i$'s equipped with its natural morphism $f=\coprod_i id_i:X\to S$, the one restricting for every $i$ to the identity $id_i:X_i=\operatorname {Spec}(\mathbb Z)\to S=\operatorname {Spec}(\mathbb Z)$.
The counterexample will simply be $\mathcal F= \mathcal O_X$: I will now show that $f_* \mathcal O_X$ is a non quasi-coherent sheaf on $S$.

b) Since for a quasi coherent sheaf $\mathcal G$ on $S$ the canonical morphism $$m_U:\mathcal G(S) \otimes _{\mathcal O_S(S)}\mathcal O_X(U)\to \mathcal G(U)$$ must be bijective for all affine $U\subset S$, it suffices to show that this property is violated for $\mathcal G =f_*(\mathcal O_X) $ and $U= D(2)=\operatorname {Spec}(\mathbb Z)\setminus \{(2)\}\subset S=\operatorname {Spec}(\mathbb Z) $.

c) In our situation we get the canonical morphism $m_U:(\prod_{i\in \mathbb N} \mathbb Z) \otimes _{\mathbb Z} \mathbb Z[{\frac 12}]\to \prod_{i\in \mathbb N}\mathbb Z[{\frac 12}]$.
This morphism is not surjective because the image $m_U(t) $ of an element $t\in (\prod_{i\in \mathbb N} \mathbb Z) \otimes _{\mathbb Z} \mathbb Z[{\frac 12}]$ is a sequence of rational numbers $(\frac {a_i}{2^M})_i\in\prod_{i\in \mathbb N}\mathbb Z[{\frac 12}]$ with some fixed denominator $2^M$.
Whereas in $\prod_{i\in \mathbb N}\mathbb Z[{\frac 12}]$ there exists sequences $(\frac {b_i}{2^{s_i}})_i $ of rational numbers with $b_i$ odd and denominators $2^{s_i}$ tending to infinity.

Conclusion: $f_* \mathcal O_X$ is a non quasi-coherent sheaf on $S=\operatorname {Spec}(\mathbb Z)$ .

Georges Elencwajg
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  • Thank you for the link! And also, yes, I don't know why I typed affine, it should be noetherian! – Stefano Jan 19 '15 at 00:38
  • You are welcome, dear Stefano . – Georges Elencwajg Jan 19 '15 at 00:41
  • @Georges The example they give must not then be in the pdf version of EGA that we can find on numdam ? Could you just give of rough idea please ? Thx – Olórin Jan 19 '15 at 01:36
  • Dear Robert: no, it is not in the Numdam version. I have given a different, detailed counterexample in an Edit . – Georges Elencwajg Jan 19 '15 at 11:25
  • @Robert Nice example, in a similar vein than Altman-Hoobler-Kleiman's, but simpler indeed. Speaking of Grothendieck's counter example that is finally not one, why is it incorrect ? – Olórin Jan 19 '15 at 16:13
  • @Robert: the incorrectness is explained on page 25 of the quoted article. – Georges Elencwajg Jan 19 '15 at 17:56
  • @GeorgesElencwajg Ok, read only their counter-example, thx. – Olórin Jan 19 '15 at 18:00
  • What a nice counterexample! – Jake Levinson Mar 28 '15 at 08:21
  • @GeorgesElencwajg Dear Georges, I am not sure if I made a mistake, however it looks like $f^{-1}(D(2))=D(2,2,2,...)$ in $X$, hence $f^*O_X(D(2))=(\prod_{i\in \mathbb{N}}\mathbb{Z})_{(2,2,2,...)}$ , while not $\prod_{i\in \mathbb{N}}\mathbb{Z_2}$ as in your example. Did I make any mistakes here? – Xin Mar 26 '21 at 03:51
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The answer by "user10000100_u", marked correct, is false. Pavel Coupek's remark is true. Perhaps to clarify where the confusion lies:

"-The category of quasi-coherent sheaves is not abelian in general, infinite direct products of quasi-coherent sheaves are not quasi-coherent in general."

The category of quasi-coherent sheaves has all limits, so these direct products exist and ARE quasi-coherent. However, one can also take the direct product as O_X-module sheaves, and this product yields a different object; in particular usually an O_X-module sheaf which is not quasi-coherent.

user258899
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To expand on Pavel's comment, the category of quasicoherent sheaves is always abelian (in fact, by Gabber's result, it is always Grothendieck). Moreover, the pullback functor $f^\ast:\operatorname{QCoh}(Y)\to \operatorname{QCoh}(X)$ for a map of schemes $f:X\to Y$ always admits a right adjoint $f_\sharp:\operatorname{QCoh}(X)\to \operatorname{QCoh}(Y).$

This follows from the following observations:

  1. The category of quasicoherent sheaves on a scheme is closed under colimits in the category of $\mathcal{O}_X$-modules.
  2. Gabber's result: Gabber's result implies that $\operatorname{QCoh}(X)$ satisfies the hypotheses of the special adjoint functor theorem, namely that $\operatorname{QCoh}(X)$ is locally presentable for any scheme $X$.
  3. The special adjoint functor theorem: Since $\operatorname{QCoh}(Y)$ is closed under colimits in $\mathcal{O}_Y$-modules, the pullback functor $f^\ast$ preserves colimits of $\mathcal{O}_Y$-modules and sends quasicoherent sheaves to quasicoherent sheaves, and both $\operatorname{QCoh}(X)$ and $\operatorname{QCoh}(Y)$ are locally presentable, we see that $f^\ast$ admits a right adjoint by the special adjoint functor theorem.

It is not true in general, however, that this right adjoint functor $f_\sharp$ agrees with the restriction of $f_\ast$ to $\operatorname{QCoh}(X)$. In general, it can be constructed using the 'associated Quasicoherent sheaf' functor.

Harry Gindi
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The category of quasi-coherent sheaves is not abelian in general, infinite direct products of quasi-coherent sheaves are not quasi-coherent in general. That's why the fact that the family of indexes is finite is crucial.

Remark for general schemes : see "Modules vs. quasi-coherent modules" of Thomason & Trobaugh in "Higher Algebraic K-theory of Schemes and Derived Categories, The Grothendieck Festschrift, Vol. III", Thomason wrote that it was as this time a priori unknown if the category of quasi-coherent sheaves over a general scheme has all limits. Since the existence of arbitrary products would imply the existence of limits...

What is the current situation on this ? Answer : as of last version of the Stacks Project, it is still unknown.

Olórin
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    In fact the exact sequence would have to use an infinite *product*! – Bruno Joyal Jan 18 '15 at 23:09
  • Thank you, now it makes sense! Is there any nice/important example to keep in mind for an infinite direct sum to fail being quasi-coherent? – Stefano Jan 18 '15 at 23:10
  • @BrunoJoyal +1 ! – Olórin Jan 18 '15 at 23:11
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    Dear Stefano, an infinite direct sum of quasi-coherent sheaves **is** quasi-coherent. – Georges Elencwajg Jan 19 '15 at 08:27
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    I don't think this is true. Firstly, the category of quasi coherent sheaves is an abelian category - it is even abelian subcategory of the cat of $O_X$-modules. Secondly, it is known that the cat of quasi coherent sheaves has all limits and colimits - it follows from Gabber's result that the category of quasi-coherent sheaves has a set of generators. It can be found on Stacks project, [here](http://stacks.math.columbia.edu/tag/077K). – Pavel Čoupek May 16 '15 at 07:00