# When is the pushforward of a quasi-coherent sheaf quasi-coherent? Hartshorne proof

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

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

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!

• 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

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 $S=\operatorname {Spec}(\mathbb Z)$ .

• 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 0:38
• You are welcome, dear Stefano . – Georges Elencwajg Jan 19 '15 at 0:41
• @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 – ujsgeyrr1f0d0d0r0h1h0j0j_juj Jan 19 '15 at 1:36
• 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 ? – ujsgeyrr1f0d0d0r0h1h0j0j_juj Jan 19 '15 at 16:13

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.

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.

• 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 ! – ujsgeyrr1f0d0d0r0h1h0j0j_juj Jan 18 '15 at 23:11
• Dear Stefano, an infinite direct sum of quasi-coherent sheaves is quasi-coherent. – Georges Elencwajg Jan 19 '15 at 8:27
• 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. – Pavel Čoupek May 16 '15 at 7:00