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I do exercise asking to show on Noetherian space $X$, any subsheaf $\mathscr{R} \subseteq \Bbb{Z}_U$ is finitely generated. $\Bbb{Z}_U$ is sheaf $i_{!}(\Bbb{Z}|_U)$ for $U \subseteq X$ open. What is the meaning of a finitely generated sheaf?

Added: Hartshorne does not even define finitely-generated sheaf. What does this even mean?

Dylan B.
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  • Where did you find that incorrect (see my answer) exercise? – Georges Elencwajg Dec 01 '13 at 17:48
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    @GeorgesElencwajg Exercise 2.6 chapter 3 of Hartshorne. – Dylan B. Dec 01 '13 at 23:07
  • Dear Dylan, although your statement is buried inside Hartshorne's exercise, I agree that what you write is what Hartshorne seems to claim. But I persist in believing that my counterexample refutes that claim. So, I hereby challenge users to prove that assertion of Hartshorne's and/or find a mistake in my answer! – Georges Elencwajg Dec 01 '13 at 23:30

3 Answers3

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The statement of the exercise you quote is incorrect : $\mathbb Z_U$ itself is not finitely generated except in trivial cases.

For example if $X=\mathbb A^1_k$ and $U=X\setminus \{O\}$ (with $O$ the origin), the stalk at $O$ of $\mathbb Z_U$ is zero: $\mathbb (Z_U)_O=0$.
If $\mathbb Z_U$ were finitely generated its stalks would be zero in a neighbourhood of $O$.
But in reality all the stalks $ (\mathbb Z_U)_x=\mathbb Z$ for $O \neq x\in X$ and thus it is false that $\mathbb Z_U$ is a finitely generated sheaf of abelian groups.

Edit
Following the friendly discussion in the comments, let me add that the last line of EGA I page 45 states that the support of a finitely generated sheaf is closed. Since the support of $\mathbb Z_U$ is $U$, this confirms that $\mathbb Z_U$ can only be finitely generated in the rare case that $U$ is both open and closed.

Georges Elencwajg
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    Dear users: this contradicts a claim in Hartshorne's Exercise III 2.6. So please scrutinize my answer and give your verdict! – Georges Elencwajg Dec 01 '13 at 23:35
  • But what do you think "finitely generated" means in this context? For my first guess at what such a notion would mean, your $\mathbb{Z}_U$ is generated by the section $1$ lying over $U$, since every section is a multiple of a restriction of this section. (or the zero global section, which is an empty linear combination) –  Dec 01 '13 at 23:50
  • I try studying your answer but I don't get why this gives contradiction to claim of Hartshorne. What is even meant by finitely generated sheaf? – Dylan B. Dec 01 '13 at 23:50
  • I think the only reasonable definition of "finitely generated" that makes the claim true is the category-theoretic notion of "finitely presentable", namely that $\mathrm{Hom}(\mathbb{Z}_U, -)$ preserves directed/filtered colimits. – Zhen Lin Dec 01 '13 at 23:52
  • Dear friends, my interpretation of "finitely generated" is exactly the one described by Martin in his answer. It is the one adopted in EGA I, Chapitre 0, (5.2.1) and that should settle the discussion :-) – Georges Elencwajg Dec 02 '13 at 00:04
  • @Hurkyl: the sheaf $\mathbb Z_U$ is defined on $X$ and cannot be generated by a section over $U$. You must realize that $\Gamma (X,\mathbb Z_U)=0$. – Georges Elencwajg Dec 02 '13 at 00:10
  • @Georges: Or just that you are using a different definition of "finitely generated". The idea I had in mind is a finite set of sections $f$ such that every $\Gamma(V, S)$ is generated by the set of all $f|_V$ where $f$ ranges over just those sections with domain containing $V$. –  Dec 02 '13 at 00:17
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    Dear @Hurkyl, I am using a definition of "finitely generated" different from whose? My definition of finitely generated is that of Grothendieck-Dieudonné, Serre, Cartan, De Jong's stack project and every algebraic geometer I have ever read. Can you name one algebraic geometer who has a different definition? – Georges Elencwajg Dec 02 '13 at 00:27
  • @Georges: Different than the one I guessed at. I have never seen a definition of "finitely generated abelian sheaf". And presumably neither has the OP, since he was asking what the phrase means. –  Dec 02 '13 at 00:38
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If $(X,\mathcal{O}_X)$ is a ringed space, then there is a well-known notion of $\mathcal{O}_X$-modules of finite type (Stacks Project, Tag 01B4). Any topological space $X$ can be endowed with the constant sheaf $\mathcal{O}_X:=\mathbb{Z}_X$ so that $\mathcal{O}_X$-modules coincide with sheaves of abelian groups on $X$.

Martin Brandenburg
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  • I am looking at http://stacks.math.columbia.edu/download/modules.pdf and in lemma 4.4 they give definition of subsheaf generated by global sections. However it seems the sheaf $\mathcal{G}$ defined there is only defined for open sets $U \subset U_i$. What if $U$ is not contained in $U_i$? – Dylan B. Dec 01 '13 at 10:23
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Based on Hartshorne's hints, I believe what he means is (a variation on) the category-theoretic notion of "finitely presentable":

A finitely presentable object in a category $\mathcal{C}$ is an object $A$ such that the functor $\mathrm{Hom}(A, -) : \mathcal{C} \to \mathbf{Set}$ preserves directed/filtered colimits.

Let us take $\mathcal{C}$ to be the category of (abelian) sheaves on a noetherian topological space $X$. Then $\mathrm{Hom}(\mathbb{Z}_U, -)$ is isomorphic (as a functor) to $\Gamma (U, -)$. But the property of being noetherian is hereditary, so we may use [Chapter II, Exercise 1.11] to deduce that $\Gamma (U, -)$ preserves directed colimits. Thus $\mathbb{Z}_U$ is indeed finitely presentable.

That said, Hartshorne speaks of subsheaves of $\mathbb{Z}_U$ as well. It is not at all clear to me that these are finitely presentable. Nor is it clear to me that a sheaf with the extension property with respect to all subsheaves $\mathscr{R} \subseteq \mathbb{Z}_U$ and all open subsets $U \subseteq X$ is necessarily injective. Certainly what is true is this: if $\mathcal{I}$ is a collection of monomorphisms such that every monomorphism in $\mathcal{C}$ can be obtained as a retract of some $\mathcal{I}$-cell complex (= a transfinite composition of pushouts of $\mathcal{I}$), then an injective object in $\mathcal{C}$ is the same thing as an object with the extension property with respect to $\mathcal{I}$; and if $\mathcal{I}$ consists of morphisms whose domain and codomain are finitely presentable, then the class of $\mathcal{I}$-injective objects is closed under directed/filtered colimits.

Zhen Lin
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  • For those who are wondering about historicity, let me point out that the notions I mentioned were invented in the 1960s and and so surely known by the 1970s. – Zhen Lin Dec 02 '13 at 09:54
  • It's been a long time since this was posted, but I got interested in this problem. Doesn't being finitely presented follow from being a Noetherian object? (I think those subsheaves are Noetherian and should follow from Noetherian-ness of the space). Furthermore I think that $\mathbb{Z}_U$ are generators of the category of abelian sheaves. So essentially we have a set of Noetherian generators so the category is locally Noetherian and hence the filtered colimit of injectives is injective. – user127776 Jun 20 '22 at 03:58