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If $0\rightarrow F\rightarrow G\rightarrow H \rightarrow 0$ is an extension of $\mathcal{O}$-modules with $F$ and $G$ locally free (each of constant finite rank, i.e. vector bundles), then is $H$ locally free? The same question can be asked with the roles of $F$, $G$, and $H$ interchanged, too.

In other words, is a quotient of a locally free sheaf by a subsheaf locally free?

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3 Answers 3

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No. A locally free sheaf of finite rank on a noetherian affine scheme is a finitely-generated projective module, but $$0 \longrightarrow 2 \mathbb{Z} \longrightarrow \mathbb{Z} \longrightarrow \mathbb{Z} / 2 \mathbb{Z} \longrightarrow 0$$ is an exact sequence of $\mathbb{Z}$-modules and $\mathbb{Z} / 2 \mathbb{Z}$ is not projective.

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    $\begingroup$ The same line of thought shows that if $G$ and $H$ (respectively $F$ and $H$) are locally free, then $F$ (respectively $G$) is locally free, since every short exact sequence ending in a projective module splits. $\endgroup$
    – RghtHndSd
    Commented Aug 17, 2014 at 5:22
  • $\begingroup$ @RghtHndSd And moreover, if $G$ and $H$ have the same rank, then $F=0$. $\endgroup$
    – Alex Youcis
    Commented Aug 17, 2014 at 6:27
  • $\begingroup$ A subsheaf of a locally free sheaf is locally free (at least on a scheme), so in RghtHndSd's comment you don't really need $H$ to be free. Also I thought the correspondence between locally free sheaves and projective modules only works in the case of smooth manifolds? Its proof uses partitions of unity so it won't work for the ring of holomorphic functions. $\endgroup$
    – Ehsaan
    Commented Aug 18, 2014 at 14:30
  • $\begingroup$ It also works for noetherian affine schemes. At any rate, you can verify by hand (if you like) that $\mathbb{Z} / 2 \mathbb{Z}$ is not locally free. $\endgroup$
    – Zhen Lin
    Commented Aug 18, 2014 at 15:37
  • $\begingroup$ @Ehsaan A subsheaf of a locally free sheaf is not always locally free on an arbitrary scheme, for example consider $X=spec(k[x]/x^2)$ , its structure sheaf has a non-locally free ideaf sheaf. But for noetherian scheme, when $G$ and $H$ are both locally fee, then $F$ is indeed locally free. It suffices to check on stalks. $\endgroup$
    – Yuan Yang
    Commented May 7, 2021 at 2:22
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Zhen Lin's answer is more than adequate, but I'll add another example. Consider the following short exact sequence of sheaves on $\Bbb{P}^1$: $$ 0\to \mathcal{O}(-1)\xrightarrow{\times f} \mathcal{O}\to k(x)\to 0. $$ Here, $f \in \Gamma(\Bbb{P}^1,\mathcal{O}(1)).$ This is the sequence realizing $\mathcal{O}(-1)$ as the ideal sheaf of a point. The skyscraper sheaf $k(x)$ is a torsion sheaf and hence not locally free.

This example indicates the geometric intuition for why a quotient of locally free sheaves is not locally free. Indeed, here is an even more basic example (basically the above example in local coordinates): consider the trivial line bundle over $\Bbb{C}$, just viewed as a product: $\pi:\Bbb{C}^2\to \Bbb{C}$ by $(x,y)\mapsto x$. Consider the bundle map given by $(x,y)\mapsto (x,xy)$. Fibre by fibre, the multiplication map $y\mapsto xy$ is an isomorphism except in the fibre $\pi^{-1}(0)$, where there is a nontrivial kernel and cokernel. This is the sense in which the quotient of a map of vector bundles could well be a skyscraper sheaf.

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The complete answer can be found in Vakil's FOAG exercise 14.2Q and 14.2R 2022 version, the result is as follows:

Let the following sequence of Quasicoherent sheaves be exact :

$$0\to \mathcal{F}' \to \mathcal{F}\to \mathcal{F}''\to 0$$

(1) if $\mathcal{F}'$ and $\mathcal{F}''$ are locally free, then so it's $\mathcal{F}$

(2) if $\mathcal{F}$ and $\mathcal{F}''$ are locally free of finite rank , then so it's $\mathcal{F}'$

(3) $\mathcal{F}'$ and $\mathcal{F}$ are locally free, $\mathcal{F}''$ needs not to be locally free.

We see the kernel of surjective morphism between locally free finite rank sheaves is locally free, but it's not true for the cokernel.

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    $\begingroup$ This answer could do with a good bit of improvement. You have a typo in 2, it would be good to specify what version of FOAG you're looking at (the numbering changes between versions), it would be better to link the appropriate version, and you have your causality wrong in the final sentence. (Finally, why do we need this answer on this question? It feels largely unrelated and it's missing the actual counterexample.) $\endgroup$
    – Hank Scorpio
    Commented May 16, 2023 at 3:54
  • $\begingroup$ @Hank Scorpio , sorry for the typo, I have corrected it, Ehsaan ask in the question:"The same question can be asked with the roles of F, G, and H interchanged, too."Therefore I provide the result exchange the role of F,G,H. $\endgroup$
    – yi li
    Commented May 16, 2023 at 4:58

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