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?


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 Aug 17 '14 at 5:22
  • $\begingroup$ @RghtHndSd And moreover, if $G$ and $H$ have the same rank, then $F=0$. $\endgroup$ – Alex Youcis Aug 17 '14 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 Aug 18 '14 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 Aug 18 '14 at 15:37

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