# Extension of Vector bundles is a Vector bundle?

I guess this is quite easy, but I don't see a counterexample:

let $X$ be a noetherian scheme (maybe with more hypotheses, but I don't think this will change much), then I have the feeling that it is not true that every extension of a vector bundle by a vector bundle is again a vector bundle, i.e. if

$0 \rightarrow F \rightarrow G \rightarrow H \rightarrow 0$

is an exact sequence of modules on $X$ with $F$ and $H$ locally free of finite rank, then is does not follow that $G$ is again locally free of finite rank?

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If $X$ is a reduced scheme, and $F, G, H$ are known in advance to be finitely-generated, then $F, H$ locally free of finite rank implies $G$ locally free of finite rank: see Hartshorne [Algebraic Geometry, Ch. II, Exercise 5.8]. –  Zhen Lin Dec 10 '11 at 13:34
Thanks, Zhen, this helps already very much! –  Cyril Dec 10 '11 at 14:41
This is true over any scheme. Any point has an open neighborhood $U$ in which $F$ and $H$ are free. Then in the same open neighborhood, $G$ is isomorphic to the direct sum of $F$ and $H$ ($F$ is projective if you want).
Over a noetherian scheme, it is true that $G$, $H$ locally free of finite rank implies the same property for $F$ (user Tor$^1$ to prove the flatness of $F$).
But $F, G$ locally free of finite rank doesn't imply the same for $H$ unless $G=0$.