TITLE: Extension of Vector bundles is a Vector bundle?
QUESTION [5 upvotes]: 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?

REPLY [4 votes]: 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 $G=0$.