# Extensions of free sheaves over a scheme

Let $X$ be scheme. Consider an extension $$0 \to \mathcal{O}_X^n \to E \to \mathcal{O}_X^m \to 0$$ with $E$ a locally free $\mathcal{O}_X$-module.

My question is : does $E$ have to be free ? I think it's false but I couldn't find a counter example.

This is related to the following question : is a locally free $\mathcal{O}_X$-module projective in the category of quasi-coherent sheaves over $X$ ?

This is true (I think) if $X$ is affine since in this case a locally free sheaf is just a locally free module over a ring which is projective. But I don't think it's true over a general scheme although I don't know a counter example.

The link with my question is that if locally free sheaves are projective it would mean that the above exact sequence has a section and thus $E$ is free.

• Since $\mathrm{Ext^1}(\mathcal{O}, \mathcal{O}) = 0$ any such exact sequence splits and $E = \mathcal{O}^{n+m}$ – qwenty May 19 '18 at 19:47
• Thanks ! Is it really true though that $Ext^1(\mathcal{O},\mathcal{O}) = 0$ if you are not affine ? – ayokointhesand May 19 '18 at 19:48
• Oops it's not true for example for elliptic curves.. – qwenty May 19 '18 at 19:51
• Ah so I guess this will provide a counter example ? – ayokointhesand May 19 '18 at 19:52
• I'am not sure that in general it is not possible that $E = \mathcal{O}^2$ for non-trivial extension. But that's true for an elliptic curve: maths.tcd.ie/pub/ims/bull60/R6005.pdf example 4.4 but may be there is much more easier proof – qwenty May 19 '18 at 20:02

No. As in the comments, the issue is precisely that $\text{Ext}^1(\mathcal{O}, \mathcal{O}) \cong H^1(X, \mathcal{O})$ need not vanish. For example, if $X$ is a smooth projective curve of genus $g$ then we have
$$\dim H^1(X, \mathcal{O}) = g$$
and so we have nontrivial extensions of $\mathcal{O}$ by $\mathcal{O}$ whenever $g \ge 1$, e.g. if $X$ is an elliptic curve as also mentioned in the comments.