# Coherent sheaf with connection is locally free?

I think it is something like a folkore result that a coherent sheaf $\mathcal F$ on a smooth algebraic variety $X$ over $k$, which is equipped with a connection

$\nabla: \mathcal F \rightarrow \mathcal F \otimes \Omega^1_{X/k}$

Maybe one may weaken the assumptions, but I think the proof wouldn't alter very much.

I thought about how to prove it, but couldn't make it rigorous. In any case one should show that the stalks $\mathcal F_x$ are free, and this somehow must follow from the existence of the connection.

Addendum: It seems that the existence of a connection in some way rigidifies the underlying sheaf. A related question in this respect is: in how much is a horizontal morphism $\phi: \mathcal F \rightarrow \mathcal F$ already rigidified by $\nabla$. E.g. if one knows that $\phi$ is the identity on $\mathcal F(x) \rightarrow \mathcal F(x)$, then is it so already around $x$?

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@Addendum: see Prop. 2. 16, 2.21 in "Notes on crystalline cohomology", by Berthelot-Ogus (Princeton Univ. Press) –  Damian Rössler Nov 19 '11 at 13:07
(late comment) Not that you have to assume that ${\rm char}(k)=0$. –  Damian Rössler Nov 27 '11 at 7:37
Yes, I realized that, thanks. –  Veen Nov 29 '11 at 8:27