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I have a simple question about coherent sheaves and line bundles:

if I have a coherent sheaf $F$ on a good scheme $X$ and I know that $F_x \otimes k(x) = k(x)$ for alle points $x$ on $X$ (where $k(x)$ is the residue field of $x$ and the tensor product goes over the local ring $\mathcal O_{X,x}$), can I then say, that $F$ is already a line bundle?

My strategy would be:

the stalk of $F$ is finitely generated over the local ring; by Nakayama you can choose a generating element $s_x$ of $F_x$ for every $x$, so you have $F_x \simeq \mathcal O_{X,x}$ and now use the statement that if a coherent sheaf is free in a point, then it is in a neighborhood. Is this the right way to see it?

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up vote 7 down vote accepted

No. You need that the scheme is reduced.

It is certainly true that if $F_x$ is a free $\mathcal{O}_{X,x}$-module of rank $n$, then there exist an open neighborhood $U$ of $x$ such that $F \vert_U$ is a free $\mathcal{O}_U$- module of rank $n$.

But from $\dim_{k(x)} F_x \otimes_{\mathcal{O}_{X,x}} k(x) = 1$ you can deduce that $F_x$ is a cyclic module and not that $F_x$ is free of rank $1$.

However on a reduced scheme the statement is true: exercise II.5.8 of Hartshorne.

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Great, thank you, Andrea! – Descartes Aug 22 '11 at 10:20

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