Hi,


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?



Sheaves with constant fibre dimension one
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.


