Fibers of a line bundle

Question

Let $C$ be a smooth curve over complex numbers and $x\in C$ be a point. Consider the line bundle $L=\mathcal{O}_C(x)$. I am confused with the following. Since $L$ is a line bundle, the fiber $L_{|_y}$ is one-dimensional vector space for any point $y\in C$ and is defined by the number $s(y)$, where $s$ is a non-zero section of $L$. But $s$ vanishes at the point $x$, which should imply the impossible consequence $L_{|_x} = 0$. Where am I wrong?

Answer 1

There is the (huge, constant, not a line bundle at all) sheaf $K$ of meromorphic functions on $C$, whose sections over any open are just the function field of your curve (= the field of fractions of the ring of functions on any affine open).

Inside of $K$ you have small sheaves like $\mathcal{O}_C(x)$. Here is what is confusing:

1. The sections of such sheaves "are meromorphic functions" because it is a subsheaf of $K$ and this is the sense in which you are getting values of $0$ (although note that your sign is opposite the more common convention, where $\mathcal{O}_C(x)$ means there is a pole of order at most $1$ at $x$.)

2. A section of this sheaf is a family of functions into the fibers (like for any sheaf) and, this sheaf being locally free of rank one, the fiber is abstractly isomorphic to the field $k$ of coefficients.

The problem is that you have to actually use the isomorphism in (2) to get the comparison. To wit, let $f$ be any meromorphic function having a pole of order exactly $1$ at $x$. Then multiplication by $f$ gives an isomorphism (in a sufficiently small neighborhood of $x$ that $f$ has no other poles or zeroes) between $\mathcal{O}_C(x)$ and $\mathcal{O}$. So the correct thing to do is multiply by $f$ and then evaluate, at which point it is clear you can get all possible values.

Note that the choice of $f$ is arbitrary; there is not a canonical way to identify the fiber of $\mathcal{O}_C(x)$ at $x$ with $k$.