2
$\begingroup$

Hi everyone.

Let $X$ be a stable curve over algebraically closed field $k$ with genus $g>1$, and $p$ is a nonsingular point of $X$. I want to prove the global section of dualizing sheaf is base point free, and I know this question is equivalent to

$dimH^{0}(X,O_{X}(p))=1$.

How to prove it?

Improve this question
$\endgroup$
0

1 Answer 1

Reset to default
7
$\begingroup$

If $X$ is a stable curve, and $X_i$ a singular component, then a section $f \in H^0(X,\mathcal{O}_X(p))$ gives a section in the blow-up (constant on the new $\mathbb{P}^1$), so we can assume that $X$ is semi-stable with non-singular components, and no $\mathbb{P}^1$ intersects in only one point.

We have two cases to consider:

  1. If the component $X_i$ of $p$ has positive genus, then a global section $f$ can be restricted to $X_i$ where it gives the isomorphism to $\mathbb{P}^1$ as usual.
  2. If $X_i$ has genus zero, then it intersects the rest of the curve in at least two points $x_1$ and $x_2$. Since $f$ can be restricted to individual components, it must be constant on the rest of the curve, so we get $f(x_1)=f(x_2)$. Now $f-f(x_1)$ has two zeroes, and thus $f$ has more than a single pole on $X_i$, so we get a contradiction.

This also shows that the result does not hold on arbitrary semi-stable curves.

Improve this answer
$\endgroup$
1
  • 2
    $\begingroup$ I just noticed a important restriction: $X \setminus X_i$ needs to be connected for this to work out. A counterexample would be a $\mathbb{P}^1$ intersecting three curves of genus bigger then 0, lets say in $0, 1$ and $\lambda$. Then the coordinate function $x/y$, extended by $0, 1$ and $\lambda$ on the rest of the curve would have a single pole at infinity. So your statement is true if and only if every rational component is part of some cycle of the dual graph. $\endgroup$
    – Sophie
    Commented Mar 13, 2012 at 21:20

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .