Can someone help me with this problem?

Let $D$ be a divisor on an algebraic curve $X$ of genus $g$, such that $\deg D = 2g-2$ and $\dim L(D) = g$. Then $D$ must be a canonical divisor.

By Riemann-Roch, I see that $\dim L(K-D) = 1$ for any canonical divisor $K$, as must certainly be the case. I don't know if this is too helpful.


A divisor of degree $0$ and dimension $1$ is principal. Hence by assumption and Riemann-Roch the divisor $K-D$ is principal, so that $D$ is linearly equivalent to the canonical divisor $K$.

  • $\begingroup$ Can you sketch a proof of the claim in your first sentence? $\endgroup$ – Tony Jan 11 '11 at 12:39
  • 6
    $\begingroup$ @Tony: Let $D$ be degree 0 with $L(D)$ non-zero. If $f$ is a non-zero function in $L(D)$ then $div(f) + D \geq 0.$ $\endgroup$ – Matt E Jan 11 '11 at 13:14

In "Algebraic Curves" by Fulton, p.212/213 it is shown that $l(K-D)$ equals the dimension of $\{ \omega\in \Omega : {\rm div} (\omega)>D\}$. In our case this dimension equals 1. Hence there exists a non-zero differential $\theta$ with ${\rm div}(\theta)>D$. But ${\rm div}(\theta)$ and $D$ have the same degree, namely $2g-2$. It follows that ${\rm div}(\theta)=D$ and $D$ is a canonical divisor.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.