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.


2 Answers 2


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 $K$.

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    $\begingroup$ Can you sketch a proof of the claim in your first sentence? $\endgroup$
    – Tony
    Jan 11, 2011 at 12:39
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    $\begingroup$ @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, 2011 at 13:14
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    $\begingroup$ Then we have $0 = deg(div(f) + D) \geq 0$ it gives that $D + div(f) = 0$ hence $D = div(1/f)$ is principal. $\endgroup$
    – Sherose
    Apr 27, 2020 at 16:07

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.


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