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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.

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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 $K$.

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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