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I am trying to solve this:

If $D$ is a divisor over a algebraic curve X of genus $g$ such that $\deg(D)=2g-2$ and $\dim L(D)=g$, then $D$ is a canonical divisor

Using Riemann-Roch theorem, being $K$ an canonical divisor over $X$, I have:

$$ \begin{array}{l} \dim L(D)-\dim L(K-D)=\deg(D)+1-g\\ \dim(K-D)=1 \end{array} $$

By Serre's duality, $$ \dim L(K-D)=\dim L^{(1)}(-D)=1 $$

This means that there exists only one 1-form $\omega \in L^{(1)}(-D)$ such that $ord_p(\omega)\geq D(p)$, for all $p\in X$.

I don't know how to finish this. Can someone help me? Thanks in advance.

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    $\begingroup$ Riemann and Roch are two people, so it's just the "Riemann–Roch theorem". $\endgroup$ – Zhen Lin Jun 11 '13 at 12:35
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You have successfully shown that $h^1(O_X(D))=1,$ i.e., that $h^0(\mathcal O_X(K-D)) = 1.$ Now try Riemann-Roch again: $$h^0(\mathcal O_X(K-D))-h^1(\mathcal O_X(K-D))=\deg(K-D)+1-g,$$ and using that $h^1(\mathcal O_X(K-D))=h^0(\mathcal O_X(D))$ we get that $\deg(K-D)=0.$

Now apply Hartshorne Lemma IV.1.2, which gives $K-D\sim 0,$ i.e., $K\sim D.$

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    $\begingroup$ Can I use the following: since $k$ is canonical then $deg(k)=2g-2$ therefore $deg (K-D)=0$? $\endgroup$ – Marra Jun 12 '13 at 13:03
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    $\begingroup$ Dear @GustavoMarra, yes, that seems reasonable, since the degree map is a homomorphism after all. $\endgroup$ – Andrew Jun 12 '13 at 13:40

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