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I have been working on Exercise IV.1.5 in Hartshorne's Algebraic Geometry, which says

For an effective divisor $D$ on a curve $X$ of genus $g$, show that $\operatorname{dim} |D| \le \operatorname{deg} D$. Furthermore, equality holds if and only if $D = 0$ or $g = 0$.

Note that by "curve" Hartshorne means complete, nonsingular curve over algebraically closed field. I was able to show the inquality using Riemann-Roch, Serre duality, and the long exact sequence of cohomology.

Regarding the equality statement, it's easy to show that $D = 0$ or $g = 0$ forces equality. However, I concluded something stronger than what Hartshorne claims in the reverse direction. I can show that

If equality holds, then $D= 0$.

As an immediate corollary, one obtains that every effective divisor on a curve of genus zero is the zero divisor, because $g = 0 \implies \deg |D| = \dim D \implies D = 0$. Have I made a mistake and concluded something false?

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    $\begingroup$ Yes, you have made a mistake. Consider the divisor given by a single point in $\Bbb P^1$. $\endgroup$ – KReiser Jun 28 '18 at 3:30

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