# How to prove that this kind of differential form exists on an algebraic curve?

The following is a problem in Miranda's Algebraic Curves and Riemann Surfaces.

Given any algebraic curve $X$ and a point $p \in X$, show that there is a meromorphic $1$-form $\omega$ on $X$ whose Laurent series at $p$ looks like $dz/z^n$ for $n > 1$, and which has no other poles on $X$.

The point of this is as a step towards the proof that the Mittag-Leffler problem can be solved for $X$.

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Why do you want to prove this? How do you know the statement you are ordering us to prove is true? –  Mariano Suárez-Alvarez Jan 12 '11 at 14:25
This is a problem in a book (Miranda's). I'll edit the question to include this. –  Tony Jan 12 '11 at 16:20
Are you allowed to use Riemann-Roch? –  Pete L. Clark Jan 12 '11 at 16:53
Yes, this is after Riemann-Roch is introduced. –  Tony Jan 13 '11 at 23:10

This is equivalent to showing that $$h^0(\Omega[n\cdot p]) > 0$$ for all n > 1.
By Riemann-Roch, we have $$h^0(O_X [-n\cdot p]) - h^0(\Omega[n\cdot p]) = -n + 1 - g$$ Since the degree of the divisor is negative, $$h^0(O_X [-n\cdot p]) = 0$$ so, when rearranging, we get: $$h^0(\Omega[n\cdot p]) = n + g - 1$$ If n > 1, then n + g - 1 > 0 for all g.