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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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    Yes, this is after Riemann-Roch is introduced.2011-01-13

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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.