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I have the following problem:

$\texttt {(1)Show that every divisor of degree ≥ p on a compact Riemann surface}$ $\texttt {of genus p is linearly equivalent to an effective divisor.}$

My initial action is to prove that: (2) $D$ be a divisor on compact Riemann surface is linearly equivalent to an effective divisor $\Leftrightarrow h^{0}(D)\ge 1$.

So if I prove (2), I prove (1).

I wonder if you're reasonable. And some idea to prove (2)?

Thank you!

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    $\begingroup$ Use Riemann-Roch, or a weak version of it $\endgroup$ – reuns Jan 29 '17 at 23:11
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As you already noticed, we need to apply Riemann-Roch for a Divisor of degree $k\ge p$. This gives $$\dim H^0(X, O_D) - \dim H^1(X,O_{D}) = 1-p + k\ge1.$$ Thus $\dim H^0(X, O_D)\ge 1$, which means there is a non-zero function $f \in O_D(X)$.
So the divisor of $f$ satisfies $(f)+D\ge 0$. But then the Divisor $\tilde D:=(f) + D$ is effective and linearly equivalent to $D$, i.e. $D$ is linearly equivalent to the effective divisor $\tilde D$.

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