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Let $X$ be a smooth projective curve in $\mathbb{P}^2(\mathbb{C})$ of degree $4$ and $p,q,r \in X$. What's $L(p+q+r)$?

With a standard computation, the genus of $X$ is $3$, so applying Riemann-Roch theorem, we obtain $\dim L(p+q+r)=1+L(K-p-q-r)$, with $K$ a canonical divisor. As $\deg(K)=4$, we have $\deg(K-p-q-r)=1$ and so $\dim L(K-p-q-r) \leq 1$. Indeed, we have two possibilities: $\lvert K-p-q-r\rvert=\lvert s\rvert$ for some $s \in X$, so that $L(K-p-q-r)=L(s)=\mathbb{C}$, because the genus of $X$ is strictly bigger than $0$; or $L(K-p-q-r)=0$.

I don't know how to conclude from this or if this the right way to proceed.

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If the points $p$, $q$, and $r$ in $\mathbb{P}^2$ are collinear, then $\dim H^0(X,K_X(-p-q-r)) = 1$, and hence by Riemann-Roch $\dim H^0(X,O_X(p+q+r)) = 2$. Otherwise, $\dim H^0(X,K_X(-p-q-r)) = 0$, and $\dim H^0(X,O_X(p+q+r)) = 1$.

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  • $\begingroup$ I'm sorry but I'm not used to this notation: as far as I've seen about Riemann-Roch,we just talked about meromorphic functions with specified poles and linear systems and never introduced homology. $\endgroup$ – Tommaso Scognamiglio Dec 17 '18 at 8:01
  • $\begingroup$ These are synonyms: $H^0 = L$. $\endgroup$ – Sasha Dec 17 '18 at 9:18

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