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I feel very stupid to ask this question but I've serached everywhere and I've not found a clear (to me) answer.

A well known result in Curve theory (over $k=\bar{k}$) states that a divisor $D$ on a curve $X$, if $\dim H^0 (D-x) = \dim H^0 (D)-1$ for all $x \in X(k)$, induces a map $\varphi : X \rightarrow \mathbb{P}^{n}_k$, where $n=\dim H^0 (D)-1$.

In particular $\varphi$ is a closed immersion if $\dim H^0 (D-x-y) = \dim H^0 (D)-2$ for all $x , y \in X(k)$.

Now my question:

What we can say about the degree of the map $\varphi$?

It's clear to me that, in the case $\varphi$ is an embeddig, we have $\deg D = \deg \varphi (X)$ but this can say someting about the degree of the map?

Are there general results (about $\deg \varphi$ and $\deg \varphi (X)$) if $\varphi$ is not a closed immersion?

Thank you

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In case $\dim H^0(D-x-y)=\dim H^0(D)-2$ for all $x,y$, the degree of $\phi=1$ as you say. In case you only know $\dim H^0(D-x)=\dim H^0(D)-1$ for all $x$, degree of $\phi$ can be any number dividing $\deg D$.

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