Let $X$ be a curve of genus $g$. Suppose $D$ is a divisor with $\dim H^0(D)\geq 2$. Then there exists a non-constant $f\in H^0(D)$. In the top answer to this question, Hartshorne 4.1.6 Gonality of a curve, it is stated that $f$ gives a finite morphism $f:X\to\mathbb{P}^1$ of degree at most $\deg D$. My question is: Why is the degree of $f$ at most $\deg D$?
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If $|D|$ is base point free, $\deg(f) = \deg(D)$. Otherwise, if $P$ is a base point of $|D|$, the map given by $|D-P|$ is the same as the map $f$, therefore by induction $\deg(f) \le \deg(D-P) = \deg(D) - 1$.
Sasha
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Why does $|D|$ base point free imply that $\deg(f)=\deg(D)$? – ponchan Jan 05 '23 at 18:43
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Because by definition of $f$ the divisor $D$ is equal (as a scheme) to the fiber of $f$ over a point. – Sasha Jan 05 '23 at 18:54
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By definition, we have that $(f)+D\geq 0$, but I don't see how that implies what you wrote. – ponchan Jan 05 '23 at 19:00
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I am talking about the definition of the morphism given by a linear system. – Sasha Jan 05 '23 at 19:38