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Let $C$ be a smooth projective curve, and let $\phi:C\rightarrow\mathbb{P}^1$ be a non-constant rational map. How to recover a divisor $D$ of the curve associated to this rational map? (I understand how to get a rational map out of a divisor, but not the other way round)

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up vote 2 down vote accepted

I'll interpret this in two different ways:

1.) If you're trying to find a divisor corresponding to a map to projective space, you can look at the pullback $\phi^*\mathcal{O}(1)$. The linear system will be base point free and correspond to the given map.

2.) I think what you probably are asking, though, is how to get a divisor from a rational function. If what you're given is a map to $\mathbb{P}^1$ with no choice of coordinates, you can't recover a divisor. However, if you are given a coordinate system on $\mathbb{P}^1$, the pullback $\phi^*(0-\infty)$ will give you the divisor you are looking for.

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I would say that $\mathbb{P}^n$ already comes with "a choice of coordinates", but I take the point that the linear equivalence class of the divisor is coordinate-independent whereas the divisor itself is coordinate-dependent. – Pete L. Clark Mar 22 '12 at 5:32

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