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Let $f:C \rightarrow C'$ be a rational map, here $C$ and $C'$ are smooth projective curves. I cannot understand how is $f$ a morphism?

(This is lemma from book by Klaus Hulek)

Thanks

chyle
  • 268

1 Answers1

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Assume $f:C\to C'$ is nonconstant.

Let us view $C'$ as embedded in some projective space $\mathbb P^N$. Since $f$ is rational, where it is defined it is the form $x\mapsto (f_0(x):\dots:f_N(x))$ for some rational functions $f_i\in k(C)$. Let $P\in C$ be any point. As $C$ is smooth, $P$ is a smooth point, thus its local ring is a DVR. So we can take a uniformizer $\pi\in \mathcal O_{C,P}\subset k(C)$. Define $$m:=\min\{\textrm{ord}_{P}(f_i)\},$$ so that $\textrm{ord}_P(\pi^{-m}f_i)\geq 0$ for every $i=0,\dots,N$, and for some $j$ we have $\textrm{ord}_P(\pi^{-m}f_j)=0$. This means that $f$ is defined at $P$.

We used no hypothesis about $C'$, except its closed embedding in $\mathbb P^N$. Indeed, the same proof says that a map from a curve to a projective variety is defined at smooth points.

Brenin
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