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I have $\phi:U⊂\mathbb{P_{\mathbb{C}}^{1}}→X$ a morphism, where U is a open dense subset and X is a projective variety in $\mathbb{P_{\mathbb{C}}^{n}}$. I want to prove that $\phi$ extends in a unique form to the morphism $\overline{\phi}:$ $\mathbb{P_{\mathbb{C}}^{1}} \rightarrow X$ such that $\overline{\phi}\restriction_U=\phi$. My question is the same as in the post https://mathoverflow.net/questions/296794/extension-of-morphism-of-quasiprojective-varieties. My problem is that I don't understand the answers there and I am looking forward to a easier one.

Laura
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    This is proved in Silverman's Arithmetic of Elliptic Curves, section II.2, Proposition 2.1. It uses the fact that the local ring at a smooth point on a curve is a DVR, which is the approach in reuns's answer below. – Viktor Vaughn Nov 27 '20 at 18:42
  • This has already been asked and answered here a couple times: here, here, probably others. – KReiser Nov 27 '20 at 20:02

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Near some $p\in U$ your morphism is given by some rational functions $f_j$ $$\phi([x:y]) = [f_0(x/y):\ldots :f_n(x/y)]$$ which is a morphism at every $[a:b]$ such that $a/b$ is not a pole of any $f_j$ nor a zero of all the $f_j$.

If $b\ne 0$ then for some $k$, replacing $f_j(x/y)$ by $f_j(x/y)(x/y-a/b)^k$ we can repair those two cases.

Can you deal with the case $b=0$ ?

reuns
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  • Because this is the definition of a morphism, given near each $p$ by some rational functions regular at $p$. – reuns Nov 27 '20 at 20:04
  • I see why replacing $f_j(x/y)$ by $f_j(x/y)(x/y-a/b)^k$ repair the case where $a/b$ is a pole of $f_j$. But in the case where $a/b$ is a zero of all the $f_j$ I don't see it because $f_j(a/b)(a/b-a/b)^k=0$ for all $k$ and $j$. Where am I wrong? – Raphaelo Nov 06 '22 at 11:45