= Solution
Let $\phi:C\setminus S\to\mathbb P^n$ be the given algebraic <morphism of algebraic varieties>. Its value at the generic point supplies homogeneous coordinates $[h_0:\cdots:h_n]$ with $h_i\in k(C)$, not all zero. Thus it defines a <rational map of projective varieties> on $C$.
Fix a missing point $p\in S$. The <local ring of a smooth algebraic curve> at $p$ is a <discrete valuation ring> $A=\mathcal O_{C,p}$, with <uniformizer> $t$ and <valuation> $v_p$. Put
$$
m=\min_{h_i\ne0}v_p(h_i),\qquad a_i=t^{-m}h_i.
$$
Zero coordinates remain zero. Every $a_i$ is in $A$, and some $a_j$ is a unit because its valuation is zero. Choose a neighborhood $U$ of $p$ on which all $a_i$ are regular and $a_j$ remains nonvanishing; this is possible because membership in the local ring means regularity on a neighborhood. The affine target-chart coordinates
$$
\frac{a_i}{a_j}\qquad(i\ne j)
$$
are regular on $U$, so they define a <morphism of algebraic varieties> $U\to\mathbb P^n$. At $p$ its homogeneous value is $[a_0(p):\cdots:a_n(p)]$, well-defined because $a_j(p)\ne0$.
Multiplication of all homogeneous coordinates by the same nonzero <rational function> does not change their projective point. Thus this local morphism agrees with the original one on the common dense open where their coordinates were chosen. They agree throughout the overlap: the target's diagonal is closed, being defined in homogeneous coordinate pairs by $T_iU_j-T_jU_i=0$ for all $i,j$. Thus the equalizer is closed and contains that dense open. Apply this construction at each point of the finite set $S$ and glue with the original morphism. The same closed-diagonal argument proves uniqueness. Therefore
$$
\boxed{\phi\text{ extends uniquely to a morphism }C\to\mathbb P^n.}
$$
This gives the <extension of a rational map from a smooth projective curve> by explicit local coordinates. The mechanism is the one-dimensional smooth local ring: it permits removal of the smallest common pole order while leaving at least one unit coordinate.
Back to article page
