Maximal domain of definition of rational morphism induced by linear system.

Question

Let $L$ be a invertible sheaf on $X$ and $s_0, \dots , s_n \in \Gamma(X,L)$ linearly independent. Then $\{s_i\}$ induces a rational morphism from $X$ to $\mathbb{P}^n$.

It is well know that if $s_i(x)$ all vanish, then the rational morphism cannot be defined at $x$. But what if $s_i = f s_i'$ for all $s_i$ locally and $s_i'$ not all vanishing at $x$. Then We can extend this morphism to $x$.

This may be impossible for $X$ a (projective) variety. But I don't know how to show it.

QUESTION: Whether or not, the maximal domain of definition of rational morphism induced by linear system is the set away from the base.

What exactly is the question here? – KReiser Dec 22 '20 at 17:46 — KReiser, Dec 22 '20 at 17:46
@KReiser I've revised. Thank you! – XT Chen Dec 22 '20 at 17:50 — XT Chen, Dec 22 '20 at 17:50

score 1 · Accepted Answer · answered Dec 22 '20 at 18:44

This isn't true. Consider $X=\Bbb P^1$, $\mathcal{L}=\mathcal{O}(2)$, and $s_0=x^2,s_1=xy$. Then this defines a nonconstant rational map $\Bbb P^1\to\Bbb P^1$, which extends to a morphism - any rational map from a regular curve to a projective variety extends to an honest morphism (ref).

Maximal domain of definition of rational morphism induced by linear system.

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