Let $D$ be a divisor on a variety $X$, and assume that $h^0(X,\mathcal O_X(D))=n+1$. So let $s_0,\dots,s_n$ be a basis of $H^0(X,\mathcal O_X(D))$. I am trying to understand the following statement :
"The map defined by $s_0,\dots,s_n$ is an embedding."
I understand that the $s_0\dots,s_n$ are rational functions on $X$, defined on the whole $X$. So I suppose that they talk about the map $X\rightarrow \mathbb P^n$ whose coordinates are the $s_i$.
- Am I correct?
- Why is this an embedding?
Thank you!
Best Answer
Yes, you are correct that one way to think about the map defined by $s_0,\cdots,s_n$ is the map which sends a point $x\in X$ to $[s_0(x):\cdots:s_n(x)]$.
Your other question requires a bit more care. This map is not always defined on the whole variety $X$ nor is it always an embedding: both of these conditions depend on $D$. The condition that the map is defined everywhere is that $\mathcal{O}_X(D)$ is base-point free, and the condition that it is an embedding is the condition that $\mathcal{O}_X(D)$ is very ample. There are plenty of varieties $X$ and divisors $D$ which do not satisfy these conditions. If this is your first exposure to these sorts of concepts, I would recommend reading the treatment of these concepts in an introductory algebraic geometry text. Hartshorne covers this in chapter II section 7, Vakil covers this in section 15.3, Gortz and Wedhorn in chapter 13, etc.