Embeddings of a surface that are given by linear systems of divisors

I am reading the proof of Castelnuovo's contractibility criterion in Beauville's Complex Algebraic Surfaces. I would like to clarify a paragraph. We have a hypeplane section $H$ of a surface $S$, a curve $E$ on $S$ isomorphic to $\mathbb P^1$, and another divisor $H':=H+kE$, where $k=H\cdot E$. We have that the (projective) dimension of the linear system $|H|$ is $n$, and that of $|H'|$ is some $q$ which is much greater than $n$.

Now the author says :

1. "$[\dots]$ $\{e_0,\dots,e_q\}$ is a basis of $H^0(S,\mathcal O_S(H'))$. Let $\phi:S\dashrightarrow\mathbb P^n$ be the rational map defined by the corresponding linear system."

I don't understand. Shouldn't this map be $S\dashrightarrow\mathbb P^q$? Then he proceeds :

2. "Since the map defined by $\{s_0,\dots,s_n\}$ is an embedding, the restriction of $\phi$ to $S-E$ is an embedding."

Here, $\{s_0,\dots,s_n\}$ is a basis of $H^0(S,\mathcal O_S(H))$, so since $H$ is very ample I understand that this map is an embedding. But how is it related to $\phi$?

I would really like to understand all the details of this proof, because it seems to be a good synthesis of the basic things I have to learn about divisors, rational geometry, intersection theory, linear systems etc... It is the Theorem II.17 of Beauville's Complex Algebraic Surfaces.

Thank you in advance!

Solution 1:

$\newcommand{\P}{\mathbb P}$ I think this is the reasoning: If you look at the projection $\P^q\dashrightarrow \P^n$ that keeps only the first $n+1$ coordinates, then this composition is the (regular) map given by $\{s_0,\ldots,s_n\}$, so it is an embedding: $$ S\setminus E \overset{{\{e_0,\ldots,e_1\}}}\longrightarrow\P^q\dashrightarrow \P^n. $$ And the only way your map composed with something else can be an embedding is if the original map is as well. I assume you can show this by the criterion about separating points and tangents, but also theorem 10.1.19 in Ravi Vakil's notes tells you that this is the case. I'm struggling to think of the easiest reason right now.