A key fact in "birational geometry" (on $\mathbb C$) is the following theorem:
Let S be a surface. Then there is a bijection between the following sets:
(i) {rational maps $\phi:S-\rightarrow\mathbb P^n $ such that $\phi(S)$ is contained in no hyperplane}.
(ii) {linear systems on $S$ without fixed part and of dimension $n$}.
The statement of the theorem is taken from the Beauville's book on algebraic surfaces, but the result is true even if we replace $S$ with a generic non-singular projective variety.
On Beauville's book there is a quick proof (only for surfaces), and unfortunately most of the details are not covered. I have understood how the bijective correspondence is constructed, but I have missed some things. For example:
$1)$ Why do we require that $\phi(S)$ is not contained in any hyperplane?
$2)$ Let $P$ be a linear system on $S$ and let $x\in S$ be a closed point. why the set $$\{D\in\text{Div}(S)\,: x\; \text{is in the support of}\;D\}$$ is a hyperplane of $P$?
So my request is the following:
I'd like to see a complete proof of the theorem for the general case (not only for surfaces). So, I need a direct proof here on ME or alternatively a reference. If there will be two distinct answers, one with a reference and one with a proof, I clearly will accept the proof.
Let $S$ be a linear system on $X$ with no fixed part. Say $S=\Bbb{P}(V)$ with $V\subset H^0(X,L)$ for some line bundle $L$ on $X$. Then the choice of a basis for $V$ gives you a rational map from $X$ to $\Bbb{P}(V)$. The points where the map is not defined is the set of base-points of $S$. Notice that if the image of this map was contained in a hyperplane it would mean that the elements of your basis are linearly dependent.
If you have a rational map $\varphi$ from $X$ to some $\Bbb{P}^n$ then define $L:=\varphi^\ast\mathcal{O}(1)$ and $V:=\varphi^\ast H^0(\Bbb{P}^n,\mathcal{O}(1))$. So you have $H^0(X,L)=V\oplus W$ (with $W$ possibly zero). Then you have a linear system $S=\Bbb{P}(V)$ and by construction its associated map, up to a change of basis of $V$, is $\varphi$.