Linear Systems And invertible Sheaf

Question

Hello Fellow Mathematicians/Algebraic Geometer ,

This is question has two parts one which is more conceptual and the other more straight forward:

i) Let $X$ be a non-singular protective variety over an algebraic closed field $k$ and $L=\mathcal{O}_X(D)$ the invertible sheaf corresponding to a Cartier Divisor $D$ in $X$ . Define the linear system $|D|$ and show there exists a bijection $$ |D|\simeq\mathbb{P}(\Gamma (X,L))$$

ii) Explain the isomorphism in the special case $X=\mathbb{P}^1 $ and $L=\mathcal{O}_{\mathbb{P}^1}(n)$ and give the isomorphism $$ \mathbb{P}(\Gamma (X,\mathcal{O}_{\mathbb{P}^1}(n))\to \mathbb{P}^n $$

now for this we need to know since $X$ is smooth/regular the Cartier divisor $Ca(X)$ and the invertible sheaves $Pic(X)$ are isomorphic - even the Weildivisors $Div(X)$.

More we now that each section in $s\in \Gamma (X,L)$ corresponds to a effective divisor (s). For $|D|$ we have

$$|D|=\left\{ D'\in Div(X):\ D'\text{ is a effictive Divisor and } D\sim D' \right\} $$ where $\sim$ means linear equivalent. I do not get the idea how construct the isomorphism or even show that there is one. And for the concrete example ii) I guess I need to get i)

Answer 1

$\newcommand{\sheaf}[1]{{\mathcal #1}}$ $\newcommand{\Ohol}{\sheaf{O}}$

If $\sheaf{K}$ is the constant sheaf of rational functions on $X$, one can assume, that $\sheaf{L} \subseteq \sheaf{K}$ for every line bundle on $X$. The embedding goes as follows: Take an open affine $U \subseteq X$ with $\sheaf{L}|_U \cong \Ohol_U$. Then for every $V$ open in $X$ and $s \in \sheaf{L}(V)$ restrict $s$ to $s|_{U\cap V}$ and consider it as an element of $\Ohol_U(U\cap V) \subseteq K(X)$. The associated element of $K(X)$ is the image of $s$ in $\sheaf{K}(V)$.

Now let $D=(f_i,U_i)$ with $f_i \in K(X)$ the Cartier-Divisor and $\Ohol_X(D) = (f_i^{-1} \Ohol_{U_i})$ the associated line-bundle.

Then a section $s \in \Ohol_X(D)$ is given as $a_i f_i^{-1} = a_j f_j^{-1}$ and the associated effective divisor is $D' = (a_i, U_i)$. But $a_i = a_i f_i^{-1} f_i$ and $a_i f_i^{-1} = a_j f_j^{-1}$ is a rational function $f$ of $K(X)$. So we have $D' = (a_i,U_i) = (a_i f_i^{-1}, U_i) + (f_i,U_i) = (f) + (D)$.

As this argument can be read forwards and backwards we have the isomorphism of i) (one has to add the note that multiplication of everything with $w \in k = \Gamma(X,\Ohol_X^*)$ does not change anything).