Let $X$ be the blow-up of $\mathbb{P}^2_\mathbb{C}$ at $P=(0:0:1)$ and $\pi:X\to\mathbb{P}^2$ the projection map. I'm trying to prove that:
$\text{Pic}(X)\simeq \mathbb{Z}\oplus\mathbb{Z}$
Here is my attempt. Define the map:
\begin{align*}
\text{Div}(X)&\to\text{Div}(\mathbb{P}^2)\\
E&\mapsto 0\\
D\neq E&\mapsto \pi(D)
\end{align*}
which induces a map $\varphi:\text{Pic}(X)\to\text{Pic}(\mathbb{P}^2)$. We then have an exact sequence:
$$\mathbb{Z}\to\text{Pic}(X)\to\text{Pic}(\mathbb{P}^2)\to 0$$
where the first map is given by $1\mapsto [E]$. Clearly the map $\pi^*:\text{Pic}(\mathbb{P}^2)\to\text{Pic}(X)$ gives a right split in the sequence.
Knowing that $\text{Pic}(\mathbb{P}^2)\simeq \mathbb{Z}$ and using the splitting lemma, the only thing left to prove is that the map $1\mapsto[E]$ is injective.
I'm stuck here. If $m[E]=0$ for some $m\geq 1$, then $mE=\text{div}(f)$ for some $f\in k(X)$. The problem is that I don't know how to actually write down rational functions in $X$.
Any ideas?
Best Answer
First, we note that $\Bbb P^2$ and $X$ are birational: the open sets $\Bbb P^2\setminus P$ and $X\setminus E$ are isomorphic (the blowup map actually restricts to the identity here). This means that the function fields are isomorphic, and a function $f$ which has $div(f)=E$ on $X$ would necessarily have $div(f)=P$ on $\Bbb P^2$, which is obviously impossible. So $E\neq 0$ in the Picard group.
Tabes Bridges' comment is also a possible approach to the problem, as the self-intersection number can be computed only by knowing the degree of the normal bundle (without assuming anything about the class of $E$). As the exceptional divisor $E$ of a blowup in a point of a smooth variety of dimension $n$ has normal bundle $\mathcal{O}_E(E) \cong \mathcal{O}_{\Bbb P^{n-1}}(-1)$ (this is standard when defining the blowup and does not require knowledge of the Picard group of the blowup), we see that the self-intersection number of the exceptional divisor is $(-1)^{n-1}$. So if one can demonstrate that the intersection product is a well-defined thing on the Picard group, this would suffice to show that $E\neq 0$.