Picard group of degree two cover of the plane Let $X$ be a smooth complex projective surface such that there is a degree two morphism $f:X\rightarrow \mathbb{P}^2$. Then I know that $X$ has to be branched over a curve of degree $2n$. I have two questions regarding $X$.
1) what is the Picard group of $X$? Is it $\mathbb{Z}$?
2) can any such $X$ be embedded in $\mathbb{P}^3$ as a closed subscheme. If this is true then $Pic\ X$ cannot be $\mathbb{Z}$ since we would need a line bundle on $X$ with four independent sections. In general what is the minimum $m$ such that $X$ can be embedded in $\mathbb{P}^m$?
 A: Assume the base field is $\mathbb{C}$.
If $n = 1$, $X \cong \mathbb{P}^1 \times \mathbb{P}^1$ (as Mohan said), so $Pic(X) \cong \mathbb{Z} \oplus \mathbb{Z}$ and $X$ can be embedded into $\mathbb{P}^3$.
If $n = 2$, $X$ is a del Pezzo surface of degree 2, it is isomorphic to $\mathbb{P}^2$ blown up at 7 points, $Pic(X) \cong \mathbb{Z}^8$, $X$ cannot be embedded into $\mathbb{P}^3$.
If $n = 3$, $X$ is a K3 surfaces. Generically, $Pic(X) \cong \mathbb{Z}$, but in the 19-dimensional moduli space of such covers there is a countable number of subspaces where the Picard rank jumps (up to 20).
If $n \ge 4$, $X$ is of general type, generically $Pic(X) \cong \mathbb{Z}$, but again it jumps on subvarieties of muduli space; this is governed by the Hodge theory.
A: For 1), there is a degree 2 map from $X=\mathbb{P}^1\times\mathbb{P}^1$ to $\mathbb{P}^2$. Notice that $X$ is a quadric surface in 3-space (and a projection from a point outside $X$ gives you your degree 2 map).
For 2), remember that any $n$-dimensional smooth projective variety can be embedded in $\mathbb{P}^{2n+1}$. 
