# An ample line bundle on a K3 surface

Let $X$ be a K3 surface obtained as a double covering of $\mathbb{P}^1 \times \mathbb{P}^1$ branching along a $(4,4)$-divisor. I think the natural line bundle $\pi^*\mathcal{O}_{\mathbb{P}^1\times \mathbb{P}^1}(1,1)$ is an ample line bundle on $X$. How can one prove that $\pi^*\mathcal{O}_{\mathbb{P}^1\times \mathbb{P}^1}(1,1)$ is ample? What is the degree of $X$ with respect to this ample line bundle?

• The pull back of an ample line bundle by a finite morphism is ample. I suggest that you use MSE for this kind of question, which is not at research level. – abx Mar 17 '14 at 10:20
• I didn't know that fact. Thanks for your advice on the use of this site. – user48202 Mar 17 '14 at 10:26

The ampleness follows from Nakai--Moichezon criterion. The degree is twice the degree of $O(1,1)$ on $P^1\times P^1$, so is $4$.
• It is. Take two divisors $D_1$ and $D_2$ in $O(1,1)$. Then $f^{-1}(D_1) \cap f^{-1}(D_2) = f^{-1}(D_1 \cap D_2)$. – Sasha Mar 17 '14 at 11:36
Let $f:X\rightarrow Y$ be a finite morphism. If $\mathcal{L}$ is an ample line bundle on $Y$ then $f^{*}\mathcal{L}$ is ample on $X$. Let $V\subseteq X$ be a subvariety. By the projection formula we have $f^{*}\mathcal{L}^{dim(V)}\cdot V = \mathcal{L}^{dim(V)}\cdot f_{*}V$. Furthermore $\mathcal{L}^{dim(V)}\cdot f_{*}V>0$ because $\mathcal{L}$ is ample, and $f$ finite implies $dim(f_{*}V) = dim(V)$. Finally $(f^*\mathcal{L})^{dim(X)} = deg(f)\mathcal{L}^{dim(Y)} >0$. By Nakai-Moishezon $f^{*}\mathcal{L}$ is ample.
In your case via the Segre embedding $s:\mathbb{P}^1\times\mathbb{P}^1\rightarrow\mathbb{P}^3$ the line bundle $\mathcal{O}(1,1)$ corresponds to the line bundle $\mathcal{O}_{Q}(1)$ on the quadric $Q = s(\mathbb{P}^1\times\mathbb{P}^1)$. Since linear sections of a quadric are plane conics and two such linear sections intersects in two points (namely two linear sections give a line and the two points are the points of intersection between the line and $Q$) we get $deg(\mathcal{O}(1,1)) = 2$. Now your $f:X\rightarrow \mathbb{P}^1\times\mathbb{P}^1$ is of degree $2$. So $deg(f^{*}\mathcal{O}(1,1)) = 2\cdot deg(\mathcal{O}(1,1)) = 4$.