# Do K3 surfaces with an Enriques involution have a polarization of bounded degree

Does there exists a real number $C$ with the following property.

For any Enriques surface $E$ over a number field $K$ with K3 cover $X\to E$, there exists an ample divisor $H$ on $X$ such that $H^2 \leq C$?

Context: A polarization of degree $d$ on a K3 surface is an ample divisor $H$ such that $H^2 =d$. I'm wondering whether the polarization degree of a K3 surface with an Enriques involution is bounded.

Yes, each K3-surface $X$ with Enriques involution has an an ample divisor $H_X$ with $deg \ H_X = 4$.

Proof: The given K3-surface $X$ is the universal covering $\pi: X \longrightarrow S$ of an Enriques surface $S$. The covering has degree $2$. Any Enriques surface has an effective divisor $D$, a "half" pencil, such that $2D$ is an elliptic pencil $|2D|$. In particular $D^2=0$. Define on $X$ the divisor $C := \pi^*D$.

Depending on divisors on $S$ distinguish two cases:

1. Either a second half pencil $D_2$ exists with $D_2^2 = 0$ and $(D,D_2)=1$ (non-special Enriques surface $S$)
2. or a $(-2)$-curve $E$ exist with $(D,E) = 1$ (special Enriques surface $S$).

In the first case, define the divisor $C_2 := \pi^*D_2$ on $X$ and set $f := f_{|C + C_2|}$. In the second case decompose $\pi^*E = E_1 + E_2$ with two $(-2)$-curves $E_i, i = 1,2,$ on $X$ and set $f := f_{|2C + E_1 + E_2|}$.

One checks: In both cases one obtains a well-defined map

$$f:X \longrightarrow \mathbb P^3.$$

It is a branched double cover $f_X:X \longrightarrow Q$ over a quadric $Q$ in $\mathbb P^3$. The quadric is non-singular in the first case and singular in the second.

The hyperplane section $H_Q$ of $Q \subset \mathbb P^3$ is very ample. It has degree $deg \ H_Q = 2$. The pull-back $H_X:=f_X^*H_Q$ along the finite map $f_X$ is an ample divisor on $X$ of degree $deg \ H_X = 2*deg \ H_Q = 4$, q.e.d.

Note. For the details of the Horikawa representation above see "Barth, W.; Hulek, K.; Peters, C.; van de Ven A.: Compact Complex Surfaces", section VIII, 18.