I'm reading the proof of Proposition VIII.15 in Beauville's "Complex algebraic surfaces", and I do not understand a part of it:
Suppose $g=3k+1$, with $k\ge 1$. Let $Q\subset\mathbb{P}^4$ be a quadric with the ordinary double point(that is $Q$ is a cone over a non-singular quadric in $\mathbb{P}^3$); let $V\subset\mathbb{P}^4$ be a cubic such that $S=Q\cap V$ is a smooth surface (by the example VIII.14 we see that $S$ is a K3 surface $S_{2,3}$). Consider one of the two pencils of planes on $Q$: it cuts out on $V$ a pencil of elliptic curves $|E|$. Then $D_k=H+(k-1)E$ is very ample on $S$, and we have: $$D_k^2=6+6(k-1)=2g-2$$
My question is why the pencil of planes on $Q$ cuts out on $V$ a pencil of elliptic curves $|E|$. And how do we get that $D_k^2=6+6(k-1)$? I see that $$D_k^2=(H+(k-1)E)^2=H^2+2(k-1)H\cdot E+E^2=6+2(k-1)H\cdot E+0$$ since $H^2=deg(S)=deg(S_{2,3})=6$ and $E^2=2g-2=0$ by Proposition VIII.13(i), but I do not understand how do we get the equality in the middle, i.e. that $2(k-1)H\cdot E=6(k-1)$ .
I will appreciate any help!
For the first question, the pencil is obtained by intersecting a plane with a cubic threefold, hence the general such intersection is a smooth plane cubic curve, i.e. an elliptic curve, as is well known.
For the second, you have computed $ D_k^2 = 6 + 2(k-1)H \cdot E $. But $ E $ has degree $ 3 $ by the first question, hence by Bezout's theorem, $ H \cdot E = 1 \cdot 3 = 3 $, so $ D_k^2 = 6 + 6(k-1) = 2g-2 $.