# K3 surface $S \subset \mathbb P^1 \times \mathbb P^2$ of bidegree $(2,3)$ has Picard rank 2?

$$\DeclareMathOperator{\Pic}{Pic}$$The following example is taken from [1, Example 5.8].

Let $$S \subset \mathbb P^1 \times \mathbb P^2$$ be a general smooth surface of bidegree $$(2,3)$$, so $$S$$ is a K3-surface. I want to understand how to compute the Picard lattice of $$S$$. First we know the Picard group of the surrounding space, $$\Pic(\mathbb P^1 \times \mathbb P^2) = \mathbb Z \cdot \mathcal O(1,0) \oplus \mathbb Z \cdot \mathcal O(0,1).$$ Choose effective divisors $$X = \{pt\} \times \mathbb P^2$$ and $$Y = \mathbb P^1 \times \mathbb P^1$$, which belong to the two linear systems of the two generators. So we may identify $$[X] = \mathcal O(1,0)$$ and $$[Y] = \mathcal O(0,1)$$. We can also calculate the intersections, namely $$[X]^2 = 0, \quad [X]\cdot[Y] = [\{pt\} \times \mathbb P^1], \quad [Y]^2 = [\mathbb P^1 \times \{pt\}],$$ and furthermore $$[X]\cdot[Y]^2 = 1, \quad [Y]^3 = 0.$$ We also have $$[S] = 2[X] + 3[Y]$$, and from this we obtain $$[X] \cdot [Y] \cdot [S] = 3, \quad [Y]^2 \cdot [S] = 2.$$ Since $$\gcd(2,3) = 1$$, this means that the restrictions $$\mathcal O(1,0)|_S$$ and $$\mathcal O(0,1)|_S$$ are $$\mathbb Z$$-linearly independent elements in $$\Pic(S)$$ and we have calculated that the rank $$2$$ lattice $$\Lambda \subset \Pic(S)$$ generated by them has intersection matrix $$\begin{pmatrix} 0 & 3 \\ 3 & 2 \end{pmatrix}.$$ However, van Geemen claims actually $$\Lambda = \Pic(S)$$, i.e. $$\Pic(\mathbb P^1 \times \mathbb P^2) \to \Pic(S)$$ is also surjective. Why is that the case?

[1] Bert van Geemen, Some remarks on Brauer groups of K3 surfaces.

• Looks like a version of Noether-Lefscetz theorem. Commented Aug 18, 2021 at 14:30