\(K3\) surfaces with Picard number one and infinitely many rational points
The author constructs infinitely many \(K3\)-surfaces \(X/\mathbb Q\) such that \(X(\mathbb Q)\) is infinite and the geometric Picard number of \(X\) equals 1. A sketch of the proof is as follows. The author starts with a family of degree 4 surfaces \(X\) in \(\mathbb P^2\), such that all members of this family reduce to the same surface modulo 6. Let \(X_p\) denote the reduction of \(X\) modulo \(p\). Using the Lefschetz trace formula the author shows that the geometric Picard number of \(X_2\) and of \(X_3\) equal two. This implies that the geometric Picard number of \(X\) is at most 2. He shows that \(X_2\) contains a line and \(X_3\) contains a conic. This implies that the discriminant of the Néron-Severi lattice of \(X_2\) and of \(X_3\) differ by a non-square. He then observes that if \(X\) would have geometric Picard number 2 then both discriminants would differ by a square, hence \(X\) has geometric Picard number 2. He considers a subfamily of \(X\) such that each member has a hyperplane section with two nodes. The normalization \(E\) of such a curve is an elliptic curve. The author finds two rational points \(O,T\) on \(E\), taking \(O\) as the identity he checks whether \(2520T=O\), since this is not the case it follows from a Theorem of Mazur that \(T\) has infinite order, hence \(X\) contains infinitely many rational points.
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