In Lazarsfeld's article "Brill Noether Petri without degenerations" he mentions the fact that for any integer $g \geq 2$, one may find a K3 surface $X$ and a curve $C$ of genus $g$ on $X$ such that the Picard group of $X$ is generated by $[C]$. How does one prove that ?
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$\begingroup$ @Youloush: What precisely are you asking? Ottem's answer proves that such surfaces exist. However, if you want to write one down in an explicit way, say over a field like $\overline{\mathbb{Q}}$ and for $g$ moderately large, that is quite difficult. For fields like $\overline{\mathbb{Q}}$, I recommend you look at the work of van Luijk. For large $g$, the results of Gritsenko, Hulek and Sankaran show that one cannot "algebraically parameterize" general K3 surfaces of large genus. $\endgroup$– Jason StarrCommented Mar 19, 2013 at 12:00
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This is equivalent to saying that there exists a K3 surface with an ample generator (a polarization) $L$, with $L^2=2g-2$ and $|L|$ has a smooth member. There are various geometric ways to construct such surfaces, e.g., by using double covers or quartic surfaces in $\mathbb P^3$ containing special curves. You will find this in VIII.15 in Beauville's book 'Complex Algebraic Surfaces". Given this, one can show the existence of a K3 where $L$ is a generator of the Picard group, using the fact that a generic K3 surface has Picard number 1.
answered Mar 18, 2013 at 17:51