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Let $X$ be a K3 surface with $U(k) \subset NS(X)$ for some $k$. Here $U$ is the hyperbolic lattice. Can one conclude that $X$ admits an elliptic fibration? If my memory serves, I read somewhere that this is true for $k=1$.

I would appreciate your help.

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1 Answer 1

In characteristic $0$, I think the answer is yes, since $U(k)$ contains elements with square zero. See Theorem 11.1 in Brendan Hassett, Potential density of rational points on algebraic varieties (pdf). It says that a K3 surface $X$ over a number field possesses an elliptic fibration iff there exists $D \in \operatorname{NS}(X)$ such that $D^2 = 0$. I see no reason that the proof should not work over any field of characteristic zero, but of course this has to be checked.

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It's also true in positive characteristic: see the proof of Proposition (1.5) of Artin's article "Supersingular K3 Surfaces". –  Christian Liedtke Jan 27 '13 at 9:37
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