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Jul 25, 2017 at 20:38 comment added user21574 As an additional comment to see why we have such surjectivity of period map for $K3$ surface .If $K_X$ the canonical bundle of $X$, is trivial, $X$ Kähler, then the monodromy on $H^2$ is finite if and only if the degeneration is trivial (i.e. $X_0$ is smooth) This implies that if the period map can be extended, then the degenerate fibre is smooth, as a result we have the surjectivity of the period map in the case of Kähler $K3 $ surfaces.
Jul 26, 2013 at 18:43 vote accept Pete L. Clark
Jul 26, 2013 at 15:17 comment added Remke Kloosterman Pete wants his elliptic curve to be semistable. Therefore you need to require that "the E you start with" has an even number of `unstable' fibers and each unstable fiber is of type $I^*_{\nu}$. Such surfaces exist, but it requires a little work to show the existence by given a Weierstrass equation for exmaple. The K3 case is easier since you can just refer to the surjectivity of the period map.
Jul 26, 2013 at 15:00 comment added Felipe Voloch More generally, I think you can just start with $E$ of rank bigger than one, find a quadratic extension in which the rank doesn't increase and twist $E$ by that quadratic extension. The twisted curve has rank zero but extending to the quadratic extension will make the rank jump by more than one.
Jul 26, 2013 at 12:36 history answered Remke Kloosterman CC BY-SA 3.0