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May 11, 2016 at 10:53 comment added Gro-Tsen @DanielLoughran Ah yes, somehow I had gotten into my mind that he was assuming $X(k_v)\neq\varnothing$ for every place.
May 11, 2016 at 10:36 comment added Daniel Loughran @Gro-Tsen: He is assuming that $X(k) \neq \emptyset$, so the Hasse principle is rather trivial here.
May 11, 2016 at 9:42 vote accept R.P.
May 11, 2016 at 9:19 answer added Martin Bright timeline score: 5
May 11, 2016 at 8:50 comment added Daniel Loughran @Gro-Tsen: The result you mention in Wittenberg's book concerns the Hasse principle. How does this give applications to rationality?
May 11, 2016 at 8:44 comment added Daniel Loughran Hi Martin. It is not quite true that for a DP4, minimal is equivalent to Picard rank $1$. You can have minimal DP4s with a conic bundle structure (the example I construct in my answer is of this type). For cubic surfaces, however, of course minimal is equivalent to Picard rank $1$.
May 11, 2016 at 8:43 comment added Gro-Tsen Daniel Loughran already pointed out that the answer is no. But if you want at least a partial result in this direction, see theorem 3.36 in Wittenberg's book Intersections de deux quadriques et pinceaux de courbes de genre 1.
May 11, 2016 at 8:33 answer added Daniel Loughran timeline score: 7
May 11, 2016 at 8:25 comment added Martin Bright You can assume that your $X$ is minimal, since otherwise blowing down reduces to the case of higher degree. For dp4s, minimal is equivalent to Picard rank 1. So you asking whether it's possible to have a surface with $\mathrm{H}^1(k,\mathrm{Pic}\,\bar{X})=0$ and $\mathrm{H}^0(k,\mathrm{Pic}\,\bar{X})=\mathbb{Z}$, and with a rational point.
May 11, 2016 at 7:38 comment added Martin Bright In the rather special but related case of diagonal cubic surfaces, what you ask follows from the calculations of Colliot-Thélène, Kanevsky and Sansuc: see Proposition 1 and the following Lemme 1 in their article.
May 11, 2016 at 1:16 history edited R.P. CC BY-SA 3.0
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May 10, 2016 at 23:32 history asked R.P. CC BY-SA 3.0