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I've been asked this question by a colleague who's not an algebraic geometer; we both feel that the answer should be "no", but I don't have a clue how to prove it. Here's the question: let $X$ be a smooth rational variety (over the complex numbers, say). Is it true that every point of $X$ has a Zariski open neighbourhood that is isomorphic to an open subset of ${\mathbb P}^n$?

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Qiaochu: every point. (I almost asked the same question, by the way.) – user5117 Jun 8 '12 at 20:40
I have discussed this over the years with several people. It is expected to be false, but it is open. – Jason Starr Jun 8 '12 at 21:16
Trivial in dimension 1. True in dimension 2 because the minimal models have this property and blow-ups preserve it. Is it true in dimension 3? – Will Sawin Jun 8 '12 at 21:52
I remember discussing this with Joe Harris many years ago. The problem seems to have been around for a while. – Angelo Jun 9 '12 at 6:59
In dimension 2 the result is true and is in fact stronger: we can assume that the open neighbourhood is isomorphic to $\mathbb{A}^2$. Do you have a counterexample in dimension $3$ of this? By the way, I would have thought that the answer to your question (not the stronger one) should be "yes", but as many people think the converse, I am now confused. If a point admits no such neighbourhood, it implies that every birational maps $X\to \mathbb{P}^n$ is either not defined at $x$ or contracts something through $x$. Do you have some candidate for $x$ and $X$? – Jérémy Blanc Oct 10 '12 at 21:08

Some partial results related to this open problem can be found in the very recent preprint by F. Bogomolov and C. Böhning On uniformly rational varieties, see arXiv:1307.0102.

According to the authors (see the Introduction) this question was first raised by M. Gromov in his paper

Oka's principle for holomorphic sections of elliptic bundles, Journal of the American Mathematical Society 2, Vol. 4 (1989).

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Grazie Francesco! – rita Jul 3 '13 at 19:45
You are welcome – Francesco Polizzi Jul 3 '13 at 20:46

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