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 ${\mathbb P}^n$?

## 2 Answers

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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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Here is a preprint of Ilya Karzhemanov constructing counterexamples in dimensions $n\geq 4$:

everypoint. (I almost asked the same question, by the way.) $\endgroup$2more comments