Local structure of rational varieties

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.) –  Artie Prendergast-Smith 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