TITLE: Local structure of rational varieties
QUESTION [47 upvotes]: 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 ${\mathbb P}^n$?

REPLY [8 votes]: Here is a preprint of Ilya Karzhemanov constructing counterexamples in dimensions $n\geq 4$:
http://research.ipmu.jp/ipmu/sysimg/ipmu/1588.pdf