## Path connectedness of varieties

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Let $X$ be a variety. Then, is $X$ path connected? And by path connected, I mean any two closed points $P, Q$ on the variety can be connected by the image of a finite number of non-singular curves.

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If $X$ is quasi-projective and of dim $\ge 2$, you can use Bertini's theorem on a sufficiently general hyperplane section through P and Q. – J.C. Ottem Apr 24 2011 at 14:36
The version in Hartshorne requires $X$ has at most a finite number of singular points and that $X$ projective (or equivalently, projective with a finite number of points removed). Do you have a more general form in mind? Also, your answer leads to another question (probably a dumb one that I cannot think of): curves are parametrizable, i.e. any segment on a curve is an image of a non-singular curve? – Brian Apr 24 2011 at 14:46

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Brian, J.C. Ottem is right. You can just use Bertini. To your question of whether every curve is the image of a non-singular one, the answer is yes, just take the normalization of the curve (see the section on curves in the first chapter of Hartshorne). I don't know what you mean by segment on a curve though. – Karl Schwede Apr 24 2011 at 15:24
Dear Karl Schwede: Thanks a lot for your answer. My question about the curve is indeed a very dumb one. – Brian Apr 24 2011 at 15:29

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We have some more than adequate answers given in the comments. Would one of the commenters be willing to step up and actually answer the question in the formal MO sense? – Pete L. Clark Apr 24 2011 at 22:29
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## 1 Answer

Given any two points on a projective variety, blow them up and re embed the blownup variety in P^N. Then by Bertini, any general linear section of the right codimension will meet the variety in an irreducible curve which also meets both exceptional divisors. Then blowing back down gives an irreducible curve connecting the original two points. Normalizing that curve gives a map from just one smooth connected curve that connects your two points. (I learned this trick from David Mumford.) – roy smith 11 hours ago

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Thanks, Roy.  – Pete L. Clark Apr 25 2011 at 4:10
This is very nice! Thanks. – QcH Apr 26 2011 at 0:06

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The above proof is attributed to C. P. Ramanujam. You can find it (and a few other such gems) in the article by Ramanan in the volume `CPR-A tribute' published by TIFR. – Mohan Apr 26 2011 at 17:21
thanks for the attribution Mohan. – roy smith Apr 26 2011 at 23:55
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