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 nonsingular curves.

4$\begingroup$ If $X$ is quasiprojective and of dim $\ge 2$, you can use Bertini's theorem on a sufficiently general hyperplane section through P and Q. $\endgroup$– J.C. OttemApr 24, 2011 at 14:36

$\begingroup$ 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 nonsingular curve? $\endgroup$– BrianApr 24, 2011 at 14:46

4$\begingroup$ Brian, J.C. Ottem is right. You can just use Bertini. To your question of whether every curve is the image of a nonsingular 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. $\endgroup$– Karl SchwedeApr 24, 2011 at 15:24

$\begingroup$ Dear Karl Schwede: Thanks a lot for your answer. My question about the curve is indeed a very dumb one. $\endgroup$– BrianApr 24, 2011 at 15:29

4$\begingroup$ 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? $\endgroup$– Pete L. ClarkApr 24, 2011 at 22:29
2 Answers
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

9$\begingroup$ 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 `CPRA tribute' published by TIFR. $\endgroup$– MohanApr 26, 2011 at 17:21
In the affine setting over $\mathbb{C}$, an algebraic set is pathconnected in the analytic topology if it is irreducible (in fact, its smooth locus is pathconnected too). Conversely, it is irreducible if and only if it contains a dense open pathconnected subset of smooth points.
See the appendix here.

5$\begingroup$ Sean: For algebraic geometers, a "curve" does not mean a "continuous path in analytic topology" (note the words "nonsingular" in the question). $\endgroup$– MishaMay 22, 2013 at 4:16

3$\begingroup$ Misha: Sorry for confusion. I was not really replying to the person who posted the question in as much as I was just offering a general piece of information that I felt was relevant to the title of the thread. My apologies. I am very inexperienced with mathoverflow and am still learning etiquette. I thought that people like me (who like to blur the categories of algebraic geometry, differential geometry and topology over C and R) might be drawn to the title of the thread and might find my comment useful or interesting. $\endgroup$ May 22, 2013 at 18:24