In Hartshorne's proof of a result of Igusa (see III, 9.13 of Hartshorne) he claims without proof that any two closed points on a variety can be connected by the image of a nonsingular curve, or by a finite number of such curves. I've seen something like this come up in other places as well, and I don't know why such a fact should be true or so obvious as not to explain. Can anyone explain it to me? Thanks
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3$\begingroup$ I think one could argue as follows: i) use Chow's Lemma (Harthshorne p.107) to reduce to the case that the variety is projective, ii) cut the variety by hyperplane sections to get a curve connecting the two points, iii) normalise to get the result. I am sure someone will correct me if I've missed something important... $\endgroup$– user5117Commented Sep 13, 2011 at 15:58
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3$\begingroup$ mathoverflow.net/questions/62843/… $\endgroup$– J.C. OttemCommented Sep 13, 2011 at 16:51
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1 Answer
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Mumford's book "Abelian Varieties" contains a proof of the following statement: given two points $x$ and $y$ on a variety, there is an irreducible curve containing both (Lemma on p. 56 in the section on the Theorem of the Cube). The normalization of the curve is non-singular, so this yields the desired result.
The proof is roughly what Artie said. Arguing by dimension, it is enough to produce an irreducible codimension $1$ subvariety containing the points. Pick a blow-up $f: X' \to X$ such that $X'$ is projective and the fibers $f^{-1}(x)$, $f^{-1}(y)$ are positive dimensional. Now fix a projective embedding of $X'$ and take a general hyperplane section $H$. This section is irreducible (Bertini) and meets the fibers $f^{-1}(x)$ and $f^{-1}(y)$ (for dimensional reasons). The image of $H$ under $f$ is the desired subvariety.
Question for the experts: What's an example where it is impossible to take the curve to be smooth?
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5$\begingroup$ To answer your question: if $X$ is itself a singular curve! $\endgroup$– M PCommented Sep 13, 2011 at 16:22
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$\begingroup$ Maybe less tautologically: if the singular locus of $X$ is the support of an ample divisor, then you are probably in trouble. $\endgroup$– M PCommented Sep 13, 2011 at 16:23
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$\begingroup$ @MP: Your remark reminded me that in my proof sketch, I should have stated at the beginning "We can assume the dimension of $X$ is at least $2$." No suitable blow-up exists if $X$ is a single point... $\endgroup$– jlkCommented Sep 13, 2011 at 19:36
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$\begingroup$ @jlk Also in order for your assertion that the general section meets the fibers $f^{-1}(x)$ and $f^{-1}(y)$ from a dimension count, I believe you also need that dim $X\geq 2$. You don't just need it to define the blow-up, unless you had some other dimensional reason in mind, which I'd like to here. Thanks for the nice answer. $\endgroup$– HNuerCommented Sep 13, 2011 at 20:06