most recent 30 from http://mathoverflow.net 2013-05-25T22:58:29Z http://mathoverflow.net/feeds/question/75320 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/75320/connecting-points-on-a-variety-by-the-image-of-a-nonsingular-curve HNuer 2011-09-13T15:34:16Z 2011-09-13T19:35:16Z

<p>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</p>

http://mathoverflow.net/questions/75320/connecting-points-on-a-variety-by-the-image-of-a-nonsingular-curve/75327#75327 jlk 2011-09-13T16:11:03Z 2011-09-13T19:35:16Z

<p>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.</p> <p>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.</p> <p>Question for the experts: What's an example where it is impossible to take the curve to be smooth?</p>