most recent 30 from http://mathoverflow.net 2013-05-23T18:41:06Z http://mathoverflow.net/feeds/question/71934 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/71934/a-little-help-with-the-unmixedness-theorem Nick Addington 2011-08-02T22:16:24Z 2011-08-03T01:57:34Z

<p>I have two smooth subvarieties $Y$ and $Z$ of a smooth variety $X$. Their intersection $Y \cap Z$ has two irreducible components, both of the expected dimension and generically reduced. I want to conclude that $Y \cap Z$ is reduced by the unmixedness theorem. Is this right?</p>

http://mathoverflow.net/questions/71934/a-little-help-with-the-unmixedness-theorem/71951#71951 Jason Starr 2011-08-03T01:57:34Z 2011-08-03T01:57:34Z

<p>Dear Nick -- First of all, if a ring satisfies Serre's criterion $S1$ and is "generically reduced", i.e., the stalk at every generic point is a field, then the ring is reduced. This is explained, for instance at the top of p. 183, Section 23 of Matsumura's "Commutative Ring Theory". Second, if $Y$, resp. $Z$ is a closed subscheme of a regular, locally Noetherian scheme which is itself regular, then it is everywhere locally cut out by a regular sequence, cf. Theorem 21.2(ii), p. 171, of Matsumura. Finally, if also $Y\cap Z$ has the "expected codimension", then $Y\cap Z$ is also locally cut out by a regular sequence, and thus Cohen-Macaulay, by Theorem 17.4, p. 135 of Matsumura. (In fact it is even LCI by Theorem 21.2 again.) A Cohen-Macaulay scheme satisfies Serre's criterion $Sn$ for every integer $n\geq 0$. Thus your scheme $Y\cap Z$ is reduced.</p>