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?
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2$\begingroup$ I am not sure what you meant by unmixed theorem, but I think generically reduced + unmixed = reduced. $\endgroup$– Hailong DaoCommented Aug 3, 2011 at 1:13
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$\begingroup$ In view of Jason's answer, it is probably worth noting that unmixed = Serre's condition $(S_1)$. $\endgroup$– Hailong DaoCommented Aug 3, 2011 at 2:13
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1 Answer
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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.
answered Aug 3, 2011 at 1:57
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$\begingroup$ Thanks, Jason (and Hailong) - I haven't really used this Cohen-Macaulay stuff before, so I wanted to make sure I had my head screwed on straight. $\endgroup$ Commented Aug 3, 2011 at 5:40