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Let $X$ be an algebraic variety over an algebraically closed field $k$. Then $X$ is said to have a stratification if one can find irreducible locally closed subsets $X_i\subset X$ such that $X=\coprod X_i$ and whenever $\overline X_i$ intersects $X_j$ one has $\overline X_i\supseteq X_j$.

Question: Does every algebraic variety $X$ has a stratification by smooth subvarieties?

To show the answer is yes (which I believe, but do not know), I tried to play with the smooth locus of the irreducible components, taking the interior and so on, but something possibly singular always seemed to pop out at the end.

If there is any hypothesis one has to add on $X$ to get an affirmative answer, or if one has to relax a bit the definition of stratification, that would also be very useful to me.

Thank you for any help.

Brenin
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  • The smooth locus is open, why do you take interior ? – Cantlog Sep 07 '13 at 11:21
  • I meant I took the interior of the components (sorry I was unclear). But this lead me nowhere as even the interior might be singular. – Brenin Sep 07 '13 at 11:47
  • Start with the smooth locus of the whole varieties, then take the smooth locus of the complent (endowed with the reduced structure) etc, is there any problem ? – Cantlog Sep 07 '13 at 14:45
  • The only nontrivial issue is why the smooth locus is nonempty. See the question and comments to my question http://math.stackexchange.com/questions/435510/generic-regularity-of-affine-varieties – Moishe Kohan Sep 07 '13 at 15:04
  • @Cantlog: Thank you. So $X=X_{sm}\coprod(X\setminus X_{sm})$, and $(X\setminus X_{sm})_{red}$ has nonempty smooth locus because the smooth locus has to be open, right? – Brenin Sep 07 '13 at 15:37
  • Yes, the smooth locus is open and dense in a reduced algebraic variety over an algebraically closed field. – Cantlog Sep 07 '13 at 15:40
  • @atricolf: The issue is that an open subset could be empty. – Moishe Kohan Sep 08 '13 at 09:29
  • @studiosus: you are right. That's why it is important that the smooth locus be dense, right? the empty set is never dense. – Brenin Sep 08 '13 at 09:54
  • @atricolf: Yes, of course, except when the ambient space is empty too! – Moishe Kohan Sep 09 '13 at 08:38
  • For a smooth projective variety you can proceed by induction on the dimension (the 0 or 1 dimensional cases being trivial), and then proving the general case using Bertini. – rfauffar Sep 09 '13 at 12:31

1 Answers1

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To remove this question from the unanswered list, here is the answer coming out of the comments.

Let $X_1=X^{sm}$ be the smooth locus, which is open and dense in $Y_0:=X$, which we assume to be reduced. Consider the closed subset $Y_1=Y_0\setminus X_{1}$, take the dense open $X_2=Y_1^{sm}\subset Y_1$, and further $Y_2=Y_1\setminus X_2$, which is closed in $X$ as well. The chain of closed subsets $(Y_i)$ stabilizes by Noetherianity of $X$, and none of the open subsets $X_i\subset Y_{i-1}$ is empty, because the smooth locus of a variety is always dense. So finally $X=\coprod X_i$.

Brenin
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