Stratifications by smooth subvarieties 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.

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 $X=\coprod X_i$.