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Nov 4, 2011 at 9:28 comment added Torsten Wedhorn I think, Greg, you are confusing the regular locus of a scheme $X$ with the smooth locus of a morphism of schemes $X \to S$. The latter one is always open, but the first one might be not open, although this can never happen for schemes of finite type over a field as Quing Liu points out. If $S$ is the spectrum of a field $k$ and $X$ is of finite type over $S$, the smooth locus is always contained in the regular locus, but in general you have equality only if $k$ is perfect. An example would be if $X$ is the spectrum of a finite inseparable extension of $k$. Then $X$ is regular but not smooth.
Nov 3, 2011 at 21:02 comment added Qing Liu If $A$ is finitely generated over any field $k$, it is excellent, so its singular locus is closed. If $A$ is not finitely generated over $k$, but $k$ has characteristic $0$, then $A$ is still excellent. If $k$ has positive characteristic, $A$ might not be excellent even if $k$ is perfect (say finite).
Nov 3, 2011 at 20:00 comment added Greg Muller Eisenbud states that, for general Noetherian rings not over a perfect field, the set of singular primes may not even be Zariski closed, so there will be no ideal which defines it.
Nov 3, 2011 at 19:54 comment added Nicolás Thanks for your answer. I am interested in the case where the base could be a general ring, not necessarily a field.
Nov 3, 2011 at 18:19 history answered Greg Muller CC BY-SA 3.0