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Let $\pi:C\to D$ be a finite surjective morphism of noetherian integral schemes. Let $x\in C$ be a nonsingular point. Does it follow that $\pi(x)$ is nonsingular?

What if we impose some conditions like normality or regularity?

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up vote 2 down vote accepted

This does not hold because the normalization of any variety is a finite branched cover of it. So, if you choose a non-smooth (and thus non-normal) curve, you'll get a finite map from its normalization to it that does not send smooth points to smooth points. In order for something like this to work, you need to assume something stronger, e.g. that the map is etale.

(Let me add that you only need flatness to make the conclusion. That is, a flat morphism of noetherian schemes $X \to Y$ sends nonsingular points to nonsingular points. This is a consequence on Serre's characterization of regular local rings as those with finite global dimension. So if $A \to B$ is a local, flat, homomorphism of local noetherian rings, then if $B$ is regular, so is $A$. To see this, let $M$ be any $A$-module, and choose an infinite free resolution $P_\bullet$ of $M$ by finitely generated modules. After tensoring with $B$, we get an infinite free resolution of $M \otimes_A B$, i.e. $P_\bullet \otimes_A B$. The kernel at some point of $P_\bullet \otimes_A B$ will be free because $B$ has finite global dimension. That means the kernel at some point of $P_\bullet $ will be flat, hence free. So $A$ has finite global dimension.)

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Yes, etaleness works. But I read somewhere that if you have a finite group acting on a smooth curve, than the quotient is smooth again. I thought, this could be generalized. What if we take a finite group acting on a smooth variety? Is the quotient smooth? – Fozad Nov 19 '11 at 15:35
@Fozad: I just saw your comment. I don't believe this is true, in general, if the action is not free. Let me see if I can find a counterexample. – Akhil Mathew Nov 19 '11 at 15:58
@Fozad: I think the reason it works for curves, incidentally, is that taking quotients preserves normality, which is equivalent to smoothness in dimension one. – Akhil Mathew Nov 19 '11 at 16:26
Ahh! That makes sense! – Fozad Nov 20 '11 at 22:57
So can we infer from your edited answer that taking the quotient by a group does not give a flat morphism in general? – Fozad Nov 20 '11 at 23:02

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