I always had trouble remembering this. Is it true that a curve over a nonalgebraicallyclosed field is normal implies that it's nonsingular? How about a 1 dimensional scheme? How about dimension 2? I think I heard once that surfaces over a nonalgebraically closed field is normal does imply that it's nonsingular. Is it true for2 dimensional schemes? What is the reason that these theorems are true for small dimensions, but fail for higher dimensions?

2$\begingroup$ "Normal implies smooth in codimension 2": rigtriv.wordpress.com/2008/07/08/… $\endgroup$ – Qiaochu Yuan Jan 22 '10 at 20:28

10$\begingroup$ You mean codimension 1? $\endgroup$ – Hailong Dao Jan 22 '10 at 20:34

$\begingroup$ Whoops. Yeah... $\endgroup$ – Qiaochu Yuan Jan 22 '10 at 20:36

7$\begingroup$ @Qiaochu: you also mean regular instead of smooth (a subtle point which is most important to the arithmetic geometers  see M. Emerton's response below). $\endgroup$ – Pete L. Clark Jan 22 '10 at 21:10

$\begingroup$ (....and mine.) $\endgroup$ – Pete L. Clark Jan 22 '10 at 21:46
For curves over a field $k$, normal implies regular. (The point is that a normal Noetherian local ring of dimension one is automatically regular, i.e. a DVR.) If $k$ is not perfect, then it might not be smooth over $k$.
The reason is that in this case it is possible to have a regular local $k$algebra of dimension one whose basechange to $\overline{k}$ is no longer regular. (On the other hand, smoothness over $k$ is preserved by basechange, since it is a determinental condition on Jacobians.)
Here is a (somewhat cheap) example: let $l$ be a nonseparable extension of $k$ (which exists since $k$ is not perfect), and let $X = \text{Spec} l[t],$ though of as a $k$scheme. This will be regular, but not smooth over $k$.
In dimension 2, even over an algebraically closed field, normal does not imply regular (and so in particular, does not imply smooth). Normal is equivalent to having the singular locus be of codimension 2 or higher (so for a surface, just a bunch of points) (this is what Serre calls R_1) together with the condition that if a rational function on some open subset has no poles in codimension one, it is in fact regular on that open set (this is Serre's condition S_2).
For a surface in ${\mathbb P}^3$, which is necessarily cut out by a single equation, the condition $S_2$ is automatic (this is true of any local complete intersection in a smooth variety), so normal is equivalent to the singular locus being 0dimensional.
For surfaces in higher dimensional projective space, $R_1$ and $S_2$ are independent conditions; either can be satisfied without the other. And certainly both together (i.e. normality) are still weaker than smoothness.
From Serre's criterion (normal is equivalent to $R_1$ and $S_2$) you can see that normality just involves conditions in codimension one or two. Thus for curves it says a lot, for surfaces it says something, but it diverges further from smoothness the higher the dimension of the variety is.
Edit: As Hailong pointed out in a comment (now removed), I shouldn't say that S_2 is a condition only in dimension 2; one must check it all points. Never the less, at some sufficiently vague level, the spirit of the preceding remark is true: $R_1$ and $S_2$ capture less and less information about the local structure of the variety, the higher the dimension of the variety.

4$\begingroup$ I don't know about cheap, but I like your example. $\endgroup$ – Georges Elencwajg Jan 22 '10 at 22:43
This is not the point of the original question, but since I am an arithmetic geometer, I do find it interesting.
Over a nonperfect ground field, one needs to distinguish between regular (all the local rings are regular) and smooth (Jacobian condition; equivalently, the base change to the algebraic closure is regular).
Emerton's example of a regular but not smooth curve is "somewhat cheap" because it is not geometrically integral. Here is a geometrically integral example which shows that the phenomenon is not so mysterious, but firmly connected to the existence of inseparable field extensions:
Let k be an imperfect field of odd characteristic p, so that there exists an inseparable field extension l/k of degree p, necessarily monogenic (since the degree is prime): say $l \cong k[t]/(P(t))$, where $P(t)$ is an inseparable polynomial. Specifically, we may take $P(t) = t^p  y$ for some element $y \in k$. Then the (unique regular projective model of) the hyperelliptic curve
$y^2 = P(x)$
is regular but not smooth, since upon base change to $l$ the polynomial $P(t)$ has multiple roots.

1$\begingroup$ This was helpful. I always wondered about the difference of smoothness and regularity. $\endgroup$ – Anweshi Jan 22 '10 at 21:26

$\begingroup$ @Pete. As of now, this answer, though addressing only a related question, got more votes than answers for the intended question. Interesting. $\endgroup$ – Anweshi Jan 22 '10 at 21:53


$\begingroup$ This is flattering, but I advise everyone who wants to understand mathematics more deeply to read all of M. Emerton's answers carefully  he really knows what he's talking about. Really. $\endgroup$ – Pete L. Clark Jan 23 '10 at 21:57

$\begingroup$ @Pete. Yes, I learned a lot from reading him. It is of course his answer that is the "best" for this question. But, you make more side remarks and comments, and they were very useful. I for instance have to clear misconceptions, etc., and it is embarrassing to ask each thing separately, and seeing this or that comment helped a lot. $\endgroup$ – Anweshi Jan 24 '10 at 12:02
In general, normality implies regular in codimension 1 (to be precise, normality is equivalent to $(R_1)$ and $(S_2)$ by Serre). So for curves, it implies regularity. For dimension 2, look at Spec ($k[x,y,z]/(x^2+y^2+z^3)$). It is normal, but not regular.
EDIT: Part of me can't resist saying something more, even after the very good answers of Emerton and Pete (warning: the following content might be disturbing to some viewers (: ). As I commented (see the last paragraph of Emerton's answer), $(S_2)$ can be rather hard to check. Let just state what it is. $X$ is said to be $(S_n)$ if:
$$ \text{depth}\ \mathcal O_{X,x} \geq min \{n, \text{dim} \mathcal O_{X,x} \} \ \forall x\in X$$
The depth of a local ring $(R,m)$ is the least index $i$ such that the local cohomology $H^i_m(R)$ is not zero.
So how come we never need to worry about checking this? Because most of the time, we already know something stronger about our variety: integrally closed, CohenMacaulay, Gorenstein, complete intersections in a smooth one, etc. But it can be quite difficult to check without prior knowledge. Let me give a somewhat famous example.
Conjecture(Hartshorne): Let $X$ be a smooth subvariety of codimension 2 in $\mathbb P^{n}_{\mathbb C}$. Assume that $n\geq 6$. Then $X$ is a complete intersection! (EDIT: the version Hartshorne stated requires $n\geq 7$, but I don't think any counterexample is known when $n=6$).
(this is closely related to the conjecture that there is no rank 2 vector bundles on $\mathbb P^{n}_{\mathbb C}$ with $n\geq 6$.)
What is relevant here is a result of EvanGriffith(theorem 2.3) which says that in the assumption of the Conjecture, if the affine cone of $X$ is $S_2$, then $X$ is complete intersection. Se we only need to prove $(S_2)$ness of the cone! But the conjecture is still open after more then 30 years.
There are two very nice proofs of "normal implies regular in codimension 1" in section 1.4 of "Lectures on Resolution of Singularities", by Kollar. One is said to be the most motivated proof Kollar can give, the other the slickest.
Dear Randy, here is an easy to remember, terse slogan:
Normal does not imply nonsingular, except in dimension one, where it does.
(Details in Hartshorne, AtiyahMacDonald Prop.9.2 and all the other fine answers to your question)