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



Nonsingular/Normal Schemes
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 base-change to $\overline{k}$ is no longer regular. (On the other hand,
smoothness over $k$ is preserved by base-change, since it is a determinental condition 
on Jacobians.)


Here is a (somewhat cheap) example: let $l$ be a non-separable 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 0-dimensional.


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.


