TITLE: Nonsingular/Normal Schemes
QUESTION [21 upvotes]: 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?

REPLY [33 votes]: 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.

REPLY [6 votes]: 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.