Let \(X\) be a variety over an algebraically closed field \(k\) of characteristic zero.
Does smoothness imply normality?
Does normality imply smoothness? If not, what happens for curves?
Solution
1. If \(X\) is smooth, then every local ring\[ \mco_{X,p} \]is regular.
The field \(k\) is perfect because it is algebraically closed. For schemes locally of finite type over a perfect field, smoothness implies regularity; indeed smoothness and regularity are equivalent in this setting by T-MORSMREG.
Therefore, if \(X\) is smooth, then for every point \(p\in X\) the local ring \(\mco_{X,p}\) is regular.
2. Every smooth variety \(X\) is normal.
By step 1, every local ring \(\mco_{X,p}\) is a regular local ring. Every regular local ring is a normal domain by D-QJ5M9. Hence every local ring of \(X\) is normal.
By definition, a variety whose local rings are all normal is normal. Therefore \[ \boxed{X\text{ smooth }\Longrightarrow X\text{ normal}.} \] This proves (a).
3. Normality does not imply smoothness in dimension at least \(2\).
Let \(n>1\) and let \[ \mu_n = \{\zeta\in\CC^\times:\zeta^n=1\} \] act on \(\AA^2_\CC\) by scalar multiplication. The quotient \[ V_n = \AA^2_\CC/\mu_n \] is a normal affine surface, and the image of the origin is a singular point of \(V_n\) P-AGXVAREXQUOTSING. Thus \[ V_n\text{ is normal but not smooth}. \] Therefore the converse to step 2 fails already in dimension \(2\).
4. For curves, normality does imply smoothness.
Let \(C\) be a normal curve over \(k\). At every closed point \(p\in C\), the local ring \(\mco_{C,p}\) is a one-dimensional Noetherian normal local domain. Such a ring is a discrete valuation ring, hence regular, by D-QJ5M9. Because \(k\) is perfect, regularity is equivalent to smoothness by T-MORSMREG.
(See also P-AGXVARNORMALCURVE and P-AGXVARNORMALPROJCURVE.) Thus \[ \boxed{ \dim C=1,\ C\text{ normal} \Longrightarrow C\text{ smooth}. } \]
5. The precise comparison is:\[ \boxed{ \text{smooth}\Longrightarrow\text{normal}, } \]the converse fails in dimension \(\geq2\), and it holds for curves.
Step 2 proves smooth implies normal. Step 3 gives a normal singular surface, so the converse fails in dimension at least two. Step 4 proves the dimension-one converse.
6. Q.E.D.
Step 5 answers both parts of the problem.