Let $k$ be a field. Let $X$ be a reduced $k$-scheme of finite type. If $X$ is geometrically reduced, then it is a basic result that $X$ has a regular point (i.e. the local ring at that point is regular). If $k$ is not perfect, then $X$ need not be geometrically reduced. I am looking for an example of a reduced variety over a non-perfect field $k$ which has no regular points at all (if such a thing exists?)
Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)
|
2
|
||||||
|
|
6
|
I can just add a reference to Angelo's comment. If $X$ is a scheme locally of finite type over a field $k$ (or more generally over a complete local noetherian ring $k$), then $Reg(X)$ is open. Cf. EGA IV.6.12.5 and IV.6.12.8. The results seem to go back to Nagata and Zariski. If $X$ is of finite type over a field $k$ and reduced, then Angelo's argument shows that $Reg(X)$ is non-empty as well. Hence $Reg(X)$ will also contain closed points in that case. |
||
|
|
You can accept an answer to one of your own questions by clicking the check mark next to it. This awards 15 reputation points to the person who answered and 2 reputation points to you.
|
4
|
No, this is impossible. The regular locus of $X$ is open, and contains the generic point of each of its irreducible components, so it is dense in $X$. |
||
|
|
