Let $S=\mathrm{Spec}(R)$ for a valuation ring $R$ and $G$ a group scheme over $S$. Is $G$ necessarily separated (maybe with some finiteness condition)?
Asked
Active
Viewed 71 times
1
-
Have you heard of the valuative criterion of properness ? – Olórin Mar 02 '15 at 20:58
-
Yes. You mean to refer to the one for separatedness? Does it hold? – User12221 Mar 03 '15 at 03:53
-
Yeah, I meant separatedness – Olórin Mar 03 '15 at 03:54
-
@RobertGreen -Still, I'm not sure why it holds... – User12221 Mar 03 '15 at 04:00
-
See this : http://stacks.math.columbia.edu/tag/06E7 it may inspire you ;-) – Olórin Mar 03 '15 at 04:01
-
Thanks, but I don't think k[x] is a valuation ring. – User12221 Mar 03 '15 at 04:09
-
Never said it was. ;-) Gave you link for inspiration, not to answer. – Olórin Mar 03 '15 at 04:11
-
General fact : let $S$ be a scheme and $G$ be a group scheme over $S$, then $G$ is separated over $S$ if and only if the identity morphism $e : S \to G$ is a closed immersion. Is this the case if $S = \textrm{Spec}(V)$ with $V$ a valuation ring which is not a field ? – Olórin Mar 03 '15 at 04:30
-
I am aware of these facts, but still couldn't find a counter example or proof... – User12221 Mar 03 '15 at 04:32
-
Have you seen the exemple of the affine line with origin over the affine line over a field, in the link I gave you ? Do you think you could produce the same example when replacing $k[T]$ by a dvr for instance ? – Olórin Mar 03 '15 at 14:15