Is there a connected $k$-group scheme $G$ such that $G_{red}$ is not a subgroup? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-19T18:02:52Z http://mathoverflow.net/feeds/question/38891 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/38891/is-there-a-connected-k-group-scheme-g-such-that-g-red-is-not-a-subgroup Is there a connected $k$-group scheme $G$ such that $G_{red}$ is not a subgroup? stankewicz 2010-09-15T22:31:36Z 2012-07-26T09:17:25Z <p>I've been trying a learn a little more about group schemes by working through a set of exercises on Brian Conrad's website. Exercise 8.3 of <a href="http://math.stanford.edu/~conrad/papers/gpschemehw1.pdf" rel="nofollow">http://math.stanford.edu/~conrad/papers/gpschemehw1.pdf</a> reads:</p> <blockquote> <p>If $k$ is a perfect field and $G$ is a locally finite type $k$-group prove that $G_{red}$ is a closed $k$-subgroup of $G$. Can you find a counterexample if $k$ is not perfect? (There's a third part to the question, but it's irrelevant here)</p> </blockquote> <p>I'll try to not give away too much for the benefit of others who want to use these exercises, but the first part of the exercise is Lemma 7.10 of <a href="http://www.math.columbia.edu/algebraic_geometry/stacks-git/groupoids.pdf" rel="nofollow">http://www.math.columbia.edu/algebraic_geometry/stacks-git/groupoids.pdf</a> in DeJong's stacks project and the hint given for the omitted proof is that $G_{red}\times_k G_{red}$ is reduced if $k$ is perfect.</p> <p>For the counterexample part, I was able to find a disconnected example over any imperfect field, but it really seemed to depend crucially on being disconnected.</p> <p>This naturally suggests the question:</p> <blockquote> <p>If $G_{red}$ is not a subgroup scheme of $G$, must $G$ be disconnected? </p> </blockquote> <hr> <p>edit: After almost a week out an answer in the zero-dimensional case was given and then retracted, but there has not been much action otherwise. Perhaps this is an open problem if $G$ is not finite.</p> <p>For any finite group scheme $H$, $H_{red}$ is a subgroup of $H$ if and only if $H/H^0$ is a subgroup of $H$ (and in fact if $k$ is perfect, $H \cong H^0 \oplus H_{red}$, see <a href="http://www.math.ethz.ch/~pink/FiniteGroupSchemes.html" rel="nofollow">Pink's notes</a> Lecture 6- note this very explicitly depends on $H$ being finite)</p> <p>For group schemes of locally finite type (as we assume here) there has been much done, for instance in <a href="http://www.msri.org/publications/books/sga/sga/3-1/3-1t_296.html" rel="nofollow">SGA 3</a> where they prove among other things that if $G$ is connected, it's actually quasi-compact and thus of finite type (Proposition 2.4). Moreso, for a locally finite type group $H$ over a field, $(H^0)_{red}$ is a group <em>in the category of reduced schemes</em>. What's unclear is whether they believed it to be a group in the category of schemes but weren't able to prove that or if they knew of a counterexample but didn't include it.</p> http://mathoverflow.net/questions/38891/is-there-a-connected-k-group-scheme-g-such-that-g-red-is-not-a-subgroup/103080#103080 Answer by Olivier Benoist for Is there a connected $k$-group scheme $G$ such that $G_{red}$ is not a subgroup? Olivier Benoist 2012-07-25T09:07:00Z 2012-07-26T09:17:25Z <p>The new edition of SGA3 by Philippe Gille et Patrick Polo provides a connected example due to Raynaud. It is in SGA3 Exposé VIA Exemples 1.3.2 (2) and you may find it at <a href="http://www.math.jussieu.fr/~polo/SGA3/Exp6A-23mai11.pdf" rel="nofollow">http://www.math.jussieu.fr/~polo/SGA3/Exp6A-23mai11.pdf</a>.</p> <p>This example is $2$-dimensional, but it is easy to modify it to get a $1$-dimensional example if the characteristic of the base field is at least $3$. More precisely, let $G$ be defined by the equations $X^p-tY^p=Y^p-tZ^p=0$ in the additive group $\mathbb{G}_a^3$ over the field $\mathbb{F}_p(t)$, $p\neq 2$. Then $G$ is a connected group scheme of dimension $1$ whose reduction is not a subgroup scheme, as may be seen by following step by step the arguments of loc. cit.</p>