Finite connected groups over a perfect field of characteristic p - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-24T23:49:30Z http://mathoverflow.net/feeds/question/83241 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/83241/finite-connected-groups-over-a-perfect-field-of-characteristic-p Finite connected groups over a perfect field of characteristic p A.E. 2011-12-12T14:02:36Z 2011-12-14T09:41:21Z <p>In 14.4 of "Introduction to Affine Group Schemes" it is proved (!) that if $A$ represents a finite connected group scheme over a perfect field $k$ of characteristic $p$ then $A$ has the form $k[X_{1}, X_{2}, ..., X_{n}] / (X_{1}^{p^{e_{1}}}, ...., X_{n}^{p^{e_{n}}})$. But what about $\mu_{p} = k[X]/(X^{p}-1)$? It is connected but not isomorphic to $k[X]/(X^{p})$ as $k$-groups. They are isomorphic as $k$-schemes. Does this theorem mean " ...... $A$ has the form $k[X_{1}, X_{2}, ..., X_{n}] / (X_{1}^{p^{e_{1}}}, ...., X_{n}^{p^{e_{n}}})$ up to isomorphism of $k$-schemes"?</p> http://mathoverflow.net/questions/83241/finite-connected-groups-over-a-perfect-field-of-characteristic-p/83261#83261 Answer by Lubin for Finite connected groups over a perfect field of characteristic p Lubin 2011-12-12T17:29:44Z 2011-12-12T18:04:00Z <p>Others can do this much better than I, but here's what's happening: to describe a group scheme of any kind, you need to talk about not only the underlying space, but also the law of composition on the group. In this case, the kernel of $[p]$ in the muliplicative group, you describe the law of composition by writing down the the comultiplication on the affine ring $k[X]/(X^p)$. This is simply $X\mapsto 1 \otimes X + X \otimes 1 + X \otimes X$.</p>