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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"?

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The book surely has an author? –  Mariano Suárez-Alvarez Dec 12 '11 at 14:30
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$A$ is an algebra, not a scheme. The statement means the only thing it can possibly mean, that $A$ is of that form as an algebra. –  Angelo Dec 12 '11 at 14:46
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The book's theorem in (14.4) states "Let A represent a finite connected group scheme ...". So A should not only be an algebra but a Hopf algebra. Moreover in the proof is used/phrased that A is a Hopf algebra. In my opinion it's not that clear, that the proof really works well. As far as I can see, the critical point is the height 1 case where Waterhouse refers to (11.4), but (11.4) doesn't reflect the Hopf algebra structure. –  Ralph Dec 12 '11 at 15:22
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Thanks, I mean Spec$A$ by saying that $A$ is a scheme. Since the categeory of affine $k$-groups are anti-equivalent to category of $k$-Hopf algebras I say like this. The author is Waterhouse. But this statement exists in many lecture notes that I can't list them now. I think what Ralph says is true, because the the critical part of the proof is the case height=1. So I think it should be understood from the theorem that " ...... A has the form $k[X_{1}, X_{2}, ..., X_{n}] / (X_{1}^{p^{e_{1}}}, ...., X_{n}^{p^{e_{n}}})$ as a $k$-algebra (not as a $k$-Hopf algebra)". –  A.E. Dec 12 '11 at 16:21
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To André: what I am saying is that, since one interpretation is obviously wrong, as the OP knows, the other must be the right one. As to a classification of finite connected group schemes, if it possible at all, it will be certainly enormously complicated. –  Angelo Dec 12 '11 at 17:37
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1 Answer

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 $X\mapsto 1 \otimes X + X \otimes 1 + X \otimes X$.

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So $\alpha_p$ and $\mu_p$ have the same underlying space, so the same affine ring, but different comultiplications. –  Lubin Dec 12 '11 at 17:33
    
Shouldn't it be $x\mapsto x\otimes x$ rather? –  darij grinberg Dec 12 '11 at 17:50
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@darij: no, not if the ring is $k[X]/(X^p)$. Yes, if the ring is $k[X]/(X^p-1)$. –  Lubin Dec 12 '11 at 18:01
    
Felipe, thanks for the edit! –  Lubin Dec 12 '11 at 18:05
    
How is this an answer ? The starting point of the OP is exactly what you describe, namely that a group scheme is classified by a Hopf algebra and not just by its coordinate algebra. As such it turned out that in the book's theorem the isomorphism has to be understood as an isomorphism of algebras. –  Ralph Dec 12 '11 at 18:34
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