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Does a connected finite locally free group scheme G over a scheme S of characteristic p>0 has degree a power of p? I know that when S is the spectrum of a field k, it is true. Someone told me that this is true by considering a generic point of $, but I don't know how to do it?

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  • $\begingroup$ I begin to think that the answer maybe negative and I don't have to know it anymore because I found another way for my original problem. But thank you for the answers. $\endgroup$
    – Qijun Yan
    Commented May 21, 2013 at 7:20

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If $f:X\rightarrow S$ is finite locally free, then formation of $f_*\mathscr{O}_X$ is compatible with arbitrary change of base on $S$. So if $X$ has constant rank $r$, i.e., $f_*\mathscr{O}_X$ is finite locally free on $S$ of constant rank $r$, the same will be true for $f^\prime:X^\prime=X\times_SS^\prime\rightarrow S^\prime$ for any $g:S^\prime\rightarrow S$. So if you know that a connected finite flat group over an algebraically closed field of characteristic $p$ has rank a power of $p$, then given $S$ of characteristic $p$ and a finite locally free $S$-group $G$ with connected geometric fibers, by base change along a geometric point lying over a given point of $S$, you find that the rank at each point of $S$ is a power of $p$. I guess if the rank on $S$ is constant, then you only need a single geometric fiber to be connected.

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  • $\begingroup$ OK if "connected" means "with connected fibers". But I suspect the OP means "connected" literally. $\endgroup$ Commented May 21, 2013 at 5:46
  • $\begingroup$ Yes, here by "connected" I mean $G$ is connected topologically. I can assume that $G/S$ has constant rank. $\endgroup$
    – Qijun Yan
    Commented May 21, 2013 at 6:11

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