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Let $\pi: G \rightarrow S$ be a finite flat group scheme over a locally noetherian connected base scheme $S$. Its degree is defined as the rank of the locally free $\mathcal O_S$-module $\pi_* \mathcal O_G$. Let $H \subset G$ be a closed subgroup scheme of $G$ which is also finite flat over $S$.

I want to show that the degree of $H$ divides the degree of $G$. How does one do this? I guess this must be easy but I'm somehow stuck.

edited May 18 2012 at 17:10

asked May 18 2012 at 7:51

Veen
417●1●6

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The proof should be the same as for Lagrange's theorem: $H$ acts freely on $G$ by translations, making $G$ an $H$-torsor over $G/H$. – Keerthi Madapusi Pera May 18 2012 at 10:28

Maybe you want to assume your $\pi$ is finite locally free instead of just finite flat, if your $S$ might not be locally Noetherian. I feel like $\pi_*\mathscr{O}_G$ might not be locally free if $\pi$ is not locally of finite presentation. – Keenan Kidwell May 18 2012 at 16:24

@Keenan: that's a good point, I simply forgot the locally noetherian assumption. – Veen May 18 2012 at 17:09

Kestutis Cesnavicius · Accepted Answer · 2012-05-18 17:44:28Z UTC

This can be seen from the existence of the quotient $G/H$ as a finite flat $S$-scheme (and an $H$-torsor). One shows that the natural map $G \rightarrow G/H$ is finite flat of order $[H : S]$; the conclusion then follows from the product formula $[G : S] = [G : G/H] [G/H : S]$. Let me give you a reference where all this is spelled out in detail (and which I am basically copying):

Tate, John. Finite flat group schemes. Modular forms and Fermat's last theorem (Boston, MA, 1995), 121–154, Springer, New York, 1997.

What you need is section (3.5) (see also (3.4) and (3.1)).

Degree of finite group schemes

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