Finite Unipotent Groups

Proof. Base-extending to k, we may assume we have a finite constant group scheme, say of order n. When we embed it as an algebraic matrix group, each g in it satisfies the separable equation X" — 1 = 0. If g is also unipotent, g — l. Thus the group is trivial. ... 

This is actually our second decomposition theorem for abelian groups in (6.8) we decomposed finite abelian group schemes into connected and etale factors. Moreover, that result is of the same type, since by (8.5) we see it is equivalent to a decomposition of the dual into unipotent and multiplicative parts. As this suggests, the theorem in fact holds for all abelian affine group schemes. To introduce the version of duality needed for this extension, we first prove separately a result of some interest in itself. 

In (11.4) we will show that all finite group schemes in characteristic 0 are etale, and hence none are unipotent. 

Let G be a finite group scheme, not necessarily commutative. Show that G is unipotent iff the augmentation ideal in the (noncommutative) algebra k[G]D is nilpotent. 

Let k be algebraically closed, G an algebraic matrix group. Show G is unipotent iff all elements of finite order have order divisible by char (k). [Use Kolchin s theorem to reduce to the abelian case, and look at diagonalizabie matrix groups.]... 

In a natural way D(G) is a 1 A-module, and D is a functor. CARTIER and GABRIEL prove that D is an anti-equivalence between the category of finitely generated A-modules annihilated by some power of V, and the category of unipotent algebraic group schemes over k (k is sup josed to be a perfect field in their theory). Examples ... 

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