# Ex. 5 From Polishcuk pg. 132

Part a

Given that

is exact sequence of finite commutative group schemes; *K* annihilated by 2. Then the sequence splits iff K’ is annihilated by 2.

Proof: If then the map is induced by the two multiplication by 2 maps to which are both the zero map, so *K’* is also annihilated by 2. Conversely, if the is annihilated by 2, then every map lands in . Now we use two results:

- Sending a finite group scheme to its cartier dual is an exact functor [ref: F.Oort, Commutative Group Schemes]
- Cartier Dual of is naturally isomorphic to [ref: pg. 123 Polischuk]

Where the denotes the subset of morphisms between the schemes that are also homomorphisms. Using these results we can say that

is exact so the natural inclusion lifts to come from a map and since is annihilated by 2, the image of the map lands in so that is actually a retraction, i.e. we get a factorization of the identity . Thus the sequence splits.

Part b

This problem involves group cohomology.

Quickly, for a group G and another group another group A, we want to define as come cocylces mod some coboundries. So we define n-cochains where the are just maps and no other special properties are asked of them (i.e. not homomorphisms). Define a boundary operator as follows: for , then

Probably its true that so you can define cocycles and coboudries. For the problem at hand, all the groups we are working with have “characteristic 2” so we ignore signs. The bilinear pairing has a chance of being a cocylce. In multiplicative notation,

So we have a cocycle. Its also a general fact that when ever you have a 2-cocycles you can twist the group structure on to get an associative group law that fits into a short exact sequence , i.e. an extension. The small tidbit, for a general exact sequence of groups and a section , then there is an isomorphism that is compatible with the inclusion and projection maps. Anyways, the twisted group law is

Although since we are in char 2, we don’t really have to express it as a fraction. Now in the case m = m’, k=k’ we have that the product is the identity since . That is we have an extension where K’ is annihilated by 2. By part a, there is an isomorphism compatible with the inclusion and surjection maps in the extension. Again, its another general fact, that all such extensions come from some 2-cocycle (the trivial map in the case of the trivial extension ). And whenever we have a compatible isomorphism between two extension it means that the two cocylces differ by a coboundry. This is said more carefully in [ref: Group Cohomology,modular theory, etc by Brunetti, Guido, Longo]. In our case it means that e(,) is in fact a coboundry, which means exactly there is a function such that .

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- May 14, 2009 / 10:24 am

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