Ex. 5 From Polishcuk pg. 132

Part a

Given that

\ 0 \to \mu_2 \to K' \to K \to 0

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

Proof: If \  K' \cong K\times \mu_2 then the map \ K' \xrightarrow{[2]} K' is induced by the two multiplication by 2 maps to \ K,\mu_2 which are both the zero map, so K’ is also annihilated by 2. Conversely, if the \ K' is annihilated by 2, then every map \ \phi \colon K' \to \mathbb{G}_m lands in \ \mu_2. Now we use two results:

  1. Sending a finite group scheme to its cartier dual is an exact functor [ref: F.Oort, Commutative Group Schemes]
  2. Cartier Dual of \ A is naturally isomorphic to \ \hom(A, \mathbb{G}_m) [ref: pg. 123 Polischuk]

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

\ 0\to \hom(K, \mathbb{G}_m) \to \hom(K',\mathbb{G}_m) \to \hom(\mu_2,\mathbb{G}_m) \to 0

is exact so the natural inclusion \ \mu_2 \to \mathbb{G}_m lifts to come from a map\ K' \xrightarrow{g} \mathbb{G}_m and since \ K' is annihilated by 2, the image of the map lands in \ \mu_2 so that \ g is actually a retraction, i.e. we get a factorization of the identity \ \mu_2 \to K' \xrightarrow{g} \mu_2.  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 \ H^*(G,A) as come cocylces mod some coboundries.  So we define n-cochains \ C^n (G,A) = \{f| \phi \colon G^n \to A \} where the \ f are just maps and no other special properties are asked of them (i.e. not homomorphisms).  Define a boundary operator \ d\colon C^n \to C^{n+1} as follows: for \ \psi = df, then

\ \psi(g_0,\dotsc,g_n) = f(g_1,\dotsc,g_n) - f(g_0g_1,g_2,\dotsc,g_n)

\ + f(g_0,g_1g_2,\dotsc,g_n) +(-1)^nf(g_0,\dotsc,g_{n-1}g_n)

\ -(-1)^nf(g_0,\dotsc,g_{n-1})

Probably its true that \ d^2 = 0 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 \ e\colon K\times K \to \mu_2 has a chance of being a cocylce.  In multiplicative notation,

\ de(x,y,z) = e(y,z)e(xy,z)e(x,yz)e(x,y)

\ =e(y,z)e(x,z)e(y,z)e(x,y)e(x,z)e(x,y)

\ = e(y,z)^2e(x,z)^2e(x,y)^2 = 1

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 \ K\times\mu_2 to get an associative group law that fits into a short exact sequence \ 1\to \mu_2 \to (K\times \mu_2)^{twisted}\to K \to 1 , i.e. an extension.  The small tidbit, for a general exact sequence of groups \ 1\to A \xrightarrow{i} G \xrightarrow{\pi} P \to 1 and a section \ P \xrightarrow{s} G, then there is an isomorphism \ g \mapsto (\frac{g}{s(\phi(g))}, \pi(g)) that is compatible with the inclusion and projection maps.  Anyways, the twisted group law is

\ (m,k)*(m',k')\mapsto (\frac{mm'}{e(k,k')}, kk')

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 \ e(k,k) \equiv 1.  That is we have an extension \ 0\to \mu_2 \to K' \to K \to 0 where K’ is annihilated by 2.  By part a, there is an isomorphism \ K'\to \mu_2 \times K 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 \ \mu_2 \times K).  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 \ \phi\colon K \to \mu_2 such that \ e(x,y) = \phi(y)\phi(xy)\phi(x).

 

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