# 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 $\ e(x,y) = \phi(y)\phi(xy)\phi(x)$.