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Let $E$ be an elliptic curve over an algebraically closed field $k$ of characteristic $p$. Is there any nice computation for the group $H^1(E,\alpha_p)$ and $H^1(E,\mathbb{G}_a)$? The cohomology is taken in the flat topology.

When these groups are trivial? and is there any way to describe them? I am not asking for the description of $H^1$ using flat torsors on $E$, I looking for a description using which I could (at least) determine the groups are trivial or not.

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See Prop. III.4.14 (p129) in Milne's book on étale cohomology for a representation of $H^1(X,\alpha_p)$ as the subspace of the closed forms in $H^0(X,\Omega^1_X)$, which are annulled by the Cartier operator. Here $X$ is any smooth scheme over a perfect field. This result can also be found in Serre'es paper "Sur la topologie des variétés..." (which I don't have at hand). –  Damian Rössler Aug 25 '12 at 21:52
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up vote 7 down vote accepted

$\mathbb G_a$ is a smooth group scheme, so the flat cohomology is the same as the etale cohomology. It is also a quasicoherent sheaf, so the etale cohomology is the same as the Zariski cohomology, which is a $1$-dimensional vector space over $k$.

For $\alpha_p$, you can use the exact sequence $0 \to \alpha_p \to \mathbb G_a \to \mathbb G_a \to 0$ and take cohomology:

$H^0(E,\mathbb G_a)\to H^0(E,\mathbb G_a) \to H^1(E,\alpha_p) \to H^1(E,\mathbb G_a)\to H^1(E,\mathbb G_a)$

The map $H^0(E,\mathbb G_a)\to H^0(E,\mathbb G_a)$ given by taking $p$th powers is surjective since $k$ is algebraically closed. If it is only separably closed and not perfect than this is false. The map $H^1(E,\mathbb G_a)\to H^1(E,\mathbb G_a)$ given by taking $p$th powers is injective because it's a field, which has no nilpotent elements. So $H^1(E,\alpha_p)$ is trivial.

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Actually, the map on $H^1(E,\mathbf{G}_a)$ induced by Frobenius on $\mathbf{G}_a$ is simply the pullback along the absolute Frobenius on $E$. Hence, $H^1(E,\alpha_p)$ is trivial if $E$ is ordinary, and is $1$-dimensional if $E$ is supersingular. –  anon Aug 25 '12 at 19:07
    
Yes, that makes sense. –  Will Sawin Aug 25 '12 at 20:55
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