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Let $R$ be a complete DVR with algebraically closed residue field $k$ and fractional field $K$ , $PGL(2)$ the automorphic group of projective line over $\overline K$.

My question is:

When $H^{1}(Gal(\overline K/K), PGL(2))=0$ ? Is this group trivial if $Char(K) \neq 2$ ?

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A more informative title wouldn't hurt, I think. –  Kestutis Cesnavicius May 3 '13 at 2:53
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1 Answer 1

up vote 8 down vote accepted

The claimed triviality holds (the nonabelian cohomology set is not a group though), and I don't think you need $Char(K) \neq 2$. To argue this, I will use the long exact nonabelian cohomology sequence of the central extension $1 \rightarrow \mathbf{G}_m \rightarrow GL_2 \rightarrow PGL_2 \rightarrow 1$, a segment of which reads $H^1(K, GL_2) \rightarrow H^1(K, PGL_2) \rightarrow H^2(K, \mathbf{G}_m)$. Firstly, $H^1(K, GL_2)$ is the one-point set because it classifies rank 2 vector bundles over $Spec(K)$, of which there is only the trivial one. Secondly, $K$ is a $C_1$ field by a theorem of Lang (see Serre "Galois cohomology", p. 80, II.3.3 c)), hence is of $dim \le 1$, so its Brauer group $H^2(K, \mathbf{G}_m)$ vanishes (loc. cit. for more details).

The same argument shows that $H^1(K, PGL_n)$ is the one-point set for any $n$ and any $C_1$ field $K$.

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Alternatively this cohomology group classifies principal homogeneous spaces for $PGl_2$ and these correspond precisely to rank 4 Azumaya algebras $A$ by descent theory. But you can lift idempotent elements in $\overline A$ over the residue field $k$ to idempotent elements in $A$ over $R$. Consequently any such algebra has non identity idempotents in it and so must be a 2×2 ring of matrices over $R$. The characteristic of $k$ or $K$ doesn't matter. –  Ray Hoobler May 4 '13 at 21:06
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