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Consider a global field $F$ and the group $\Gamma =GL(2,F)$. An element $\gamma \in \Gamma$ is called elliptic, if its eigenvalues do not lie in $F$. Now consider a completion $F_v$ of $F$ and $G_v = GL(2,F_v)$. Now the eigenvalues of $\gamma$ may or may not lie in $F_v$. What is the centralizer $$ C_{G_v}(\gamma) = \{ g \in G_v : g\gamma = \gamma g \}?$$

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up vote 3 down vote accepted

The centralizer of $\gamma$ is a torus of the form $E^\times$, where $E/F$ is a quadratic field extension. Assume moreover that $\gamma$ is regular (so that $E/F$ is separable). The centralizer of $\gamma$ in $G_v$ is $(E\otimes_F F_v )^\times$. The algebra $E\otimes F_v$ is either a field (in that case $\gamma$ is elliptic in $G_v)$) or a sum of two fields isomorphic to $F_v$, according to whether the prime $v$ splits in $E$ or not. In the latter case $\gamma$ is a regular element lying in a split torus of $G_v$.

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Thanks for the fast response. I guess that answers my question completely, but I have some questions. You start with $\gamma$ semi simple, and I guess you embed $E^\times \subset \Gamma$, but how? Is $E$ the splitting field of the characteristic polynomial? I have read that semi simple implies that the centralizer is a torus, and I guess also that for $GL(n)$, you will need extensions of degree $n$? What is the standard reference for these facts? – Marc Palm Oct 17 '11 at 8:25
Identify $GL_2(F)$ with $GL_F(E)$, the group of $F$-linear automorphisms of $E$, where $E=F(\gamma)$. Under such an identification, $E^\times$ becomes a subgroup of $GL_2(F)$; $x\in E^\times$ acts on $E$ as multiplication by $x$, which is $F$-linear. – Amritanshu Prasad Oct 17 '11 at 8:49
And yes, you can do something similar for $GL(n)$. Take a matrix with irreducible characteristic polynomial. Identify $GL_n(F)$ with $G_F(E)$, where $E$ is obtained from $F$ by adjoining the roots of the characteristic polynomial. The centralizer of the matrix will be identified with $E^\times$. – Amritanshu Prasad Oct 17 '11 at 8:55

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