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Let $k$ be any field. Suppose $D$ is a central division algebra over $k$ of degree $n^2$, then we can understand its multiplicative group $D^{\times}$ as an algebraic group (defined over $k$). I wonder how to show $D^{\times}$ is anisotropic (modulo the centre $G_{m}$). In other words, the maximal $k$-split torus is the centre $G_{m}$.

According to Milne's Algebraic Group book, we can argue by looking at the conjugation representation on $D$ of a maximal $k$-split torus. But I was stuck at the last implication: how can one see $S \subset Z(G)$? In my understanding, $se_{i}s^{-1}e_{i}^{-1} \in k$ only implies it should be an $n$-root of unity, by looking at its determinant after base change to $M_{n}(\bar{k})$. Not sure why it has to be $1$.

Milne

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    $\begingroup$ The set $\{ se_i s^{-1} e_i^{-1}\ ; \ s \in S(k^a )\}$ is finite and connected in $(k^a )^{\times}$. $\endgroup$ May 15, 2022 at 7:26
  • $\begingroup$ @PaulBroussous I appreciate your comment. Could you explain a bit why the set is connected? $\endgroup$
    – Rigid AOE2
    May 15, 2022 at 10:14

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We consider the map $f$ : $S(k^a ) \longrightarrow (k^a )^{\times}$, $s\mapsto se_i s^{-1}e_i^{-1}$.It is a morphism of algebraic varieties, with $S (k^a )$ connected. So its image is connected. On the other hand it is finite so it must be reduced to one element. Since it contains $1$, it is reduced to $\{ 1\}$.

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