# Why is the multiplicative group of a central division algebra anisotropic?

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 $$1$$.

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

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 $$\{ 1\}$$.