## In a division ring, every centralizer is a division ring

Prove that if $D$ is a division ring, then $C_D(a)$ is a division ring for all $a \in D$.

We saw in the previous exercise that $C_D(a)$ is a subring which contains 1. It remains to be shown that every nonzero element of $C_D(a)$ has an inverse in $C_D(a)$. To that end, let $x \in C_D(a)$ be nonzero. Then $xa = ax$. Now $x^{-1}$ exists in $D$, and we have $x^{-1}xa = x^{-1}ax$, so that $a = x^{-1}ax$. Similarly, right multiplying by $x^{-1}$ yields $ax^{-1} = x^{-1}a$. Thus $x^{-1} \in C_D(a)$. Thus $C_D(a)$ is a division ring.