I am looking at the proof of Wedderburn's theorem and I am bothered by the following fact:
Let $K$ be a division ring (it is supposed finite, but I don't think it is important for what I ask) and $x\in K$. Let $K_x$ be the set of elements of $K$ which commute with $x$.
In the proofs I have been looking at, it is supposed clear that $K_x$ is a division subring of $K$. I see why it is a subring but what I don't understand is why we have this implication:
$y\in K_x \;\mathrm{and}\; y\ne 0 \Rightarrow y^{-1}\in K_x$.
I'm probably missing an easy trick with the inversions and multiplications...