# Why do the elements which commute with a given element in a division ring form a division subring?

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...

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What you're calling a "field" is usually called a "division ring" or "skew field" in current English usage, the term "field" being reserved for a division ring that is commutative. – Robert Israel Nov 12 '12 at 8:53

So fix $x\in K$ and let $0\neq y\in K$ such that $xy=yx$. You want to show that $$y^{-1}x=xy^{-1}.$$ The latter is equivalent to $$y^{-1}x(xy^{-1})^{-1}=y^{-1}xyx^{-1}=1,$$ and this is indeed so: switch $x$ and $y$ and everything cancels out.
$$y^{-1}x=y^{-1}xyy^{-1}=y^{-1}yxy^{-1}=xy^{-1}.$$
Let $y\neq 0\in K_{x}$. Then $xy = yx$, and because $K$ is a field, $y^{-1}\in K$. Therefore, $y^{-1}xy = x$, and $y^{-1}x = xy^{-1}$. This means that $y^{-1}\in K_{x}$