Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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...

share|improve this question
2  
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

3 Answers 3

up vote 3 down vote accepted

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.

share|improve this answer

$$y^{-1}x=y^{-1}xyy^{-1}=y^{-1}yxy^{-1}=xy^{-1}.$$

share|improve this answer

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 $y^{-1}\in K_{x}$

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.