Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Can someone please explain me the following proof?

Proposition: If $D$ is a division ring, then the division ring $R$ generated by $Z(D)$ and all additive commutators (elements of the form $xy-yx$) is the whole of $D$.

Proof: Let $x$ be a member of $Z(D)$. Then there exists $y$ in $D$, such that $xy\neq yx$, therefore $x(xy - yx)$ is in $R^*$, and $xy - yx$ is in $R^*$, thus $x$ is in $R^*$.

Thanks! G.

share|cite|improve this question
Maybe you want to say $x\notin Z(D)$. – user26857 Jan 12 '13 at 23:32
even if x is not in Z(D), I still don't understand the proof – cruvadom Jan 12 '13 at 23:41
what is [x,xy]? anyway, the way I look at it, we need to take an element of R, and show it is an element of D. I don't see where is it done here. – cruvadom Jan 12 '13 at 23:48
yep, the converse. just confused, sorry. so any idea? – cruvadom Jan 13 '13 at 0:02
up vote 2 down vote accepted

Let $x\in D$. If $x\in Z(D)$, then $x\in R$. If $x\notin Z(D)$ then there is $y\in D$ such that $xy\neq yx$. But $xy-yx\in R^*$ and $x(xy-yx)=x(xy)-(xy)x\in R^*$, so $x\in R$.

share|cite|improve this answer
how do we know that xy-yx is invertible in R? – cruvadom Jan 13 '13 at 0:48
you are correct, I said that R is a division ring. Thanks you for your help. my problem is a bit different, the following: show that as a Z(D) algebra, D is generated by the commutators – cruvadom Jan 13 '13 at 0:55

Your Answer


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.