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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.

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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

1 Answer 1

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 $x\in R$.

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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

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