# A Condition for division ring being a field.

I'm reading a paper of Maurice Chacron and I.N. Herstein entitled Powers of skew and symmetric elements in division rings.
At the first page of the paper, I got stuck in a problem that:

"if in a division ring $R$, $a^{m}b^{n}=b^{n}a^{m}$, for all $a, b \in R$, and appropriate $m,n >0$ depending on $a,b$, then $R$ is a field".

Is this a right statement? Thanks for helping me!

• IOW, $\forall a,b\in R ~\exists m,n>0~a^mb^n=b^na^m$ – arctic tern Mar 16 '17 at 5:53
• Do you think that statement is right? @arctictern – Minh Nam Mar 16 '17 at 10:04
• A commutative division ring is a field, so it's true if you always take $m=n=1$. Is there some context in that paper for the appearance of $m, n$? – Kimball Mar 16 '17 at 19:54
• @Kimball What do you mean by "take" $m=n=1$? While such $m$ and $n$ exist, they may depend on $a,b$. – arctic tern Mar 16 '17 at 21:35
• @arctictern It seems I interpreted the word "appropriate" in the quote differently than you did. I thought there was some context for it so the quote meant something different from your comment without words. But probably your interpretation is the intended one. – Kimball Mar 16 '17 at 21:46