I was reading a paper recently concerning a non-commutative version of the matrix determinant. On the third page, it stated a fact without providing a proof or a reference:

If $D$ is a division ring, let $D^\times$ be its multiplicative group, then the abelianisation of $D^\times$, $\frac{D^\times}{[D^\times,D^\times]}$, is isomorphic to $Z(D)^\times$ - the centre of $D^\times$.

This result doesn't seem at all trivial, although of course it holds for any commutative field. I had considered that perhaps the natural map $Z(D)^\times\to \frac{D^\times}{[D^\times,D^\times]},c\mapsto c [D^\times,D^\times]$ could be proved to be an isomorphism. But I can see no reason why this should be be the case.

Is this result really true? If so, does anyone have any idea why? Is the natural map an isomorphism, or is there some other non-canonical isomorphism? Otherwise, can anyone think of a counterexample?

  • $\begingroup$ Perhaps this is mentioned in the article Brenner, J. L. "Corrections to “Applications of the Dieudonné determinant”." Linear Algebra and its Applications 13.3 (1976): 289. $\endgroup$ – rschwieb Oct 20 '17 at 16:17
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    $\begingroup$ Out of curiosity, I followed up on this by obtaining the correction article. Sadly, the error you found here was not among the corrections offered. I think the others are important if you are still reading the article. They are (paraphrasing slightly the article above): (C) 7.9 and 7.10 are false. 7.11 should be rephrased by changing the words "proper value" to "right [left] proper value" throughout. (D) If $K$ is not commutative, the material on page 520 is needed to establish 3.8. So to date it seems 3.5 serious errors have been found. $\endgroup$ – rschwieb Nov 2 '17 at 13:13
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    $\begingroup$ For the sake of completeness the other two points offered were (A) It was an unintentional oversight to omit reference to Artin's book (Geometric algebra. Interscience, 1957.) which includes a complete definition of the Dieudonne determinant. (B) Certain results of [the corrected article] have been complemented by P. M. Cohn in "The similarity reduction of matrices over a skew field." Mathematische Zeitschrift 132.2 (1973): 151-163. $\endgroup$ – rschwieb Nov 2 '17 at 13:17

This seems to be false.

An counter-example would be the division algebra $Q_\mathbb{R}$ of real quaternions. The center of $Q_\mathbb{R}$ is the field of real numbers $\mathbb{R}$, and the derived group $[Q_\mathbb{R}^\times, Q_\mathbb{R}^\times]$ is the group of elements $w=a+bi+cj+dk$ with $n(w):=a^2+b^2+c^2+d^2 = 1$ (a proof of this can be found here). Since $n$ is multiplicative, we get that $$ Q_\mathbb{R}^\times/[Q_\mathbb{R}^\times, Q_\mathbb{R}^\times] \cong \mathbb{R}^\times_{>0}. $$

This isomorphism is proved on page 12 of this Intelligencer article of Helmer Aslaksen (with thanks to @rschwieb for pointing out this reference).

However, $Z(Q_\mathbb{R})^\times = \mathbb{R}^\times$. But $\mathbb{R}^\times$ is not isomorphic to $\mathbb{R}^\times_{>0}$ (the former contains two elements that are their own inverses, and the latter only one).

Therefore $Q_\mathbb{R}^\times/[Q_\mathbb{R}^\times, Q_\mathbb{R}^\times]$ is not isomorphic to $Z(Q_\mathbb{R})^\times$.

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    $\begingroup$ +1 Indeed: precisely that is shown in this Intelligencer article. $\endgroup$ – rschwieb Oct 20 '17 at 16:18
  • $\begingroup$ @rschwieb Nice, thanks! I was unable to find a suitable reference. I'll incorporate it to the answer. $\endgroup$ – Pierre-Guy Plamondon Oct 20 '17 at 16:57

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