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I would like to find an example of principal ideal domain $R$, such that there exists a square matrix $A\in \mathfrak{M}_n(R)$ with zero trace that is not a commutator (i.e. for all $B,C \in \mathfrak{M}_n(R)$, $A\neq BC-CB$).

I know that such a PID (if it can be found) cannot be a field, or $\mathbb{Z}$.

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  • $\begingroup$ Why can't you take $R = \mathbb{Z}$, n=2, and $A = \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix}$? $\endgroup$
    – Mike Skirvin
    Commented Apr 18, 2010 at 16:08
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    $\begingroup$ Because that A can be written as BC-CB! $\endgroup$
    – Kevin Buzzard
    Commented Apr 18, 2010 at 16:16
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    $\begingroup$ @Mike: $A=[B,C]$ with $B=\begin{pmatrix}1&0\\0&0\end{pmatrix}$, $C=A$. $\endgroup$
    – Vladimir Dotsenko
    Commented Apr 18, 2010 at 16:20
  • $\begingroup$ I see, I was thinking that $B,C$ also had to be trace zero. $\endgroup$
    – Mike Skirvin
    Commented Apr 18, 2010 at 18:16
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    $\begingroup$ Is there a NATURAL proof for $R$ being a field? I have only seen ones involving a lot of hacking. $\endgroup$
    – darij grinberg
    Commented Sep 23, 2010 at 16:43

4 Answers 4

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It is not difficult to see that Rosset & Rosset's result for $2\times2$ matrices is equivalent to the surjectivity of the bilinear map $(X,Y)\mapsto X\times Y$ (called vector product when $A={\mathbb R}^3$) over $A^3$. For this, just search $B$ and $C$ such that $b_{22}=c_{22}=0$.

To prove it, let $Z=(a,b,c)\in A^3$ be given. One can choose a primitive vector $X=(x,y,z)$ such that $ax+by+cz=0$. By primitive, I mean that $gcd(x,y,z)=1$. Bézout tells that there exist a vector $U=(u,v,w)$ such that $ux+vy+wz=1$. Set $Y=Z\times U$. Then $Z=X\times Y$.

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Every matrix with trace zero over a PID is a commutator, according to the MR review of

Rosset, Myriam(IL-BILN); Rosset, Shmuel(IL-TLAV) Elements of trace zero that are not commutators. Comm. Algebra 28 (2000), no. 6, 3059--3072.

From the Math Review:

Although Shoda's method fails when $C$ is a PID, the authors do prove the result in this case, and give counterexamples for $C$ of dimension $\ge 2$.

However, I just took a look at the paper, and as far as I can see the authors only claim the result for 2x2 matrices!

Can anyone resolve this conundrum?

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    $\begingroup$ The paper of Rosset and Rosset contains an explicit calculation for 2x2 matrices over a PID. The math review, and to some extent the paper, claim that it solves this for general square matrices over PIDs, but I do not see that this is the case. $\endgroup$
    – Jack Schmidt
    Commented Apr 18, 2010 at 17:48
  • $\begingroup$ @Jack: I just independently spotted this! I thought it was a bit disingenuous to give a MR review for a reference, and when I checked it out I came to the same conclusion as you! I edited accordingly. $\endgroup$
    – Kevin Buzzard
    Commented Apr 18, 2010 at 17:49
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    $\begingroup$ @Kevin: this confirms my feeling - this paper does not claim it for matrices bigger than 2x2. Moreover, on the second page of the article, they claim explicitly that they do not know the answer for the 3x3 case! $\endgroup$
    – Vladimir Dotsenko
    Commented Apr 18, 2010 at 18:05
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I'm a bit (?) late on that one, but the theorem for general PIDs has been proven after this question was asked.

The reference is (see here for the ArXiv version):

Stasinski, Alexander, Similarity and commutators of matrices over principal ideal rings, Trans. Amer. Math. Soc. 368 (2016), no. 4, 2333–2354.

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Here (see the very last paragraph) it is stated that every matrix with trace zero over a PID is a commutator. However, I can't come up with a proof right away; the only proof for matrices over a field that I remember (due to Albert?) does not immediately generalize.

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  • $\begingroup$ If it helps anyone, the proof for matrices over a field (Albert and Muckenhoupt) is available via the following link: projecteuclid.org/… $\endgroup$
    – Vladimir Dotsenko
    Commented Apr 18, 2010 at 17:14
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    $\begingroup$ Vladimir, that is my .pdf file you linked to and I actually do not have a good citation! I may have just read the MathSciNet review as Jack did without looking at the Rosset--Rosset article itself. So I have now removed that claim from the .pdf file and will see if I can find someone to settle this question. ой.... $\endgroup$
    – KConrad
    Commented Apr 18, 2010 at 20:22
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    $\begingroup$ I have written to someone who has worked on these kinds of problems and will post a reply when I get it. $\endgroup$
    – KConrad
    Commented Apr 18, 2010 at 20:29
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    $\begingroup$ Vladimir, I wrote to Zak Mesyan (who wrote a paper on trace 0 and commutators) and he told me that whether an arbitrary nxn matrix over a PID with trace 0 is a commutator is still open as far as he knows. $\endgroup$
    – KConrad
    Commented Apr 19, 2010 at 14:04
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    $\begingroup$ Keith, thanks! Why don't you post it as an answer here? I think it's a bit misleading that the accepted answer to this question states that every nxn traceless matrix over a PID is a commutator! $\endgroup$
    – Vladimir Dotsenko
    Commented Apr 19, 2010 at 21:13

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