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Any traceless $n\times n$ matrix with coefficients in a field of caracteristic $0$ is a commutator (or Lie bracket) of two matrices. What happens when the field has positive caracteristic?

When trying to reproduce the proof I have in the caracteristic $0$ case for the positive caractersitic case, I run into two problems:

  1. multiples of the identity may have trace $=0$.
  2. a matrix may have a spectrum equal to the whole field.

Are all traceless matrices commutators? If not, for which $n\in\mathbb{N}\setminus\lbrace 0 \rbrace$ and fields $k$ does it still hold? EDIT given a traceless matrix $M$, I want to know wether there are two matrices $A,B$ with $M=AB-BA$.

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up vote 4 down vote accepted

If $k$ is any field, the $k$-algebra $M_n(k)$ is Morita equivalent (as a $k$-algebra) to $k$. It follows that $M_n(k)$ and $k$ have isomorphic Hochschild homologies. In particular, they have isomorphic $0$th Hochschild homology.

In general, if $A$ is a $k$-algebra, the zeroth homology is $HH_0(A)=A/[A,A]$, the quotient of $A$ by the subspace generated by commutators. Since $k$ is a commutative $k$-algebra, it is obvious that $HH_0(k)=k$. The first paragraph, then, tells us that $$M_n(k)/[M_n(k),M_n(k)]\cong k.$$

Now, the trace is a non-zero linear map $\mathrm{tr}:M_n(k)\to k$ which vanishes on $[M_n(k),M_n(k)]$. It follows by the above isomorphism that $[M_n(k),M_n(k)]$ is precisely the kernel of the trace.

N.B.: The details underlying my first paragraph above are explain in Jean-Louis Loday's book on cyclic homology or in Chuck Weibel's one on homological algebra, among other places.

Later. Olivier wanted matrices to be actual commutators, not sums thereof. It is a result of Albert and Muckenhoupt that this is always possible over a field. See

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I see, but wouldn't this show that traceless matrices are sums of commutators? I want to know wether any traceless matrix is a commutator. I'll edit my question to make it clearer. – Olivier Bégassat Mar 27 '12 at 21:57
Yes. That's what I wrote :) – Mariano Suárez-Alvarez Mar 27 '12 at 22:01
But are traceless matrices always a particular commutator? – Olivier Bégassat Mar 27 '12 at 22:08
Thank you for your last edit. Unfortunately I don't have access to the whole article but only to the first page. I think it might be improper for me to ask you to send me the file via email, but if it isn't and you'd do it I could give you my email address. Thanks in any case! – Olivier Bégassat Apr 29 '12 at 10:57

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