Traceless matrices and commutators 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:


*

*multiples of the identity may have trace $=0$.

*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 $M=AB-BA$.

 A: 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 http://projecteuclid.org/euclid.mmj/1028990168
A: Not all traceless matrix are commutators given a finite field with order $p^n$. Note that $M^{p^n} - M$ will have trace zero. Use the fact that the trace of $M$ is the sum of its eigenvalues and that $M^k$ has trace as sum of the eigenvalues of $M$ raised to $k$. 
