A central division algebra is not its commutator In looking at old qualifying exam questions, I've come upon a question that has me stumped.

Let $A$ be a central division algebra (of finite dimension) over a field $k$. Let $[A,A]$ be the $k$-subspace of $A$ spanned by the elements $ab-ba$ with $a, b \in A$. Show that $[A,A] \neq A$.

If $\text{char } k \nmid [A:k]$, there is a very pleasant argument at hand: fix a basis of $A$ and define $\text{tr}(r)$ to be the trace of the matrix associated to the vector space endomorphism $a \mapsto ra$. It's easily seen that $\text{tr}(ab-ba) = 0$; but by our assumption on the dimension, $\text{tr}(1) \neq 0$, so that $[A,A] \neq A$. This argument fails spectacularly in general. I'd like to salvage it, but I'm not convinced that's possible.
I would appreciate hints - if you give a full answer, I would appreciate it if you put it in spoiler boxes.
 A: I googled the question verbatim and the first result had a solution to the problem. As it turns out, one should perhaps look at more than just StackExchange when hunting for solutions. Here's a slightly expanded version of the proof (under Problem 2 here near the start of the document):
Let $\bar k$ be the algebraic closure of $k$. Then $\bar k$ is trivially a simple division algebra over $k$. Consider $B = A \otimes_k \bar k$; there is a theorem (see, for instance, Lorenz and Levy, Algebra II, ch 29 theorem 7) that, if $A$ and $B$ are simple $k$-algebras, and at least one is central, then $A \otimes_k B$ is a simple $k$-algebra. In particular, this means that (because $A$ is central) $B$ is a simple $k$-algebra, thus a simple $\bar k$-algebra. In addition, because $A$ is finite-dimensional over $k$, so is $B$ over $\bar k$.
It is not hard to see that the only division algebra over an algebraically closed field is the field itself; so by the Artin-Wedderburn theorem $B \cong M_n(\bar k)$ for some $n$. Then $[A,A] \otimes_k \bar k = [B,B]$, and because in $M_n(\bar k),$ $[B,B]$ consists of the traceless matrices, $[B, B] \neq B$; so $[A, A] \neq A$. 
