# Let $A$ be a central division algebra (of finite dimension) over a field $k$. Show that $[A,A] \neq A$.

I am looking at the post A central division algebra is not its commutator and I have a few questions regarding the proof that was provided in the answer.

1. Why is $$A$$ a simple $$k$$-algebra?

My first observation is that $$A$$ is Artinian because it is finite-dimensional. Then, I normally think about the Jacobson Radical being trivial to arrive at $$A$$ being semisimple. Are all central division algebras simple? Every time I Google central division algebra, it leads me to the Wikipedia page for central simple algebras but there is no discussion on my question so I think I am misunderstanding something fundamental.

1. Given that $$B =A \otimes_k \overline{k}$$ how do I see that $$[B,B] = [A,A] \otimes_k \overline{k}$$? I usually view adding and subtracting pure tensors to be completely abstract.
• About the title question: Have you heard of the reduced trace? Apr 19, 2023 at 4:58

As a hint for the first question, consider what a two sided ideal might look like in your division algebra, is there anything in the setup that forces this to be a trivial ideal? (Let $$x$$ be in this ideal…)
For the second part, imagine you had picked a basis over $$k$$. Can you use this to build a generating set of this commutator subalgebra over both $$k$$ and $$\bar{k}$$?
• So for (1), if I take an element of an ideal of our division algebra $x\in I \subset A$, we get that $xx^{-1} = x^{-1}x = 1 \in I$ so $I = A$. Thus, division algebras don't have any proper nontrivial ideals so it must be a field. Since fields are central simple algebras over itself we have proven (1). Jan 2, 2023 at 5:33
• For (2), if we say that $\{ a_1, \dots, a_n \}$ is a basis for $A$ over $k$, then $[A,A]$ is generated by $\{ a_i a_j - a_j a_i \}_{i,j=1}^n$ over $k$. I guess here is where I begin to lose intuition. I am thinking that $B = A \otimes_k \overline{k}$ has basis $\{ a_i \otimes 1 \}_{i=1}^n$ and so $[B,B]$ has basis $\{ a_ia_j \otimes 1 - a_j a_i \otimes 1 \}_{i,j=1}^n$ over $\overline{k}$ which exactly also generates $[A,A] \otimes_k \overline{k}$? Jan 2, 2023 at 5:38
• Yep, that’s correct for the first question! For the second, that’s nearly it, but these commutators need not be a basis, but they do generate. A general algebra lemma that might help is that if $\{e_i\}$ generate a left $R$ module $M$ for any ring, then $\{1\otimes e_i\}$ generate $S\otimes_R M$ as an $S$ module. You can prove this “by hand”, but it’s also a great exercise to utilise the left exactness of tensor product to prove this generation statement. Jan 2, 2023 at 8:44