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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.
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  • $\begingroup$ About the title question: Have you heard of the reduced trace? $\endgroup$ Apr 19, 2023 at 4:58

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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 $\bar{k}$?

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    $\begingroup$ 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). $\endgroup$ Jan 2, 2023 at 5:33
  • $\begingroup$ 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 $[A,A] \otimes_k \overline{k}$? $\endgroup$ Jan 2, 2023 at 5:38
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    $\begingroup$ 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. $\endgroup$
    – Chris H
    Jan 2, 2023 at 8:44
  • $\begingroup$ This makes a lot of sense. Thank you! $\endgroup$ Jan 2, 2023 at 18:02

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