Is it really necessary to go to all this trouble to split division algebras?

Question

Let $D$ be a finite-dimensional central simple algebra over a field $K$. $D$ is said to split over a field extension $L/K$ if we have $D_L \cong D \otimes_K L \cong M_n(L)$ for some $n$. In a few places (e.g. the Stacks Project) people seem to go to a lot of trouble to prove that $D$ always splits over a finite separable extension; one first shows that $D$ always splits over a maximal subfield of $D$ (this argument is nice), then that such a subfield can be chosen to be separable (this argument is a little fiddly).

The point of this, as far as I can tell, is to show the Brauer group $\text{Br}(K)$ can be computed as a filtered colimit of relative Brauer groups $\text{Br}(L/K)$ as $L$ runs over finite separable extensions of $K$.

However, it seems to me like we can prove that $D$ always splits over a finite separable extension via a general spreading out argument, without knowing anything about the subfields of $D$ at all, as follows:

1. First show that the extension of scalars $D_L$ remains a finite central simple algebra over $L$.

2. Next, the only finite central simple algebras over a separably closed field are matrix algebras, so $D$ splits over the separable closure $K^s$ at worst, meaning we have an isomorphism $D_{K^s} \cong M_n(K^s)$.

3. Now the spreading out argument: this isomorphism and its inverse, when expressed as matrices with respect to any basis of $D$ and the standard matrix basis of $M_n$, only involve finitely many elements of $K^s$, so $D$ in fact splits over the subfield generated by these elements, which is a finite separable extension of $K^s$.

Does this argument work? If so, why go to all the trouble that the usual line of argument goes to? I admit you do learn more from the usual line of argument but I'm surprised I haven't seen the spreading out argument anywhere yet.

Answer 1

Okay, I think I see what the issue is. It's the second step:

2. Next, the only finite central simple algebras over a separably closed field are matrix algebras, so $D$ splits over the separable closure $K^s$ at worst, meaning we have an isomorphism $D_{K^s} \cong M_n(K^s)$.

This is clear for the algebraic closure (we just need to show that there are no nontrivial finite-dimensional division algebras over $\overline{K}$ which is standard); however, to prove this result for the separable closure we need to rule out the possibility that there is a finite central division algebra $D$ over $K^s$ all of whose elements are purely inseparable over $K^s$. This can be done but the argument is exactly the argument used by the Stacks project to show that we can find a separable maximal subfield so we haven't actually avoided that argument. And also, now that I've taken a second look at it it's a pretty nice argument anyway!