A problem in division rings and Brauer group

Question

Suppose that $D, E$ are two division rings in the Brauer group of $F$ ($Br(F)$), where $F$ is local field. Show that $D\otimes_FE$ is a division ring iff $([D:F],[E:F])=1$.

Answer 1

Some hints:

1) The implication $[D:F], [E:F]$ relatively prime implies $D \otimes_F E$ is a division algebra has nothing to do with $F$ being local: it is true (and is a very standard fact) for finite dimensional central division algebras over an arbitrary field.

Without knowing what your background in this subject is it will be hard to sketch out an answer that will necessarily be helpful to you, but here is one way to go that may or may not be accessible to you: you want to show that the index of $A := D \otimes_F E$ is equal to the product of the indices of $D$ and $E$. Thus if $d = \sqrt{[D:F]}$ and $e = \sqrt{[E:F]}$, you want to show that the index of $A$ is $de$. For this, it suffices to show that $A$ is not split by any field extension of degree prime to $d$ and is not split by any field extension of degree prime to $e$. And that's easy: after making a field extension $L/K$ of degree prime to $d$, the period of $D_{/L}$ is still divisible by $d$, and the period of $E_{/L}$ is still prime to $d$...

2) Here you do need to use that the Brauer group of a local field is $\mathbb{Q}/\mathbb{Z}$. I would also like to use that the period and index are equal in the Brauer group of a local field (which follows from the "standard fact" that the restriction map on Brauer groups induced by a degree $d$ extension of local fields is multiplication by $d$). And if you know these things, the result is not hard: really we are saying something like: since $6$ and $4$ are not relatively prime, the order of $\frac{1}{6} + \frac{1}{4}$ in $\mathbb{Q}/\mathbb{Z}$ is smaller than $24$.