Let $K\subset L$ be finite extensions of $\Bbb{Q}$.
Background. Let $D$ be a finite dimensional division algebra with center $K$. Its class in the Brauer group $Br(K)$ then maps injectively into the direct sum of the Brauer groups $Br(K_v)$ with $v$ ranging over completions of $K$. Each of the groups $Br(K_v)\simeq \Bbb{Q}/\Bbb{Z}$, and these fit into the fundamental short exact sequence (see e.g. Chapter VIII of Milne's notes on CFT) $$ 0\to Br(K)\to\bigoplus_{v}Br(K_v)\to\Bbb{Q}/\Bbb{Z}.\qquad(*) $$ Here the class $[D]\in Br(K)$ is mapped to $\bigoplus_v[D_v], D_v:=D\otimes_KK_v$, and we know that $[D_v]$ is non-trivial only finitely many $v$. The non-trivial components of the image are the Hasse invariants $inv(D_v)$ (viewed as elements of $\Bbb{Q}/\Bbb{Z}$) of $D$, and by exactness of $(*)$ they sum up to an integer.
Question. We can also extend the scalars from $K$ to $L$ and form the central simple $L$-algebra $D_L:=D\otimes_KL$. We have a natural commutative diagram (correct me if I'm wrong) $$ \begin{array}{ccc} Br(K)&\longrightarrow&Br(K_v)\\ \downarrow&&\downarrow\\ Br(L)&\longrightarrow&\bigoplus_{w\mid v}Br(L_w), \end{array} $$ where all the mappings are more or less extensions of scalars, and $w$ ranges over the extensions of $v$.
Is there a description of the mapping of the right column in terms of the behavior of $v$ in the extension $L/K$. In other words, if $v$ comes from a prime $\mathfrak{p}$ of $K$, and we know the inertia degree and ramification index of each prime $\mathfrak{P}$ of $L$ above $\mathfrak{p}$, is there a simple formula for the Hasse invariant $inv({D_L}_w)$ in terms of $inv(D_v)$, $e(\mathfrak{P}\mid \mathfrak{p})$ and $f(\mathfrak{P}\mid\mathfrak{p})$ (obviously $w$ corresponding to $\mathfrak{P}$)?
Example. Consider the case of $K=\Bbb{Q}$ and $D=\Bbb{H}$, the ring of Hamiltonian quaternions with rational coefficients. Then $D$ has two non-trivial Hasse invariants, both $=1/2$. One at the infinite place (corresponding to the fact that $D\otimes\Bbb{R}$ is a division algebra) and the other at the prime $v=2$ ($2$ is a factor of the discriminant of a maximal order of $D$, or $-1$ is not in the image of the norm map $N:\Bbb{Q}_2(i)\to\Bbb{Q}_2$). If we use the extension field $L=\Bbb{Q}(\sqrt5)$ we get the division algebra of Icosians. The Icosian ring (a maximal order of $D_{\Bbb{Q}(\sqrt5)}$) has unit discriminant, so the non-trivial Hasse invariant at $v=2$ became trivialized. This is probably related to some combination of facts
- $2$ is inert in $L/K$,
- $-1$ is a norm in the extension $L_2(i)/L_2$ because $L_2$ contains primitive third roots of unity $\omega$ and $\omega^2$ and $N^{L_2(i)}_{L_2}(\omega+i\omega^2)=\omega^2+\omega^4=-1$.
On the other hand $D_L$ has two non-trivial Hasse invariants at the two infinite places.
I gave the Example to help the answerers gauge my level of familiarity with the relevant basic results. I am familiar with the construction of cyclic division algebras using non-norm elements. I waded thru Chapters 1-7 of Milne's notes in our study group 12-15 years ago, and familiarized myself with some of his extras but I'm a bit rusty. Pointers to relevant literature are welcome, but as CFT can be given in many different languages I may need help in reinterpreting the results in another.
