# Corestriction of Brauer group of local field is an isomorphism

Brauer group of any field is the set of finite dimensional central simple algebras over that field modulo an equivalence relation, called the Brauer equivalence. It turns out to be an abelian group under the operation $\otimes$.  Every equivalence class has a unique central division algebra as a representative. So if we can classify all central division algebras over the field, we know its Brauer group.

Another way to describe the Brauer group of a field is via the profinite group cohomology of the Galois group of $K$:

$H^2(K, \mathbb{Q}/\mathbb{Z})$

Thus for (finite) field extensions $L/K$, we have the following morphisms :

1. Restriction : $Res : Br(K)\to Br(L)$, which at the central simple algebra level is just a base change  from K to L
2. Corestriction : $Cor : Br(L)\to Br(K)$

If K is an algebraically closed field, then there are no nontrivial finite dimensional central division algebras over it (If D is one such, then it must contain a nontrivial field extension of finite degree over K, which is not possible). Thus Br(K) = (0).

If K is a finite field, then there are still no nontrivial finite dimensional central division algebras over it (This is not as simple as the case of an algebraically closed field though, but the proof is still very elementary. One elegant way of showing this is to show K is a $C_1$ field. A $C_1$ field is any field such that any homogeneous polynomial with degree strictly less than the number of variables used has a nontrivial zero. And it is easy to see that there are no nontrivial central division algebras over such a field by simply observing that the reduced norm polynomial is a homogeneous polynomial with degree = $\sqrt{[D:K]}$ in $[D:K]$ number of variables where $D$ is the central division algebra over $C_1$ field $K$) Thus Br(K)=(0) here also.

The field $\mathbb{R}$ has exactly one nontrivial central division algebra over it. It is the famous Hamiltonian quaternions. Thus $Br(\mathbb{R})\cong \mathbb{Z}/{2\mathbb{Z}}$. I have forgotten how to prove this ..

We now come to $K$, a local field (examples are $\mathbb{Q}_p$ and $k((t))$ where $k$ is a finite field). Any finite extension of a local field is also a local field. Local class field theory tells us that $Br(K)\cong \mathbb{Q}/{\mathbb{Z}}$. In fact, it gives us the following very useful commutative diagram

$\begin{array}{ccc} Br(K) & \to^{Res} & Br(L) \\ \downarrow & . &\downarrow \\ \frac{\mathbb{Q}}{\mathbb{Z}} & \to ^{[L:K]} & \frac{\mathbb{Q}}{\mathbb{Z}} \end{array}$

The vertical arrows are isomorphisms. These maps which give you the isomorphism of the Brauer groups of local fields with $\frac{\mathbb{Q}}{\mathbb{Z}}$ are called the Hasse-invariants or simply $inv$.

To see the usefulness of this diagram, here is a nice application :

If $K$ is a local field and central division algebra $D/K$ has exponent $r$ (that is $r[D]=0\in BR(K)$), then any field $L/K$ of degree $r$ splits $D$.

Thus checking whether a field is a splitting field of a division algebra becomes as trivial as checking the degree of the field extension. The proof easily follows from the diagram :

$\begin{array}{ccc} D & \to^{Res} & D\otimes L \\ \downarrow & . &\downarrow \\ a & \to ^{[L:K]} & {ra=0} \end{array}$

Thus the restriction map $Res : Br(K)\to Br(L)$ can be simply described as multiplication map $\frac{\mathbb{Q}}{\mathbb{Z}} \to \frac{\mathbb{Q}}{\mathbb{Z}}$, sending $p/q \to np/q$ where $n=[L:K]$. This is clearly surjective.

Thus restriction map between Brauer groups of local fields is a surjection

What about the corestriction $Cor : Br(L)\to Br(K)$ ? It corresponds to the identity map of $\frac{\mathbb{Q}}{\mathbb{Z}}$ and in fact fits into the following commutative diagram :

$\begin{array}{ccc} Br(L) & \to^{Cor} & Br(K) \\ \downarrow & &\downarrow \\ \frac{\mathbb{Q}}{\mathbb{Z}} & \to ^{id} & \frac{\mathbb{Q}}{\mathbb{Z}} \end{array}$

Here is a proof :

Once we show that the above diagram commutes, then since the vertical maps are isomorphisms and so is the bottom map, $Cor$ will be an isomorphism.

Let $[D]\in Br(L)$. Since retsriction map is surjective, it comes from $[C]\in BR(K)$. The following is a well-known and very useful fact from finite group cohomology, namely if $H$ is a subgroup of $G$ of index $n$, then $Cor\circ Res : H^i(G,-)\to H^i(H,-)\to H^i(G,-)$ is simply multiplication by $n$.

Thus $Cor([D]) = Cor(Res([C])) = n[C]$ where $n=[L:K]$. Let $inv([C])=a$ and $inv([D])=b$. The following diagram will make things clear !

$\begin{array}{ccccc} C & \to^{Res} & D & \to^{Cor} & nC\\| & . & | & . & |\\a & \to & b & \to & na \end{array}$