# About the isomorphism $\operatorname{Br}(\mathbb{Q}_p)\cong \mathbb{Q}/\mathbb{Z}$

I've been reading the section about Brauer groups in Introduction to Modern Number Theory, and I couldn't quite understand how this isomorphism is defined.

We start with a central simple algebra $A$, over $\mathbb{Q}_p$. Then, the book says, we take a maximal unramified extension'' $L$ of $\mathbb{Q}_p$ in $A$. The book informs me that there are several of those, and that they are all conjugate in $A$. Then the book says the the valuation on $\mathbb{Q}_p$ extends to a valuation $v_A$ on $A$, and that the Frobenius automorphism of $L$ over $\mathbb{Q}_p$ extends to an inner automorphism of $A$ by some element $\gamma \in A$. Thus $v_A:A^{\times}\rightarrow \frac{1}{n}\mathbb{Z}$, and we define the map that is designed to be an isomorphism from $Br(\mathbb{Q}_p)$ to $\mathbb{Q}/\mathbb{Z}$ by $A\mapsto v_A(\gamma)$. The book informs me that this is well defined and in fact an isomophism.

But I'm not sure I understand what $L$ being a maximal unramified extension'' means. Is it: let $L$ be a field extension of $\mathbb{Q}_p$, contained in $A$, and unramified over $K$? I can't think of any examples where $L$ wouldn't be $K$ itself, since the center of $A$ is $\mathbb{Q}_p$? Is this because my intuition is bad (if so please give me an example), or is it because I have the wrong definition in mind?

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Maybe the stuff in $L$ commutes with stuff in $L \supset \mathbf{Q}_p$ but not with other stuff in $A$? In a way analogous to how $\{\pm1, \pm i\}$ is not in the center of the quaternion group but is yet abelian. Just thinking aloud. –  Dylan Moreland Oct 31 '11 at 0:12
Yes, that's exactly how this works. Just think of the subalgebra of diagonal matrices inside the CSA $M_{n \times n}(K)$ of $n \times n$ matrices; this subalgebra is commutative, but not central. (It's not a field, of course.) –  David Loeffler Oct 31 '11 at 9:34
You can assume that $A$ is a division algebra. If you adjoin any element $z$ of a division algebra to its center $K$, you get an extension field $K[z]$ of the center, because everything in sight commutes. –  Jyrki Lahtonen Oct 31 '11 at 14:40
In Jacobson's Basic Algebra II he shows a fin-dim CSA $A$ over ${\mathbf Q}_p$ (or any local field) is a cyclic algebra $(E_n/{\mathbf Q}_p,\sigma,a)$, where ${\rm dim}_{{\mathbf Q}_p}(A) = n^2$, $E_n/{\mathbf Q}_p$ is unram. of deg. $n$, $\sigma$ is the Frob. in ${\rm Gal}(E_n/{\mathbf Q}_p)$, and $a \in {\mathbf Q}_p^\times$. We can write $a=p^iu$ for $u \in {\mathbf Z}_p^\times$. From properties of cyclic algebras, $(E/{\mathbf Q}_p,\sigma,p^iu) \cong (E/{\mathbf Q}_p,\sigma,p^jv)$ as ${\mathbf Q}_p$-algebras iff $i\equiv j \bmod n$. Then $A \mapsto i/n \bmod {\mathbf Z}$ is the isom. –  KCd Apr 28 '12 at 3:19
Your confusion over how there could be multiple copies of $L$ inside $A$ can be removed by looking at an actual example. Take $A = {\mathbf H}({\mathbf Q}_2)$, the quaternions over the 2-adics. Convince yourself this is a division ring. There are a lot of solutions in $A$ of $q^2 = -3$ (one example is $q = i+j+k$) and for any of them, ${\mathbf Q}_2[q]$ is a copy of ${\mathbf Q}_2(\sqrt{-3})$ inside $A$, and this is a quad. unram. extn. of ${\mathbf Q}_2$. Do you know there are infinitely many copies of ${\mathbf C}$ in ${\mathbf H}({\mathbf R})$? –  KCd Apr 28 '12 at 3:28