What does $(a,b)_{\zeta}$ correspond to in $\mathrm{Br}(\mathbb{Q}_p)=\mathbb{Q}/\mathbb{Z}$

Let $$p$$ be a prime number, let $$\mathbb{Q}_p$$ be the local field, by Hensel's lemma, we know it has $$p-1$$-th roots of unity, let $$\zeta$$ be a fixed primitive $$p-1$$-th root of unity in $$\mathbb{Q}_p$$.

Let $$a,b\in\mathbb{Q}_p^*$$, let $$(a,b)_{\zeta}$$ be the cyclic algebra defined by $$\mathbb{Q}_p\langle x,y|x^{p-1}=a,y^{p-1}=b, xy=\zeta yx\rangle$$.

Is there an explicit way to tell which class $$[(a,b)_\zeta]\in\mathrm{Br}(\mathbb{Q}_p)=\mathbb{Q}/\mathbb{Z}$$ is? (For example, when $$a,b\in\mathbb{Q}$$, how does the two rational numbers determine an element in $$\mathbb{Q}/\mathbb{Z}$$?)

• en.wikipedia.org/wiki/Hilbert_symbol – Qiaochu Yuan Nov 13 '18 at 22:49
• @QiaochuYuan Thanks! But for example how could one calculate $(2,3)_5$ explicitly? – Qixiao Nov 14 '18 at 17:15
• @QiaochuYuan Sorry this may be a stupid question, but I feel confused: from the wikipedia description, the Hilbert symbol is expressed by Legendre symbols, $(a,b)_p=(-1)^{\alpha\beta\epsilon(p)}(\frac{u}{p})^{\beta}(\frac{v}{p})^{\alpha}$, but it seems this expression only take $\pm1$, but $Br(\mathbb{Q}_p)=\mathbb{Q}/\mathbb{Z}$. – Qixiao Nov 14 '18 at 17:37
• That section is about the quadratic Hilbert symbol, which corresponds to $2$-torsion in the Brauer group. Scroll down further for the general Hilbert symbol – Qiaochu Yuan Nov 14 '18 at 19:56
• @QiaochuYuan Thanks! But is the Artin symbol easier to calculate explicitly? For example we have two integers $2,3\in\mathbb{Z}$ and a $4$-th root of unity $\zeta$ in $\mathbb{Q}_5$, is there an explicit way to determine what the class of $(2,3)_{\zeta}$ in $\mathbb{Q}/\mathbb{Z}$ is?(As for quadratic symbols, this can be determined by solution of a degenerate conic) – Qixiao Nov 14 '18 at 20:12

I should clarify going into this that I don't know any class field theory, so I might say some silly things. The number you're trying to compute is the Hasse invariant of the cyclic algebra, and it should also correspond to a Hilbert symbol as in the comments. I found a nice formula for a special case of the Hilbert symbol in Section 3.3 of these notes (Lemma 3.22). Specialized to this case it gives

$$(a, b)_{\xi} = \omega \left( (-1)^{\nu(a) \nu(b)} \frac{b^{\nu(a)}}{a^{\nu(b)}} \bmod p \right)$$

where $$\nu : \mathbb{Q}_p \to \mathbb{Z}$$ is the valuation and $$\omega : \mathbb{F}_p^{\times} \to \mu_{p-1}(\mathbb{Q}_p)$$ is the Teichmüller character.

You asked in the comments about $$p = 5, a = 2, b = 3$$. In this case $$\nu(a) = \nu(b) = 0$$ so the Hilbert symbol is trivial. In order for it to be nontrivial at least one of $$\nu(a)$$ and $$\nu(b)$$ must be nonzero. For example, sticking to $$p = 5$$, if we take $$a = 2, b = 5$$ then we get

$$(1, 5)_{\xi} = \omega \left( \frac{1}{2} \bmod 5 \right) = \omega(3)$$

which is the unique $$4^{th}$$ root of unity in $$\mathbb{Z}_5$$ congruent to $$3 \bmod 5$$, and in particular primitive. Unfortunately I don't know exactly how this is identified with an element of $$\mathbb{Q}/\mathbb{Z}$$, although it must be either $$\frac{1}{4}$$ or $$\frac{3}{4}$$.

• I've tried to decide the last question in my answer, please check if you agree. – Torsten Schoeneberg Jan 26 at 6:47

Qiaochu Yuan's answer gives good insight into computing the Hilbert symbol here, giving the result as an element of $$\mu_{p-1}(\Bbb Q_p)$$; but it avoids the machinery of cyclic (division) algebras which are sort of the original setting to define the Hasse invariant as element of $$Br(\Bbb Q_p) \simeq \Bbb Q/\Bbb Z$$. I want to amend that. It turns out that in the example at the end of his answer, whether the result is $$\frac14$$ or $$\frac34$$ depends on (some conventions, and:) which primitive $$(p-1)$$-th root of unity $$\zeta$$ you choose in the original question.

References: I learned a lot of this stuff from R. Pierce: Associative Algebras (GTM 88, Springer, 1982) and I. Reiner: Maximal Orders (LMS Monographs 5, Academic Press, 1975) some years ago, although I don't have them at hand by now, so I cannot even guarantee that I follow the exact same conventions.

Namely, let us restrict to $$p \neq 2$$ for simplicity, and to the basic interesting case $$\nu(a) = 0, \nu(b) =1$$, i.e. $$a\in \Bbb Z_p^\times \simeq \mu_{p-1} \times (1+\Bbb Z_p)$$. Since elements in the second factor have $$(p-1)$$-th roots, we can easily restrict to the case $$a \in \mu_{p-1}$$; however, for your definition of the cyclic algebra to work, I think we should further restrict to the case that $$a$$ is a primitive $$(p-1)$$-th root of unity. Namely, now one would usually define the cyclic algebra as follows: Let $$r \in \overline{\Bbb Q_p}$$ satisfy $$r^{p-1}=a$$. The extension $$L= \Bbb Q_p(r) \vert \Bbb Q_p$$ is the unique unramified extension of degree $$p-1$$; its Galois group is cyclic and has a distinct generator, namely the (lift of the) Frobenius, let's call it $$\sigma$$. Then the cyclic algebra you define can be realised as a subalgebra of $$M_{(p-1) \times (p-1)}(L)$$, generated by

$$x:= \pmatrix{r&0& &\\ 0&\zeta^{-1} r& &\\ && \ddots &0\\ &&0&\zeta^{2-p}r}$$ and $$y:= \pmatrix{0&1&0& &\\ 0&0&1 &&\\ &&\ddots& \ddots &\\ &&& &1\\ b&&&0&0}.$$

Note that we have $$x^{p-1}=a$$ and $$y^{p-1}=b$$ as well as $$yxy^{-1}= \zeta^{-1}x$$, as demanded. One checks that this is a division algebra, which contains as a subfield the diagonal matrices

$$\pmatrix{z&0& &\\ 0&\tau(z)& &\\ && \ddots &0\\ &&0&\tau^{p-1}(z)}$$

where $$z \in L$$ and $$\tau \in Gal(L\vert \Bbb Q_p)$$ is the automorphism induced by $$r\mapsto \zeta^{-1}r.$$ We identify this subfield with $$L$$. This division algebra, which in this case corresponds exactly to your definition of $$(a,b)_\zeta$$, is usually denoted by something like $$(L\vert \Bbb Q_p, \tau, b)$$.

Now the original definition of the Hasse invariant goes like this: One can extend the $$p$$-adic valuation $$\nu$$ to a valuation of the division algebra. Note that e.g. $$\nu(x) = 0$$ and $$\nu(y) = \dfrac{1}{p-1}$$. In the division algebra, there exist elements $$\gamma$$ such that the Frobenius $$\sigma$$ on $$L$$ is induced by conjugation with $$\gamma$$, i.e.

$$\gamma z \gamma^{-1} = \sigma(z) \text{ for all } z \in L . (*)$$

Then the Hasse invariant is (well-)defined as $$\nu(\gamma) + \Bbb Z \in \Bbb Q/ \Bbb Z$$.

A quick calculation shows that $$\sigma(r) = ar$$. Now find $$1\le k \le p-1$$ such that $$\tau^k = \sigma$$, or in other words, $$\zeta^{-k} = a$$. Then one checks that $$\gamma := y^k$$ satisfies $$(*)$$ and hence

$$(a,b)_\zeta = \dfrac{k}{n} + \Bbb Z \in \Bbb Q / \Bbb Z.$$

In the example $$a=2, b=p=5$$ of the other answer, it depends: There are two primitive $$4$$-th roots of unity in $$\Bbb Q_5$$, namely $$\omega(2)$$ and $$\omega(3)$$ ($$\omega$$ denoting the Teichmüller map). With your definitions we have

$$(2,5)_{\omega(2)} = \dfrac{3}{4} + \Bbb Z$$ $$(2,5)_{\omega(3)} = \dfrac{1}{4} + \Bbb Z$$

For $$a \in \Bbb Z_p^\times, b \in p \Bbb Z_p$$, we have $$(a,b)_{\zeta} = \dfrac{k}{p-1} + \Bbb Z \in \Bbb Q/\Bbb Z \simeq Br(\Bbb Q_p)$$ where $$\zeta^{-k} \equiv a$$ mod $$p$$.
For $$a \in \Bbb Z_p^\times, b \in \Bbb Q_p$$, we have $$(a,b)_{\zeta} = \nu(b)\dfrac{k}{p-1} + \Bbb Z \in \Bbb Q/\Bbb Z \simeq Br(\Bbb Q_p)$$ where $$\zeta^{-k} \equiv a$$ mod $$p$$.
(The only class field theory that goes into this case is basic Kummer theory, relating $$\sigma$$ and $$\tau$$ and $$\mu_{p-1}$$.)
Note that if one rewrote conjugation in $$(*)$$ as $$\gamma^{-1}z\gamma$$, everything would be multiplied by $$(-1)$$ in $$\Bbb Q/\Bbb Z$$ which in Qiaochu's example would exactly flip the two values $$\frac14$$ and $$\frac34$$. So in a way, it's really a convention which one is which. However, if one looks at the full Brauer group instead of just the $$p-1$$-torsion part as here, one indeed has to fix such a convention.