Ramification of quaternion algebras over $\mathbb Q$

Question

I'm reading on the classification of quaternion algebras over $\mathbb{Q}$. In the most common definition, we have a quaternion algebra $Q = \left(\frac{a,b}{\mathbb{Q}} \right)$ splits at a finite prime $p$ if $$ Q\otimes_\mathbb{Q}\mathbb{Q}_p\cong M_2(\mathbb{Q}_p)$$ and $Q$ ramifies at $p$ if $ Q\otimes_\mathbb{Q}\mathbb{Q}_p$ is a division algebra. Is there a way to phrase this without using $p$-adics. How was this done classically?

Now if $\mathcal{O}$ is a maximal order of $Q =\left(\frac{a,b}{\mathbb{Q}} \right)$, then the discriminant of $\mathcal{O}$ is divisible by the primes that ramify in $Q$. It seems this correlates with the structure of $\mathcal{O}/p\mathcal{O}$. Is there a precise formulation of this?

Answer 1

I'm not sure how to define "classically" here (Hensel's $p$-adics hail from around 1890s), but the Hilbert symbol gives a way to do this without $p$-adics. (See Chapter 12 of my book, http://quatalg.org.) In a nutshell, multiplying by an even power of $p$, we may assume without loss of generality that $a,b$ are integers not divisible by $p^2$. We may suppose also that $b$ is not divisible by $p$: if necessary, either swap $a,b$ or replace $b$ by $-ab/p^2$ (taking the standard generators $i,ij$ in place of $i,j$). Then $(a,b\,|\,\mathbb{Q})$ ramifies at an odd prime $p$ if and only if:

• $p \nmid ab$, or
• $p \mid a$ and ($b$ is not a square modulo $p$).

For $p=2$, there is something more elaborate modulo $8$ (see (12.4.16)), and of course there is a simple criterion at the infinite place (over $\mathbb{R}$).

If you like, you an also interpret this as whether or not the conic $ax^2+by^2=1$ has a solution modulo an arbitrary power of $p$ (hiding the $p$-adics, but not very well yet). With the above normalizations, using Hensel's lemma you can see that this is equivalent to the above.

This isn't a very conceptual criterion! But it is useful.

To your second question: yes, if $\mathcal{O}$ has basis $1,i,j,k$ (not necessarily 'standard' generators), then it has a reduced discriminant (square root of the discriminant) $\mathrm{discrd}(\mathcal{O})=\pm \mathrm{trd}((ij-ji)k)$ and this is exactly the product of the primes that ramify (chapter 15). For example, in the algebra $B := (-1,3\,|\,\mathbb{Q})$, the order $\mathbb{Z} \oplus \mathbb{Z} i \oplus \mathbb{Z} j \oplus \mathbb{Z} k$ where $k=(1+i+j+ij)/2$ is maximal, and $ \mathrm{trd}((ij-ji)k)=6 $ showing that $B$ is ramified at $2,3$.

Of course, there's much more to say here!

Answer 2

The concept of a quaternion algebra over a field, going beyond the case of Hamilton's quaternions, is due to Dickson in 1906 and he introduced the concept of maximal orders in finite-dimensional algebras over a number field in his 1923 book Algebras and their Arithmetics. (The term "arithmetic", as a noun, was Dickson's wacky label for what we'd call an order in an algebra -- at that time some people used the word "order" to refer to the degree of a number field over $\mathbf Q$.) While the $p$-adics were introduced by Hensel in the 1890s and Hasse used them extensively in the 1920s to understand central simple algebras over number fields (culminating in the Albert-Brauer-Hasse-Noether theorem around 1930), Dickson's work on algebras over number fields in the 1920s was closer to Dededkind-style ideal theory. So it is legitimate to ask for a way to describe ramification of a prime in a quaternion algebra over a number field without using $p$-adics directly even if it sounds at first like an anachronistic question based on when $p$-adics and quaternion algebras were each first created as general concepts.

Because quaternion algebras over a number field are noncommutative and maximal orders in them are not unique (unlike maximal orders in number fields), rather than describe ramification in a quaternion algebra in terms of something analogous to factorization I'll use extensions of valuations. I don't know if this was the first "classical" approach, but it has the virtue of not relying directly on maximal orders.

In algebraic number theory, you can define ramification of a prime $p$ in a number field $K$ without using $p$-adics in two ways: (i) look at how $p\mathcal O_K$ decomposes in $\mathcal O_K$ and see if some exponent in a prime ideal factor is bigger than $1$ or (ii) look at all the extensions $v$ of the valuation ${\rm ord}_p$ from $\mathbf Q$ to $K$ and see if there is some $v$ where $[v(K^\times):{\rm ord}_p(\mathbf Q^\times)] = [v(K^\times):\mathbf Z]$ is bigger than $1$. The connection between viewpoints (i) and (ii) is that each $v$ is $(1/e(\mathfrak p|p)){\rm ord}_{\mathfrak p}$ for a unique prime ideal $\mathfrak p$ dividing $p$ in $\mathcal O_K$ and $v(K^\times) = (1/e(\mathfrak p|p))\mathbf Z$, so $[v(K^\times):\mathbf Z] = e(\mathfrak p|p)$. That connects the interpretation of ramification in terms of prime ideal exponents in the factorization of $p\mathcal O_K$ and the interpretation of ramification in terms of subgroup indices $[v(K^\times):{\rm ord}_p(\mathbf Q^\times)]$ as $v$ varies. We can carry over method (ii) to quaternion algebras fairly easily in the following way.

For a quaternion algebra $Q$ over $\mathbf Q$ that is not ${\rm M}_2(\mathbf Q)$, $Q$ is a division algebra and we can ask how ${\rm ord}_p$ extends from a valuation on $\mathbf Q$ to a valuation $v$ on $Q$. There are finitely many valuations $v$ on $Q$ that extend ${\rm ord}_p$ on $\mathbf Q$. The group $v(Q^\times)$ contains ${\rm ord}_p(\mathbf Q^\times) = \mathbf Z$ with finite index, and we define $[v(Q^\times):\mathbf Z]$ to be the ramification index of $v$ over $p$. Call $p$ unramified in $Q$ if $[v(Q^\times):\mathbf Z] = 1$ for every $v$ extending ${\rm ord}_p$ from $\mathbf Q$ to $Q$. Call $p$ ramified in $Q$ if there is a valuation $v$ on $Q$ extending ${\rm ord}_p$ on $\mathbf Q$ such that $[v(Q^\times):\mathbf Z] > 1$. To include the case of the trivial/split quaternion algebra ${\rm M}_2(\mathbf Q)$, we declare all primes to be unramified in it. That's reasonable since, in terms of quaternion algebras over the $p$-adic numbers, a nonsplit quaternion algebra $Q$ over $\mathbf Q$ is unramified at $p$ if and only if $\mathbf Q_p \otimes_\mathbf Q Q \cong {\rm M}_2(\mathbf Q_p)$ and such an isomorphism is automatically true for all $p$ if we use $Q = {\rm M}_2(\mathbf Q)$.

Answer 3

Another answer to your second question is as follows:

• if $Q$ splits at $p$, then $\mathcal{O}/p\mathcal{O} \cong M_2(\mathbb{F}_p)$, and
• if $Q$ ramifies at $p$, then $\mathcal{O}/p\mathcal{O} \cong A$ where $A$ is the $\mathbb{F}_p$ algebra $A = \mathbb{F}_{p^2} + \mathbb{F}_{p^2}j$ where $j^2=0$ and $jx = x^pj$ for all $x\in \mathbb{F}_{p^2}$.

This can be proved, for instance, by using the standard form of maximal orders over complete DVRs with finite residue field. If you wanted to use this as a definition of ramification (which nobody does) avoiding all $p$-adics, you would have to prove that this dichotomy holds without using the local theory, which might be possible by constructing explicit overorders depending on the structure of $\mathcal{O}/p\mathcal{O}$ for an arbitrary order $\mathcal{O}$; if you are given the dichotomy then it is clear that the second case can only hold for finitely many $p$ by using discriminants (the first one is a simple algebra while the second one has a nontrivial Jacobson radical).