# Ramification of quaternion algebras over $\mathbb Q$

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?

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!

• This is the sort of criterion I was looking for and I found your book very helpful for Hilbert's criterion. The problem I had was proving that the existence of a solution $ax^2+by^2=1$ modulo $p$ was sufficient to conclude $(a,b|\mathbb{Q})$ is a division algebra without using Hensel's lemma. This may be a bit much to ask since most of the natural answers do seem to flirt with the $p$-adics the very least
– Rdrr
Aug 9, 2021 at 13:27

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 $$Q = {\rm M}_2(\mathbf Q)$$.

• I like the term arithmetic, at least as much as order anyway, and I think it makes more sense, given that it's an arithmetic structure inside an algebraic structure called an algebra. Aug 7, 2021 at 13:57
• @Kimball I was initially bothered by the use of the term order in ring theory until I realized it was just a borrowing from the classification system in biology for a "type of thing" in the same way as a genus (of a quadratic form) or a class or a family. Aug 7, 2021 at 20:29
• I like this criterion but I don't see a natural way to prove a classification of quaternion algebras over $\mathbb{Q}$ as in Voight's answer using this definition.
– Rdrr
Aug 9, 2021 at 13:31
• @Rdrr your post asks how to interpret ramification in quat. algebras over $\mathbf Q$, while your comment here asks about classifying such quaternion algebras. You can classify quaternion algebras over $\mathbf Q$ by their set $S$ of ramified places: $|S|$ has to be even, and conversely for each even number of places $S$ on $\mathbf Q$ there is a unique (up to isomorphism) quat. alg. over $\mathbf Q$ ramified precisely at $S$. For example, ${\rm M}_2(\mathbf Q)$ is the only quat. alg. over $\mathbf Q$ ramified nowhere (here $S = \emptyset$). I don't know how to prove this without $p$-adics. Aug 9, 2021 at 16:53
• Two quaternion algebras $(a,b)_\mathbf Q$ and $(c,d)_\mathbf Q$ are isomorphic if and only if the ternary quadratic forms $-ax^2 - by^2 + abz^2$ and $-cx^2 - dy^2 + cdz^2$ over $\mathbf Q$ are equivalent. The proof of this "if and only if" only uses very basic algebra, no $p$-adics, but to decide when two ternary quadratic forms are equivalent (special case of the Hasse-Minkowski theorem) normally proceeds today by the use of $p$-adics. Do you want to think about HM theorem without using the $p$-adics? I don't think that's been done in over $100$ years. :) Aug 9, 2021 at 16:58

• 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).
• That reminds me that no worldwide notation expresses "$a$ is a multiple of $b$" with $a$ on the left. We can write $b \mid a$, as if we had $y < x$ but nobody had ever created $x > y$. Since the divides relation $\mid$ is symmetric, we can't directly use it to "reverse" the relation, but in Eastern Europe there is a notation for this that is basically unknown elsewhere: see math.stackexchange.com/questions/3906711/… and Georges Elencwajg's answer to math.stackexchange.com/questions/135253/… Oct 17, 2021 at 4:38