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?


3 Answers 3


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!

  • $\begingroup$ 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 $\endgroup$
    – 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 $Q = {\rm M}_2(\mathbf Q)$.

  • $\begingroup$ 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. $\endgroup$
    – Kimball
    Aug 7, 2021 at 13:57
  • $\begingroup$ @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. $\endgroup$
    – KConrad
    Aug 7, 2021 at 20:29
  • $\begingroup$ 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. $\endgroup$
    – Rdrr
    Aug 9, 2021 at 13:31
  • $\begingroup$ @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. $\endgroup$
    – KConrad
    Aug 9, 2021 at 16:53
  • $\begingroup$ 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. :) $\endgroup$
    – KConrad
    Aug 9, 2021 at 16:58

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).

  • $\begingroup$ overorders? Sorry, I'm not familiar with that term. $\endgroup$
    – Rdrr
    Aug 9, 2021 at 13:29
  • $\begingroup$ I simply meant an order properly containing the given one. $\endgroup$
    – Aurel
    Aug 9, 2021 at 19:57
  • 1
    $\begingroup$ It is unfortunate that English never acquired a standard label for the concept complementary to a subgroup or subring (i.e., a name for a group that contains a specific group or a ring that contains a specific ring). In field theory we have the term "extension field" but the terms "extension group" and "extension ring" with analogous meanings are not standard. $\endgroup$
    – KConrad
    Oct 17, 2021 at 4:30
  • $\begingroup$ 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/… $\endgroup$
    – KConrad
    Oct 17, 2021 at 4:38

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