# ramified quaternion algebras

I'm trying to better understand the connection between the concepts of ramification of a field extension, and ramification of a quaternion algebra. I'm also trying to build a better understanding of the motivation behind this term. Bear with me as I summarize my understanding of the stuff...

If $K$ is a field with valuation $v$, and $L$ is a finite extension of it, the definition of ramification depends on whether $v$ is Archimedean. If $v$ is Archimedean, then it corresponds to an embedding $\sigma:K\rightarrow\mathbb{C}$, and also $v$ extends to a valuation $w$ on $L$. We say the extension is ramified here iff there is a nontrivial $\tau\in Gal(L,K)$ such that $w\tau=w$ (turns out to be independent of choice of $w$, hence well-defined). If $v$ is not Archimedean, then $v$ corresponds to a prime ideal $\mathcal{P}$ of $R_K$, and there are prime ideals $\mathcal{P}_i$ of $R_L$ such that $\mathcal{P}R_L=\Pi_{i=1}^r\mathcal{P}_i^{e_i}$, and we have that the extension is ramified here iff $\exists i: e_i>1$.

If $A$ is a quaternion algebra over the field $K$, and $v$ is a valuation on $K$, let $A_v:=A\otimes_KK_v$, and then we have (after a good amount of work) that $A_v\cong M_2(K_v)$ or $A_v$ is the unique division algebra over $K_v$. And the definition of $A$ being ramified at $v$ is that $A_v$ is the division algebra and not the other one.

Now, behind these definitions, there is a lot of other theory one would use to actually determine these things. Let's stick to number fields for simplicity. For a number field as an extension of $\mathbb{Q}$, the infinite places are very simple, just have to do with whether an embedding is totally real. And for the finite places you've got Kummer's theorem: look at when the minimal polynomial of the primitive element has roots modulo $p$, which I find very enlightening because it gives you an explicit description of how $(p)$ splits in the number field. In the case of the quaternion algebras, any $A$ has a Hilbert symbol $(\frac{a,b}{K})$, and you can figure out where it ramifies by looking at $a$ and $b$. The infinite places depend on the sign of the images of $a$ and $b$ under the embedding, and the finite places depend on whether or not $a$ and $b$ (and sometimes $-a^{-1}b$) are in $\mathcal{P}$ and $\mathcal{P}^2$.

So I get all that, but what exactly are we trying to measure here? The infinite places seem more obvious: we are looking at whether or not there are embeddings in the bigger field that look the same in the smaller field. But why should it be so important that the prime factorization of a prime in the bigger ring has repetition? Why is it not called "ramified" just for the prime to split apart into several primes? Also, for $A$ to ramify at a finite place $v$, does this imply some kind of factorization of lifts of prime ideals? $A$ doesn't have a "ring of integers," but it does have an order, does this happen there?

I apologize for the vagueness of my questions, I'm just looking for some good intuition about this from someone who knows it very well. My intention is to understand enough of this for now to use it for something else, but I would like to come back to it periodically until I've mastered it.

If $F/{\mathbf Q}$ is a quadratic field then all but finitely many places $v$ of ${\mathbf Q}$ are unramified in $F$, and we could interpret what that means in a couple of ways: prime ideal factorization (for nonarchimedean $v$), extensions of absolute values (any $v$), or base extension by ${\mathbf Q}_v$ (any $v$). Let's use the last way: we look at ${\mathbf Q}_v \otimes_{\mathbf Q} F$. First suppose $v$ is nonarchimedean. For unramified $v$, the tensor product is ${\mathbf Q}_v \times {\mathbf Q}_v$ (if $v$ splits) or it is an unramified quadratic extension of ${\mathbf Q}_v$ (if $v$ is inert). For ramified $v$, the tensor product is a ramified quadratic extension of ${\mathbf Q}_v$ in the sense of ramified extensions of complete discretely valued fields. Next suppose $v$ is the archimedean place of ${\mathbf Q}$. We call $v$ unramified in $F$ when ${\mathbf Q}_v \otimes F$ is ${\mathbf R} \times {\mathbf R}$ ($F$ is real quadratic), and call it ramified in $F$ when ${\mathbf Q}_v \otimes F$ is $\mathbf C$ ($F$ is imaginary quadratic).

When $B$ is a quaternion algebra over ${\mathbf Q}$ and $v$ is a place of ${\mathbf Q}$, the base extension ${\mathbf Q}_v \otimes_{\mathbf Q} B$ is the matrix algebra ${\rm M}_2({\mathbf Q}_v)$ except for finitely many $v$, when it is a division algebra (something subtle). Think of ${\rm M}_2({\mathbf Q}_v)$ as being a noncommutative analogue of ${\mathbf Q}_v \times {\mathbf Q}_v$, and this is a reason for calling ${\rm M}_2({\mathbf Q}_v)$ the unramified case while the division algebra case is called ramified since those are the finitely many peculiar cases. The situation with quaternion algebras doesn't have anything like the inert case from quadratic fields, so "split" and "unramified" are used as synonyms (we say $v$ is "split" or "unramified" in $B$ when ${\mathbf Q}_v \otimes_{\mathbf Q} B$ is a matrix algebra, and "nonsplit" or "ramified" otherwise).

An analogy with quadratic extensions $L$ of $K :={\mathbf C}(X)$, where all residue fields are algebraically closed, is useful to bear in mind. For a discrete valuation $v$ on $K$ that is trivial on the constants, there is a notion of ramified or unramified for $v$ in $L$. It could be defined in terms of how the prime ideal associated to $v$ in the localization ${\mathcal O}_v$ of $K$ decomposes in the integral closure of ${\mathcal O}_v$ in $L$. Or it could be defined in terms of the base extension of $L$ by the completion of $K$ at $v$,$K_v \otimes_{K} L$: if this is $K_v \times K_v$ then we call $v$ unramified or split in $L$ (the two words are synonyms here) and otherwise this base extension is a quadratic ramified extension of $K_v$ (in the sense of ramified extension of a complete discretely valued field) and we then call $v$ ramified or nonsplit in $L$.

Near as I can tell, you're confusing things because you're trying to consider a general number field as an analogue to a quaternion algebra when the right analogue is a quadratic field.

In a quadratic field $F$, a prime $p$ of $\mathbf{Z}$ is ramified if and only if there is a prime $\mathfrak{p}$ of $\mathbf{Z}_F$ such that $\mathfrak{p}^2 = p\mathbf{Z}_F$.

Similarly in a quaternion algebra $B$ over $\mathbf{Q}$, you will have many orders $\mathcal{O}$. We will say that a prime $p$ of $\mathbf{Z}$ is ramified in $\mathcal{O}$ if there is a two-sided ideal $\mathfrak{P}$ of norm $p$ in $\mathcal{O}$ such that $\mathfrak{P}^2 = p\mathcal{O}$. If $\mathcal{O}$ is maximal, then the ramified primes of $\mathcal{O}$ are exactly the primes at which $B$ is not division.

Of course historically, these two notions probably had no more connection than "these are the finite set of primes at which something funny happens." In general, just because two things in mathematics have the same name does not necessarily mean there is a deep connection between the two. See for instance What are the most overloaded words in mathematics?

• As a reference for this ramification stuff, see the book of Vigneras, especially section 2.1 – stankewicz Mar 20 '13 at 8:33
• That is a much clearer analogy. I'm reading Maclachlan/Ried, but they cover quaternion orders in 4 pages and I definitely needed more detail. I was aware of the Vigneras book but had not looked for it because I thought it was only available in French. Then a quick Google search shows a free download of an English translation from Nanjing University: math.nju.edu.cn/~guoxj/notes/qa.pdf I checked this against the French version and it's indeed the same, plus modern typesetting. Very cool. I think the intuition I'm looking for will just take time, so thanks for the tip and the reference. – j0equ1nn Mar 20 '13 at 19:25
• I neglected to mention that later in the book I'm reading Maclachlan/Ried actually do devote a whole chapter to orders in quaternion algebras, they just give an overview of it early on. I'm reading the book carefully from the beginning and was dissatisfied with this overview but in fairness they probably chose wisely to ask the reader to accept some stuff on faith. I think this speaks to the fact that the answer to this question is really pretty nontrivial ... and also the fact that I'm really bad at accepting anything on faith. – j0equ1nn Mar 21 '13 at 2:12