Suppose $H$ is a indefinite quaternion algebra over $\mathbb{Q}$. Are there infinitely many quadratic fields that can be embedded into $H$?

  • $\begingroup$ Since a finite field has characteristic $p$, you certainly can't embed in in a $\mathbb{Q}$ algebra. Did you mean to ask something else? $\endgroup$ – David E Speyer Sep 6 '10 at 15:17
  • $\begingroup$ I mean some quadratic field can be embeded into H, my question is how many such field ,is ther finite or infinite many? $\endgroup$ – TOM Sep 6 '10 at 15:24
  • $\begingroup$ OK, I see what you're asking. Infinitely many, as explained below... $\endgroup$ – David E Speyer Sep 6 '10 at 15:37
  • $\begingroup$ Fixed up the language a bit. Feel-free to undo if this was not a faithful rewording. $\endgroup$ – Cam McLeman Sep 6 '10 at 15:55
  • 4
    $\begingroup$ Inf. many: if $A$ is c.s.a. of deg. $n^2$ over field $L$ then deg-$n$ ext'n field $L'/L$ embeds in $A$ if and only if it splits $A$. For global $L$, get a finite collection of local conditions (by global & local CFT), and so satisfied for inf. many $L'$. For any field $L$ the $L$-isom. classes of finite \'etale $L$-subalgebras of rank $n$ in $A$ correspond to $L$-rat'l conj. classes of max'l $L$-tori in $L$-group $G$ of units of $A$. Probably any non-comm. conn'd reductive group over any finitely generated infinite field has infinitely many rational conj. class of max'l tori; is it known? $\endgroup$ – BCnrd Sep 6 '10 at 15:58

There are infinitely many. Let $V$ be the subspace of $H$ where the trace is zero. Then norm gives an nondegenerate quadratic form on $V$. For any $v \in V$, the field $\mathbb{Q}(v)$ is isomorphic to $\mathbb{Q}(\sqrt{-N(v)})$. Recall that the fields $\mathbb{Q}(\sqrt{D_1})$ and $\mathbb{Q}(\sqrt{D_2})$ are isomorphic if and only if $D_1/D_2$ is a square.

So we need to show that $V$ takes infinitely many values in $\mathbb{Q}^*/(\mathbb{Q}^*)^2$. For example, if we are dealing with the standard quaternion algebra, we need to show that $p^2+q^2+r^2$ takes infinitely many values in $\mathbb{Q}^*/(\mathbb{Q}^*)^2$. This is easy enough that probably any method you think of will work. Here is what I came up with: Take $u$ and $v$ linearly independent members of $V$. Let $a=u^{-1} v$.

First, suppose that $-N(a)$ is a square, say $k^2$. Then, for $s$ and $t \in \mathbb{Q}$, we have $N(su+tv) = N(u) N(s+t a)= N(u) (s-kt)(s+kt)$, and this expression clearly takes infinitely many values in $\mathbb{Q}^*/(\mathbb{Q}^*)^2$ as we vary $s$ and $t$.

If $N(a)$ is not a square, let $K=\mathbb{Q}(a)$, this is a subfield of $H$ isomorphic to $\mathbb{Q}[\sqrt{N(a)}]$. For $b \in K$, we have $N(ub) = N_{K/\mathbb{Q}}(b) N(u)$, and $ub$ is in $V$. Since there are infinitely many primes that split principally in $K$, there are infinitely primes that occur as norms $ N_{K/\mathbb{Q}}(b)$, and thus we get an infinite subgroup of $\mathbb{Q}^*/(\mathbb{Q}^*)^2$.


To supplement David's answer: there is a standard "local-global" criterion for determining whether a quadratic field $K/\mathbb{Q}$ can be embedded in a rational quaternion algebra $B/\mathbb{Q}$.

For this recall that $B$ is said to be ramified at a prime number $p$ if $B_p = B \otimes_{\mathbb{Q}} \mathbb{Q}_p$ is a division algebra. Moreover, we say that $B$ ramifies "at infinity" if $B_{\infty} = B \otimes_{\mathbb{Q}} \mathbb{R}$ is a division algebra.

It is known that a rational quaternion algebra $B$ is determined up to isomorphism by the set of ramified places $p \leq \infty$, that this set of places is finite and of even cardinality, and conversely for any finite set of even cardinality there is a rational quaternion algebra ramifying at these places.

Now, let $B$ be a rational quaternion algebra, $K$ a quadratic field, and $p \leq \infty$ a ramified place of $B$. Suppose that we have an embedding $K \hookrightarrow B$. Then tensoring with $\mathbb{Q}_p$ we get $K_p := K \otimes \mathbb{Q}_p \hookrightarrow B_p$.

Now, if $p$ is inert or ramified in $K$, then $K_p$ is a quadratic field extension of $\mathbb{Q}_p$, and it turns out that every such quadratic extension does indeed embed in $B_p$. However, if $p$ is split in $K$, then $K_p \cong \mathbb{Q}_p \times \mathbb{Q}_p$ has nontrivial idempotent elements, so cannot embed in $B_p$ if the latter is a division algebra. (If $p = \infty$, then we say that $p$ is split in $K$ iff $K$ is a real quadratic field.)

In summary, this gives a necessary local criterion for the embeddability of $K$ into $B$: each ramified prime $p \leq \infty$ of $B$ is nonsplit in $K$. By the local-global theory of quaternion algebras over $\mathbb{Q}$, it turns out that this necessary condition is also sufficient. In particular, the quadratic fields which embed into a given quaternion algebra are precisely those which are determined by finitely many splitting conditions.

It follows easily from this that there are infinitely many quadratic fields which embed in $B$, for instance any imaginary quadratic field $\mathbb{Q}(\sqrt{D})$ where $D$ is divisible by each finite ramified prime $p$ of $B$. Moreover, one can see that the set of such quadratic fields has, in some natural sense, positive density (as does its complement, unless $B \cong M_2(\mathbb{Q})$ in which case we recover the result that every quadratic field embeds, as one sees much more easily by a Cayley's Theorem / regular representation style argument).

  • $\begingroup$ Note that this is a fleshing out of part of BCnrd's comment above, phrased in a somewhat more middlebrow way. $\endgroup$ – Pete L. Clark Sep 6 '10 at 17:07

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.