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

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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? –  David Speyer Sep 6 '10 at 15:17
    
I mean some quadratic field can be embeded into H, my question is how many such field ,is ther finite or infinite many? –  TOM Sep 6 '10 at 15:24
    
OK, I see what you're asking. Infinitely many, as explained below... –  David Speyer Sep 6 '10 at 15:37
    
Fixed up the language a bit. Feel-free to undo if this was not a faithful rewording. –  Cam McLeman Sep 6 '10 at 15:55
4  
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? –  BCnrd Sep 6 '10 at 15:58

2 Answers 2

up vote 10 down vote accepted

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 $\mathbb{Q}^*/(\mathbb{Q}^*)^2$.

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Thanks you for your great help! –  TOM Sep 6 '10 at 15:58

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).

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

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