Which number fields can appear as subfields of a finite-dimensional division algebra over Q with center Q?

I have some idle questions about what's known about finite-dimensional division algebras over $\mathbb{Q}$ (thought of as "noncommutative number fields"). To keep the discussion focused, let's concentrate on these:

1. Which number fields $K$ occur as subfields of a finite-dimensional division algebra over $\mathbb{Q}$ with center $\mathbb{Q}$?

2. Which pairs of number fields $K, L$ occur as subfields of the same finite-dimensional division algebra over $\mathbb{Q}$ with center $\mathbb{Q}$?

There are some easy examples involving quaternions but I am curious how completely these kinds of questions are understood. Some preliminary Googling on my part was not successful.

• I think you want $\mathbb{Q}$ in the title (instead of Q) – Belgi Aug 5 '12 at 4:58
• @Belgi: I don't know. $\LaTeX$ in titles takes time to display and I'd prefer not to slow down the main page. I am also not sure what using the $\LaTeX$ would do to the searchability of the title. – Qiaochu Yuan Aug 5 '12 at 4:59
• +1 A very good question. May be class field theory offers an answer? By the construction here all the cyclic extensions occur, but that is nowhere near a complete answer to your 1st question. Undoubtedly you found that quaternions contain all the fields $\mathbb{Q}(\sqrt{-d})$, where $d$ is a sum of three squares. – Jyrki Lahtonen Aug 5 '12 at 5:23
• @JyrkiLahtonen : could you please expand your last sentence? Given any quadratic extension $K/\Bbb Q$ the quaternion algebra $D=K\oplus Ku$ with $ku=u\bar k$ for all $k\in K$ and $u^2\in\Bbb Q$ not a norm from $K$ is not split and contains $K$ as a subfield. – Andrea Mori Aug 5 '12 at 9:44
• I asked a colleague. A degree $n$ extension field splits the the division algebra $D$, iff it is isomorphic to a maximal subfield of $D$. Also the question, whether a field splits $D$, can be decided locally by studying the behavior of Hasse invariants under extension of scalars. I am still working to understand the description of that in a way that I could communicate. Hopefully one of the resident class field theorists shows up soon. – Jyrki Lahtonen Aug 6 '12 at 11:31

$\def\Q{\mathbf{Q}}$ $\def\Z{\mathbf{Z}}$ $\def\Br{\mathrm{Br}}$ $\def\inv{\mathrm{inv}}$ $\def\Gal{\mathrm{Gal}}$

I'm going to begin by stating a bunch of very standard results without either proofs or references. Any standard text on class field theory (including Milne's online notes) will have all the details.

Lemma: If $K/\Q$ can be centrally embedded into $D/\Q$ of dimension $n^2$, then there is a finite extension $L/K$ where $L$ can be centrally embedded into $D/\Q$ and $[L:\Q] = n$.

This is another way of saying that all maximal commutative subfields have degree $n$.

Lemma: $L/\Q$ of dimension $n$ can be centrally embedded into $D/\Q$ of dimension $n^2$ if and only if $L$ splits $D$.

Standard.

Theorem: There is an exact sequence:

$$0 \rightarrow \Br(\Q) \rightarrow \bigoplus_{p,\infty} \Br(\Q_p) \rightarrow \Q/\Z \rightarrow 0.$$

This is a key result of global class field theory.

Lemma: $\Br(\Q_p) = \Q/\Z$, $\Br(\mathbf{R}) = \Z/2\Z$.

This follows from local class field theory.

Lemma: Let $L_v/\Q_p$ be a finite extension. Then $D_p/\Q_p$ splits over $L_v$ if and only if $[L_v:\Q_p]$ annihilates the class $D_p \in \Br(\Q_p)$, or equivalently $$[L_v:\Q_p] \inv(D) = 0,$$ where $\inv_p(D) \in \Q/\Z$ or $\Z/2\Z$ depending on whether $p$ is finite or not.

Standard.

Now, simply by connecting the dots, one obtains the following:

Lemma: A field $L/\Q$ of degree $n$ embeds centrally and maximally into $D/\Q$ of dimension $n^2$ if and only if, for all primes $p$ and for all $v|p$ in $p$, $$[L_v:\Q_p] \inv_p(D) = 0.$$

There is a lot of freedom in constructing $D$. We can make $\inv_p(D)$ anything we want, as long as it is zero for all but finitely many primes $p$, has order dividing $2$ at infinity, and satisfies

$$\sum_p \inv_p(D) = 0.$$

Hence we also deduce:

Lemma: A field $L/\Q$ of degree $n$ embeds centrally and maximally into $D/\Q$ of dimension $n^2$ if and only if, for all primes $q | n$ with $q^m \| n$, we can find two distinct primes $p_i$ for $i = 1,2$ (one possibly infinite) such that, for any prime $v|p_i$, we have $$q^m | [L_v:\Q_{p_i}].$$

Proof: Since $\sum \inv_p(D) = 0$ and $D$ has order $n$, there exist at least two (possibly one is infinite if $q^m = 2$) primes $p_i$ such that $\inv_p(D)$ has order divisible by $q^m$. This gives one direction.

For the converse, take the invariants to be the sum of the vector over $q^m | n$ whose contribution is $1/q^m$ at $p_1$ and $-1/q^m$ at $p_2$ and zero elsewhere. Now we deduce:

Claim: A field $K/\Q$ of degree $n$ embeds into $D/\Q$ of dimension $N^2$ for some $N$ if and only if, for all primes $q | n$ with $q^m \| n$, we can find two distinct primes $p_i$ (one possibly infinite) such that, for any prime $v|p_i$, we have $$q^m | [L_v:\Q_{p_i}].$$

Proof: If there is such an embedding, there is a maximal embedding of an $L$ which contains $K$. Suppose that $N = dn$, and let $q^M \| d$ so $q^{m+M} \| N$. We deduce from the previous Lemma that there is a prime $p_i$ such that $$q^{m+M} | [L_w:\Q_{p_i}] = [L_w:K_v][K_v:\Q_{p_i}]$$ for all $v|p_i$ and all $w|v$. Hence it follows that $$q^{m+M} | \sum_{w|v} [L_w:K_v] [K_v:\Q_{p_i}] = [L:K][K_v:\Q_{p_i}] = d [K_v:\Q_{p_i}],$$ and thus $$q^m | [K_v:\Q_{p_i}],$$ as required. For the converse, one can easily construct a totally ramified $L/K$ which has $L_w = K_v$ for all $v|p_i$ for any finite set of $p_i$, and then use the previous Lemma.

The claim basically gives an answer to your first question, note that one also deduces that $K$ is included inside some $D$ if and only if it is included maximally for some $D$ of dimension $[K:\Q]^2$. The second question presents no real extra difficulties.

Let me give some examples.

Example: $K/\Q$ has prime degree $q$. Let $L$ be the Galois closure of $K$. We have $G = \Gal(L/\Q) \subset S_q$ has order divisible by $q$, and hence contains a $q$-cycle. Thus by Chebotarev there are infinitely many $p_i$ such that Frobenius at $p_i$ is a $q$-cycle. This implies that $[K_v:\Q_{p_i}] = q$ for all $v|q$, and so $K$ embeds.

Example: $K/\Q$ is Galois and cyclic. In this case, $G = \Gal(K/\Q) = \Z/n \Z \subset S_n$ also has an $n$ cycle, so as in the last question one is done by Cebotarev, and $K$ embeds.

Example: $K/\Q$ is bi-quadratic, and $[K_v:\Q_p] \le 2$ for all ramified primes. In this case, $K$ can never be embedded inside a $D/\Q$, because there are no primes $p_i$ such that $4 | [K_v:\Q_p]$.

Remark In the last example, one does require the hypothesis at the ramified primes, since it could happen that $[K_v:\Q_p] = 4$ for two primes dividing $\Delta_K$. Thus the answer depends not only on the Galois group but also at the inertia groups at primes of bad reduction.

Example: $K = \Q(\sqrt{-1},\sqrt{17})$ does not embed centrally into any $D/\Q$. This is a special case of the last example.

Example: $K = \Q(\sqrt{-1},\sqrt{3})$ does embed centrally into a $D/\Q$ (for example with invariants $\inv_2(D) = 1/4$ and $\inv_3(D) = -1/4$ and $\inv_p(D) = 0$ otherwise). This is a special non-case of the last example, because $[K_3:\Q_3] = 4$ and $[K_2:\Q_2] = 4$.

Example: $K/\Q$ is totally ramified at two primes $p_1$ and $p_2$. In this case, $[K_v:\Q_{p_i}] = [K:\Q]$, so we can aways embed $K$.

I think these examples are enough to get the point.