Why is every quadratic subfield of a Galois extension of the rationals with the quaternions as Galois group real?

Question

Suppose that L is a field extension of the rationals with Galois group the quaternions Q={1,-1,i,-i,j,-j,k,-k}. Furthermore assume that L contains a quadratic subfield K. I have learned from this link

http://www.math.princeton.edu/generals/bhargava_manjul

that K is real. How can we prove this result? Of course one should use Galois theory There are three subgroups inside Q of order four. Each of them is cyclic and generated by i, j, or k, resp. But I don't know how to move on. Any hint is appreciated.

Answer 1

To expand slightly my brief comment. Regard $L$ as a subfield of $\mathbb{C}$. Let $L'=\mathbb{R}\cap L$. Then the index $|L:L'|=1$ or $2$. In the former case all quadratic subfields of $L$ are real. In the latter case $L'$ is the unique degree $4$ subfield of $L$; by Galois theory and the structure of the quaternion group this is the compositum of its quadratic subfields. Again all quadratic subfields are real.

Answer 2

Here's another proof:

Let $L/F$ be a cyclic extension of degree $4$, in characteristic $\neq 2$. Let $K$ be the intermediate extension of degree $2$. Then I claim there is an element $u$ of $K$ with $N_{K/F}(u)=-1$.

Proof of Claim: Let $\sigma$ be a generator for $\mathrm{Gal}(L/F)$. Let $L=K(\sqrt{b})$; choose a particular square root of $b$, in $L$, to denote by $\sqrt{b}$. Let $\sigma(\sqrt{b}) = u \sqrt{b}$. Note that $- \sqrt{b} = \sigma^2 (\sqrt{b}) = u \sigma(u) \sqrt{b}$, so $u \sigma(u) = -1$. Applying $\sigma$ to this last relation, we also have $\sigma(u) \sigma^2(u) = -1$ so $u = \sigma^2(u)$ and $u$ is in $K$. Then $u \sigma(u) = N_{K/F}(u) = -1$. QED

Why is this relevant? It shows that we can't have a real place of $F$ become complex in $K$ as, otherwise, all norms from $K$ to $F$ would be positive. In particular, let $L/\mathbb{Q}$ has Galois group the quaternion $8$ group. Let $K$ be the biquadratic subfield of $L$, with $F_1$, $F_2$ and $F_3$ the quadratic subfields of $K$. At least one of the $F_i$ must be real. If $K$ is not totally real, then the tower $L/K/F_i$ violates the claim.

Answer 3

Here's a more general question which should give you a hint how to solve your problem: prove that the given assertion is true whenever the finite group Gal(L/K) has the property that every involution in Gal(L/K) is in the kernel of every homomorphism Gal(L/K) -> Z/2Z (a condition which is easily seen to hold for the quaternion group.)

Answer 4

Here is a solution using some representation theory of the quaternions and Dirichlet's theorem on the units:
the quadratic subfields of our quaternion extension (call it $F$) are of the form $\mathbb{Q}[\sqrt{a}]$, $\mathbb{Q}[\sqrt{b}]$, $\mathbb{Q}[\sqrt{ab}]$, corresponding to, say, $i$,$j$,$k$, respectively, in $Q_8$. So either all three are real or exactly two are complex and one is real. Assume, the latter is the case and wlog $a,b<0$. Then in the following table I list the intermediate fields, the number of real embeddings $r_1$, the number of pairs of complex embeddings $r_2$ and the rank of the unit group, which is $r_1+r_2 -1$:

• $F$: 0, 4, 3
• $\mathbb{Q}[\sqrt{a},\sqrt{b}]$: 0, 2, 1
• $\mathbb{Q}[\sqrt{a}]$: 0, 1, 0
• $\mathbb{Q}[\sqrt{b}]$: 0, 1, 0
• $\mathbb{Q}[\sqrt{ab}]$: 2, 0, 1
• $\mathbb{Q}$: 1, 0, 0

The units of $F$ (thought of as an abelian group) tensored with $\mathbb{Q}$ therefore yield a three dimensional rational representation of our Galois group $Q_8$, call it $V$. Now, the irreducible rational representations of $Q_8$ are the trivial representation, three sign representations and a 4-dimensional one, namely twice the complex irreducible 2-dimensional representation. So $V$ is a sum of a certain number of trivial and sign representations, 3 altogether. Now, there can be no copies of the trivial representation in $V$, since the rank of the units over $\mathbb{Q}$ is 0. But by the same argument, there can be no copy of the sign representations that factor through $i$ and $j$, so the only one occuring in $V$ is the one factoring through $k$. But then the rank of the fixed subspace of $V$ under $k$ would have to be the same as the rank of $V$ itself, which is a contradiction.

Note that to write out all the details was slightly lenghty, but the idea is simple: look at the rank of the units in all the subfields and show that this cannot correspond to a rational representation of $Q_8$.

Answer 5

A theorem of Witt asserts that a biquadratic extension $F(\sqrt{a},\sqrt{b}),\; a,b\in F^{\times}$ can be embedded in a quaternion Galois extension of $F$ iff the quadratic form $ax^2+by^2+abz^2$ is isomorphic over $F$ to $x^2+y^2+z^2$. For $F=\mathbb{Q}$ this forces $a,b,ab>0$.