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This is a soft but very mathematically hands-on question. Hopefully it will be interesting to more than just me. Thanks in advance for your help in thinking clearly about what follows.

I have been trying to build a quaternion extension of $\mathbb{Q}$. I have been pursuing an overly optimistic method, which has not quite been working, and I just learned from this MO question that it’s impossible for it to work. I would like your help in thinking about the way it is failing; more precisely, the extent to which it is/is not working. I have obtained a family of degree 8 number fields that are not quaternion extensions, and are not even normal, but due to the construction there is (perhaps??) something *quaternion-ish* about them. I guess my question is really: what is the relationship of $Q_8$ to the galois group of their normal closure?

Let $L=\mathbb{Q}(\alpha)$ be a $Q_8$-extension of $\mathbb{Q}$. $L$ has a unique biquadratic subfield $K=\mathbb{Q}(\theta,\psi)$ where $\theta^2,\psi^2\in\mathbb{Q}$, and thus quadratic subfields $\mathbb{Q}(\theta),\mathbb{Q}(\psi),\mathbb{Q}(\theta\psi)$ fixed by $\mathbf{i},\mathbf{j},\mathbf{k}$ respectively. From the MO question linked above I learned that $\theta^2,\psi^2>0$, but before I knew that, I reasoned as follows:

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$\mathbf{i}$ must fix $\theta$ and reverse the sign of $\psi$, and have order $4$ cyclic action on $\alpha$. $\alpha$ can be taken to be the square root of an element of $K$; then we must have $\mathbf{i}^2(\alpha)=-\alpha$. So, optimistically, suppose $\mathbf{i}(\alpha)= \psi/\alpha$. Then $\mathbf{i}^2(\alpha)=-\psi/(\psi/\alpha)=-\alpha$; this is promising! Suppose likewise that $\mathbf{j}(\alpha)=\theta/\alpha$. Then $\mathbf{k}(\alpha)=\mathbf{ij}(\alpha)=\theta/(\psi/\alpha)=(\theta/\psi)\alpha$, and

$$\mathbf{k}^2(\alpha) = \left(\frac{-\theta}{-\psi}\right)\left(\frac{\theta}{\psi}\right)\alpha=-\alpha$$

Thus $\theta^2,\psi^2$ are equal magnitude and opposite sign. This tells us, a propos of the linked MO question, that this optimistic construction cannot actually work. However, bear with me:

If all of this were actually working, we would have $\theta/\psi = i$ (up to sign), and thus the conjugates of $\alpha$ would be

$$\pm \alpha, \pm i\alpha, \pm \psi/\alpha, \pm i\psi/\alpha$$

A straightforward calculation shows that any polynomial of the form

$$f_{\psi^2}(x) = x^8 – 2r x^4 + \psi^4$$

with $r,\psi^2\in\mathbb{Q}$ has these as roots. If we choose $r,\psi^2$ so that $f$ is irreducible, then $\mathbb{Q}[x]/(f)$ is a field of degree 8 over $\mathbb{Q}$, so even if it is not the originally-desired quaternion extension of $\mathbb{Q}$, it is *something* with a kind of promising-looking bunch of roots.

So, consider $L=\mathbb{Q}[x]/(f)$ and let $\alpha$ be the residue of $x$ in this field. If it were really the desired extension, it would contain $i, \psi$. We can force it to contain either one of these because we have control over the discriminant of $\alpha^4$ since we can control $r,\psi^2$. For example, choosing them so that $r^2+1=\psi^4$, e.g. $r=4/3, \psi^2=5/3$, we get

$$\left(\alpha^4-r\right)^2 = (2r\alpha^4-\psi^4) -2r\alpha^4 + r^2 = r^2-\psi^4= -1$$

so that $L\supset \mathbb{Q}(i)$; in fact, $\mathbb{Q}(\alpha^4)=\mathbb{Q}(i)$ because it contains it and is degree $2$.

Before I realized that $L$ couldn’t possibly be my desired quaternion extension, I was hoping I could put a further fullfillable constraint on $r,\psi^2$ to force it to contain $\psi$; this was of course in vain. However, the search did lead me to realize that

$$\left[\alpha^2((\psi^2-2r)+\alpha^4)\right]^2=\alpha^4\left(-4r(\psi^2-r)+2(\psi^2-r)\alpha^4\right)$$

$$=2(\psi^2-r)\left(-2r\alpha^4+\alpha^8\right)=2(r-\psi^2)\psi^4$$

Thus (because $\psi^4$ is a rational square) $L$, in fact $\mathbb{Q}(\alpha^2)$, contains a square root of $2(r-\psi^2)$. I found this square root by searching for elements of $L$ that were invariant under the substitution $\alpha\mapsto i\psi/\alpha$, which I wanted to correspond to $\mathbf{j}$. I was hoping for $\psi$ but instead I got a different rational square root!

At any rate, we conclude that $\mathbb{Q}(\alpha^2)\subset L$ contains both $i$ and $\sqrt{2(r-\psi^2)}$. (For example with $r=4/3, \psi^2=5/3$, it contains $\sqrt{-2/3}$, thus $\mathbb{Q}(\sqrt{6})$.) Thus (under the condition $r^2+1=\phi^4$), $\mathbb{Q}(\alpha^2)$ is generically a biquadratic extension of $\mathbb{Q}$, and thus normal.

On the other hand, $L$ is not normal, because it can’t contain $\psi$ unless it is generated over $\mathbb{Q}$ by $i$, $\psi$, and $\sqrt{2(r-\psi^2)}$, and this is not plausible. On the other hand, $f$ splits in $L(\psi)$, so $L$’s normal closure is a quadratic extension of it. $L$ ought to have one conjugate $L’=\mathbb{Q}(\psi/\alpha)$ corresponding to the fact that $\alpha$ and $\psi/\alpha$ are both roots of $f$ and all other roots of $f$ are obtained from these by multiplying by $i$, which $L$ and thus any conjugate contains.

So, we’ve obained a normal degree 16 extension $L(\psi)$ of $\mathbb{Q}$. I guess my question amounts to, what’s the relationship between its Galois group and $Q_8$?

(It seems to me that, at least, $Q_8$ should be a subgroup, with $\mathbf{i}$ sending $\psi\mapsto -\psi$, $\alpha\mapsto \psi/\alpha$, and $\mathbf{j}$ fixing $\psi$ and sending $\alpha\mapsto i\psi/\alpha$. Is this right?)

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I figured it out.

First of all, we can think of $\alpha$ as $\sqrt{\sqrt{p}+\sqrt{q}}$, with $p,q$ rationals such that $-4pq=1$. Then $r=p+q$, $\psi = \sqrt{p-q}$. Also, $i=2\sqrt{pq}$, and $\sqrt{2(r-\psi^2)}=2\sqrt{q}$, explaining why $i,\sqrt{2(r-\psi^2)}$ are in $\mathbb{Q}(\alpha^2)$.

This makes it clear that, in general, $\psi\notin L$.

Then, $f$’s splitting field is $L(\psi)$; an automorphism of this field is determined by its action on $\alpha,\psi$. Defining $\mathbf{i},\mathbf{j}$ as in the question, and setting $\mathbf{k}=\mathbf{ij}$, we find that $\mathbf{k}$ sends $\alpha\mapsto i\alpha$ and $\psi\mapsto -\psi$. Let $\mathbf{-1}$ denote the automorphism that fixes $\psi$ and sends $\alpha\mapsto -\alpha$; it commutes with the others. Then $\mathbf{i},\mathbf{j},\mathbf{k}$ satisfy the famous $\mathbf{i}^2=\mathbf{j}^2=\mathbf{k}^2=\mathbf{ijk}=\mathbf{-1}$. Thus $G=\operatorname{Gal}(L(\psi)/\mathbb{Q})$ does contain $Q_8$ as a subgroup.

I was secretly hoping that $G$ would also have $Q_8$ as a quotient, so I could still pull a quaternion extension of $\mathbb{Q}$ out of all this, but alas, this is not to be. $G$ is generated over $Q_8$ by a central element $\mathbf{l}$, that, tragically, has order 4. (For a moment, I though it might have order $2$ and thus imply that $G=Q_8\times C_2$. Alas…) We can find $\mathbf{l}$ as follows:

Let $\sigma$ be the automorphism of $L(\psi)$ that sends $\psi\mapsto -\psi$ and fixes $\alpha$, i.e. the nontrivial automorphism of $L(\psi)$ over $L$. It is clear that $\sigma$ generates $G$ over $Q_8$ because $Q_8$ is index $2$ and $\sigma\notin Q_8$.

Now $L(\psi)$ has $\mathbb{Q}(\sqrt{p},\sqrt{q},\psi)$ as a subfield. Restriction to this subfield gives us a surjective homomorphism $G\rightarrow \operatorname{Gal}(\mathbb{Q}(\sqrt{p},\sqrt{q},\psi))$; thus we have an exact sequence

$$1\rightarrow C_2\rightarrow G\rightarrow C_2\times C_2\times C_2\rightarrow 1$$

A normal subgroup of order $2$ is automatically central, thus $G$ is a central extension of $C_2\times C_2\times C_2$ by $C_2$. It was pointed out to me by Fedor Bogomolov that whenever $A,B$ are abelian groups and $G$ is a central extension of $A$ by $B$, the commutator $[\cdot,\cdot]$ induces a $\mathbb{Z}$-bilinear form on $A$ with values in $B$. (This is *awesome*! This must be a standard fact, right? Where would one normally learn this?) This is because commutators $[x,y]$ in $G$ are necessarily in $B$ because they are knocked out by homomorphing to $A$ since it is abelian, and furthermore they are not affected by changing $x,y$ by central elements (in particular, by elements of $B\subset Z(G)$), thus $[\cdot,\cdot]$ is still unambiguous when it is seen as accepting values from $A$ rather than $G$.

In the present case, $A=C_2\times C_2\times C_2$ is an $\mathbb{F}_2$ vector space and $B=C_2=\mathbb{F}_2$, so we can see the commutator as a skew form on $\mathbb{F}_2^3$!

Any skew form on an odd-dimensional space has a nontrivial null space. The preimage in $G$ of this nullspace is the center of $G$. This gives us a way of finding a central element of $G$ that generates it over $Q_8$.

Let $\bar{\mathbf{i}},\bar{\mathbf{k}},\bar{\sigma}$ be the images of $\mathbf{i},\mathbf{k},\sigma$ in $C_2\times C_2\times C_2$. $\mathbf{i}$ reverses $\psi$ and $\sqrt{q}$ while fixing $\sqrt{p}$; $\mathbf{k}$ reverses all three; while $\sigma$ just reverses $\psi$. Thus $\bar{\mathbf{i}},\bar{\mathbf{k}},\bar{\sigma}$ are a basis of $C_2\times C_2 \times C_2$. Since $\mathbf{k},\sigma$ commute while $\mathbf{i}$ doesn’t commute with either, the matrix of the commutator form with respect to this basis is

$$\begin{pmatrix}0&1&1\\1&0&0\\1&0&0\end{pmatrix}$$

and, by inspection, the nullspace is spanned by $\begin{pmatrix}0&1&1\end{pmatrix}^T$. It follows that $\mathbf{k}\sigma$, the automorphism of $L(\psi)$ that fixes $\psi$ and sends $\alpha\mapsto i\alpha$, is in the center of $G$! Let $\mathbf{l}=\mathbf{k}\sigma$.

Then $G=Q_8\cup\mathbf{l}Q_8$, $\mathbf{l}^2=\mathbf{-1}$, and $\mathbf{l}$ commutes with everything. This is enough information to calculate in $G$.

About building a quaternion extension of **Q** (or more generally of any number field) : the problem can be solved completely and explicitly starting from a biquadratic field, using techniques of embedding problem. Actually, given a field K containing a primitive *p*-th root of unity (where *p* is any prime), the more general problem of embedding a Galois extension L/K of type (*p, p*) into a non abelian Galois extension M/L/K of degree $p^3$ of any possible type can be solved via Kummer theory and cohomology. Allow me to refer to R. Massy & T. Nguyen Quang Do , *J. reine angew. Math.,* 291 (1977), 149-161. Beware : the last result there (thm. 15) contains an error by omission, which is repaired in R. Massy, *J. of Algebra*, 109 (1987), 508-535, thm. 3 B .

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