The Hurwitz order

Abstract

With the preceding chapters on lattices and orders in hand, we are now prepared to embark on a general treatment of quaternion algebras over number fields and the arithmetic of their orders. Before we do so, for motivation and pure enjoyment, in this chapter we consider the special case of the Hurwitz order. Not only is this appropriate in a historical spirit, it is also instructive for what follows; moreover, the Hurwitz order has certain exceptional symmetries that make it worthy of specific investigation.

With the preceding chapters on lattices and orders in hand, we are now prepared to embark on a general treatment of quaternion algebras over number fields and the arithmetic of their orders. Before we do so, for motivation and pure enjoyment, in this chapter we consider the special case of the Hurwitz order. Not only is this appropriate in a historical spirit, it is also instructive for what follows; moreover, the Hurwitz order has certain exceptional symmetries that make it worthy of specific investigation.

1 $\triangleright$ The Hurwitz order

Hurwitz developed the theory of integral quaternions in a treatise [Hur19] in 1919. A more modern treasure trove of detail about quaternion groups and the Hurwitz order (as well as many other things) can be found in the book by Conway–Smith [CSm2003]; the review by Baez [Bae2005] also provides an accessible overview.

We consider in this chapter the restriction of the Hamiltonians from $\mathbb R$ to $\mathbb Q$, namely, the quaternion algebra . We further restrict to those elements with integer coordinates

$$\begin{aligned} \mathbb Z \langle i,j \rangle =\mathbb Z + \mathbb Z i + \mathbb Z j + \mathbb Z k, \end{aligned}$$

(11.1.1)

where $k :=ij$. By Example 10.2.4, this is an order in B, called the Lipschitz order. In the rest of this chapter, we will work over $\mathbb Z$ and so we will simply refer to lattices and orders.

The Lipschitz order is not a maximal order, and maximal orders have better properties. This is analogous to the fact that the ring $\mathbb Z [\sqrt{-3}]$ is an order in $\mathbb Q (\sqrt{-3})$ but is not maximal (not integrally closed), properly contained in the better-behaved maximal order $\mathbb Z [(-1+\sqrt{-3})/2]$ of Eisenstein integers. The comparison with the Eisenstein integers is more than incidental: the element $\alpha =i+j+k$ satisfies $\alpha ^2+3=0$, so it is natural to consider

$$\begin{aligned} \omega :=\frac{-1+i+j+k}{2} \end{aligned}$$

which satisfies $\omega ^2+\omega +1=0$. We can enlarge the Lipschitz order to include $\omega$—indeed, this is the only possibility.

Lemma 11.1.2

The lattice

$$\begin{aligned} \mathcal {O}= \mathbb Z + \mathbb Z i + \mathbb Z j + \mathbb Z \omega = \mathbb Z \langle i,j \rangle + \mathbb Z \langle i,j\rangle \omega \end{aligned}$$

(11.1.3)

in B is the unique order that properly contains $\mathbb Z \langle i,j \rangle$, and $\mathcal {O}$ is maximal.

The order $\mathcal {O}$ in (11.1.3) is called the Hurwitz order, and it contains $\mathbb Z \langle i,j \rangle$ with index 2. Note that if $\alpha \in \mathcal {O}$, then $\alpha \in \mathbb Z \langle i,j \rangle$ if and only if ${{\,\mathrm{trd}\,}}(\alpha ) \in 2\mathbb Z$.

Proof. By Exercise 11.1, the lattice $\mathcal {O}$ is an order.

Suppose that $\mathcal {O}' \supsetneq \mathbb Z \langle i,j \rangle$ and let $\alpha = t+xi+yj+z k \in \mathcal {O}'$ with $t,x,y,z \in \mathbb Q$. Then ${{\,\mathrm{trd}\,}}(\alpha )=2t \in \mathbb Z$, so by Corollary 10.3.6 we have $t \in \frac{1}{2}\mathbb Z$. Similarly, $\alpha i \in \mathcal {O}'$ therefore ${{\,\mathrm{trd}\,}}(\alpha i) = -2x \in \mathbb Z$ and $x \in \frac{1}{2}\mathbb Z$, and in the same way $y,z \in \frac{1}{2}\mathbb Z$. Finally, ${{\,\mathrm{nrd}\,}}(\alpha )=t^2+x^2+y^2+z^2 \in \mathbb Z$, and considerations modulo 4 imply that t, x, y, z either all belong to $\mathbb Z$ or to $\frac{1}{2}+\mathbb Z$; thus $\alpha \in \mathcal {O}$ and so $\mathcal {O}'=\mathcal {O}$. $\square$

11.1.4

We can recast this calculation in terms of the local-global dictionary for lattices (Theorem 9.1.1). Since $\mathcal {O}[\frac{1}{2}]=\mathbb Z \langle i,j \rangle [\frac{1}{2}]$, for every odd prime p we have $\mathcal {O}_{(p)} = \mathbb Z \langle i,j \rangle _{(p)}$, and $\mathcal {O}_{(2)} \supsetneq \mathbb Z \langle i,j \rangle _{(2)}$.

2 $\triangleright$ Hurwitz units

We now consider unit groups; in this section, we take $k :=ij$. An element $\gamma =t+xi+yj+zk \in \mathbb Z \langle i,j\rangle$ is a unit if and only if ${{\,\mathrm{nrd}\,}}(\gamma )=t^2+x^2+y^2+z^2 \in \mathbb Z ^{\times }$, i.e. ${{\,\mathrm{nrd}\,}}(\gamma )=1$, and since $t,x,y,z \in \mathbb Z$ we immediately have

$$\begin{aligned} \mathbb Z \langle i,j \rangle ^\times = \{\pm 1,\pm i, \pm j, \pm k\} \simeq Q_8 \end{aligned}$$

is the quaternion group of order 8. In a similar way, taking $\gamma \in \mathcal {O}$ in the Hurwitz order and allowing $t,x,y,z \in \frac{1}{2}\mathbb Z$ so that 2t, 2x, 2y, 2z all have the same parity, we find that

$$\begin{aligned} \mathcal {O}^\times = Q_8 \cup (\pm 1 \pm i \pm j \pm k)/2 \end{aligned}$$

is a group of order 24.

We have $\mathcal {O}^\times \not \simeq S_4$ (the symmetric group on 4 letters) because there is no embedding $Q_8 \hookrightarrow S_4$. (The permutation representation $Q_8 \rightarrow S_4$ obtained by the action on the cosets of the unique subgroup $\langle -1 \rangle$ of index 4 factors through the quotient $Q_8 \rightarrow Q_8/\{\pm 1\} \simeq V_4 \hookrightarrow S_4$, where $V_4$ is the Klein 4-group.) There are 15 groups of order 24 up to isomorphism! We identify the right one as follows.

Lemma 11.2.1

We have $\mathcal {O}^\times \simeq {{\,\mathrm{SL}\,}}_2(\mathbb F _3)$.

Proof. We reduce modulo 3. There is a ring homomorphism

Any quaternion algebra over a finite field is isomorphic to the matrix ring by Wedderburn’s little theorem (Exercises 3.16, 6.16, and 7.30). Specifically, the element $\epsilon =i+j+k$ has $\epsilon ^2 = 0 \in \mathcal {O}/3\mathcal {O}$. The left ideal I generated by $\epsilon$ is an $\mathbb F _3$-vector space, and we compute that it has basis $\epsilon$ and $i\epsilon = -1 - j + k$. As in (7.6.3) (Proposition 7.6.2) this yields an isomorphism

$$\begin{aligned} \mathcal {O}/3\mathcal {O}&\rightarrow {{\,\mathrm{M}\,}}_2(\mathbb F _3) \\ i, j&\mapsto \begin{pmatrix} 0 &{} -1\\ 1 &{} 0 \end{pmatrix}, \begin{pmatrix} 1 &{} 1 \\ 1 &{} -1 \end{pmatrix} \end{aligned}$$

(a statement that can be explicitly and independently verified in Exercise 11.3). We obtain a group homomorphism $\mathcal {O}^\times \rightarrow {{\,\mathrm{SL}\,}}_2(\mathbb F _3)$, since the reduced norm corresponds to the determinant and ${{\,\mathrm{nrd}\,}}(\mathcal {O}^\times )=\{1\}$, and this homomorphism is injective because if $\gamma \in \mathcal {O}^\times$ has $\gamma -1 \in 3\mathcal {O}$ then $\gamma =1$, by inspection. Since $\#\mathcal {O}^\times =\#{{\,\mathrm{SL}\,}}_2(\mathbb F _3)=24$, the map $\mathcal {O}^\times \hookrightarrow {{\,\mathrm{SL}\,}}_2(\mathbb F _3)$ is an isomorphism. $\square$

11.2.2

The group ${{\,\mathrm{SL}\,}}_2(\mathbb F _3)$ acts on the left on the set of nonzero column vectors $\mathbb F _3^2$ up to sign, a set of cardinality $(9-1)/2=4$. (More generally, ${{\,\mathrm{SL}\,}}_2(\mathbb F _p)$ acts on $\mathbb P ^1(\mathbb F _p) :=(\mathbb F _p^2 \smallsetminus \{(0,0)\})/\mathbb F _p^\times$, a set of cardinality $p+1$.) This action yields a permutation representation ${{\,\mathrm{SL}\,}}_2(\mathbb F _3) \rightarrow S_4$; the kernel of this map is the subgroup generated by the scalar matrix $-1$ and so the representation gives an injective group homomorphism from ${{\,\mathrm{PSL}\,}}_2(\mathbb F _3) :={{\,\mathrm{SL}\,}}_2(\mathbb F _3)/\{\pm 1\}$ into $S_4$. Since $A_4 \le S_4$ is the unique subgroup of size $24/2=12$, we must have ${{\,\mathrm{PSL}\,}}_2(\mathbb F _3) \simeq A_4$, giving an exact sequence

$$\begin{aligned} 1 \rightarrow \{\pm 1\} \rightarrow \mathcal {O}^\times \rightarrow A_4 \rightarrow 1. \end{aligned}$$

(11.2.3)

11.2.4

We can also visualize the group $\mathcal {O}^\times$ and the exact sequence (11.2.3), thinking of the Hamiltonians as acting by rotations (section 2.4). Recall there is an exact sequence (Corollary 2.4.21)

$$\begin{aligned} 1 \rightarrow \{\pm 1 \} \rightarrow \mathbb H ^1 \rightarrow {{\,\mathrm{SO}\,}}(3) \rightarrow 1 \end{aligned}$$

(11.2.5)

obtained by the left action $\alpha \mapsto \alpha v \alpha ^{-1}$ for $\alpha \in \mathbb H ^1$ and $v \in \mathbb H ^0 \simeq \mathbb R ^3$; specifically, by Proposition 2.4.18, a quaternion $\alpha =\cos \theta + I(\alpha ) \sin \theta$ acts by rotation through the angle $2\theta$ about the axis $I(\alpha )$.

We have been considering

(11.2.6)

and we now consider the corresponding embedding of groups $\mathcal {O}^1=\mathcal {O}^\times \hookrightarrow \mathbb H ^1$. We are led to think of the group $\mathcal {O}^\times /\{\pm 1 \} \simeq A_4$ as the group of symmetries (rigid motions) of a tetrahedron (or rather, a tetrahedron and its dual), as in Figure 11.2.7.

Figure 11.2.7:

Symmetries of a tetrahedron, viewed quaternionically

Inside the cube in $\mathbb R ^3$ with vertices $(\pm 1, \pm 1, \pm 1) = \pm i \pm j \pm k$, we can find four inscribed tetrahedra, for example, the tetrahedron T with vertices

$$\begin{aligned} i+j+k, i-j-k, -i+j-k, -i-j+k. \end{aligned}$$

Then the elements $\pm i, \pm j, \pm k$ act by rotation about the x, y, z axes by an angle $\pi$ (so interchanging points with the same x, y, z coordinate). The element $\pm \omega =\pm (-1+i+j+k)/2$ rotates by the angle $2\pi /3$ fixing the point (1, 1, 1) and cyclically permuting the other three points, and by symmetry we understand the action of the other elements of $\mathcal {O}^\times$. We therefore call $\mathcal {O}^\times$ the binary tetrahedral group. Following Conway–Smith [CSm2003, §3.3], we also write $2T=\mathcal {O}^\times$ for this group; the notation $\widetilde{A}_4$ is also used.

The subgroup $Q_8 \trianglelefteq 2T$ is normal (as it is characteristic, consisting of all elements of $\mathcal {O}$ of order dividing 4), and so we can write $2T = Q_8 \rtimes \langle \omega \rangle$ where $\langle \omega \rangle \simeq \mathbb Z /3\mathbb Z$ acts on $Q_8$ by conjugation, cyclically rotating the elements i, j, k. Finally, the group 2T has a presentation (Exercise 11.6)

$$\begin{aligned} 2T \simeq \langle r,s,t \mid r^2 = s^3 = t^3 = rst = -1 \rangle \end{aligned}$$

(11.2.8)

via $r=i$, $s=\omega =(-1+i+j+k)/2$, and $t=(-1+i+j-k)/2$.

We conclude by noting that the difference between the Lipschitz and Hurwitz orders is “covered” by the extra units.

Lemma 11.2.9

For every $\beta \in \mathcal {O}$, there exists $\gamma \in \mathcal {O}^\times$ such that $\beta \gamma \in \mathbb Z \langle i,j \rangle$.

Proof. If $\beta \in \mathbb Z \langle i,j \rangle$ already, then we are done. Otherwise, $2\beta =t+xi+yj+zk$ with all $t,x,y,z \in \mathbb Z$ odd. Choosing matching signs, there exists $\gamma \in \mathcal {O}^\times$ such that $2\beta \equiv 2\gamma \pmod {4\mathcal {O}}$. Thus

$$\begin{aligned} (2\beta )\gamma ^{-1} \equiv 2 \pmod {4\mathcal {O}} \end{aligned}$$

so $\beta \gamma ^{-1} \in \mathbb Z + 2\mathcal {O}= \mathbb Z \langle i,j \rangle$, so we may take $\gamma ^{-1}$ for the statement of the lemma. $\square$

3 $\triangleright$ Euclidean algorithm

The Eisenstein order $\mathbb Z [(-1+\sqrt{-3})/2]$ has several nice properties. Perhaps nicest of all is that it is a Euclidean domain, so in particular it is a PID and UFD. (Alas, the ring $\mathbb Z [\sqrt{-3}]$ just fails to be Euclidean.)

11.3.1

The Hurwitz order also has a left (or right) Euclidean algorithm generalizing the commutative case, as follows. There is an embedding $B \hookrightarrow B \otimes _\mathbb{Q } \mathbb R \simeq \mathbb H$, and inside $\mathbb H \simeq \mathbb R ^4$ the Hurwitz order sits as a ($\mathbb Z$-)lattice equipped with the Euclidean inner product, so we can think of the reduced norm by instead thinking of distance. In the Lipschitz order, we see by rounding coordinates that for all $\gamma \in B$ there exists $\mu \in \mathbb Z \langle i,j \rangle$ such that ${{\,\mathrm{nrd}\,}}(\gamma -\mu ) \le 4\cdot (1/2)^2 = 1$—a farthest point occurs at the center (1/2, 1/2, 1/2, 1/2) of a unit cube. But this is precisely the point where the Hurwitz quaternions occur, and it follows that for all $\gamma \in B$, there exists $\mu \in \mathcal {O}$ such that ${{\,\mathrm{nrd}\,}}(\gamma -\mu )<1$. (In fact, we can take ${{\,\mathrm{nrd}\,}}(\gamma -\mu ) \le 1/2$; see Exercise 11.7.)

Paragraph 11.3.1 becomes a right Euclidean algorithm as in the commutative case.

Lemma 11.3.2

(Hurwitz order is right norm Euclidean). For all $\alpha ,\beta \in \mathcal {O}$ with $\beta \ne 0$, there exists $\mu ,\rho \in \mathcal {O}$ such that

$$\begin{aligned} \alpha =\beta \mu +\rho \end{aligned}$$

(11.3.3)

and ${{\,\mathrm{nrd}\,}}(\rho ) < {{\,\mathrm{nrd}\,}}(\beta )$.

Proof. If ${{\,\mathrm{nrd}\,}}(\alpha )<{{\,\mathrm{nrd}\,}}(\beta )$, we may take $\mu =0$ and $\rho =\alpha$, so suppose ${{\,\mathrm{nrd}\,}}(\alpha ) \ge {{\,\mathrm{nrd}\,}}(\beta )>0$. Let $\gamma =\beta ^{-1}\alpha \in B$. Then by 11.3.1, there exists $\mu \in \mathcal {O}$ such that ${{\,\mathrm{nrd}\,}}(\gamma -\mu )<1$. Let $\rho =\alpha -\beta \mu$. Then by multiplicativity of the norm,

$$ {{\,\mathrm{nrd}\,}}(\rho )={{\,\mathrm{nrd}\,}}(\alpha -\beta \mu )<{{\,\mathrm{nrd}\,}}(\beta ). $$

$\square$

A similar statement to Lemma 11.3.2 holds for division on the left, i.e., in (11.3.3) we may take $\alpha =\mu \beta +\rho$ (with possibly different elements $\mu ,\rho \in \mathcal {O}$, of course).

Proposition 11.3.4

Every right ideal $I \subseteq \mathcal {O}$ is right principal, i.e., there exists $\beta \in I$ such that $I=\beta \mathcal {O}$.

Proof. Let $I \subseteq \mathcal {O}$ be a right ideal. If $I=\{0\}$, we are done. Otherwise, there exists an element $0 \ne \beta \in I$ with minimal reduced norm ${{\,\mathrm{nrd}\,}}(\beta ) \in \mathbb Z _{>0}$. We claim that $I=\beta \mathcal {O}$. For all $\alpha \in I$, by the left Euclidean algorithm in Lemma 11.3.2, there exists $\mu \in \mathcal {O}$ such that $\alpha =\mu \beta +\rho$ with ${{\,\mathrm{nrd}\,}}(\rho )<{{\,\mathrm{nrd}\,}}(\beta )$; but $\rho =\alpha -\beta \mu \in I$, so by minimality, ${{\,\mathrm{nrd}\,}}(\rho )=0$ and $\rho =0$, hence $\alpha =\beta \mu \in \beta \mathcal {O}$ as claimed. $\square$

Definition 11.3.5

Let $\alpha ,\beta \in \mathcal {O}$. We say $\beta$ right divides $\alpha$ (or $\alpha$ is a right multiple of $\beta$) and write $\beta \mid \!\!{}_{\textsf {\tiny {R}} }\, \alpha$ if there exists $\gamma \in \mathcal {O}$ such that $\alpha =\beta \gamma$.

A right common divisor of $\alpha ,\beta \in \mathcal {O}$ is an element $\gamma \in \mathcal {O}$ such that $\gamma \mid \!\!{}_{\textsf {\tiny {R}} }\,\alpha ,\beta$. A right greatest common divisor of $\alpha ,\beta$ is a common divisor $\gamma$ such that $\delta \mid \!\!{}_{\textsf {\tiny {R}} }\,\gamma$ for all common divisors $\delta$ of $\alpha ,\beta$.

It follows from Lemma 11.3.2 in the same way as in the commutative case that if $\alpha ,\beta$ are not both zero, then there exists a greatest common divisor of $\alpha ,\beta$, taking the last nonzero remainder in the right Euclidean algorithm.

Corollary 11.3.6

(Bézout’s theorem). For all $\alpha ,\beta \in \mathcal {O}$ not both zero, there exists $\mu ,\nu \in \mathcal {O}$ such that $\alpha \mu +\beta \nu =\gamma$ where $\gamma$ is a right greatest common divisor of $\alpha ,\beta$.

Proof. By Proposition 11.3.4, we may write $\alpha \mathcal {O}+ \beta \mathcal {O}= \gamma \mathcal {O}$ for some $\gamma \in \mathcal {O}$, and then $\gamma \in \alpha \mathcal {O}+ \beta \mathcal {O}$ implies there exists $\mu ,\nu \in \mathcal {O}$ such that $\alpha \mu +\beta \nu =\gamma$.

Proposition 11.3.7

Let $\mathcal {O}' \subset B$ be a maximal order. Then there exists $\alpha \in B^\times$ such that $\mathcal {O}'=\alpha ^{-1} \mathcal {O}\alpha$, and in particular $\mathcal {O}' \simeq \mathcal {O}$ as rings.

Proof. By clearing denominators, there exists nonzero $a \in \mathbb Z$ such that $a\mathcal {O}' \subseteq \mathcal {O}$. Let $I=a\mathcal {O}'\mathcal {O}$ be the right ideal of $\mathcal {O}$ generated by $a\mathcal {O}'$. Then $\mathcal {O}' \subseteq \mathcal {O}{}_{\textsf {\tiny {L}} }(I)$, and equality holds since $\mathcal {O}'$ is maximal. By Proposition 11.3.4, we have $I=\beta \mathcal {O}$ for some $\beta \in B^\times$. We have $\mathcal {O}{}_{\textsf {\tiny {L}} }(I)=\beta \mathcal {O}\beta ^{-1}$ by Exercise 10.14, so $\mathcal {O}'=\beta \mathcal {O}\beta ^{-1}$ and we may take $\alpha =\beta ^{-1}$. $\square$

Example 11.3.8

The Lipschitz order $\mathbb Z \langle i,j \rangle$ does not enjoy the property that every right ideal is principal, as the following example shows.

Let $I=2\mathcal {O}=2\mathbb Z + 2i \mathbb Z + 2j \mathbb Z + (1+i+j+k)\mathbb Z$. Then $I \subseteq \mathbb Z \langle i,j \rangle$ and I has the structure of a right $\mathbb Z \langle i,j \rangle$-ideal, in fact $I=2\mathbb Z \langle i,j \rangle + (1+i+j+k)\mathbb Z \langle i,j \rangle$. We claim that I is not principal as a right $\mathbb Z \langle i,j \rangle$-ideal. Indeed, suppose $I=\alpha \mathbb Z \langle i,j \rangle$ with $\alpha \in I$. Since $\alpha \in 2\mathcal {O}$, we have $4 \mid {{\,\mathrm{nrd}\,}}(\alpha )$. But $2 \in I$ so $2=\alpha \beta$ with $\beta \in \mathbb Z \langle i,j \rangle$, so $4={{\,\mathrm{nrd}\,}}(2)={{\,\mathrm{nrd}\,}}(\alpha ){{\,\mathrm{nrd}\,}}(\beta )$, whence ${{\,\mathrm{nrd}\,}}(\alpha )=4$ and ${{\,\mathrm{nrd}\,}}(\beta )=1$ so $\beta \in \mathbb Z \langle i,j \rangle ^\times$ and so $2\mathcal {O}= I=\alpha \mathbb Z \langle i,j \rangle = 2 \mathbb Z \langle i,j \rangle$. Cancelling the factor 2, we conclude $\mathcal {O}=\mathbb Z \langle i,j \rangle$, a contradiction.

For more, see Exercise 11.10.

4 $\triangleright$ Unique factorization

It does not follow that there is unique factorization in $\mathcal {O}$ in the traditional sense, as the order of multiplication matters. Nevertheless, there is a theory of prime factorization in $\mathcal {O}$ as follows.

Lemma 11.4.1

Let p be prime. Then there exists $\pi \in \mathcal {O}$ such that $\pi \overline{\pi }={{\,\mathrm{nrd}\,}}(\pi )=p$.

Proof. We have ${{\,\mathrm{nrd}\,}}(1+i)=1^2+1^2=2$, so we may suppose $p \ge 3$ is odd. Then $\mathcal {O}/p\mathcal {O}\simeq ({-1,-1} \mid \mathbb{F _p}) \simeq {{\,\mathrm{M}\,}}_2(\mathbb F _p)$ by Wedderburn’s little theorem, and there exists a right ideal $I \bmod {p} \subset \mathcal {O}/p\mathcal {O}$ with $\dim _\mathbb{F _p}(I \bmod {p})=2$. Let

$$\begin{aligned} I=\{\alpha \in \mathcal {O}: \alpha \bmod {p} \in I \bmod {p}\} \end{aligned}$$

be the preimage of $I \bmod p$ in the map $\mathcal {O}\rightarrow \mathcal {O}/p\mathcal {O}$. Then $p\mathcal {O}\subsetneq I \subsetneq \mathcal {O}$. Then $I \subset \mathcal {O}$ is a right ideal, and $I \ne \mathcal {O}$. But $I=\beta \mathcal {O}$ is right principal by Proposition 11.3.4.

We claim that ${{\,\mathrm{nrd}\,}}(\beta )=p$. Since $p \in I$, we have $p=\beta \mu$ for some $\mu \in \mathcal {O}$, whence ${{\,\mathrm{nrd}\,}}(p)=p^2={{\,\mathrm{nrd}\,}}(\beta ){{\,\mathrm{nrd}\,}}(\mu )$ so ${{\,\mathrm{nrd}\,}}(\beta ) \mid p^2$. We cannot have ${{\,\mathrm{nrd}\,}}(\beta )=1$ or ${{\,\mathrm{nrd}\,}}(\beta )=p^2$, as these would imply $I=\mathcal {O}$ or $I=p\mathcal {O}$, impossible. We conclude that ${{\,\mathrm{nrd}\,}}(\beta )=p$. $\square$

Remark 11.4.2. Once we have developed a suitable theory of norms, the proof that ${{\,\mathrm{nrd}\,}}(\beta )=p$ above will be immediate: if we define $\mathsf{N }(I):=\#(\mathcal {O}/I)$ then $\mathsf{N }(I)=p^2$ by construction, and it turns out that $\mathsf{N }(I)={{\,\mathrm{nrd}\,}}(\beta )^2$.

Theorem 11.4.3

(Lagrange). Every integer $n \ge 0$ is the sum of four squares, i.e., there exist $t,x,y,z \in \mathbb Z$ such that $n=t^2+x^2+y^2+z^2$.

Proof. We seek an element $\beta \in \mathbb Z \langle i,j \rangle$ such that ${{\,\mathrm{nrd}\,}}(\beta )=n$. By multiplicativity of the reduced norm, it is sufficient to treat the case where $n=p$ is prime. We obtain $\pi \in \mathcal {O}$ such that ${{\,\mathrm{nrd}\,}}(\pi )=p$ by Lemma 11.4.1. But now the result follows from Lemma 11.2.9, as there exists $\gamma \in \mathcal {O}^\times$ such that $\pi \gamma \in \mathbb Z \langle i,j \rangle$. $\square$

Remark 11.4.4. A counterpart to Lagrange’s theorem (Theorem 11.4.3) is the following theorem of Legendre and Gauss on sums of three squares: Every integer n that is not of the form $n=4^am$ with $m \equiv 7~(\text{ mod } ~{8})$ can be written as the sum of three squares $n=x^2+y^2+z^2$. We will revisit this classical theorem in Chapter 30 as motivation for the study of embedding numbers, and the number of such representations will be given in terms of class numbers, following Gauss. A direct proof of the three square theorem is given by Mordell [Mor69, §20, Theorem 1], but he notes that “no really elementary treatment [of this theorem] is known”.

We finish this section with a discussion of ‘unique factorization’ in the Hurwitz order.

Definition 11.4.5

An element $\pi \in \mathcal {O}$ is irreducible if whenever $\pi =\alpha \beta$ with $\alpha ,\beta \in \mathcal {O}$ then either $\alpha \in \mathcal {O}^\times$ or $\beta \in \mathcal {O}^\times$.

Lemma 11.4.6

Let $\pi \in \mathcal {O}$. Then $\pi$ is irreducible if and only if ${{\,\mathrm{nrd}\,}}(\pi )=p \in \mathbb Z$ is prime.

Proof. If ${{\,\mathrm{nrd}\,}}(\pi )=p$ is prime and $\pi =\alpha \beta$ then ${{\,\mathrm{nrd}\,}}(\pi )=p={{\,\mathrm{nrd}\,}}(\alpha ){{\,\mathrm{nrd}\,}}(\beta )$ so either ${{\,\mathrm{nrd}\,}}(\alpha )=1$ or ${{\,\mathrm{nrd}\,}}(\beta )=1$, thus $\alpha \in \mathcal {O}^\times$ or $\beta \in \mathcal {O}^\times$. Conversely, suppose $\pi$ is irreducible and let $p \mid {{\,\mathrm{nrd}\,}}(\pi )$. Let $I=\pi \mathcal {O}+p\mathcal {O}=\alpha \mathcal {O}$. Then ${{\,\mathrm{nrd}\,}}(\alpha ) \mid {{\,\mathrm{nrd}\,}}(p)=p^2$. We cannot have ${{\,\mathrm{nrd}\,}}(\alpha )=1$, as every element of I has reduced norm divisible by p. We similarly cannot have ${{\,\mathrm{nrd}\,}}(\alpha )=p^2$, since this would imply $\pi \in p\mathcal {O}$; but by Lemma 11.4.1, p is reducible, a contradiction. We conclude that ${{\,\mathrm{nrd}\,}}(\alpha )=p$. From $\pi \in I=\alpha \mathcal {O}$ we obtain $\pi =\alpha \beta$ with $\beta \in \mathcal {O}$; by irreducibility, $\beta \in \mathcal {O}^\times$ and ${{\,\mathrm{nrd}\,}}(\pi )={{\,\mathrm{nrd}\,}}(\alpha )=p$. $\square$

Definition 11.4.7

An element $\alpha \in \mathcal {O}$ is primitive if $\alpha \not \in n\mathcal {O}$ for all $n \in \mathbb Z _{\ge 2}$.

Theorem 11.4.8

(Conway–Smith). Let $\alpha \in \mathcal {O}$ be primitive and let $a={{\,\mathrm{nrd}\,}}(\alpha )$. Factor $a=p_1 p_2 \cdots p_r$ into a product of primes. Then there exists $\pi _1,\pi _2,\dots ,\pi _r \in \mathcal {O}$ such that

$$\begin{aligned} \alpha = \pi _1 \pi _2 \cdots \pi _r, \quad \text {and } {{\,\mathrm{nrd}\,}}(\pi _i)=p_i\, {for\, all}\, i. \end{aligned}$$

(11.4.9)

Moreover, every other such factorization is of the form

$$\begin{aligned} \alpha = (\pi _1 \gamma _1)(\gamma _1^{-1} \pi _2 \gamma _2) \cdots (\gamma _{r-1}^{-1} \pi _r) \end{aligned}$$

(11.4.10)

where $\gamma _1,\dots ,\gamma _r \in \mathcal {O}^\times$.

Proof. Let $I=\alpha \mathcal {O}+ p_1\mathcal {O}$; as in the proof of Lemma 11.4.6, we find $I=\pi _1\mathcal {O}$ with ${{\,\mathrm{nrd}\,}}(\pi _1)=p_1$, arguing that ${{\,\mathrm{nrd}\,}}(\pi _1)\ne p_1^2$ since $\alpha \in p_1\mathcal {O}$ is in contradiction to $\alpha$ being primitive. Then $\pi _1$ is unique up to right multiplication by a unit and $\alpha =\pi _1\alpha _2$. The result then follows by induction. $\square$

The factorization (11.4.10) is said to be obtained from $\alpha =\pi _1 \cdots \pi _r$ by unit migration.

Remark 11.4.11. To look at all possible prime factorizations of $\alpha$ as in (11.4.9), it is necessary to consider the possible factorizations $a=p_1 \cdots p_r$. Conway–Smith call this process metacommutation [CSm2003, Chapter 5]; metacommutation is analyzed by Cohn–Kumar [CK2015], Forsyth–Gurev–Shrima FGS, and in a very general context by Chari [Cha2020].

5 Finite quaternionic unit groups

We conclude this section by a discussion of quaternion unit groups extending the discussion 11.2: we classify finite subgroups of $\mathbb H ^\times$ and realize the possible subgroups as coming from quaternionic unit groups.

11.5.1

To begin with the classification, suppose that $\Gamma \subseteq \mathbb H ^\times$ is a finite subgroup. Then ${{\,\mathrm{nrd}\,}}(\Gamma )$ is a finite subgroup of $\mathbb R _{>0}^\times$, hence identically 1, so $\Gamma \subseteq \mathbb H ^1$.

Similarly, if $\Gamma \subseteq \mathbb H ^\times /\mathbb R ^\times \simeq \mathbb H ^1/\{\pm 1\}$ is a finite subgroup, then it lifts via the projection $\mathbb H ^1 \rightarrow \mathbb H ^1/\{\pm 1\}$ to a finite subgroup of $\mathbb H ^1$.

So let $\Gamma \subseteq \mathbb H ^1$ be a finite subgroup. Then

$$\begin{aligned} \Gamma /\{\pm 1\} \hookrightarrow \mathbb H ^1/\{\pm 1\} \simeq {{\,\mathrm{SO}\,}}(3) \end{aligned}$$

the latter isomorphism by Hamilton’s original (!) motivation for quaternion algebras (Corollary 2.4.21). Therefore $\Gamma /\{\pm 1\} \subseteq {{\,\mathrm{SO}\,}}(3)$ is a finite rotation group, and these groups have been known since antiquity.

Proposition 11.5.2

A finite subgroup of ${{\,\mathrm{SO}\,}}(3)$ is one of the following:

• (i) a cyclic group;

• (ii) a dihedral group;

• (iii) the tetrahedral group $A_4$ of order 12;

• (iv) the octahedral group $S_4$ of order 24; or

• (v) the icosahedral group $A_5$ of order 60.

Cases (iii)–(v) are the symmetry groups of the corresponding Platonic solids and are called exceptional rotation groups.

Proof. Let $G \le {{\,\mathrm{SO}\,}}(3)$ be a finite subgroup with $\#G=n>1$; then G must consist of rotations about a common fixed point (its center of gravity), which we may take to be the origin. The group G then acts on the unit sphere, and every nonidentity element of G acts by rotation about an axis, fixing the poles of its axis on the sphere. Let V be the subset of these poles in the unit sphere; the set V will soon be the vertices of our (possibly degenerate) polyhedron. Let

$$\begin{aligned} X=\{(g,v) : g \in G\smallsetminus \{1\}\text { and } v \text { is a pole of } g \}. \end{aligned}$$

Since each $g \in G \smallsetminus \{1\}$ has exactly two poles, we have $\#X = 2(n -1)$. On the other hand, we can also count organizing by orbits. Choose a representative set $v_1,\dots ,v_r$ of poles, one from each orbit of G on V, and let

$$\begin{aligned} n_i=\#{{\,\mathrm{Stab}\,}}_G(v_i)=\#\{g \in G : gv_i=v_i\} \end{aligned}$$

be the order of the stabilizer: this group is a cyclic subgroup about a common axis. Then

$$\begin{aligned} 2n-2=\# X = \sum _{i=1}^r \#(G v_i)(n_i-1) = \sum _{i=1}^r \frac{n}{n_i}(n_i-1) = n\sum _{i=1}^r \left( 1-\frac{1}{n_i}\right) , \end{aligned}$$

by the orbit–stabilizer theorem. Dividing both sides by n gives

$$\begin{aligned} 2-\frac{2}{n}=\sum _{i=1}^r \left( 1-\frac{1}{n_i}\right) . \end{aligned}$$

(11.5.3)

Since $n>1$, we have $1 \le 2-2/n<2$; and since each $n_i \ge 2$, we have $1/2 \le 1-1/n_i < 1$. Putting these together, we must have $r=2,3$.

If $r=2$, then (11.5.3) becomes $2=n/n_1+n/n_2$, with $n/n_i=\#(G v_i) \ge 1$, so $n_1=n_2=n$, there is only one axis of rotation, and G is cyclic.

If $r=3$, then the only possibilities for $(n_1,n_2,n_3)$ with $n_1 \le n_2 \le n_3$ are (2, 2, c), (2, 3, 3), (2, 3, 4), (2, 3, 5); the corresponding groups have sizes 2c, 12, 24, 60, respectively, and can be identified with $D_{2c},A_4,S_4,A_5$ by a careful but classical analysis of orbits. See Armstrong [Arm88, Chapter 19], Grove–Benson [GB2008, §2.4], or Conway–Smith [CSm2003, §3.3]. $\square$

In 11.2.4, we gave a quaternionic visualization of the binary tetrahedral group (lifting the tetrahedral group to $\mathbb H ^1$); we repeat this with the two other exceptional rotation groups, taking again $k :=ij$.

11.5.4

The octahedral group $S_4$ pulls back to the binary octahedral group $2O \subseteq \mathbb H ^1$ of order $24\cdot 2 = 48$, whose elements act by rigid motions of the octahedron (or dually, the cube). We make identifications following 11.2.4, shown in Figure 11.5.5.

Figure 11.5.5:

Symmetries of an octahedron and a cube, viewed quaternionically

The binary tetrahedral group $2T \trianglelefteq 2O$ of order 24 acts as a subgroup of rigid motions; the group 2O is generated by an element which maps to a rotation of order 4 around the 6 faces, i.e., one of the 12 elements

$$\begin{aligned} \frac{\pm 1\pm i}{\sqrt{2}}, \frac{\pm 1\pm j}{\sqrt{2}}, \frac{\pm 1\pm k}{\sqrt{2}}. \end{aligned}$$

The group 2O has a Coxeter presentation

$$\begin{aligned} 2O \simeq \langle r,s,t \mid r^2 = s^3 = t^4 = rst = -1 \rangle \end{aligned}$$

(with $-1$ central and $(-1)^2=1$). One also writes $2O \simeq \widetilde{S}_4$.

Let $F=\mathbb Q (\sqrt{2})$ and $R=\mathbb Z [\sqrt{2}]$. If we consider the Hamiltonians restricted to F as , then the group $2O \subseteq \mathbb H ^1$ generates an R-order: letting i, j be the standard generators and still $k :=ij$, and letting $\alpha =(1+i)/\sqrt{2}$ and $\beta =(1+j)/\sqrt{2}$, then

$$\begin{aligned} \mathcal {O}_{2O} = R + R\alpha + R\beta + R\alpha \beta ; \end{aligned}$$

(11.5.6)

this order contains the scalar extension of the Hurwitz order to R and is in fact a maximal R-order. (The extension of scalars is necessary: $S_4$ contains an element of order 4 which lifts to an element of order 8 in 2O; such an element has trace $\pm (\zeta _8+\zeta _8^{-1})=\pm \sqrt{2}$.)

11.5.7

Finally, we treat the binary icosahedral group $2I \subseteq \mathbb H ^1$ of order $60\cdot 2 = 120$, acting by rigid motions of the icosahedron (or dually, the dodecahedron). We choose the regular icosahedron to have vertices at

$$\begin{aligned} \pm i \pm j \pm k,\ \pm \tau i \pm \tau ^{-1} j,\ \pm \tau j \pm \tau ^{-1} k,\ \pm \tau k \pm \tau ^{-1}i \end{aligned}$$

where $\tau =(1+\sqrt{5})/2$ is the golden ratio. The elements of order 5 are given by conjugates and powers of the element $\zeta =(\tau + \tau ^{-1} i + j)/2$, which acts by rotation about a face. The group 2I can be presented as

$$\begin{aligned} 2I \simeq \langle r,s,t \mid r^2 = s^3 = t^5 = rst = -1 \rangle \end{aligned}$$

and we have $2I \simeq \widetilde{A}_5 \simeq {{\,\mathrm{SL}\,}}_2(\mathbb F _5)$. Letting now $F=\mathbb Q (\sqrt{5})$ and $R=\mathbb Z [\tau ]$, the R-algebra generated by 2I is the maximal order

$$\begin{aligned} \mathcal {O}_{2I} = R + Ri + R\zeta + Ri\zeta . \end{aligned}$$

(11.5.8)

For further references, see Conway–Sloane [CSl88, §8.2], who describe the binary icosahedral group in detail, calling it the icosian group.

We now consider the related possibilities over $\mathbb Q$. (We will return to a general classification in section 32.4.) To put ourselves in a situation like (11.2.6), let be a quaternion algebra over $\mathbb Q$ such that : in this case, we say that B is definite . By Exercise 2.4, B is definite if and only if $a,b<0$. Let $\mathcal {O}\subseteq B$ be an order in B; we would like to understand its unit group.

Lemma 11.5.9

The group $\mathcal {O}^\times =\mathcal {O}^1$ is finite.

Proof. We may take $B = ({a,b} \mid \mathbb{Q })$ with $a,b<0$. Consider the reduced norm ${{\,\mathrm{nrd}\,}}:B \rightarrow \mathbb Q$, given by ${{\,\mathrm{nrd}\,}}(t+xi+yj+zij)=t^2+|a \,|x^2+|b \,|y^2 + |ab \,|z^2$, so ${{\,\mathrm{nrd}\,}}(B^\times ) \subseteq \mathbb Q _{>0}^\times$. At the same time, ${{\,\mathrm{nrd}\,}}(\mathcal {O}^\times ) \subseteq \mathbb Z ^\times =\{\pm 1\}$, so we conclude $\mathcal {O}^\times =\mathcal {O}^1$. This group is finite because the restriction ${{\,\mathrm{nrd}\,}}|_{\mathcal {O}}$ of the reduced norm to $\mathcal {O}\simeq \mathbb Z ^4$ defines a (still) positive definite quadratic form, so there are only finitely many elements of $\mathcal {O}$ of any fixed reduced norm. (For a geometric perspective, viewing the elements of $\mathcal {O}^1$ as lattice points on an ellipsoid in $\mathbb R ^4$), see Proposition 17.5.6.) $\square$

In view of Lemma 11.5.9, the classification of finite rotation groups (Proposition 11.5.2) applies. We consider each case in turn.

11.5.10

Among the (nontrivial) cyclic groups, only subgroups of order 2, 4, 6 are possible over $\mathbb Q$. Indeed, a generator satisfies a quadratic equation with integer coefficients and so belongs to the ring of integers of an imaginary quadratic field; and only two imaginary quadratic fields have units other than $\pm 1$, namely, the Eisenstein order $\mathbb Z [(-1+\sqrt{-3})/2]$ of discriminant $-3$ and the Gaussian order $\mathbb Z [\sqrt{-1}]$ of discriminant $-4$ with groups of size 4, 6, respectively. (The more precise question of whether or not there is a unit of specified order is a question of embedding numbers, the subject of Chapter 30.)

11.5.11

Next, suppose that $\mathcal {O}^\times /\{\pm 1\}$ is dihedral, and let $j \in \mathcal {O}^\times \smallsetminus \{\pm 1\}$ act by inversion (equivalently, conjugation) on a cyclic group (of order 2, 3, by 11.5.10), generated by an element i. Let $K=\mathbb Q (i)$. Since j acts by inversion, we have $j^2 \in \mathbb Q$, and since $j \in \mathcal {O}^\times$ we have $j^2=-1$. It follows that $j\alpha = \overline{\alpha } j$ for all $\alpha \in K$. Thus , and we have two possibilities:

• (i) If i has order 4, then $B \simeq ({-1,-1} \mid \mathbb{Q })$ and $\mathcal {O}$ contains the order generated by i, j. This is the case treated in section 11.2: $\mathcal {O}$ is the Lipschitz order, and $\mathcal {O}^\times \simeq Q_8$ is the quaternion group of order 8.

• (ii) Otherwise, $i=\omega$ has order 6, and $B \simeq ({-3,-1} \mid \mathbb{Q })$. By Exercise 11.11(a), we have $({-3,-1} \mid \mathbb{Q }) \not \simeq ({-1,-1} \mid \mathbb{Q })$. By an argument similar to Lemma 11.1.2—and boy, there is more of this to come in Chapter 32—we see that

  $$\begin{aligned} \mathcal {O}=\mathbb Z + \mathbb Z \omega + \mathbb Z j + \mathbb Z \omega j \end{aligned}$$

  (11.5.12)

  is maximal. The group $\mathcal {O}^\times /\{\pm 1\} \simeq D_6$ is a dihedral group of order 6, and the group $\mathcal {O}^\times$ is generated by $\omega ,j$ with relations $\omega ^3=j^2=-1$ and $j\omega = \omega ^{-1} j$; in other words, $\mathcal {O}^\times \simeq C_3 \rtimes C_4$ is the semidirect product of the cyclic group $C_3$ of order 3 by the action of the cyclic group $C_4$ with a generator acting by inversion on $C_3$. Because $i^2=-1$ is central, we also have an exact sequence

  $$\begin{aligned} 1 \rightarrow C_6 \rightarrow \mathcal {O}^\times \rightarrow C_2 \rightarrow 1 \end{aligned}$$

  where $C_6 \simeq \langle \omega \rangle$ and $C_2 \simeq \langle j \rangle / \{\pm 1\}$. This group is also called the binary dihedral or dicyclic group of order 12, denoted $2D_6$.

11.5.13

To conclude, suppose that $\mathcal {O}^\times /\{\pm 1\}$ is exceptional. Each of these groups contain a dihedral group, so the argument from 11.5.11 applies: the only new group we see is the (binary) tetrahedral group obtained from the Hurwitz units (section 11.2). Here is another proof: the group $S_4$ contains an element of order 4 and $A_5$ an element of order 5, and these lift to elements of order 8, 10 in $\mathcal {O}^\times$, impossible.

We have proven the following theorem.

Theorem 11.5.14

Let $B=({a,b} \mid \mathbb{Q })$ be a quaternion algebra over $\mathbb Q$ with $a,b<0$, and let $\mathcal {O}\subseteq B$ be an order. Then $\mathcal {O}^\times$ is either cyclic of order 2, 4, 6, quaternion of order 8, binary dihedral of order 12, or binary tetrahedral of order 24.

Moreover, $\mathcal {O}^\times$ is quaternion, binary dihedral, or binary tetrahedral if and only if $\mathcal {O}$ is isomorphic to the Lipschitz order, the order (11.5.12), or the Hurwitz order, respectively.

Proof. Combine 11.5.10, 11.5.11, and 11.5.13. $\square$

Exercises

$\triangleright$ 1.:

Check directly that the Hurwitz order

$$\begin{aligned} \mathcal {O}= \mathbb Z + \mathbb Z i + \mathbb Z j + \mathbb Z \left( \frac{1+i+j+k}{2}\right) \end{aligned}$$

is indeed an order in .

2.:

Let , and let $\mathcal {O}\subseteq B$ be the Hurwitz order. For the normalizer

$$\begin{aligned} N_{B^\times }(\mathcal {O}) :=\{\alpha \in B^\times : \alpha ^{-1} \mathcal {O}\alpha = \mathcal {O}\} \end{aligned}$$

show the equality $N_{B^\times }(\mathbb Z \langle i,j \rangle ) = N_{B^\times }(\mathcal {O})$. [Hint: consider units and their traces.]

3.:

(a):

Show that the Lipschitz order $\mathbb Z \langle i,j \rangle$ is the unique suborder of the Hurwitz order $\mathcal {O}$ with index 2 (as abelian groups).

(b):

Show that

$$\begin{aligned} \mathbb Z \langle i,j \rangle = \{\alpha \in \mathcal {O}: {{\,\mathrm{trd}\,}}(\alpha ) \text { is even}\}. \end{aligned}$$

Check that the map

$$\begin{aligned} \mathcal {O}/3\mathcal {O}&\rightarrow {{\,\mathrm{M}\,}}_2(\mathbb F _3) \\ i, j&\mapsto \begin{pmatrix} 0 &{} -1\\ 1 &{} 0 \end{pmatrix}, \begin{pmatrix} 1 &{} 1 \\ 1 &{} -1 \end{pmatrix} \end{aligned}$$

from Lemma 11.2.1 is an $\mathbb F _3$-algebra isomorphism.

4.:

Generalizing the previous exercise, show that for an odd prime p that $\mathcal {O}/p\mathcal {O}\simeq {{\,\mathrm{M}\,}}_2(\mathbb F _p)$.

5.:

Draw the subgroup lattice for ${{\,\mathrm{SL}\,}}_2(\mathbb F _3)$, indicating normal subgroups (and their quotients).

$\triangleright$ 6.:

Show explicitly that

$$\begin{aligned} 2T \simeq \langle r,s,t \mid r^2 = s^3 = t^3 = rst = -1 \rangle \end{aligned}$$

(cf. (11.2.8)).

$\triangleright$ 7.:

Let

$$\begin{aligned} \Lambda =\mathbb Z ^4 + \mathbb Z (\tfrac{1}{2},\tfrac{1}{2},\tfrac{1}{2},\tfrac{1}{2}) \subset \mathbb R ^4 \end{aligned}$$

be the image of the Hurwitz order $\mathcal {O}$ under the natural embedding $\mathcal {O}\hookrightarrow \mathbb H \simeq \mathbb R ^4$. Show that for every $x \in \mathbb R ^4$, there exists $\lambda \in \Lambda$ such that $\Vert \lambda \Vert ^2 \le 1/2$. [Hint: without loss of generality we may take $0\le x_i \le 1/2$ for all i; then show we may take $x_1+x_2+x_3+x_4 \le 1$; conclude that the maximum value of $\Vert x\Vert ^2$ with these conditions occurs at the point $(\frac{1}{2},\frac{1}{2},0,0)$.]

8.:

Let B be a definite quaternion algebra over $\mathbb Q$ and let $\mathcal {O}\subseteq B$ be an order. Show that $\mathcal {O}$ is left Euclidean if and only if $\mathcal {O}$ is right Euclidean (with respect to a norm N).

9.:

Let $\mathcal {O}\subset B:=({-1,-1} \mid \mathbb{Q })$ be the Hurwitz order.

(a):

Consider the natural ring homomorphism $\mathcal {O}\rightarrow \mathcal {O}/2\mathcal {O}= \mathcal {O}\otimes _\mathbb{Z } \mathbb F _2$ giving the reduction of the algebra $\mathcal {O}$ modulo 2. Show that $\mathcal {O}/2\mathcal {O}$ is an $\mathbb F _2$-algebra, that $\#(\mathcal {O}/2\mathcal {O})=16$, and that $(\mathcal {O}/2\mathcal {O})^\times \simeq A_4$ is isomorphic to the alternating group on 4 elements. Conclude that $\mathcal {O}/2\mathcal {O}\not \simeq {{\,\mathrm{M}\,}}_2(\mathbb F _2)$ and hence that $\mathcal {O}/2\mathcal {O}$ is not a quaternion algebra over $\mathbb F _2$.

(b):

Show that the group of ring automorphisms of $\mathcal {O}/2\mathcal {O}$ is

$$\begin{aligned} {{\,\mathrm{Aut}\,}}(\mathcal {O}/2\mathcal {O}) \simeq S_4. \end{aligned}$$

(c):

More generally, if F is a field of characteristic 2 show that there is an exact sequence

$$\begin{aligned} 1 \rightarrow F^2 \rightarrow {{\,\mathrm{Aut}\,}}(\mathcal {O}\otimes _\mathbb{Z } F) \rightarrow {{\,\mathrm{SL}\,}}_2(F) \rightarrow 1 \end{aligned}$$

where $F^2$ is an additive group. [Hint: let $J={{\,\mathrm{rad}\,}}(\mathcal {O}\otimes _\mathbb{Z } F)$ be the Jacobson radical of the algebra, and show that the sequence is induced by F-linear automorphisms of J and the automorphisms $\omega \mapsto \omega +\epsilon$ with $\epsilon \in J$.] [This kind of construction, considered instead over the octonions, arises when constructing the exceptional group $G_2$ in characteristic 2 [Wils2009, §4.4.1].]

10.:

Although the Lipschitz order just misses being Euclidean with respect to the norm (see Example 11.3.8), bootstrapping from the Hurwitz order we still obtain a result on principality by restricting the set of ideals, as follows. Let $I \subseteq \mathbb Z \langle i,j \rangle$ be a right ideal.

(a):

Show that $I\mathcal {O}=\beta \mathcal {O}$ for some $\beta \in I\mathcal {O}\cap \mathbb Z \langle i,j \rangle$.

(b):

Prove that $I_{(2)}=\mathbb Z _{(2)}\langle i,j \rangle$ if and only if I is generated by elements of odd reduced norm.

(c):

If $I_{(2)}=\mathbb Z _{(2)}\langle i,j \rangle$, show that $I\mathcal {O}\cap \mathbb Z \langle i,j \rangle = I$ and conclude that I is right principal. [Hint: Argue locally.]

11.:

Let $B :=({-1,-3} \mid \mathbb{Q })$, and let

$$\begin{aligned} \mathcal {O}:=\mathbb Z \langle i,(1+j)/2 \rangle = \mathbb Z + \mathbb Z i + \mathbb Z \frac{1+j}{2} + \mathbb Z i \frac{1+j}{2}. \end{aligned}$$

(a):

Show that $B \not \simeq ({-1,-1} \mid \mathbb{Q })$.

(b):

Show that $\mathcal {O}$ is a maximal order in B.

(c):

Show that $\mathcal {O}$ is Euclidean with respect to the reduced norm

$$\begin{aligned} {{\,\mathrm{nrd}\,}}(t + xi + y(1+j)/2 + zi(1+j)/2) = t^2+ty+x^2+xz+y^2+z^2. \end{aligned}$$

(d):

Show that every maximal order in B is conjugate to $\mathcal {O}$.

12.:

Let $G \le {{\,\mathrm{O}\,}}(2)$ be a finite subgroup such that ${{\,\mathrm{tr}\,}}(g),\det (g) \in \mathbb Q$ for all $g \in G$. Show that G is conjugate in ${{\,\mathrm{O}\,}}(2)$ to one of the following: (i) a cyclic group of order 1, 2, 3, 4, 6 that is a subgroup of ${{\,\mathrm{SO}\,}}(2)$, or (ii) a dihedral group of order 2, 4, 6, 8, 12, not contained in ${{\,\mathrm{SO}\,}}(2)$.

13.:

Let p be an odd prime.

(a):

The group ${{\,\mathrm{GL}\,}}_2(\mathbb Z _{(p)})$ acts by right multiplication on the set of matrices $\pi \in {{\,\mathrm{M}\,}}_2(\mathbb Z _{(p)})$ with $p \parallel \det (\pi )$ (i.e., p exactly divides the numerator of $\det (\pi )$, written in lowest terms). Show that there are precisely $p+1$ orbits, represented by

$$\begin{aligned} \pi = \begin{pmatrix} p &{} 0 \\ 0 &{} 1 \end{pmatrix} \end{aligned}$$

and

$$\begin{aligned} \pi = \begin{pmatrix} 1 &{} 0 \\ a &{} p \end{pmatrix}, \quad a = 0,1,\dots ,p-1. \end{aligned}$$

[Hint: use column operations.]

(b):

Repeat (a) but with ${{\,\mathrm{SL}\,}}_2(\mathbb Z _{(p)})$ acting on the set of matrices $\pi \in {{\,\mathrm{M}\,}}_2(\mathbb Z _{(p)})$ with $\det (\pi )=p$, with the same conclusion.

(c):

Show that the number of (left or) right ideals of $\mathcal {O}$ of reduced norm p is equal to $p+1$.

(d):

Accounting for units, conclude that the number of ways of writing an odd prime p as the sum of four squares is equal to $8(p+1)$.

14.:

In the following exercise, we consider a computational problem, suitable for those with some background in number theory algorithms (see e.g. Cohen [Coh93]).

(a):

Show that one can find $x,y,z \in \mathbb Z$ such that $x^2+y^2+z^2=pm$ with $p \not \mid m$ in probabilistic polynomial time in $\log p$.

(b):

Describe the right Euclidean algorithm as applied to $\alpha =xi+yj+zk$ and p to obtain $\pi \in \mathcal {O}$ with ${{\,\mathrm{nrd}\,}}(\pi )=p$. Adjust as in Lemma 11.2.9 to find a solution to $x^2+y^2+z^2=p$ with $x,y,z \in \mathbb Z$. Estimate the running time of this algorithm.