Generalized Quaternion Rings over $\mathbb{Z}/n\mathbb{Z}$ for an odd $n$

TL;DR

This paper explores a broad class of quaternion rings over the integers modulo an odd number, providing a classification and counting the distinct isomorphism types for these generalized structures.

Contribution

It introduces a generalization of quaternion rings over rac{a,b}{R} that includes non-unit elements and computes the number of non-isomorphic rings over rac{a,b}{\u00bb} for odd n.

Findings

Derived formulas for counting non-isomorphic quaternion rings

Extended the theory to include non-unit elements in quaternion rings

Provided explicit classifications for odd modulus cases

Abstract

We consider a generalization of the quaternion ring $\Big(\frac{a,b}{R}\Big)$ over a commutative unital ring $R$ that includes the case when $a$ and $b$ are not units of $R$. In this paper, we focus on the case $R=\mathbb{Z}/n\mathbb{Z}$ for and odd $n$. In particular, for every odd integer $n$ we compute the number of non-isomorphic generalized quaternion rings $\Big(\frac{a,b}{\mathbb{Z}/n\mathbb{Z}}\Big)$

Equations

$$\Big{(}\frac{a,b}{\mathbb{Z}/n\mathbb{Z}}\Big{)}\cong\mathbb{M}_{2}(\mathbb{Z}/n\mathbb{Z}).$$

$$\Big{(}\frac{a,b}{\mathbb{Z}/n\mathbb{Z}}\Big{)}\cong\begin{cases}\mathbb{M}_{2}(\mathbb{Z}/m\mathbb{Z})\times(\frac{-1,-1}{\mathbb{Z}/2^{s}\mathbb{Z}}),&\textrm{if $s=1$ or $a\equiv b\equiv-1\pmod{4}$};\\ \mathbb{M}_{2}(\mathbb{Z}/m\mathbb{Z})\times(\frac{1,1}{\mathbb{Z}/2^{s}\mathbb{Z}}),&\textrm{otherwise}.\end{cases}$$

$$H(R)=\{x_0+x_{1}i+x_{2}j+x_{3}k:x_0,x_1,x_2,x_3\in R\}.$$

$$i^2$$

$$j^2$$

$$ij$$

$$I^2$$

$$J^2$$

$$IJ$$

$$\overline{f(q_0)}=A\overline{f(i)}+B\overline{f(j)}+C\overline{f(k)}=-Af(i)-Bf(j)-Cf(k)=-f(q_0),$$

$$\left(\begin{array}{cccc}1& 0& 0& 0\\ 0& 1& {\alpha }_1& {\alpha }_2\\ 0& 0& {\beta }_1& {\beta }_2\\ 0& 0& {\gamma }_1& {\gamma }_2\end{array}\right),$$

$$\Big{(}\frac{a,b}{R}\Big{)}\cong\Big{(}\frac{b,a}{R}\Big{)}.$$

$$\Big{(}\frac{a,b}{R}\Big{)}\cong\Big{(}\frac{a,-ab}{R}\Big{)}\cong\Big{(}\frac{b,-ab}{R}\Big{)}$$

$$\Big{(}\frac{a,bp^{s}}{\mathbb{Z}/p^{k}\mathbb{Z}}\Big{)}\cong\Big{(}\frac{a,-abp^{s}}{\mathbb{Z}/p^{k}\mathbb{Z}}\Big{)}.$$

$$\left(\begin{array}{cccc}1& 0& 0& 0\\ 0& 1& 0& 0\\ 0& 0& 0& a\\ 0& 0& 1& 0\end{array}\right)$$

$$\Big{(}\frac{a_{1},a_{2}}{\mathbb{Z}/p^{k}\mathbb{Z}}\Big{)}\cong\Big{(}\frac{a_{3},a_{4}}{\mathbb{Z}/p^{k}\mathbb{Z}}\Big{)}$$

$$\Big{(}\frac{a^{\prime}_{1},a^{\prime}_{2}}{\mathbb{Z}/p^{s}\mathbb{Z}}\Big{)}\cong\Big{(}\frac{a^{\prime}_{3},a^{\prime}_{4}}{\mathbb{Z}/p^{s}\mathbb{Z}}\Big{)}$$

$$R_{1}=\Big{(}\frac{ap^{s},bp^{r}}{\mathbb{Z}/p^{k}\mathbb{Z}}\Big{)},\ R_{2}=\Big{(}\frac{cp^{s},0}{\mathbb{Z}/p^{k}\mathbb{Z}}\Big{)},\ R_{3}=\Big{(}\frac{0,0}{\mathbb{Z}/p^{k}\mathbb{Z}}\Big{)}$$

$$\Big{(}\frac{ap^{s_{1}},bp^{s_{2}}}{\mathbb{Z}/p^{k}\mathbb{Z}}\Big{)}\ncong\Big{(}\frac{cp^{s_{3}},dp^{s_{4}}}{\mathbb{Z}/p^{k}\mathbb{Z}}\Big{)}$$

$$\Big{(}\frac{ap^{s_{1}},bp^{s_{1}}}{\mathbb{Z}/p^{s_{4}}\mathbb{Z}}\Big{)}\cong\Big{(}\frac{cp^{s_{1}},0}{\mathbb{Z}/p^{s_{4}}\mathbb{Z}}\Big{)},$$

$$\Big{(}\frac{ap^{s_{1}},bp^{s_{2}}}{\mathbb{Z}/p^{s_{4}}\mathbb{Z}}\Big{)}\cong\Big{(}\frac{cp^{s_{1}},0}{\mathbb{Z}/p^{s_{4}}\mathbb{Z}}\Big{)},$$

$$\Big{(}\frac{ap^{s_{1}},bp^{s_{1}}}{\mathbb{Z}/p^{s_{3}}\mathbb{Z}}\Big{)}\cong\Big{(}\frac{0,0}{\mathbb{Z}/p^{s_{3}}\mathbb{Z}}\Big{)},$$

$$\Big{(}\frac{ap^{s_{1}},0}{\mathbb{Z}/p^{s_{2}}\mathbb{Z}}\Big{)}\cong\Big{(}\frac{0,0}{\mathbb{Z}/p^{s_{2}}\mathbb{Z}}\Big{)},$$

$$\Big{(}\frac{ap^{s_{1}},0}{\mathbb{Z}_{p^{s_{3}}}}\Big{)}\cong\Big{(}\frac{0,0}{\mathbb{Z}/p^{s_{3}}\mathbb{Z}}\Big{)},$$

$$R=\Big{(}\frac{tp^{r},m}{\mathbb{Z}/p^{k}\mathbb{Z}}\Big{)}\cong\Big{(}\frac{sp^{r},m}{\mathbb{Z}/p^{k}\mathbb{Z}}\Big{)}=S.$$

$$A=\left(\begin{array}{cccc}1& 0& 0& 0\\ 0& x& 0& 0\\ 0& 0& 1& 0\\ 0& 0& 0& x\end{array}\right)$$

$$R=\Big{(}\frac{p^{r},p^{r}}{\mathbb{Z}/p^{k}\mathbb{Z}}\Big{)}\cong\Big{(}\frac{sp^{r},sp^{r}}{\mathbb{Z}/p^{k}\mathbb{Z}}\Big{)}=S$$

$$A=\left(\begin{array}{cccc}1& 0& 0& 0\\ 0& x& -y& 0\\ 0& y& x& 0\\ 0& 0& 0& s^{-1}\end{array}\right)$$

$$(xI+yJ{)}^2=(x^2+y^2)s{p}^r\equiv p^r(\mathrm{mod}{p}^k),$$

$$(-yI+xJ{)}^2=(x^2+y^2)s{p}^r\equiv p^r(\mathrm{mod}{p}^k),$$

$$(xI+yJ)(-yI+xJ)\equiv (x^2+y^2)K\equiv s^{-1}K(\mathrm{mod}{p}^k)$$

$$\Big{(}\frac{1,ap^{s}}{\mathbb{Z}/p^{k}\mathbb{Z}}\Big{)}\cong\Big{(}\frac{1,p^{s}}{\mathbb{Z}/p^{k}\mathbb{Z}}\Big{)}\ \textrm{and}\ \Big{(}\frac{u,p^{s}}{\mathbb{Z}/p^{k}\mathbb{Z}}\Big{)}\cong\Big{(}\frac{u,bp^{s}}{\mathbb{Z}/p^{k}\mathbb{Z}}\Big{)}$$

Full text

AMS 2010 Mathematics Subject Classification 11R52,16-99

1. Introduction

The origin of quaternions dates back to 1843, when William Rowan Hamilton considered a $4$-dimensional vector space over $\mathbb{R}$ with basis $\{1,i,j,k\}$ and defined an associative product given by the now classical rules $i^{2}=j^{2}=-1$ and $ij=-ji=k$. These so-called “Hamilton quaternions” turned out to be the only division algebra over $\mathbb{R}$ with dimension greater than 2. Later on, this idea was extended to define quaternion algebras over arbitrary fields. Thus, if $F$ is a field and $a,b\in F\setminus\{0\}$ we can define a unital, associative, $4$-dimensional algebra over $F$ just considering a basis $\{1,i,j,k\}$ and the product given by $i^{2}=a$, $j^{2}=b$ and $ij=-ji=k$. The structure of quaternion algebras over fields of characteristic different from two is well-known. Indeed, such a quaternion algebra is either a division ring or isomorphic to the matrix ring $\mathbb{M}_{2}(F)$ [11, p.19]. This is no longer true if $F$ is of characteristic 2, since quaternions over $\mathbb{Z}/2\mathbb{Z}$ are not a division ring but they form a commutative ring, while $\mathbb{M}_{2}(\mathbb{Z}/2\mathbb{Z})$ is not commutative. Nevertheless, some authors consider a different product in the characteristic 2 case given by $i^{2}+i=a$, $j^{2}=b$, and $ji=(i+1)j=k$. The algebra defined by this product is isomorphic to the corresponding matrix ring.

Generalizations of the notion of quaternion algebra to other commutative base rings $R$ have been considered by Kanzaki [5], Hahn [4], Knus [6], Gross and Lucianovic [3], Tuganbaev [15], and most recently by John Voight [16, 17]. Quaternions over finite rings have attracted significant attention since they have applications in coding theory see, [9, 10, 14]. In [2] the case $R=\mathbb{Z}/n\mathbb{Z}$ was studied proving the following result.

Theorem 1** ([2], Theorem 4).**

Let $n$ be an integer and let $a,b$ be such that $\gcd(a,n)=\gcd(b,n)=1$. The following hold:

• i)

If $n$ is odd, then

[TABLE]

• ii)

If $n=2^{s}m$ with $s>0$ and $m$ odd, then

[TABLE]

In this paper, we extend the concept of quaternion rings over commutative, associative, unital rings to the case when $i^{2}$ and $j^{2}$ are not necessarily units of the ring $R$. In particular, we will focus on the case $R=\mathbb{Z}/n\mathbb{Z}$ for an odd $n$.

2. Basic concepts

Let $R$ be a commutative and associative ring with identity and let $H(R)$ denote the free $R$-module of rank $4$ with basis $\{1,i,j,k\}$. That is,

[TABLE]

Now, let $a,b\in R$ and define an associative multiplication in $H(R)$ according to the following rules:

[TABLE]

Thus, we obtain an associative, unital ring called a quaternion ring over $R$ which is denoted by \Big{(}\frac{a,b}{R}\Big{)}.

Definition 1**.**

A standard basis of \Big{(}\frac{a,b}{R}\Big{)} is any base $\mathcal{B}=\{1,I,J,K\}$ of the free $R$-module $H(R)$ such that

[TABLE]

Given the standard basis $\{1,i,j,k\}$, the elements of the submodule $R\langle i,j,k\rangle$ are called pure quaternions. Note that the square of a pure quaternion always lays on $R$.

Remark 1**.**

Given q\in\Big{(}\frac{a,b}{R}\Big{)} and a fixed standard basis, there exist $x_{0}\in R$ and a pure quaternion $q_{0}$ such that $q=x_{0}+q_{0}$. Observe that both $x_{0}$ and $q_{0}$ are uniquely determined and also that the only pure quaternion in $R$ is 0.

The following classical concepts are not altered by the fact that $a$ and $b$ are not necessarily units.

Definition 2**.**

Consider the standard basis $\{1,i,j,k\}$ and let q\in\Big{(}\frac{a,b}{R}\Big{)}. Put $q=x_{0}+q_{0}$ with $x_{0}\in R$ and $q_{0}=x_{1}i+x_{2}j+x_{3}k$ a pure quaternion. Then,

• i)

The conjugate of $q$ is: $\overline{q}=x_{0}-q_{0}=x_{0}+x_{1}i-x_{2}j-x_{3}k$.

• ii)

The trace of $q$ is $\textrm{tr}(q)=q+\overline{q}=2x_{0}.$

• iii)

The norm of $q$ is $\textrm{n}(q)=q\overline{q}=x_{0}^{2}-q_{0}^{2}=x_{0}^{2}-ax_{1}^{2}-bx_{2}^{2}+abx_{3}^{2}$.

Note that $\textrm{n}(q),\textrm{tr}(q)\in R$ and $\textrm{n}(q_{1}q_{2})=\textrm{n}(q_{1})\textrm{n}(q_{2})$.

Remark 2**.**

Observe that, if $q$ is a pure quaternion, then $\overline{q}=-q$ and $\textrm{tr}(q)=0$. The converse also holds only if $R$ has odd characteristic.

In the following result we will see that isomorphisms preserve conjugation. The classical proof in the case when $a$ and $b$ are units (see [1, Theorem 5.6] for instance) is no longer valid in our setting and it must be slightly modified.

Theorem 2**.**

Let f:\Big{(}\frac{a,b}{R}\Big{)}\to\Big{(}\frac{c,d}{R}\Big{)} be a ring isomorphism. Then, for every q\in\Big{(}\frac{a,b}{R}\Big{)} it holds that $f(\overline{q})=\overline{f(q)}$.

Proof.

Let q\in\Big{(}\frac{a,b}{R}\Big{)} and put $q=x_{0}+q_{0}$ with $x_{0}\in R$ and $q_{0}$ a pure quaternion. Then, $\overline{q}=x_{0}-q_{0}$ and $f(\overline{q})=f(x_{0})-f(q_{0})=x_{0}-f(q_{0})$. On the other hand, $\overline{f(q)}=\overline{f(x_{0}+q_{0})}=\overline{f(x_{0})+f(q_{0})}=\overline{x_{0}+f(q_{0})}=x_{0}+\overline{f(q_{0})}$. Hence, in order to prove the result, it is enough to prove that $\overline{f(q_{0})}=-f(q_{0})$ for every pure quaternion $q_{0}$.

Let us consider the standard basis $\{1,i,j,k\}$ of \Big{(}\frac{a,b}{R}\Big{)}. Then, $f(i)=\alpha_{1}+q_{1}$ with $\alpha_{1}\in R$ and $q_{1}$ a pure quaternion in \Big{(}\frac{c,d}{R}\Big{)}. Now, since $i^{2}\in R$ and taking into account that $f$ fixes $R$, we have that $f(i^{2})=f(i)^{2}=(\alpha_{1}+q_{1})^{2}=\alpha_{1}^{2}+q_{1}^{2}+2\alpha_{1}q_{1}\in R$. Consequently, $2\alpha_{1}q_{1}\in R$ (because both $\alpha_{1}^{2}$ and $q_{1}^{2}$ are in $R$) and since $2\alpha_{1}q_{1}$ is a pure quaternion, it must be $2\alpha_{1}q_{1}=0$. Thus, $f(2\alpha_{1}i)=2\alpha_{1}f(i)=2\alpha_{i}^{2}$ and, since $f$ fixes $R$, it follows that $2\alpha_{1}i=0$ and also that $2\alpha_{1}=0$. Equivalently, $\alpha_{1}=-\alpha_{1}$ and then, $\overline{f(i)}=\alpha_{1}-q_{1}=-\alpha_{1}-q_{1}=-f(i)$.

In the same way, it can be seen that $\overline{f(j)}=-f(j)$ and $\overline{f(k)}=-f(k)$. Thus, if $q_{0}=Ai+Bj+Ck$ is a pure quaternion in \Big{(}\frac{a,b}{R}\Big{)} we have that:

[TABLE]

and the result follows. ∎

Since both the trace and the norm are defined in terms of the conjugation, the following result easily follows from Theorem 2.

Corollary 1**.**

Let f:\Big{(}\frac{a,b}{R}\Big{)}\to\Big{(}\frac{c,d}{R}\Big{)} be a ring isomorphism. Then, for every q\in\Big{(}\frac{a,b}{R}\Big{)} the following hold.

• i)

{\rm tr}\Big{(}f(q)\Big{)}={\rm tr}(q).

• ii)

{\rm n}\Big{(}f(q)\Big{)}={\rm n}(q).

Remark 3**.**

Theorem 2 and Corollary 1 imply in particular that the conjugate, the trace and the norm of an element are independent from the standard basis of \Big{(}\frac{a,b}{R}\Big{)} used to compute them. Moreover, according to Remark 2, Theorem 2 implies that (in the odd characteristic case) every isomorphism preserves pure quaternions.

Proposition 1**.**

Let $R$ be a ring with odd characteristic and Let f:\Big{(}\frac{a,b}{R}\Big{)}\to\Big{(}\frac{a,c}{R}\Big{)} be a ring isomorphism. Then, for some pair of standard bases the matrix of $f$ has the form

[TABLE]

with $\alpha_{1}a=\alpha_{2}a=0$.

Proof.

Let $\{1,i,j,k\}$ be any standard basis in \Big{(}\frac{a,b}{R}\Big{)}. Since $f(i)^{2}=f(i^{2})=a$, let us consider $S$ the subring of \Big{(}\frac{a,c}{R}\Big{)} generated by $\{1,f(i)\}$. Now, we apply the Cayley-Dickson process [13] to $S$ and $c$ so that we extend $\{1,f(i)\}$ to an standard basis $\{1,I:=f(i),J,K\}$ of \Big{(}\frac{a,c}{R}\Big{)}.

Now, since $R$ has odd characteristic, $f$ preserves pure quaternions. Thus, $f(j)=\alpha_{1}I+\beta_{1}J+\gamma_{1}K$ and $f(k)=\alpha_{2}I+\beta_{2}J+\gamma_{2}K$.

Finally, $f(k)=f(ij)=f(i)f(j)=I(\alpha_{1}I+\beta_{1}J+\gamma_{1}K)=\alpha_{1}a+\beta_{1}K+\gamma_{1}aJ$ must be a pure quaternion and hence $\alpha_{1}a=0$. In the same way it can be seen that $\alpha_{2}a=$ and the result follows. ∎

In what follows, we will be interested in determining whether two different quaternion rings are isomorphic or not. The following isomorphism, which is well-known if $a$ and $b$ are units, also holds in our setting. The proof is straightforward.

Lemma 1**.**

Let $a,b\in R$. Then,

[TABLE]

Nevertheless, some other easy isomorphisms that hold in the case when $a$ and $b$ are units, like

[TABLE]

are, as we will see, no longer generally true in our setting.

3. Some results regarding \Big{(}\frac{a,b}{\mathbb{Z}/p^{k}\mathbb{Z}}\Big{)} for a prime $p$

Throughout this section $p$ will denote any prime. The next two results present some isomorphisms that will be useful in forthcoming sections. The first one (Lemma 2) is, in some sense, an analogue to the classical isomorphism (1). The second one (Lemma 3) presents some kind of descent principle.

Lemma 2**.**

Let $a$ and $b$ be integers with $\gcd(a,p)=1$. Then,

[TABLE]

Proof.

Let us consider standard bases $\{1,i,j,k\}$ and $\{1,I,J,K\}$ of \Big{(}\frac{a,bp^{s}}{\mathbb{Z}_{p^{k}}}\Big{)} and \Big{(}\frac{a,-abp^{s}}{\mathbb{Z}_{p^{k}}}\Big{)}, respectively. Then it is enough to consider the ring homomorphism $f$ induced by $f(1)=1$, $f(I)=i$, $f(J)=k$ and $f(K)=aj$; which is bijective because its coordinate matrix

[TABLE]

is regular over $\mathbb{Z}/p^{k}\mathbb{Z}$. ∎

Lemma 3**.**

Let $a_{i}$ ($1\leq i\leq 4$) and $k\geq 1$ be integers such that

[TABLE]

and let $s\leq k$. If $a_{i}\equiv a^{\prime}_{i}\pmod{p^{s}}$ for every $1\leq i\leq 4$, then

[TABLE]

Proof.

Let $f$ be an isomorphism between \Big{(}\frac{a_{1},a_{2}}{\mathbb{Z}/p^{k}\mathbb{Z}}\Big{)} and \Big{(}\frac{a_{3},a_{4}}{\mathbb{Z}/p^{k}\mathbb{Z}}\Big{)}. If $A$ is the coordinate matrix of $f$ with respect to some standard bases, it is obvious that $A$ is regular over $\mathbb{Z}/p^{k}\mathbb{Z}$ and, consequently, also over $\mathbb{Z}/p^{s}\mathbb{Z}$.

Then, the linear map $g$ between \Big{(}\frac{a^{\prime}_{1},a^{\prime}_{2}}{\mathbb{Z}/p^{s}\mathbb{Z}}\Big{)} and \Big{(}\frac{a^{\prime}_{3},a^{\prime}_{4}}{\mathbb{Z}/p^{s}\mathbb{Z}}\Big{)} defined by the matrix $A$ with respect to some standard bases, induces an ring isomorphism because $a_{i}\equiv a^{\prime}_{i}\pmod{p^{s}}$ for every $i$. ∎

It is also interesting, and often harder, to determine whether two quaternion rings are not isomorphic. The following results go in this direction.

Lemma 4**.**

Let $p$ be a prime and consider integers $a$, $b$ and $c$ coprime to $p$. Also, let $0\leq s\leq r<k$. Then, the quaternion rings $R_{1}$, $R_{2}$ and $R_{3}$ defined by

[TABLE]

are pairwise non-isomorphic.

Proof.

For each $i\in\{1,2,3\}$ let us define the set $\mathbb{P}_{i}:=\{q\in R_{i}:{\rm tr}(q)=0\}$.

Note that, if $p$ is odd, then $\mathbb{P}_{i}$ is precisely the set of pure quaternions and is hence preserved by isomorphisms. Now, for every element $q\in\mathbb{P}_{3}$ it holds that $q^{2}=0$, while $\mathbb{P}_{1}$ and $\mathbb{P}_{2}$ contain elements whose square is non-zero. This implies that $R_{3}$ is not isomorphic to $R_{1}$ or $R_{2}$. On the other hand, the set $\mathbb{P}_{2}\setminus p\mathbb{P}_{2}$ contains elements with zero square while this is not the case for $\mathbb{P}_{1}\setminus p\mathbb{P}_{1}$. This implies that $R_{1}$ and $R_{2}$ are not isomorphic.

Finally, if $p=2$, then $\mathbb{P}_{i}=\{\alpha 2^{k-1}+q_{0}:\alpha\textrm{ is odd and }q_{0}\textrm{ is a pure quaternion}\}$ and we can reason in the same way. ∎

Lemma 5**.**

Let $p$ be a prime and consider integers $a$, $b$, $c$ and $d$ coprime to $p$. Also, let $s_{1}\leq s_{2}\leq k$ and $s_{3}\leq s_{4}\leq k$ and assume that either $s_{1}\neq s_{3}$ or $s_{2}\neq s_{4}$. Then

[TABLE]

Proof.

Let us assume that both rings are isomorphic. Without loss of generality, we can also assume that $s_{1}\leq s_{3}$. Five different situations arise:

• i)

If $s_{1}=s_{3}=s_{2}<s_{4}$, then Lemma 3 implies that

[TABLE]

which contradicts Lemma 4.

• ii)

If $s_{1}=s_{3}<s_{2}<s_{4}$, then due to Lemma 3 we have that

[TABLE]

which contradicts Lemma 4.

• iii)

If $s_{1}=s_{2}<s_{3}$, by Lemma 3 we have that

[TABLE]

which contradicts Lemma 4 again.

• iv)

If $s_{1}<s_{2}\leq s_{3}$, Lemma 3 implies that

[TABLE]

contradicting Lemma 4.

• v)

If $s_{1}<s_{3}\leq s_{2}$, Lemma 3 leads to

[TABLE]

which is a contradiction due to Lemma 4.

Hence, in any case we reach a contradiction and the result follows. ∎

4. Quaternions over $\mathbb{Z}/p^{k}\mathbb{Z}$ for an odd prime $p$

This section is devoted to determine the number of non-isomorphic generalized quaternion rings over $\mathbb{Z}/p^{k}\mathbb{Z}$ for an odd prime $p$. Hence, throughout this section $p$ will be assumed to be an odd prime.

Lemma 6**.**

Let $s$ and $t$ be integers coprime to $p$ such that $st$ is a quadratic residue modulo $p$ and let $m$ be any integer. Then, for every $r\geq 0$,

[TABLE]

Proof.

Since $\gcd(st,p)=1$, it follows that $st$ is also a quadratic residue modulo $p^{k}$ so let $x$ be an integer such that $x^{2}\equiv ts^{-1}\pmod{p^{k}}$. Let us consider $\{1,i,j,k\}$ and $\{1,I,J,K\}$ standard bases of $R$ and $S$, respectively. Then, the linear map whose matrix with respect to these bases is

[TABLE]

induces an isomorphism because $(xI)^{2}=x^{2}I^{2}\equiv ts^{-1}sp^{r}\equiv tp^{r}\pmod{p^{k}}$ and $A$ is regular over $\mathbb{Z}/p^{k}\mathbb{Z}$. ∎

Lemma 7**.**

Let $s$ be an integer such that $\gcd(p,s)=1$. Then, for every $r\geq 0$,

[TABLE]

Proof.

Let $x,y\in\mathbb{Z}/p^{k}\mathbb{Z}^{*}$ such that $x^{2}+y^{2}\equiv s^{-1}\pmod{p^{k}}$ (the exist due to [2, Proposition 1]). Now let us consider $\{1,i,j,k\}$ and $\{1,I,J,K\}$ standard bases of $R$ and $S$, respectively. Then, the linear map whose matrix with respect to these bases is

[TABLE]

induces an isomorphism because

[TABLE]

[TABLE]

[TABLE]

and $A$ is regular over $\mathbb{Z}/p^{k}\mathbb{Z}$. ∎

Lemma 8**.**

Let $u$ be a quadratic nonresidue modulo $p$ with $p\nmid u$ and consider integers $a$ and $b$ coprime to $p$ and let $0\leq s$. Then,

• i)

[TABLE]

• ii)

The isomorphism

[TABLE]

holds if and only if $s=0$.

Proof.

• i)

To see that \displaystyle R=\Big{(}\frac{1,p^{s}}{\mathbb{Z}/p^{k}\mathbb{Z}}\Big{)}\cong\Big{(}\frac{1,ap^{s}}{\mathbb{Z}/p^{k}\mathbb{Z}}\Big{)}=S, let us consider $\{1,i,j,k\}$ and $\{1,I,J,K\}$ standard bases of $R$ and $S$, respectively. It is obvious that there exist $x,y\in\mathbb{Z}/p^{k}\mathbb{Z}$ with $\gcd(p,y)=1$ such that $x^{y}-y^{2}\equiv a^{-1}\pmod{p^{k}}$. Then, the linear map whose matrix with respect to these bases is

[TABLE]

induces an isomorphism because

[TABLE]

and $A$ is regular over $\mathbb{Z}/p^{k}\mathbb{Z}$.

The remaining isomorphism can be proved in a similar way.

• ii)

Assume that $s>0$. To see that \displaystyle\Big{(}\frac{1,p^{s}}{\mathbb{Z}/p^{k}\mathbb{Z}}\Big{)}\not\cong\Big{(}\frac{u,p^{s}}{\mathbb{Z}/p^{k}\mathbb{Z}}\Big{)} it is enough to observe that \displaystyle\Big{(}\frac{1,p^{s}}{\mathbb{Z}/p^{k}\mathbb{Z}}\Big{)} does not contain any pure quaternion $q$ with $q^{2}=u$. In fact, if $\{1,i,j,k\}$ is a standard basis, $q=ai+bj+ck$ and $q^{2}=a^{2}+(b^{2}-c^{2})p^{s}$. Hence, if $q^{2}\equiv u\pmod{p^{k}}$ if follows that $u$ is a quadratic residue modulo $p$, which is a contradiction.

On the other hand, if $s=0$, we know that \Big{(}\frac{1,1}{\mathbb{Z}/p^{k}\mathbb{Z}}\Big{)}\cong\Big{(}\frac{u,1}{\mathbb{Z}/p^{k}\mathbb{Z}}\Big{)} using [2, Theorem 4].

∎

Lemma 9**.**

Let $u$ be a quadratic nonresidue modulo $p$ with $p\nmid u$ and let $0<s<k$. Then,

• i)

\displaystyle R_{1}=\Big{(}\frac{up^{s},p^{s}}{\mathbb{Z}_{p^{k}}}\Big{)}\not\cong\Big{(}\frac{p^{s},p^{s}}{\mathbb{Z}_{p^{k}}}\Big{)}=R_{2}.**

• ii)

\displaystyle S_{1}=\Big{(}\frac{up^{s},0}{\mathbb{Z}_{p^{k}}}\Big{)}\ncong\Big{(}\frac{p^{s},0}{\mathbb{Z}_{p^{k}}}\Big{)}=S_{2}.**

Proof.

• i)

Let us consider the following sets:

[TABLE]

[TABLE]

In order to prove that $R_{1}\ncong R_{2}$ we will see that $\textrm{card}(N_{1})\neq\textrm{card}(N_{2})$.

To do so, let $\{1,i,j,k\}$ and $\{1,I,J,K\}$ be standard bases of $R_{1}$ and $R_{2}$, respectively. Then, if $q_{1}\in N_{1}$, it must be $q_{1}=x_{1}i+x_{2}j+x_{3}k$ with $x_{1}^{2}up^{s}+x_{2}^{2}p^{s}-x_{3}^{s}up^{2s}\equiv 0\pmod{p^{k}}$ and $p\nmid x_{l}$ for some $l\in\{1,2,3\}$. On the other hand, if $q_{2}\in N_{2}$, it must be $q_{2}=y_{1}I+y_{2}J+y_{3}K$ with $y_{1}^{2}p^{s}+y_{2}^{2}p^{s}-y_{3}^{s}p^{2s}\equiv 0\pmod{p^{k}}$ and $p\nmid y_{l}$ for some $l\in\{1,2,3\}$.

Thus, $\textrm{card}(R_{1})$ is the number of non-zero solutions of the congruence

[TABLE]

while $\textrm{card}(R_{2})$ is the number of non-zero solutions of the congruence

[TABLE]

Now, reducing modulo $p$, we can see that:

• –

If $-1$ is a quadratic residue $\pmod{p}$ (i.e., if $p\equiv 1\pmod{4}$), then the congruence (3) has non-zero solutions while the congruence (2) has not.

• –

If $-1$ is a quadratic nonresidue $\pmod{p}$ (i.e., if $p\equiv 3\pmod{4}$), then the congruence (2) has non-zero solutions while the congruence (3) has not.

In any case, it follows that $\textrm{card}(N_{1})\neq\textrm{card}(N_{2})$ as claimed.

• ii)

For this case, it is enough to observe that $S_{2}$ does not contain pure quaternions $q$ such that $q^{2}=up^{s}$, while $S_{1}$ obviously does contain such type of elements. To do so, just note that the congruence $x^{2}p^{s}\equiv up^{s}\pmod{p^{k}}$ ha no solutions because $u$ is a quadratic nonresidue modulo $p$.

∎

Lemma 10**.**

Let $u$ be a quadratic nonresidue $\pmod{p}$ with $p\nmid u$ and let $0<s<r<k$. Then, the quaternion rings \displaystyle R_{1}=\Big{(}\frac{up^{s},up^{r}}{\mathbb{Z}/p^{k}\mathbb{Z}}\Big{)}, \displaystyle R_{2}=\Big{(}\frac{p^{s},up^{r}}{\mathbb{Z}/p^{k}\mathbb{Z}}\Big{)}, \displaystyle R_{3}=\Big{(}\frac{up^{s},p^{r}}{\mathbb{Z}/p^{k}\mathbb{Z}}\Big{)} and \displaystyle R_{4}=\Big{(}\frac{p^{s},p^{r}}{\mathbb{Z}/p^{k}\mathbb{Z}}\Big{)} are pairwise non-isomorphic.

Proof.

Let us see that $R_{1}\ncong R_{2}$, $R_{1}\ncong R_{4}$, $R_{2}\ncong R_{3}$ and $R_{3}\ncong R_{4}$. If they were isomorphic, the due to Lemma 3 we would have (reducing modulo $p^{r}$) that \displaystyle\Big{(}\frac{up^{s},0}{\mathbb{Z}/p^{r}\mathbb{Z}}\Big{)}\cong\Big{(}\frac{p^{s},0}{\mathbb{Z}/p^{r}\mathbb{Z}}\Big{)}, which contradicts Lemma 9.

Now, let us see that $R_{1}\ncong R_{3}$. Assume that $R_{1}\cong R_{3}$. Then, due to Proposition 1, we can consider $\{1,i,j,k\}$ and $\{1,I,J,K\}$ standard bases of $R_{1}$ and $R_{3}$, respectively such that the matrix of the isomorphism with respect to these bases is

[TABLE]

with $\alpha_{1}up^{s}=0$.

In particular, $up^{r}=j^{2}=f(j^{2})=f(j)^{2}=(\alpha_{1}I+\beta_{1}J+\gamma_{1}K)^{2}=\alpha_{1}^{2}up^{s}+\beta_{1}^{2}p^{r}-\gamma_{1}^{2}up^{r+s}=\beta_{1}^{2}p^{r}-\gamma_{1}^{2}u^{2}p^{r+s}$. In other words, $\beta_{1}^{2}p^{r}-\gamma_{1}^{2}up^{r+s}\equiv up^{r}\pmod{p^{k}}$ but this implies that $\beta_{1}^{2}-\gamma_{1}^{2}up^{s}\equiv u\pmod{p^{k-r}}$ and, consequently, that $\beta_{1}^{2}\equiv u\pmod{p}$ which is a contradiction because $u$ is a quadratic nonresidue.

The remaining case, namely $R_{2}\ncong R_{4}$ can be proved in the exact same way. ∎

Corollary 2**.**

Let $u$ be a quadratic nonresidue modulo $p$ with $p\nmid u$. Consider integers $a$ and $b$ coprime to $p$ and let $0<r$. Then,

[TABLE]

Proof.

If $a$ is a quadratic nonresidue:

[TABLE]

Now, if $a$ is a quadratic residue:

[TABLE]

Finally, \Big{(}\frac{u,p^{r}}{\mathbb{Z}/p^{k}\mathbb{Z}}\Big{)} and \Big{(}\frac{1,p^{r}}{\mathbb{Z}/p^{k}\mathbb{Z}}\Big{)} are not isomorphic due to Lemma 8. ∎

Corollary 3**.**

Let $u$ be a quadratic nonresidue modulo $p$ with $p\nmid u$. Consider integers $a$ and $b$ coprime to $p$ and let $0<r$. Then,

[TABLE]

Proof.

If $ab$ is a quadratic nonresidue, only one among $a$ and $b$ is a quadratic residue. We can assume without loss of generality that $a$ is a quadratic residue and that $b$ is a quadratic nonresidue (so $ub$ is a quadratic residue) and then:

[TABLE]

Now, if $ab$ is a quadratic residue:

[TABLE]

Finally, \Big{(}\frac{p^{r},p^{r}}{\mathbb{Z}/p^{k}\mathbb{Z}}\Big{)} and \Big{(}\frac{up^{r},p^{r}}{\mathbb{Z}/p^{k}\mathbb{Z}}\Big{)} are not isomorphic due to Lemma 9. ∎

Corollary 4**.**

Let $u$ be a quadratic nonresidue modulo $p$ with $p\nmid u$. Consider integers $a$ and $b$ coprime to $p$ and let $0<s<r$. Then,

[TABLE]

Proof.

Like in the previous results, it is enough to apply Lemma 6 repeatedly. The four different cases that arise are non-isomorphic due to Lemma 10. ∎

Now, we can prove the main result of this section.

Theorem 1**.**

Let $p$ be an odd prime and let $k$ be a positive integer. Then, there exist exactly $2k^{2}+2$ non-isomorphic generalized quaternion rings over $\mathbb{Z}/p^{k}\mathbb{Z}$.

Proof.

Taking into account the previous results, any generalized quaternion ring over $\mathbb{Z}/p^{k}\mathbb{Z}$ is isomorphic to one of the following:

[TABLE]

where $u$ is a quadratic nonresidue $\pmod{p}$ with $p\nmid u$ and $0\leq s\leq r\leq k$.

• •

If $0=s=r$, due to Lemmata 1, 7 and 8, there is only one ring to consider, namely \Big{(}\frac{1,1}{\mathbb{Z}/p^{k}\mathbb{Z}}\Big{)}.

• •

If $0=s<r<k$, we must consider the rings

[TABLE]

Due to Lemma 8 we know that \Big{(}\frac{u,up^{r}}{\mathbb{Z}/p^{k}\mathbb{Z}}\Big{)}\cong\Big{(}\frac{u,p^{r}}{\mathbb{Z}/p^{k}\mathbb{Z}}\Big{)}, \Big{(}\frac{1,up^{r}}{\mathbb{Z}/p^{k}\mathbb{Z}}\Big{)}\cong\Big{(}\frac{1,p^{r}}{\mathbb{Z}/p^{k}\mathbb{Z}}\Big{)} and \Big{(}\frac{u,p^{r}}{\mathbb{Z}/p^{k}\mathbb{Z}}\Big{)}\ncong\Big{(}\frac{1,p^{r}}{\mathbb{Z}/p^{k}\mathbb{Z}}\Big{)}. Hence, in this case we have 2 non-isomorphic generalized quaternion rings for each $1\leq r\leq k-1$. A total of $2(k-1)$.

• •

If $0=s$ and $k=r$ we must only consider the rings

[TABLE]

which are non-isomorphic due to Lemma 8. Thus, in this case we have 2 non-isomorphic generalized quaternion rings.

• •

If $0<s=r<k$, we must consider the rings

[TABLE]

Using Lemma 1, Lemma 7 and Lemma 9 we know that \Big{(}\frac{up^{r},up^{r}}{\mathbb{Z}/p^{k}\mathbb{Z}}\Big{)}\cong\Big{(}\frac{p^{r},p^{r}}{\mathbb{Z}/p^{k}\mathbb{Z}}\Big{)}\ncong\Big{(}\frac{up^{r},p^{r}}{\mathbb{Z}/p^{k}\mathbb{Z}}\Big{)}\cong\Big{(}\frac{p^{r},up^{r}}{\mathbb{Z}/p^{k}\mathbb{Z}}\Big{)}. Hence, in this case we have 2 non-isomorphic generalized quaternion rings for each $1\leq r\leq k-1$ for a total of $2(k-1)$.

• •

If $0<s<r<k$, Lemma 10 implies that the four rings are non-isomorphic. Hence, in this case we have 2 non-isomorphic generalized quaternion rings for each $1\leq s\leq k-2$ and each $s+1\leq k\leq k-1$. A total of $2(k-2)(k-1)$.

• •

If $0<s<r=k$, we must only consider the rings

[TABLE]

which are non-isomorphic due to Lemma 9. Thus, in this case we have 2 non-isomorphic generalized quaternion rings for each $1\leq s\leq k-1$. A total of $2(k-1)$.

• •

If $s=r=k$ there is only one ring to consider, namely \Big{(}\frac{0,0}{\mathbb{Z}/p^{k}\mathbb{Z}}\Big{)}.

Finally, taking into consideration all the previous information, we conclude that there exist

[TABLE]

non-isomorphic generalized quaternion rings over $\mathbb{Z}/p^{k}\mathbb{Z}$. ∎

Remark 4**.**

The sequence $a_{k}=2k^{2}+2$ is sequence A005893 in the OEIS.

5. Quaternions over $\mathbb{Z}/n\mathbb{Z}$ for an odd $n$

Note that if $n=p_{1}^{r_{1}}\ldots p_{k}^{r_{k}}$ is the prime factorization of $n$, then by the Chinese Remainder Theorem we have that

[TABLE]

Decomposition (4) induces a natural isomorphism

[TABLE]

Consequently, if we denote by $\omega(n)$ the number of different primes dividing $n$ and by $\nu_{p}(n)$ the $p$-adic order of $n$ we obtain the following corollary to Theorem 1.

Corollary 5**.**

Let $n$ be an odd integer. Then, the number of non-isomorphic generalized quaternion rings over $\mathbb{Z}/n\mathbb{Z}$ is

[TABLE]