\documentclass[10pt]{studiamnew} \usepackage{amsmath,amsthm,amsfonts} \usepackage{mathtools} \usepackage{graphicx} \usepackage{placeins} \frenchspacing \sloppy \newtheorem{theorem}{Theorem}[section] \newtheorem{lemma}[theorem]{Lemma} \newtheorem{corollary}[theorem]{Corollary} \newtheorem{proposition}[theorem]{Proposition} \newtheorem{axiom}[theorem]{Axiom} \newtheorem{case}[theorem]{Case} \newtheorem{condition}[theorem]{Condition} \newtheorem{conjecture}[theorem]{Conjecture} \newtheorem{criterion}[theorem]{Criterion} \newtheorem{definition}[theorem]{Definition} \theoremstyle{definition} \newtheorem{remark}[theorem]{Remark} \newtheorem{algorithm}[theorem]{Algorithm} \newtheorem{problem}[theorem]{Problem} \newtheorem{example}[theorem]{Example} \newtheorem{exercise}[theorem]{Exercise} \newtheorem{solution}[theorem]{Solution} \newtheorem{notation}[theorem]{Notation} \renewcommand{\theequation}{\thesection.\arabic{equation}} \numberwithin{equation}{section} \newcommand{\eqdefa}{\overset{comm.}{=}} \newcommand{\eqdefb}{\overset{Mouf.}{=}} \newcommand{\eqdefc}{\overset{altern.}{=}} \newcommand{\eqdefd}{\overset{flex.}{=}} \begin{document} % \setcounter{page}{1} \setcounter{firstpage}{1} \setcounter{lastpage}{4} \renewcommand{\currentvolume}{??} \renewcommand{\currentyear}{??} \renewcommand{\currentissue}{??} % \title{Idempotent and Nilpotent Elements in Octonion Rings over ${\mathbb Z}_{\mathbf{p}}$} \author{Michael Aristidou} \address{``Texas A\&M University at Galveston" Galveston, TX\\ USA} \email{maristidou@tamu.edu} % \author{Philip R. Brown} \address{``Texas A\&M University at Galveston" Galveston, TX\\ USA} \email{brownp@tamug.edu} % \author{George Chailos} \address{``University of Nicosia", Nicosia\\ Cyprus} \email{chailos.g@unic.ac.cy} % \subjclass{15A33, 15A30, 20H25, 15A03} \keywords{quaternion, octonion, ring, skew field, idempotent, nilpotent} \begin{abstract} \noindent In this paper, we show that the set $\mathbb{O}\slash\mathbb{Z}_{p}$, where $p$ is a prime number, does not form a skew field and discuss idempotent and nilpotent elements in the (finite) ring $\mathbb{O}\slash\mathbb{Z}_{p}$. We provide examples and establish conditions for idempotency and nilpotency. \end{abstract} \maketitle \section{Introduction} Quaternions, denoted by $\mathbb{H}$, were first discovered by William. R. Hamilton in 1843 as an extension of complex numbers into four dimensions \cite{Remmert}. Namely, a quaternion is of the form $ x=a_0+a_1 i+a_2 j+a_3 k$, where $a_i$ are reals and $i , j, k$ are such that $i^2 = j^2 = k^2 = ijk = -1$. Algebraically speaking, $\mathbb{H}$ forms a division algebra (skew field) over $\mathbb{R}$ of dimension 4 (\cite{Remmert}, p.195-196).About the same time, John T. Graves discovered the octonions, denoted by $\mathbb{O}$, which are 8-dimensional numbers of the form $ x=a_0+a_1 e_1+a_2 e_2+a_3 e_3+a_4e_4+a_5e_5+a_6e_6+a_7e_7$ where $a_i$ are reals and $e_i$'s are mutually anti-commuting roots of unity. (i.e. $e_i^2 = -1$ and $e_ie_j=e_k$, $e_je_i = -e_k$, $i \neq j$) \cite{Baez}. Algebraically speaking, $\mathbb{O}$ forms a normed division algebra (skew field) over $\mathbb{R}$ of dimension 8 \cite{Baez}. It is the largest of the (only) four$^1$ normed division algebras and it is nonassociative. A study of the structure and some of its properties of the finite ring$^2$ $\mathbb{H}\slash\mathbb{Z}_{p}$, where $p$ is a prime number, was done in \cite{Aristidou09}. A more detailed description of the structure $\mathbb{H}\slash\mathbb{Z}_{p}$ was given by Miguel and Serodio in \cite{Miguel}. Among others, they found the number of zero-divisors, the number of idempotent elements, and provided an interesting description of the zero-divisor graph. In particularly, they showed that the number of idempotent elements in $\mathbb{H}\slash\mathbb{Z}_{p}$ is $p^2 +p +2$, for $p$ odd prime. As discussed in \cite{Aristidou12}, the only scalar idempotents in $\mathbb{H}\slash\mathbb{Z}_{p}$ are $a_0 =0$, $1$. Furthermore, there are no purely imaginary idempotents in $\mathbb{H}\slash\mathbb{Z}_{p}$. On the other hand, in \cite{Aristidou13}, it was shown that nilpotents $x$ in $\mathbb{H}\slash\mathbb{Z}_{p}$ are purely imaginary with norm $N(x) = 0$ and $x^2 = 0$. In the sections that follow, we look at the structure of the finite ring $\mathbb{O}\slash\mathbb{Z}_{p}$. The multiplication of octonions followed the Fano Plane and it was programmed in Maple$^3$. We give examples of idempotent and nilpotent elements in $\mathbb{O}\slash\mathbb{Z}_{p}$ and provide conditions for idempotency and nilpotency in $\mathbb{O}\slash\mathbb{Z}_{p}$. \vspace*{-14.4mm} \section{Is $\mathbb{O}\slash\mathbb{Z}_{p}$ a finite skew field? A counterexample} In \cite{Aristidou09} we saw that since ${\mathbb Z}_{\mathbf{p}}$ is a field, then $\mathbb{H}\slash\mathbb{Z}_{p}$ is a quaternion algebra. The theory of quaternion algebras over a field $\mathbb{K}$ (char$\mathbb{K}\ne2$) tells us that a quaternion algebra $Q$ is either a division ring or $Q = \mathbb{M}_{2 \times 2}(\mathbb{K})$ ([16], p.16, 19). Since $\mathbb{H}\slash\mathbb{Z}_{p}$ is not a division ring (see \cite{Aristidou09}), then $\mathbb{H}\slash\mathbb{Z}_{p}\cong \mathbb{M}_{2 \times 2}(\mathbb{Z}_{p})$ if $p\ne2$. The real matrix representation of $\mathbb{H}\slash\mathbb{Z}_{p}$, where $x=a_0+a_1 i+a_2 j+a_3 k$ $\in$ $\mathbb{H}\slash\mathbb{Z}_{p}$, is achieved by the $4\times 4$ left or right Hamilton Operators as follows: \[ H_x^L = \begin{bmatrix} a_0 & -a_1 & -a_2 & -a_3 \\ a_1 & a_0 & -a_3 & a_2 \\ a_2 & a_3 & a_0 & -a_1 \\ a_3 & -a_2 & -a_1 & -a_0 \end{bmatrix} \,\,H_x^R = \begin{bmatrix} a_0 & -a_1 & -a_2 & -a_3 \\ a_1 & a_0 & a_3 & -a_2 \\ a_2 & -a_3 & a_0 & a_1 \\ a_3 & a_2 & -a_1 & a_0 \end{bmatrix} \] But is the finite ring $\mathbb{O}\slash\mathbb{Z}_{p}$ a skew field? Consider the elements $x_1 =2e_2-e_3, x_2 = e_4+3e_5$ in $\mathbb{O}\slash\mathbb{Z}_{5}$. Multiplying the two, we get: \\ \\ \hspace*{25mm}$x_1\cdot x_2 = (2e_2-e_3)(e_4+3e_5) = 0 (mod\,5)$ \\ \\ This shows that $\mathbb{O}\slash\mathbb{Z}_{5}$ has zero-divisors, and hence $\mathbb{O}\slash\mathbb{Z}_{5}$ is not a skew field. This was also anticipated by some well-known theorem in algebra, by Wedderburn in 1905 ([11], p.361), which says that: ``Every finite skew field is a field". Since $\mathbb{O}\slash\mathbb{Z}_{p}$ is not commutative, then it is not a field, and so it is not a skew-field. So, what is the structure of $\mathbb{O}\slash\mathbb{Z}_{p}$? Since $\mathbb{Z}_{p}$ is a field, then $\mathbb{O}\slash\mathbb{Z}_{p}$ is a non-associative octonion algebra. As a matter of fact, is it an alternative, flexible and power associative algebra$^4$. It is well known that ${\mathbb O}$ is a skew field, yet it has no ‘proper’ matrix representation due to the non-associativity. Nevertheless, as ${\mathbb O}$ is an extension of ${\mathbb H}$, by the Cayley-Dickson process, some non-proper $8 \times 8$ real matrix representations were introduced, by Tian in \cite{Tian}, through the left and right Hamilton Operators of quaternions analogous to the one above. Namely: \vspace*{5mm} \[ H_x^L = \begin{bmatrix} a_0 & -a_1 & -a_2 & -a_3 &-a_4 & -a_5 & -a_6 & -a_7 \\ a_1 & a_0 & -a_3 & a_2 & -a_5 & a_4 & a_7&-a_6 \\ a_2 & a_3 & a_0 & -a_1 &-a_6 & -a_7 & a_4 & a_5 \\ a_3 & -a_2 & a_1 & a_0 &-a_7 & a_6 & -a_5 & a_4 \\ a_4 & a_5 & a_6 & a_7 &a_0 & -a_1 & -a_2 & -a_3 \\ a_5 & -a_4 & a_7 & -a_6 & a_1 & a_0 & a_3 & a_2 \\ a_6 & -a_7 & -a_4 & a_5 &a_2 & -a_3 & a_0 & a_1 \\ a_7& a_6 & -a_5 & -a_4 &a_3 & a_2 & -a_1 & a_0 \\ \end{bmatrix} \] \[H_x^R = \begin{bmatrix} a_0 & -a_1 & -a_2 & -a_3 &-a_4 & -a_5 & -a_6 & -a_7 \\ a_1 & a_0 & a_3 & -a_2 & a_5 & -a_4 & -a_7 & a_6 \\ a_2 & -a_3 & a_0 & a_1 &a_6 & a_7 & -a_4 & -a_5 \\ a_3 & a_2 & -a_1 & a_0 &a_7 & -a_6 & a_5 & -a_4 \\ a_4 & -a_5 & -a_6 & -a_7 & a_0 & a_1 & a_2 & a_3 \\ a_5 & a_4 & -a_7 & a_6 & -a_1 & a_0 & -a_3 & a_2 \\ a_6 & a_7 & a_4 & -a_5 &-a_2 & a_3 & a_0 & -a_1 \\ a_7& -a_6 & a_5 & a_4 &-a_3 & -a_2 & a_1 & a_0 \\ \end{bmatrix} \] Modifying the above over ${\mathbb Z}_{p}$, one could easily get the left and right $8 \times 8$ real representations of $\mathbb{O}\slash\mathbb{Z}_{p}$ as follows$^5$: \[ H_x^L = \begin{bmatrix} a_0 & p-a_1 & p-a_2 & p-a_3 &p-a_4 &p-a_5 & p-a_6 &p-a_7 \\ a_1 & a_0 & p-a_3 & a_2 & p-a_5 & a_4 & a_7&p-a_6 \\ a_2 & a_3 & a_0 & p-a_1 &p-a_6 & p-a_7 & a_4 & a_5 \\ a_3 & p-a_2 & a_1 & a_0 &p-a_7 & a_6 & p-a_5 & a_4 \\ a_4 & a_5 & a_6 & a_7 &a_0 & p-a_1 & p-a_2 & p-a_3 \\ a_5 & p-a_4 & a_7 & -a_6 & a_1 & a_0 & a_3 & a_2 \\ a_6 & p-a_7 & p-a_4 & a_5 &a_2 & p-a_3 & a_0 & a_1 \\ a_7& a_6 & p-a_5 & -a_4 &a_3 & a_2 & p-a_1 & a_0 \\ \end{bmatrix} \] \[H_x^R = \begin{bmatrix} a_0 & p-a_1 & p-a_2 & p-a_3 &p-a_4 & p-a_5 & p-a_6 & p-a_7 \\ a_1 & a_0 & a_3 & p-a_2 & a_5 & p-a_4 & p-a_7 & a_6 \\ a_2 & p-a_3 & a_0 & a_1 &a_6 & a_7 &p-a_4 & p-a_5 \\ a_3 & a_2 & p-a_1 & a_0 &a_7 &p-a_6 & a_5 & p-a_4 \\ a_4 & p-a_5 & p-a_6 & p-a_7 & a_0 & a_1 & a_2 & a_3 \\ a_5 & a_4 & p-a_7 & a_6 & p-a_1 & a_0 & p-a_3 & a_2 \\ a_6 & a_7 & a_4 & p-a_5 &p-a_2 & a_3 & a_0 & p-a_1 \\ a_7& p-a_6 & a_5 & a_4 &p-a_3 &p-a_2 & a_1 & a_0 \\ \end{bmatrix} \] Notice that for the octonionic cases ${\mathbb O}$ and $\mathbb{O}\slash\mathbb{Z}_{p}$, we have that $H_{xy}^L \ne H_x^LH_y^L$ because of the non-associativity. \section{Idempotent and Nilpotents Elements in $\mathbb{O}\slash\mathbb{Z}_{p}$} Recall that an element $x$ in a ring $R$ is called idempotent if $x^2 = x$. In the ring $\mathbb{H}\slash\mathbb{Z}_{p}$, $p$ prime, in the special case where $x = a_0, a_0 \ne 0$ (i.e., $x$ is a non-zero scalar in $\mathbb{H}\slash\mathbb{Z}_{p}$) one quickly observes that if $x$ is idempotent then $x = 1$, for $x$ in { 1, 2, ..., $p$-1}, since $(x, p) = 1$. Therefore, the only scalar idempotent in $\mathbb{H}\slash\mathbb{Z}_{p}$ is 1 (we omit the case $x = 0$ as trivial). Another simple case is the case where $x = ai, aj$ or $ak, a \ne 0$ (i.e., a non-zero scalar multiple of the imaginary units). Then, $x^2 = (ai)^2 = -a^2 i^2 = -a^2 \ne ai =x$, which shows that there are no idempotents of the form $ai, aj$ or $ak$. (Again, we omitted the case $x = 0$ as trivial). Examples of proper idempotents$^6$ and conditions for idempotency in $\mathbb{H}\slash\mathbb{Z}_{p}$ were given in \cite{Aristidou12}. Due to the isomorphism $\mathbb{H}\slash\mathbb{Z}_{p}$ $\cong$ $\mathbb{O}[e_i, e_j, e_ie_j]$ (where $e_i \neq e_j$) idempotents in $\mathbb{H}\slash\mathbb{Z}_{p}$ will transfer in some subalgebras$^7$ of $\mathbb{O}\slash\mathbb{Z}_{p}$. For example, $x = 4+i + 3j + 4k$ is idempotent in $\mathbb{H}\slash\mathbb{Z}_{7}$ and therefore $x = 4+e_1 + 3e_2 + 4e_3$ is idempotent in $\mathbb{O}\slash\mathbb{Z}_{7}$ . Nevertheless, $x = 4+e_1 + 3e_3 + 4e_5$ is a non-``quaternionic" idempotent in $\mathbb{O}\slash\mathbb{Z}_{7}$. Notice that $x = 7i+4j$ is nilpotent in $\mathbb{H}\slash\mathbb{Z}_{13}$ and so $x = 7e_1+4e_2$ is also nilpotent in $\mathbb{O}\slash\mathbb{Z}_{13}$ . Nevertheless, $x=4e_1 + e_2 + 3e_3 + 4e_5$ is a non-``quaternionic" nilpotent in $\mathbb{O}\slash\mathbb{Z}_{7}$. As we will show below, purely imaginary octonions in $\mathbb{O}\slash\mathbb{Z}_{p}$ cannot be idempotents, just as in $\mathbb{H}\slash\mathbb{Z}_{p}$ \cite{Aristidou12}. And nilpotents in $\mathbb{O}\slash\mathbb{Z}_{p}$ are purely imaginary, just as in $\mathbb{H}\slash\mathbb{Z}_{p}$ \cite{Aristidou13}. \begin{theorem} Let $x$ $\in$ $\mathbb{O}\slash\mathbb{Z}_{p}$ be an octonion of the form $x = a_0 + \sum_{i=1}^7 a_ie_i $. Then $x$ is idempotent if and only if $a_0 = \frac{1+p}{2}$ and $ \sum_{i=1}^7 a_i^2 = \frac{p^2 -1}{4}$. \end{theorem} \begin{proof} We follow the steps given in the proof for the quaternion case in \cite{Aristidou12}. Since $x$ is idempotent, we have: \begin{align*} x^2 = x &\Rightarrow (a_0 + \sum_{i=1}^7 a_ie_i)(a_0 + \sum_{i=1}^7 a_ie_i) = a_0 + \sum_{i=1}^7 a_ie_i \\ &\Rightarrow a_0^2 + 2a_0\sum_{i=1}^7 a_ie_i + (\sum_{i=1}^7 a_ie_i)(\sum_{i=1}^7 a_ie_i) = a_0 + \sum_{i=1}^7 a_ie_i \\ & \xRightarrow[\text{Fano}]{\text{distr.}} \, a_0^2 - \sum_{i=1}^7 a_i^2 = a_0 \quad \textrm{and} \quad 2a_0a_i = a_i \end{align*} From the 2$^{nd}$ equation, we have that either $a_i = 0$ or $2a_0 = 1$. That is $a_0 = \frac{1+p}{2}$, as $p =0(mod\,p)$. Substituting the latter in the 1$^{st}$ equation, we get $\sum_{i=1}^7 a_i^2 = \frac{p^2 -1}{4}$. \end{proof} \begin{corollary} Let $x$ $\in$ $\mathbb{O}\slash\mathbb{Z}_{p}$ be a purely imaginary octonion of the form $x = \sum_{i=1}^7 a_ie_i$. Then $x$ is not idempotent. \end{corollary} \begin{proof} If $x$ is purely imaginary then $a_0 =0$. Then from Theorem $3.1$, $0 = \frac{1+p}{4}$ which is a contradiction. \end{proof} \begin{example} Consider $x =4+e_1 + 3e_3 +4e_5$ in $\mathbb{O}\slash\mathbb{Z}_{7}$. Then $x$ is idempotent. Notice that $4 = \frac{1+7}{2}$ and $1^2+3^2+4^2 =26= \frac{49-1}{4}\, mod(7)$. \end{example} \begin{remark} To find the number of idempotents in $\mathbb{O}\slash\mathbb{Z}_{p}$, one could naturally find how many ways $\frac{P^2 -1}{4}$ can be written as a sum of seven or fewer squares. The equation $\sum_{i=1}^7 a_i^2 = \frac{p^2 -1}{4}$ in Theorem $3.1$ brings to mind the ‘sum of seven squares problem’, which is to find the different values $r_7(n)$ for which $n = \sum_{i=1}^7 x_i^2$, $n$ $\in$ $\mathbb{N}$. A formula for square-free values of $n$ were stated without proof by Eisenstein in 1847, and those were extended to all positive integers $n$ by Smith in 1864, also without a proof. Hardy in 1920 developed a method in deriving the proof for $r_k(n)$, where $k$ is odd, but he explicitly showed only the $r_5(n)$ case in \cite{Hardy20,Hardy18}. More general results for r$_7(n)$ were given by Cooper in 2001 \cite{Cooper2} and Cooper and Hirschhorn in 2007 \cite{Cooper7}. \end{remark} Recall that an element $x$ in a ring $R$ is called nilpotent if $x^k =0$ for some $k \in \mathbb{N}$. In \cite{Aristidou13}, it was shown that if $x$ in $\mathbb{H}\slash\mathbb{Z}_{p}$ is nilpotent then the norm $N(x) =0$ (where $N(x) =xx^*=\sum_{i=0}^3 a_i^2$) and, furthermore, that $x$ is purely imaginary and $x^2 =0$. If $x$ $\in$ $\mathbb{O}\slash\mathbb{Z}_{p}$, we have similar results. First, consider the following Lemmas: \begin{lemma} For any $x$ $\in$ $\mathbb{O}\slash\mathbb{Z}_{p}$, we have that $x^2 - 2a_0x+N(x) =0$. \end{lemma} \begin{proof} Let $x = a_0 + \sum_{i=1}^7 a_ie_i$. Then the left-hand side of the equation becomes: \begin{align*} x^2 - &2a_0x + N(x) =(a_0 + \sum_{i=1}^7 a_ie_i)(a_0 + \sum_{i=1}^7 a_ie_i) -2a_0x + N(x) \\ &= a_0^2 + 2a_0\sum_{i=1}^7 a_ie_i+(\sum_{i=1}^7 a_ie_i)(\sum_{i=1}^7 a_ie_i) - 2a_0(a_0+ \sum_{i=1}^7 a_ie_i) + \sum_{i=0}^7 a_i^2 \\ & = a_0^2 + \sum_{i=1}^7 2a_0a_ie_i - \sum_{i=1}^7 a_i^2 - 2a_0(a_0 + \sum_{i=1}^7 a_ie_i) + \sum_{i=0}^7 a_i^2 \\ &= a_0^2 + 2a_0\sum_{i=1}^7 a_ie_i - \sum_{i=1}^7 a_i^2 - 2a_0^2 - 2a_0 \sum_{i=1}^7 a_ie_i + a_0^2 + \sum_{i=1}^7 a_i^2 \\ & =0 \end{align*} \end{proof} \begin{lemma} Let $x$ $\in$ $\mathbb{O}\slash\mathbb{Z}_{p}$. If $x$ is nilpotent, then $N(x) =0$. \end{lemma} \begin{proof} We follow the steps given in the proof for the quaternion case in \cite{Aristidou13}. If $x$ is nilpotent, then $x^k = 0$ for some $k$. From Lemma $3.5$ above, we have: \begin{align*} x^2 - 2a_0x + N(x) =0 &\Rightarrow x(x -2a_0) = - N(x) \\ &\Rightarrow (x(x - 2a_0))^k = (- N(x))^k \\ &\Rightarrow x^k(x - 2a_0)^k = (- N(x))^k \mbox{ (see Remark $3.7$ below)} \\ &\Rightarrow 0 = (N(x))^k \\ &\Rightarrow N(x) = 0 , \mbox{ because } {\mathbb Z}_{\mathbf{p}} \mbox{ is a field.} \end{align*} \end{proof} \begin{remark} We discuss the statement $(x(x - 2a_0))^k = x^k(x - 2a_0)^k$ in the proof in the Lemma $3.6$ above: The statement is taken as obvious, without a proof, in \cite{Aristidou13} (in Lemma 2.1) for the quaternionic case $\mathbb{H}\slash\mathbb{Z}_{p}$, but it deserves a bit more explanation in our case here considering the non-commutativity and non-associativity of $\mathbb{O}\slash\mathbb{Z}_{p}$. As we mentioned in Sec.2, $\mathbb{O}\slash\mathbb{Z}_{p}$ is an alternative algebra (and flexible). Therefore, it also satisfies the \textit{Moufang Identities}, in particularly the identity $(xy)(zx) = (x(yz))x$. Given this, it is not hard to show the following: \begin{proposition} If $A$ is an alternative algebra such that $xy = yx$, $x$, $y$ $\in$ $A$, then $(xy)^k = x^ky^k$. \end{proposition} \begin{proof} We show this for $k = 2$ (the general case follows by iteration). Indeed:%\\ \begin{equation*} \begin{split} (xy)^2 = (xy)(xy) \eqdefa (yx)(xy) &\eqdefb (y(xx))y \\ &\eqdefc ((yx)x)y \\ &\eqdefa ((xy)x)y \\ &\eqdefd (x(yx))y\\ &\eqdefa (x(xy))y\\ &\eqdefc((xx)y)y\\ &\eqdefb (xx)(yy) \end{split} \end{equation*} \end{proof} \noindent Hence, the statement $(x(x -2a_0))^k = x^k(x - 2a_0)^k$ is also true in our particular case here, because $\mathbb{O}\slash\mathbb{Z}_{p}$ is alternative (and flexible) and $x(x-2a_0) = (x-2a_0)x$. It is also clear now why the statement is easy to prove in $\mathbb{H}\slash\mathbb{Z}_{p}$, considering that $\mathbb{H}\slash\mathbb{Z}_{p}$ is actually associative. Finally, given the above result, one could also obtain the binomial formula $(x+y)^k = \sum_{j=0}^k \begin{pmatrix} k \\ j \end{pmatrix} x^jy^{k-j}$, which could also be used to prove the statement in question. That is: \begin{align*} (x(x-2a_0))^k = (x^2-2a_0x)^k &= \sum_{j=0}^k \begin{pmatrix} k \\ j \end{pmatrix} (x^2)^j(-2a_0x)^{k-j} \\ &= x^k \sum_{j=0}^k \begin{pmatrix} k \\ j \end{pmatrix} x^j(-2a_0)^{k-j} \\ &= x^k(x-2a_0)^k \end{align*} \end{remark} \begin{theorem} Let $x$ $\in$ $\mathbb{O}\slash\mathbb{Z}_{p}$. Then $x$ is nilpotent if and only if $x$ is purely imaginary and $N(x) =0$. Furthermore, if $x$ is nilpotent, then $x^2 = 0$. \end{theorem} \begin{proof} If $x$ is nilpotent, then $x^k =0$ for some $k >1$ (where $k$ is the least such natural number). From Lemma $3.6$ above, we have that $N(x) =0$. Combining Lemmas $3.5$ and $3.6$, we get $x^2 = 2a_0x$. Following the steps given in the proof for the quaternion case in \cite{Aristidou13}, we have: \begin{align*} \textrm{If\ } k\ \textrm{is even}:\quad x^2 = 2a_0x & \Rightarrow (x^2)^{k/2} = (2a_0)^{k/2}x^{k/2}\\ &\Rightarrow x^k = (2a_0)^{k/2}x^{k/2}\\ &\Rightarrow 0= (2a_0)^{k/2}x^{k/2}\\ &\Rightarrow a_0 = 0\, \end{align*} \begin{align*} \hspace*{1.3cm}\textrm{If\ } k\ \textrm{is odd}:\quad x^2 = 2a_0x &\Rightarrow (x^2)^{(k+1)/2} = (2a_0)^{(k+1)/2}x^{(k+1)/2} \\ &\Rightarrow (x)^{(k+1)/2} = (2a_0)^{(k+1)/2}x^{(k+1)/2} \\ &\Rightarrow 0 =(2a_0)^{(k+1)/2}x^{k/2} \\ & \Rightarrow a_0 =0 \end{align*} Hence, $a_0 = 0$ and therefore $x$ is imaginary. Furthermore, since $a_0 = 0$, from $x^2 = 2a_0x$ we have that $x^2 = 0$. For the converse, since $N(x) =0$, Lemma $3.5$ gives $x^2 =2a_0x$. Since also $x$ is imaginary $(a_0 = 0)$ the equation $x^2 = 2a_0x$ gives $x^2 = 0$. Then for any $k >1$ we have: $x^k = x^{k-2}x^2 = x^{k-2}\cdot 0 = 0$, so $x$ is nilpotent. \end{proof} \begin{example} Consider $x = 4e_1 + e_2 + 3e_3 +4e_5$ in $\mathbb{O}\slash\mathbb{Z}_{7}$. Then $x$ is nilpotent. Notice that $N(x) =0^2+4^2+1^2+3^2+ 0^2+4^2+0^2 +0^2= 0(mod\,7)$. \end{example} \section{Connection to General Rings and Applications} There is a lot in the literature on idempotents, nilpotents and k-potents in general, in more general rings $R$. It would be interesting to see if and how some of these results relate to the `special', in a sense, ring $\mathbb{O}\slash\mathbb{Z}_{p}$. In \cite{Hirano}, Hirano and Tominaga proved that in a ring $R$ the following are equivalent: (i) Every element of R is a sum of two commuting idempotents; (ii) R is commutative and every element of R is a sum of two idempotents; (iii) $x^3 =x$, for all $x$ in $R$.$^8$ As $\mathbb{O}\slash\mathbb{Z}_{p}$ is not commutative, the above fails. For example, consider the idempotents $a=3+e_1$ and $b = 3+e_2$ in $\mathbb{O}\slash\mathbb{Z}_{5}$. Then, $x=a+b=(3+e_1)+(3+e_2)=6+e_1+e_2=1+e_1+e_2$, but $x$ is not tripotent (indeed, $(1+e_1+e_2)^3 = e_1+e_2 \neq 1+e_1+e_2)$. The above fails even when the idempotents commute. Take, for example, $a = b = 3 + e_1$ in $\mathbb{O}\slash\mathbb{Z}_{5}$. Also, Mosic in \cite{Mosic} gives the relation between idempotent and tripotent elements in any associative ring $R$, generalizing the result on matrices by Trenkler and Baksalary \cite{Trenkler}. Namely, for any $x$ $\in$ $R$, where 2, 3 are invertible, $x$ is idempotent if and only if $x$ is tripotent and $1-x$ is tripotent or $1+x$ is invertible. Notice that even though $\mathbb{O}\slash\mathbb{Z}_{p}$ is not associative, the result does hold in some cases. Take for example the tripotent $x = 4 +3e_1 + e_2+ 4e_3$ in $\mathbb{O}\slash\mathbb{Z}_{7}$, which is also an idempotent. It is not hard to check that directly or using the conditions for idempotency in Theorem $3.1$ above. Notice also that $1 - x$ is tripotent and $1 + x$ is invertible as $N(x) =2\neq 0$. So, we conjecture that Mosic’s result may extend to (some) non-associative rings. Finally, it is interesting to note any possible applications of rings related to the ring $\mathbb{O}\slash\mathbb{Z}_{p}$. Malekian and Zakerolhosseini in \cite{Malekian} use octonionic algebras to construct a high speed public key cryptosystem. More specifically, they consider the convolution polynomial rings $R = \mathbb{Z}[x]/(x^N -1)$, $R_{p} =\mathbb{Z}_{p}[x]/(x^N-1)$ and $R_{q} = \mathbb{Z}_{q}[x]/(x^N -1)$, where $p$, $q$ are primes such as $q \gg p$. From these they construct the octonionic algebras: \begin{align*} \mathbb{A} &= \{\,a_0(x)+ \sum_{i=1}^7 a_i(x)e_i\ |\ a_i(x) \in R\,\}\, , \\ \mathbb{A}_{p} &= \{\,a_0(x) + \sum_{i=1}^7a_i(x)e_i\ |\ a_i(x)\ \in R_{p}\,\}\, , \\ \mathbb{A}_{q} &= \{\,a_0(x)+ \sum_{i=1}^7 a_i(x)e_i\ |\ a_i(x) \in R_{q}\,\}\, , \end{align*} respectively. Then, the public (and private) key is generated as follows: initially two small octonions $F\in\mathbb{L}_{f}$ and $G\in\mathbb{L}_{g}$, where $\mathbb{L}_{f}, \mathbb{L}_{g}$ are some specifically constructed subspaces of $\mathbb{A}$, are randomly generated. Namely, \begin{align*} F &= f_0 + \sum_{i=1}^7f_ie_i\ |\ f_i\in \mathbb{L}_{f}\, , \\ G &= g_0 + \sum_{i=1}^7g_ie_i\ |\ g_i\in \mathbb{L}_{g}\, . \end{align*} The octonion $F$ must be invertible in $\mathbb{A}_{p}$ and $\mathbb{A}_{q}$ , otherwise a new octionion $F$ is generated. The inverses of $F$ in $\mathbb{A}_{p}$ and $\mathbb{A}_{q}$ are denoted by in $F_p^{-1}$ and $F_q^{-1}$, respectively. The public key, which is an octonion, is then given by $H = F_p^{-1} o\,G \in \mathbb{A}_{q}$ , where $o$ is a multiplication defined on $\mathbb{A}_{q}$, in terms of the convolution product. Encryption and decryption are done with similar calculations. \section*{\noindent Notes} 1. $\mathbb{R}$, $\mathbb{C}$, $\mathbb{H}$ and $\mathbb{O}$ are the only normed division algebras. This was proved by Hurwitz in 1898 \cite{Hurwitz}. \\ \\ 2. ``+" and ``." on $\mathbb{H}$ are defined in {\cite[p. 124]{Herstein}. As $p=0 (mod\,p)$ on $\mathbb{H}\slash\mathbb{Z}_{p}$ they are defined as follows:\\ \hspace*{2cm}$x+y = (a_0 + a_1i + a_2j+a_3k) + (b_0 + b_1i + b_2j+b_3k) $\\ \hspace*{3.6cm} $= (a_0 +b_0) +(a_1+b_1)i + (a_2+b_2)j + (a_3+b_3)k$\\ \hspace*{2.7cm} $ x\cdot y \hspace*{0.15cm}= (a_0 + a_1i + a_2j+a_3k)\cdot(b_0 + b_1i + b_2j+b_3k) $\\ \hspace*{3.6cm}$ =a_0b_0 + (p-1)a_1b_1 + (p-1)a_2b_2 + (p-1) a_3b_3 + \\ \hspace*{3.8cm} (a_0b_0 + a_1b_0 + a_2b_3 + (p-1)a_3b_2)j + \\ \hspace*{3.8cm} (a_0b_2 + (p-1)a_1b_3 + a_2b_0 + a_3b_1)j+\\ \hspace*{3.8cm}(a_0b_3 + a_1b_2 + (p-1)a_2b_1 +a_3b_0)k$\\ \\ 3. Fano Plane (Figure~\ref{fig:fano}); Multiplication table (Figure~\ref{fig:octmult}); Program in Maple (Figure~\ref{fig:maple}): \begin{figure}[h]\centering \includegraphics[scale=0.5]{fanoplane.jpg} \caption{Fano Plane} \label{fig:fano} \end{figure} \begin{figure}[h]\centering \includegraphics[scale=0.26]{octonion-multiplication-table.jpg} \caption{Multiplication table} \label{fig:octmult} \end{figure} \begin{figure}[h]\centering \includegraphics[scale=0.22]{mapleextract.jpg} \caption{Maple program} \label{fig:maple} \end{figure} \FloatBarrier % \noindent 4. Accordingly, the following hold: $(xx)y = x(xy)$ (alternative), $x(yx) = (xy)x$ (flexible), $\left$ is power associative for all $x$. \\ \\ 5. These representations are given in \cite{Halici} without a proof. The proof for $\mathbb{O}\slash\mathbb{Z}_{p}$ is actually straightforward, following the exact steps in the proofs of Theorems 2.1 and 2.3 in \cite{Tian} for the case of $\mathbb{O}$. \\ \\ 6. In Herstein \cite[p. 130]{Herstein}, we have as an exercise that: In a ring $R$, if $x^2 =x$, for all $x$ in $R$, then $R$ is commutative. It is not hard to show that the converse is not true. (e.g. $\mathbb{F} = \mathbb{Z}_{3}$, 2 is not idempotent). Actually, a field $\mathbb{F}$ has only trivial idempotents. Hence, in $\mathbb{H}\slash\mathbb{Z}_{p}$ some elements are non-trivial idempotents and they were described in \cite{Aristidou12}. \\ \\ 7. Namely, the seven quaternionic subalgebras of $\mathbb{O}$ each generated by the seven “lines” (including the circle) in the Fano Plane.\\ \\ 8. A ring $R$ is called a \emph{tripotent ring} if $x^3 =x$, for all $x$ in $R$. The fact that a tripotent ring is commutative is found as an exercise in Hernstein \cite[p. 136]{Herstein}. Several proofs of this fact have been given since the 60's \cite{Ayoub}. In Bourbaki, we find it also as an exercise with guided steps/hints for the proof \cite[p. 176]{Bourbaki}. See also \cite{Poorten}. 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