Generalized Projections in Zn

TL;DR

This paper investigates the conditions under which the set of integers modulo n, with a specific partial order, forms a lattice structure, expanding understanding of algebraic properties in modular arithmetic.

Contribution

It provides necessary and sufficient conditions for the poset of integers modulo n to be a lattice under a particular partial order.

Findings

Characterization of when (Z_n, ≤) is a lattice.

Conditions depend on the structure of n and the partial order.

Enhances understanding of lattice structures in modular arithmetic.

Abstract

We consider the ring $\mathbb Z_n$ (integers modulo $n$) with the partial order `$\leq$' given by `$a \leq b$ if either $a=b$ or $a\equiv ab~(mod~n)$'. In this paper, we obtain necessary and sufficient conditions for the poset ($\mathbb Z_n,~\leq$) to be a lattice.

Equations

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Full text

Abstract: We consider the ring $\mathbb{Z}_{n}$ (integers modulo $n$) with the partial order ‘$\leq$’ given by ‘$a\leq b$ if either $a=b$ or $a\equiv ab~{}(mod~{}n)$’. In this paper, we obtain necessary and sufficient conditions for the poset ($\mathbb{Z}_{n},~{}\leq$) to be a lattice.

Keywords: generalized projections, regular elements, nilpotent elements.

1. Introduction

An element $a$ in a commutative ring $R$ is said to be a generalized projection if $a^{k}=a$ for some $k\in\mathbb{N}$ with $k\geq 2$ (see [1]); an element $a$ is called a regular element if $a=aba(=a^{2}b)$ for some element $b$ in the ring. It is proved by L$\acute{\text{a}}$szl$\acute{\text{o}}$ T$\acute{\text{o}}$th [2] that in $\mathbb{Z}_{n}$, an element is a generalized projection if and only if it is regular; in-fact the following result is proved.

Theorem 1.1** ([2], Theorem 1).**

*Let $n=\displaystyle\prod_{i=1}^{k}p_{i}^{\alpha_{i}}$ be the prime factorization of $n\in\mathbb{N}$ with $\alpha_{i}>0$, for all $i$. For an integer $a\geq 1$, the following assertions are equivalent:

i) $a$ is regular (mod $n$);           ii) for every $i\in\{1,2,\dots,k\}$, either $p_{i}^{\alpha_{i}}|a$ or $p_{i}\nmid a$;

iii) $gcd(a,n)=gcd(a^{2},n)$;       iv) $gcd(a,n)|n$ and $gcd(gcd(a,n),\frac{n}{gcd(a,b)})=1$ ;

v) $a^{\varphi(n)+1}\equiv a~{}(mod~{}n)$;            vi) there exists an integer $m\geq 1$ such that $a^{m+1}\equiv a~{}(mod~{}n)$.*

We denote by $GP(\mathbb{Z}_{n})$, the set of generalized projections (i.e. the set of regular elements) and $P(\mathbb{Z}_{n})=\{a\in\mathbb{Z}_{n}~{}|~{}a^{2}=a\}$, the set of projections in $\mathbb{Z}_{n}$. Let $R$ be a commutative ring, the relation ‘$\leq$’ defined by: for $a,b\in R$, ‘$a\leq b$ if and only if either $a=b$ or $a=ab$’ is a partial order on $R$ (see [1]). In particular, $(\mathbb{Z}_{n},~{}\leq)$ is a poset with the smallest element [math] and the largest element $1$. It is known that $(P(\mathbb{Z}_{n}),~{}\leq)$ is a lattice. However, Khairnar and Waphare [1] proved that for any finite commutative ring $R$, $(GP(R),~{}\leq)$ is a lattice, hence in particular, $(GP(\mathbb{Z}_{n}),~{}\leq)$ is a lattice for every $n$. Whenever $n$ is a square-free integer, we get that $GP(\mathbb{Z}_{n})=\mathbb{Z}_{n}$. In general, $(\mathbb{Z}_{n},~{}\leq)$ is not a lattice, for example $(\mathbb{Z}_{9},~{}\leq)$ (see Figure 2). In this paper, we give a necessary and sufficient conditions for the poset ($\mathbb{Z}_{n},~{}\leq$) to be a lattice.

We denote by $U(\mathbb{Z}_{n})$, the set of units in $\mathbb{Z}_{n}$ and $N(\mathbb{Z}_{n})$, the set of nilpotents in $\mathbb{Z}_{n}$. In the following remark, we list observations required in a sequel.

Remark 1.2**.**

Let $n=\displaystyle\prod_{i=1}^{k}p_{i}^{\alpha_{i}}$ be the prime factorization of $n\in\mathbb{N}$ with $\alpha_{i}>0$, for all $i$, and let $a\in\mathbb{Z}_{n}$. Then,

(i) $a\in U(\mathbb{Z}_{n})$ if and only if $a\equiv$ unit (mod $p_{i}$) for all $i\in\{1,2,\cdots,k\}$.

(ii) $a\in N(\mathbb{Z}_{n})$ if and only if $a\equiv$ zero (mod $p_{i}$) for all $i\in\{1,2,\cdots,k\}$.

(iii) $a\in GP(\mathbb{Z}_{n})$ if and only if $a\equiv$ zero or unit (mod $p_{i}^{\alpha_{i}}$) for all $i\in\{1,2,\cdots,k\}$.

2. Upper covering projection and lower covering projection

In a poset $(P,\leq)$, $a<b$ denotes $a\leq b$ with $a\neq b$. We say that $b$ is an upper cover of $a$ or $a$ is a lower cover of $b$ (denoted by $a\prec b$), if $a<b$ and there is no $c\in P$ such that $a<c<b$.

The following theorem gives an existence of the unique lower cover and the unique upper cover of any element $a\in GP(R)\backslash P(R)$ in the poset $GP(R)$.

Theorem 2.1** ([1], Theorem 2.7).**

Let $R$ be a finite commutative ring. If $a\in GP(R)\backslash P(R)$, then there exist a unique $a_{u}\in GP(R)$ and a unique $a_{l}\in GP(R)$ such that $a_{l}\prec a\prec a_{u}$. Further, these unique elements are projections and for $b\in GP(R)$, $a<b$ if and only if $a_{u}\leq b$; and $b<a$ if and only if $b\leq a_{l}$.

With notations as in Theorem 2.1, the unique projection $a_{l}$ is called the lower covering projection of $a$ and the unique projection $a_{u}$ is called the upper covering projection of $a$. If $a$ is a projection, then we assume that the lower covering projection of $a$ and the upper covering projection of $a$ is $a$ itself.

Theorem 2.1 gives an existence of the upper covering projection and the lower covering projection of any element in $GP(\mathbb{Z}_{n})\backslash P(\mathbb{Z}_{n})$. In this section, we determine the upper covering projection and the lower covering projection of elements in $GP(\mathbb{Z}_{n})\backslash P(\mathbb{Z}_{n})$.

The following lemma gives the conditions for strict comparability of a projection and a generalized projection.

Lemma 2.2**.**

*Let $n=\displaystyle\prod_{i=1}^{k}p_{i}^{\alpha_{i}}$ be the prime factorization of $n\in\mathbb{N}$ with $\alpha_{i}>0$, for all $i$. Let $a\in GP(\mathbb{Z}_{n})$ and $e,f\in P(\mathbb{Z}_{n})\backslash\{1\}$. Then,

(1) $a<e$ if and only if for $i\in\{1,2,\cdots,k\}$, $e\equiv 0~{}(mod~{}p_{i}^{\alpha_{i}})$ implies that $a\equiv 0~{}(mod~{}p_{i}^{\alpha_{i}})$.

(2) $f<a$ if and only if for $j\in\{1,2,\cdots,k\}$, $f\nequiv 0~{}(mod~{}p_{j}^{\alpha_{j}})$ implies that $a\equiv 1~{}(mod~{}p_{j}^{\alpha_{j}})$.*

Proof.

$(1)$ Let $a<e$ and $i\in\{1,2,\cdots,k\}$ be such that $e\equiv 0~{}(mod~{}p_{i}^{\alpha_{i}})$. Then $a(1-e)\equiv 0~{}(mod~{}n)$ and $1-e\nequiv 0~{}(mod~{}p_{i}^{\alpha_{i}})$. Therefore $a\equiv 0~{}(mod~{}p_{i}^{\alpha_{i}})$. Conversely, suppose that for $i\in\{1,2,\cdots,k\}$, $e\equiv 0~{}(mod~{}p_{i}^{\alpha_{i}})$ implies that $a\equiv 0~{}(mod~{}p_{i}^{\alpha_{i}})$. Then $a(1-e)\equiv 0~{}(mod~{}n)$. Thus $a<e$.

$(2)$ Let $f<a$ and $j\in\{1,2,\cdots,k\}$ be such that $f\nequiv 0~{}(mod~{}p_{j}^{\alpha_{j}})$. Then $1-a\equiv 0~{}(mod~{}p_{j}^{\alpha_{j}})$. Therefore $a\equiv 1~{}(mod~{}p_{j}^{\alpha_{j}})$. Conversely, suppose that for $j\in\{1,2,\cdots,k\}$, $f\nequiv 0~{}(mod~{}p_{j}^{\alpha_{j}})$ implies that $a\equiv 1~{}(mod~{}p_{j}^{\alpha_{j}})$. This gives $f(1-a)\equiv 0~{}(mod~{}n)$. Thus $f<a$. ∎

Remark 2.3** ([1], Remark 3).**

Let $a\in GP(\mathbb{Z}_{n})$. Suppose $k\geq 2$ be the smallest integer such that $a^{k}=a$. Then $(a^{k-1})^{2}=a^{2k-2}=a^{k}a^{k-2}=aa^{k-2}=a^{k-1}$. Therefore $a^{k-1}\in P(\mathbb{Z}_{n})$, and $a^{k-1}=a_{u}$. Clearly, $a\leq a_{u}$; and $a=a_{u}$ if and only if $a\in P(\mathbb{Z}_{n})$.

For any $a\in GP(\mathbb{Z}_{n})\backslash P(\mathbb{Z}_{n})$ the following theorem gives a construction for $a_{u}$.

Theorem 2.4**.**

Let $n=\displaystyle\prod_{i=1}^{k}p_{i}^{\alpha_{i}}$ be the prime factorization of $n\in\mathbb{N}$ with $\alpha_{i}>0$, for all $i$, and $a\in GP(\mathbb{Z}_{n})\backslash P(\mathbb{Z}_{n})$. If $a\in U(\mathbb{Z}_{n})$, then $a_{u}=1$. If $a\notin U(\mathbb{Z}_{n})$, then $a_{u}=b^{\varphi(\frac{n}{b})}$ where $b=\displaystyle\prod_{\begin{subarray}{c}j=1\\ a~{}\equiv~{}0~{}(mod~{}p_{j}^{\alpha_{j}})\end{subarray}}^{k}p_{j}^{\alpha_{j}}$.

Proof.

Let $a\in GP(\mathbb{Z}_{n})\backslash P(\mathbb{Z}_{n})$. If $a\in U(\mathbb{Z}_{n})$ then by Remark 2.3, $a_{u}=1$. Suppose $a\notin U(\mathbb{Z}_{n})$. By Remark 1.2, there exists $j\in\{1,2,\cdots,k\}$ such that $a\equiv 0~{}(mod~{}p_{j}^{\alpha_{j}})$, and $a\equiv 0~{}or~{}unit~{}(mod~{}p_{i}^{\alpha_{i}})$ for all $i\neq j$. Let $b=\displaystyle\prod_{\begin{subarray}{c}j=1\\ a~{}\equiv~{}0~{}(mod~{}p_{j}^{\alpha_{j}})\end{subarray}}^{k}p_{j}^{\alpha_{j}}$ and $a_{u}=b^{\varphi(\frac{n}{b})}$. Then $a_{u}\in P(\mathbb{Z}_{n})$ and by Lemma 2.2, $a\leq a_{u}$. Let $e\in P(\mathbb{Z}_{n})$ be such that $a<e$. If $e=1$, then $a_{u}\leq e$. Suppose $e\neq 1$. Again by Lemma 2.2, for $i\in\{1,2,\cdots,k\}$, $e\equiv 0~{}(mod~{}p_{i}^{\alpha_{i}})$ implies that $a\equiv 0~{}(mod~{}p_{i}^{\alpha_{i}})$. Hence, for $i\in\{1,2,\cdots,k\}$, $e\equiv 0~{}(mod~{}p_{i}^{\alpha_{i}})$ implies that $a_{u}\equiv 0~{}(mod~{}p_{i}^{\alpha_{i}})$. Thus $a_{u}\leq e$. ∎

For any $a\in GP(\mathbb{Z}_{n})\backslash P(\mathbb{Z}_{n})$ the following theorem gives a construction for $a_{l}$.

Theorem 2.5**.**

Let $n=\displaystyle\prod_{i=1}^{k}p_{i}^{\alpha_{i}}$ be the prime factorization of $n\in\mathbb{N}$ with $\alpha_{i}>0$, for all $i$, and $a\in GP(\mathbb{Z}_{n})\backslash P(\mathbb{Z}_{n})$. Then $a_{l}=b^{\varphi(\frac{n}{b})}$ where $b=\displaystyle\prod_{\begin{subarray}{c}j=1\\ a~{}\nequiv~{}1~{}(mod~{}p_{j}^{\alpha_{j}})\end{subarray}}^{k}p_{j}^{\alpha_{j}}$.

Proof.

Let $a\in GP(\mathbb{Z}_{n})\backslash P(\mathbb{Z}_{n})$. If $a\equiv 1~{}(mod~{}p_{j}^{\alpha_{j}})$ for all $j\in\{1,2,\cdots,k\}$ then $a-1\equiv 0~{}(mod~{}n)$. Hence $a=1\in P(\mathbb{Z}_{n})$, a contradiction. Therefore there exists $j\in\{1,2,\cdots,k\}$ such that $a\nequiv~{}1~{}(mod~{}p_{j}^{\alpha_{j}})$. Let $b=\displaystyle\prod_{\begin{subarray}{c}j=1\\ a~{}\nequiv~{}1~{}(mod~{}p_{j}^{\alpha_{j}})\end{subarray}}^{k}p_{j}^{\alpha_{j}}$ and $a_{l}=b^{\varphi(\frac{n}{b})}$. We prove that $a_{l}\leq a$. Let $i\in\{1,2,\cdots,k\}$ be such that $a_{l}\nequiv 0~{}(mod~{}p_{i}^{\alpha_{i}})$. Then $b\nequiv 0~{}(mod~{}p_{i}^{\alpha_{i}})$ and hence $a\equiv 1~{}(mod~{}p_{i}^{\alpha_{i}})$. This yields, $a_{l}(a-1)\equiv 0~{}(mod~{}n)$. Therefore $a_{l}\leq a$. Let $f\in P(\mathbb{Z}_{n})$ be such that $f<a$, and $j\in\{1,2,\cdots,k\}$ be such that $f\nequiv 0~{}(mod~{}p_{j}^{\alpha_{j}})$. Then by Lemma 2.2, we get $a\equiv 1~{}(mod~{}p_{j}^{\alpha_{j}})$. Consequently, $b\nequiv 0~{}(mod~{}p_{j}^{\alpha_{j}})$ and hence $a_{l}\nequiv 0~{}(mod~{}p_{j}^{\alpha_{j}})$. This implies that $a_{l}\equiv 1~{}(mod~{}p_{j}^{\alpha_{j}})$. Therefore $f(1-a_{l})\equiv 0~{}(mod~{}n)$. Thus $f\leq a_{l}$. ∎

Let $P$ be a poset and $a,b\in P$. The join of $a$ and $b$, denoted by $a\vee b$, is defined as $a\vee b=\sup~{}\{a,b\}$. The meet of $a$ and $b$, denoted by $a\wedge b$, is defined as $a\wedge b=inf~{}\{a,b\}$.

We conclude this section with the following examples.

In the following example, $n\in\mathbb{N}$ is not square-free but the poset $\mathbb{Z}_{n}$ is a lattice.

Example 2.6**.**

Consider the ring $\mathbb{Z}_{4}$. Then $GP(\mathbb{Z}_{4})=\{0,1,3\}$ and $N(\mathbb{Z}_{4})=\{0,2\}$. Note that $4$ is not square-free but the poset $\mathbb{Z}_{4}$ is a lattice (see Figure 1). Also, the nilpotent element $2$ possess unique upper cover.

Example 2.7**.**

Consider the ring $\mathbb{Z}_{8}$. Then $GP(\mathbb{Z}_{8})=\{0,1,3,5,7\}$ and $N(\mathbb{Z}_{8})=\{0,2,4,6\}$. Note that $8$ is not square-free but the poset $\mathbb{Z}_{8}$ is a lattice (see Figure 1). Also, each of $2$ and $6$ possess unique upper covers but $4$ does not possess unique upper cover.

In the following example, the poset $\mathbb{Z}_{n}$ is not a lattice. Also, none of the nilpotent elements possess unique upper cover.

Example 2.8**.**

Consider the ring $\mathbb{Z}_{9}$. Then $GP(\mathbb{Z}_{9})=\{0,1,2,4,5,7,8\}$ and $N(\mathbb{Z}_{9})=\{0,3,6\}$. By Figure 2, the poset $\mathbb{Z}_{9}$ is not a lattice. Note that $3\vee 6$ and $4\wedge 7$ do not exist. Also, each of $3$ and $6$ do not possess unique upper covers and each of $4$ and $7$ do not possess unique lower covers.

In the following example, $n$ is not square-free but the poset $\mathbb{Z}_{n}$ is a lattice.

Example 2.9**.**

Consider the ring $\mathbb{Z}_{12}$. Then $GP(\mathbb{Z}_{12})=\{0,1,3,4,5,7,8,9,11\}$ and $N(\mathbb{Z}_{12})=\{0,6\}$. Note that $12$ is not square-free but the poset $\mathbb{Z}_{12}$ is a lattice (see Figure 2). Observe that, the nilpotent element $6$ does not possess an unique upper cover.

In the next section, we give a necessary and sufficient condition for the existence of supremum and infimum of any two elements of the poset $\mathbb{Z}_{n}$.

3. Existence of $a\vee b$ and $a\wedge b$ for $a,b\in\mathbb{Z}_{n}$

For $x\in\mathbb{Z}_{n}$, the ideal generated by $x$ is denoted by $(x)$.

The following theorem characterizes the existence of $a\vee b$ for $a,b\in\mathbb{Z}_{n}$.

Theorem 3.1**.**

Let $n=\displaystyle\prod_{i=1}^{k}p_{i}^{\alpha_{i}}$ be the prime factorization of $n\in\mathbb{N}$ with $\alpha_{i}>0$, for all $i$. Let $a,b\in\mathbb{Z}_{n}$ be incomparable and $d=gcd(gcd(a,b),n)$. Then, $a\vee b$ exists if and only if the coset $(\frac{n}{d})+1$ has the smallest element.

Proof.

For each $i\in\{1,2,\cdots,k\}$, let $\beta_{i},\gamma_{i}\in\mathbb{W}=\mathbb{N}\bigcup\{0\}$ be the largest powers of prime $p_{i}$ such that $a\equiv 0~{}(mod~{}p_{i}^{\beta_{i}})$ and $b\equiv 0~{}(mod~{}p_{i}^{\gamma_{i}})$ respectively. Let $f_{i}=max\{(\alpha_{i}-\beta_{i}),(\alpha_{i}-\gamma_{i}),0\}$ and $m=\displaystyle\prod_{i=1}^{k}p_{i}^{f_{i}}$. Then $m=\frac{n}{d}$. Let $c\in\mathbb{Z}_{n}$ and for each $i\in\{1,2,\cdots,k\}$, let $t_{i}\in\mathbb{N}$ be the largest powers of prime $p_{i}$ such that $c\equiv 1~{}(mod~{}p_{i}^{t_{i}})$. Then, $a<c$ and $b<c$, if and only if $a(c-1)\equiv 0~{}(mod~{}n)$ and $b(c-1)\equiv 0~{}(mod~{}n)$, if and only if $t_{i}\geq(\alpha_{i}-\beta_{i}),(\alpha_{i}-\gamma_{i})$ for all $i$, if and only if $c-1\in(\frac{n}{d})$, if and only if $c\in(\frac{n}{d})+1$.

Suppose $a\vee b$ exists. Since $a$ and $b$ are incomparable, we have $a<a\vee b$ and $b<a\vee b$. This yields $a\vee b\in(\frac{n}{d})+1$. Let $x\in(\frac{n}{d})+1$. Then $a<x$ and $b<x$. Therefore $a\vee b\leq x$. Thus $a\vee b$ is the smallest element of the coset $(\frac{n}{d})+1$. Conversely, suppose that the coset $(\frac{n}{d})+1$ has the smallest element, say $e\in(\frac{n}{d})+1$. This yields $a<e$ and $b<e$. We claim that $a\vee b=e$. Let $f\in\mathbb{Z}_{n}$ be such that $a<f$ and $b<f$. Then $f\in(\frac{n}{d})+1$. Therefore $e\leq f$. Thus $a\vee b=e$. ∎

From the proof of Theorem 3.1, it is clear that, if $a\vee b$ exists, then $a\vee b$ is the least element of the coset $(\frac{n}{d})+1$. Also, if the coset $(\frac{n}{d})+1$ has the smallest element $e$, then $a\vee b=e$.

The following corollary is an immediate consequence of Theorem 3.1.

Corollary 3.2**.**

Let $n\in\mathbb{N},~{}n>1$ and $S=\{d\in\mathbb{N}~{}|~{}d=gcd(gcd(a,b),n),$ for some incomparable elements $a,b\in\mathbb{Z}_{n}\}$. Then, $\mathbb{Z}_{n}$ is a lattice if and only if every coset in $\{(\frac{n}{d})+1~{}|~{}d\in S\}$ has smallest element.

In the following theorem, we characterize the existence of $a\wedge b$ for $a,b\in\mathbb{Z}_{n}$.

Theorem 3.3**.**

Let $n=\displaystyle\prod_{i=1}^{k}p_{i}^{\alpha_{i}}$ be the prime factorization of $n\in\mathbb{N}$ with $\alpha_{i}>0$, for all $i$. Let $a,b\in\mathbb{Z}_{n}$ be incomparable and $d=gcd(gcd(a-1,b-1),n)$. Then, $a\wedge b$ exists if and only if the ideal $(\frac{n}{d})$ has the largest element.

Proof.

For each $i\in\{1,2,\cdots,k\}$, let $\beta_{i},\gamma_{i}\in\mathbb{W}$ be the largest powers of prime $p_{i}$ such that $a\equiv 1~{}(mod~{}p_{i}^{\beta_{i}})$ and $b\equiv 1~{}(mod~{}p_{i}^{\gamma_{i}})$ respectively. Let $f_{i}=max\{(\alpha_{i}-\beta_{i}),(\alpha_{i}-\gamma_{i}),0\}$ and $m=\displaystyle\prod_{i=1}^{k}p_{i}^{f_{i}}$. Then $m=\frac{n}{d}$. Let $c\in\mathbb{Z}_{n}$ and for each $i\in\{1,2,\cdots,k\}$, let $s_{i}\in\mathbb{N}$ be the largest powers of prime $p_{i}$ such that $c\equiv 0~{}(mod~{}p_{i}^{s_{i}})$. Then, $c<a$ and $c<b$ if and only if $c(a-1)\equiv 0~{}(mod~{}n)$ and $c(b-1)\equiv 0~{}(mod~{}n)$ if and only if $s_{i}\geq(\alpha_{i}-\beta_{i}),(\alpha_{i}-\gamma_{i})$ for all $i$ if and only if $c\in(\frac{n}{d})$.

Suppose $a\wedge b$ exists. Since $a$ and $b$ are incomparable, we have $a\wedge b<a$ and $a\wedge b<b$. This yields $a\wedge b\in(\frac{n}{d})$. Let $x\in(\frac{n}{d})$. Then $x<a$ and $x<b$. Therefore $x<a\wedge b$. Thus $a\wedge b$ is the largest element of the ideal $(\frac{n}{d})$. Conversely, suppose that the ideal $(\frac{n}{d})$ has the largest element, say $e\in(\frac{n}{d})$. This yields $e<a$ and $e<b$. We claim that $a\wedge b=e$. Let $f\in\mathbb{Z}_{n}$ be such that $f<a$ and $f<b$. Then $f\in(\frac{n}{d})$. Therefore $f\leq e$. Thus $a\wedge b=e$. ∎

From the proof of Theorem 3.3, it is clear that if $a\wedge b$ exists, then $a\wedge b$ is the largest element of the ideal $(\frac{n}{d})$. Also, if the ideal $(\frac{n}{d})$ has the largest element $e$, then $a\wedge b=e$.

The following corollary is an immediate consequence of Theorem 3.3.

Corollary 3.4**.**

Let $n\in\mathbb{N},~{}n>1$ and $S^{\prime}=\{d\in\mathbb{N}~{}|~{}d=gcd(gcd(a-1,b-1),n),$ for some incomparable elements $a,b\in\mathbb{Z}_{n}\}$. Then, $\mathbb{Z}_{n}$ is a lattice if and only if every ideal in $\{(\frac{n}{d})~{}|~{}d\in S^{\prime}\}$ has largest element.

Corollary 3.5**.**

Let $n\in\mathbb{N},~{}n>1$, $S=\{d\in\mathbb{N}~{}|~{}d=gcd(gcd(a,b),n),$ for some incomparable elements $a,b\in\mathbb{Z}_{n}\}$ and $S^{\prime}=\{d\in\mathbb{N}~{}|~{}d=gcd(gcd(a-1,b-1),n),$ for some incomparable elements $a,b\in\mathbb{Z}_{n}\}$. Then, every ideal in $\{(\frac{n}{d})~{}|~{}d\in S^{\prime}\}$ has largest element if and only if every coset in $\{(\frac{n}{d})+1~{}|~{}d\in S\}$ has smallest element.

Proof.

Follows from Corollaries 3.2 and 3.4. ∎

The following two lemmas relate the largest element of an ideal with the smallest element of a coset and vice versa.

Lemma 3.6**.**

Let $n=n_{1}n_{2}$ with $n_{1}\geq 1,n_{2}\geq 3$ and $I=(n_{1})$, $J=(n_{2})$. Then, the largest element of the ideal $I$ becomes the smallest element of the coset $J+1$.

Proof.

Since $|I|=n_{2}\geq 3$, we have $n_{1}\nequiv-n_{1}~{}(mod~{}n)$. Therefore $n_{1}$ and $-n_{1}$ are distinct elements in $I$. Let $e_{1}n_{1}\in I$ be the largest element of $I$. Then $x_{1}n_{1}\leq e_{1}n_{1}$ for all $x_{1}\in\mathbb{Z}$. This yields $n_{1}\leq e_{1}n_{1}$ and $-n_{1}\leq e_{1}n_{1}$. That is $n_{1}\equiv e_{1}n_{1}~{}(mod~{}n)$ or $n_{1}\equiv n_{1}e_{1}n_{1}~{}(mod~{}n)$; and $-n_{1}\equiv e_{1}n_{1}~{}(mod~{}n)$ or $-n_{1}\equiv-n_{1}e_{1}n_{1}~{}(mod~{}n)$. If $n_{1}\equiv e_{1}n_{1}~{}(mod~{}n)$ and $-n_{1}\equiv e_{1}n_{1}~{}(mod~{}n)$, then $n_{1}\equiv-n_{1}~{}(mod~{}n)$, a contradiction to the fact that $n_{2}\geq 3$. Thus, either $n_{1}\equiv n_{1}e_{1}n_{1}~{}(mod~{}n)$ or $-n_{1}\equiv-n_{1}e_{1}n_{1}~{}(mod~{}n)$. Suppose $n_{1}\equiv n_{1}e_{1}n_{1}~{}(mod~{}n)$. That is $n_{1}(e_{1}n_{1}-1)\equiv 0~{}(mod~{}n_{1}n_{2})$. This implies that $e_{1}n_{1}-1\equiv 0~{}(mod~{}n_{2})$, hence $e_{1}n_{1}\in J+1$. Similarly, $-n_{1}\equiv-n_{1}e_{1}n_{1}~{}(mod~{}n)$ implies that $e_{1}n_{1}\in J+1$. Thus, in any case, $e_{1}n_{1}\in J+1$. Let $y_{2}n_{2}+1\in J+1$ be any element. Then $(y_{2}n_{2}+1)e_{1}n_{1}=y_{2}n_{2}e_{1}n_{1}+e_{1}n_{1}\equiv e_{1}n_{1}~{}(mod~{}n)$. Thus $e_{1}n_{1}$ is the smallest element of $J+1$. ∎

Lemma 3.7**.**

Let $n=n_{1}n_{2}$ with $n_{1}\geq 3,n_{2}\geq 1$ and $I=(n_{1})$, $J=(n_{2})$. Then the smallest element of the coset $J+1$ becomes the largest element of the ideal $I$.

Proof.

Since $|J|=n_{1}\geq 3$, we have $n_{2}\nequiv-n_{2}~{}(mod~{}n)$. Therefore $n_{2}+1$ and $-n_{2}+1$ are distinct elements in the coset $J+1$. Let $e_{2}n_{2}+1\in J+1$ be the smallest element of $J+1$. Then $e_{2}n_{2}+1\leq x_{2}n_{2}+1$ for all $x_{2}\in\mathbb{Z}$. This yields $e_{2}n_{2}+1\leq n_{2}+1$ and $e_{2}n_{2}+1\leq-n_{2}+1$. That is $e_{2}n_{2}+1\equiv n_{2}+1~{}(mod~{}n)$ or $e_{2}n_{2}+1\equiv(e_{2}n_{2}+1)(n_{2}+1)~{}(mod~{}n)$; and $e_{2}n_{2}+1\equiv-n_{2}+1~{}(mod~{}n)$ or $e_{2}n_{2}+1\equiv(e_{2}n_{2}+1)(-n_{2}+1)~{}(mod~{}n)$. If $e_{2}n_{2}+1\equiv n_{2}+1~{}(mod~{}n)$ and $e_{2}n_{2}+1\equiv-n_{2}+1~{}(mod~{}n)$, then $n_{2}+1\equiv-n_{2}+1~{}(mod~{}n)$, a contradiction to the fact that $n_{1}\geq 3$. Thus, either $e_{2}n_{2}+1\equiv(e_{2}n_{2}+1)(n_{2}+1)~{}(mod~{}n)$ or $e_{2}n_{2}+1\equiv(e_{2}n_{2}+1)(-n_{2}+1)~{}(mod~{}n)$. Suppose $e_{2}n_{2}+1\equiv(e_{2}n_{2}+1)(n_{2}+1)~{}(mod~{}n)$. That is $(e_{2}n_{2}+1)(n_{2})\equiv 0~{}(mod~{}n_{1}n_{2})$. This implies that $e_{2}n_{2}+1\equiv 0~{}(mod~{}n_{1})$, hence $e_{2}n_{2}+1\in I$. Similarly, $e_{2}n_{2}+1\equiv(e_{2}n_{2}+1)(-n_{2}+1)~{}(mod~{}n)$ implies that $e_{2}n_{2}+1\in I$. Thus, in any case, $e_{2}n_{2}+1\in I$. Let $y_{1}n_{1}\in I$ be any element. Then $(y_{1}n_{1})(e_{2}n_{2}+1)=y_{1}n_{1}e_{2}n_{2}+y_{1}n_{1}\equiv y_{1}n_{1}~{}(mod~{}n)$. Thus $e_{2}n_{2}+1$ is the largest element of $I$. ∎

Remark 3.8**.**

Let $a\in GP(\mathbb{Z}_{n})$ and $I$ be the ideal generated by $a$. Then $a_{u}$ is the largest element of $I$. For $ma\in I$, $maa_{u}=ma$, hence $ma\leq a_{u}$.

If $a,b\in GP(\mathbb{Z}_{n})$ then $a\vee b$ and $a\wedge b$ both exists in the poset $GP(\mathbb{Z}_{n})$ (see [1]). The following two theorems gives the existence of $a\vee b$ and $a\wedge b$ in the poset $\mathbb{Z}_{n}$ where $a,b\in GP(\mathbb{Z}_{n})$.

Theorem 3.9**.**

If $a,b\in GP(\mathbb{Z}_{n})$ then $a\vee b$ exists in the poset $\mathbb{Z}_{n}$. Further, $a\vee b\in GP(\mathbb{Z}_{n})$.

Proof.

If $a$ and $b$ are comparable then clearly $a\vee b$ exists and $a\vee b\in GP(\mathbb{Z}_{n})$. Suppose $a$ and $b$ are incomparable. Let $d=gcd(gcd(a,b),n)$, $I$ be the ideal generated by $d$ and $J$ be the ideal generated by $\frac{n}{d}$. As, $a,b\in GP(\mathbb{Z}_{n})$, by Remark 1.2(iii), $gcd(a,b)\in GP(\mathbb{Z}_{n})$. Therefore $d\in GP(\mathbb{Z}_{n})$. By Remark 3.8, the ideal $I$ possesses the largest element, say $e$ and $e\in GP(\mathbb{Z}_{n})$. Since $d|a$ and $d|b$, we have $a,b\in I$. As, $a$ and $b$ are incomparable, we have $|I|=\frac{n}{d}\geq 3$. By Lemma 3.6, $e$ becomes the smallest element of the coset $J+1$. By Theorem 3.1, $a\vee b$ exists and $a\vee b=e$. Thus, $a\vee b=e\in GP(\mathbb{Z}_{n})$. ∎

Theorem 3.10**.**

Let $a,b\in GP(\mathbb{Z}_{n})$ be such that $a-1,b-1\in GP(\mathbb{Z}_{n})$. Then $a\wedge b$ exists in the poset $\mathbb{Z}_{n}$ and $a\wedge b\in GP(\mathbb{Z}_{n})$.

Proof.

If $a$ and $b$ are comparable then clearly $a\wedge b$ exists and $a\wedge b\in GP(\mathbb{Z}_{n})$. Suppose $a$ and $b$ are incomparable. Let $d=gcd(gcd(a-1,b-1),n)$ and $I$ be the ideal generated by $\frac{n}{d}$. As, $a-1,b-1\in GP(\mathbb{Z}_{n})$, by Remark 1.2(iii), $gcd(a-1,b-1)\in GP(\mathbb{Z}_{n})$. Therefore $d\in GP(\mathbb{Z}_{n})$, and hence $\frac{n}{d}\in GP(\mathbb{Z}_{n})$. By Remark 3.8, the ideal $I$ possesses the largest element, say $e$ and $e\in GP(\mathbb{Z}_{n})$. By Theorem 3.3, $a\wedge b$ exists and $a\wedge b=e$. Thus $a\wedge b=e\in GP(\mathbb{Z}_{n})$. ∎

In the following theorem, we give a necessary and sufficient condition for the poset $\mathbb{Z}_{n}$ to be a lattice.

Theorem 3.11**.**

Let $n=\displaystyle\prod_{i=1}^{k}p_{i}^{\alpha_{i}}$ be the prime factorization of $n\in\mathbb{N}$ with $\alpha_{i}>0$, for all $i$. Then, $\mathbb{Z}_{n}$ is a lattice if and only if for every $n_{1}\geq 3$ with $n=n_{1}n_{2}$, $(n_{1})$ possess the largest element.

Proof.

Suppose $\mathbb{Z}_{n}$ is a lattice. Let $n=n_{1}n_{2}$ with $n_{1}\geq 3$. If $(n_{1})$ does not possess the largest element, then $|(n_{1})|=n_{2}\geq 3$. Let $a=n_{1}$; and $b=pn_{1}$, where $p$ is a prime such that $gcd(p,n_{2})=1$ and $n_{2}\nmid(p-1)$. Then $a,b\nequiv 0~{}(mod~{}n)$ and $a\nequiv b~{}(mod~{}n)$. If $a<b$, then $a\equiv ab~{}(mod~{}n)$. That is $n_{1}\equiv n_{1}pn_{1}~{}(mod~{}n)$. This yields $n_{1}(pn_{1}-1)\equiv 0~{}(mod~{}n)$. This implies that $pn_{1}\equiv 1~{}(mod~{}n_{2})$. Hence $gcd(n_{1},n_{2})=1$. Consequently $n_{1}\in GP(\mathbb{Z}_{n})$. By Remark 3.8, $(n_{1})$ possess the largest element, a contradiction. Therefore $a\nleq b$. Similarly, $b\nleq a$. Thus $a$ and $b$ are incomparable. Observe that $gcd(gcd(a,b),n)=n_{1}$. As, $(\frac{n}{n_{2}})=(n_{1})$ does not possess the largest element. By Lemma 3.7, the coset $(n_{2})+1$ does not possess the smallest element. By Corollary 3.2, $\mathbb{Z}_{n}$ is not a lattice, a contradiction. Therefore $(n_{1})$ possess the largest element. Conversely, suppose for every $n_{1}\geq 3$ with $n=n_{1}n_{2}$, $(n_{1})$ possess the largest element. If $\mathbb{Z}_{n}$ is not a lattice, then by Corollary 3.2, there exists incomparable elements $a^{\prime},b^{\prime}\in\mathbb{Z}_{n}$ such that $d^{\prime}=gcd(gcd(a^{\prime},b^{\prime}),n)$ and $(\frac{n}{d^{\prime}})+1$ does not possess the smallest element. Therefore $\frac{n}{d^{\prime}}=|(d^{\prime})|\geq 3$ and $|(\frac{n}{d^{\prime}})+1|\geq 3$. Hence $|(\frac{n}{d^{\prime}})|=d^{\prime}\geq 3$. Let $n_{1}^{\prime}=d^{\prime}$ and $n_{2}^{\prime}=\frac{n}{d^{\prime}}$. Then $n_{1}^{\prime}\geq 3$ and $n=n_{1}^{\prime}n_{2}^{\prime}$. By Lemma 3.6, $(n_{1}^{\prime})$ does not possess the largest element, a contradiction. Thus $\mathbb{Z}_{n}$ is a lattice. ∎

Remark 3.12**.**

Let $a\in N(\mathbb{Z}_{n})\backslash\{0\}$. If $b\in\mathbb{Z}_{n}$ be such that $b<a$ then $b=ba=ba^{m}$ for any $m\in\mathbb{N}$. Since $a\in N(\mathbb{Z}_{n})$, we have $b=0$. Hence $a_{l}=0$. From this, it follows that, if $I$ is an ideal generated by a nilpotent element of $\mathbb{Z}_{n}$ such that $|I|\geq 3$, then $I$ does not possess the largest element. Thus $\mathbb{Z}_{n}$ is not a lattice.