Full article: Some Results on Filters and Congruence Relations of Hoops

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ABSTRACT

This manuscript presents some results on the conditions and characterizations of filters, congruence relations and homomorphisms in algebraic hoops. The paper investigates the structural properties of filters that are generated by a set in various ways and establishes several descriptions for such filters. It is demonstrated in this paper that the class of filters in hoops forms an algebraic lattice. This finding contributes to our understanding of the structural properties of filters and their relationship within hoops. Moreover, this manuscript explores congruence relations in hoops, revealing a fascinating connection between the lattice of filters and the lattice of congruences. In particular, it is shown that the lattice of filters is in one-to-one correspondence with the lattice of congruences in hoops, confirming that the variety of hoops is an ideal determined. Additionally, the manuscript delves into the properties of hoop homomorphisms in relation to congruences and filters, as well as to quotient structures. This paper presents several homomorphism and correspondence theorems for hoops.

KEYWORDS:

MATHEMATICS SUBJECT CLASSIFICATION:

1. Introduction

Hoops, originated by Bosbach (Bosbach, Citation1969, Citation1970), are algebras with a binary operation satisfying a set of axioms that have been an area of significant interest and research in algebraic logics. In recent years, there have been significant advancements in hoop theory, as evidenced by the introduction of deep structure theorems (see (Aaly Kologani & Borzooei, Citation2019; Aglianó et al., Citation2007; Borzooei & Aaly Kologani, Citation2014b; Namdar et al., Citation2017)). Many of these findings have had a significant impact on fuzzy logic. One noteworthy result is the structure theorem of finite basic hoops (as stated in (Aglianó et al., Citation2007) Corollary 2.10), which provides an elegant and concise proof of the completeness theorem for propositional basic logic (refer to (Aaly Kologani & Borzooei, Citation2019), Theorem 3.8). This completeness theorem was originally introduced by Hajek in (Hájek, Citation1998). Currently, the study of filter theory in hoops has gained considerable attention, leading to the discovery of important results. Notably, various types of filters, such as (positive) implicative filters and fantastic filters (as discussed in (Borzooei & Aaly Kologani, Citation2014a)), have been introduced, and their characteristics have been presented in (Alavi et al., Citation2017; Borzooei & Kologani, Citation2014; Kondo & Dudek, Citation2008; Namdar et al., Citation2017).

In this manuscript, we delve into the study of filters in hoops and derive several important characterizations and properties related to them. The concept of filters in hoops has been a subject of investigation due to its relevance in understanding the algebraic structure and properties of hoops. Filters are subsets of hoops that possess certain characteristics and play a fundamental role in the algebraic formulation and analysis of hoops. By understanding the conditions under which a set in a hoop can be classified as a filter, we can gain insights into the fundamental properties and structures within a hoop.

The primary focus of this manuscript is to obtain various characterizations of filters generated by a set in different ways. By deriving these conditions, we aim to provide a comprehensive understanding of the structures and properties of filters. This exploration leads us to establish that the class of filters in hoops forms an algebraic lattice. This result not only deepens our understanding of the relationships between filters but also lays the foundation for further investigation into the lattice structures within a hoop. We further investigate the structural properties of those implicative filters in hoops. Interestingly, we obtain the smallest implicative filter $I_{0} (R)$ in a hoop $R$ and we give an algebraic description for implicative filters generated by a set in general and by a filter in particular. It is also proved that the collection of all implicative filters in a hoop $R$ is complete and is a filter in the lattice of all filters of $R$ .

The other important aspect of our study involves the examination of congruence relations on hoops. Congruence relations, which are equivalence relations on a hoop preserving its algebraic operations, provide valuable insights into the structure and behavior of hoops. We show that the lattice of filters in hoops is in one-to-one correspondence with the lattice of congruences, establishing a strong connection between these two fundamental aspects of hoop theory. This correspondence further solidifies the idea that the variety of hoops is ideally determined.

In addition to exploring the relationships between filters and congruences, we thoroughly investigate the properties of hoop homomorphisms in connection with congruences and quotient structures. Homomorphisms are morphisms that preserve the algebraic structure between hoops, and by studying their interactions with congruences and filters, we can uncover additional insights into the relationships and correlations within hoops.

2. Preliminaries

In this section, we present the necessary background and introduce the fundamental concepts that form the basis of our study. A hoop is an algebraic structure that combines the properties of groups and semigroups, providing a framework for studying operations that are both associative and partially invertible.

Definition 2.1.

(Bosbach, Citation1969, Citation1970) A hoop is a system $⟨ R, \circ, \to, 1 ⟩$ ; consisting of a non-empty set $R$ , binary operations $\circ$ and $\to$ and a constant 1 in $R$ satisfying the following conditions:

(1)	$R$ together with $\circ$ is a commutative semigroup with identity 1;
(2)	$h \to h = 1$ ;
(3)	$(h \circ g) \to s = h \to (g \to s)$
(4)	$h \circ (h \to g) = g \circ (g \to h)$

for all $h, g, s \in R$ . A hoop $R$ can be made into a partially ordered set together with a natural ordering ≤ on $R$ given by $h \leq g$ if and only if $h \to g = 1$ .

It is observed in Bosbach (Citation1970) that infimum of any two elements (with respect to the natural order) in a hoop $R$ always exists, and it is given as:

\begin{array}{l} h \land g = h \circ (h \to g) for all h, g \in R \end{array}

So that $R$ together with the natural ordering can be viewed as a meet semi-lattice. Moreover, the binary operations $\circ$ and $\to$ satisfies the adjointness property; i.e. for any $h, g, s \in R$ it holds that

\begin{array}{l} h \leq g \to s if~and only if h \circ g \leq s \end{array}

Theorem 2.2.

Every hoop $R$ satisfies the following properties:

(1)	$h \circ g \leq h, g$ and $s \leq g \to s$ ;
(2)	$h \to 1 = 1$ and $1 \to h = h$ ;
(3)	$h \leq g \to (h \circ g)$ ;
(4)	$h \to g \leq (g \to s) \to (h \to s)$ ;
(5)	$h \to g \leq (s \to h) \to (s \to g)$ ;
(6)	$h \leq g \Rightarrow h \circ s \leq g \circ s$ , $s \to h \leq s \to g$ and $g \to s \leq h \to s$

Lemma 2.3.

(Aglianó et al., Citation2007) Given a hoop $R$ , the pseudo-join of h and g in $R$ , denoted $h \underset{―}{\lor} g$ , is defined by:

\begin{array}{l} h \underset{―}{\lor} g = ((h \to g) \to g) \land ((g \to h) \to h) . \end{array}

Then for any, $h, g, s \in H$ ;

(1)	$h \underset{―}{\lor} g = g \underset{―}{\lor} h$ and $h, g \leq h \underset{―}{\lor} g$ ;
(2)	$h \leq g$ if and only if $h \underset{―}{\lor} g = g$ ;
(3)	$(h \to g) \underset{―}{\lor} (g \to h) = 1$ .

Lemma 2.4

(Aglianó et al., Citation2007). The following are equivalent in a hoop $R$ :

(1)	$\underset{―}{\lor}$ is associative;
(2)	for all $h, g, s \in R$ ; $h \leq g \Rightarrow h \underset{―}{\lor} s \leq g \underset{―}{\lor} s$ ;
(3)	for all $h, g, s \in R$ ; $h \underset{―}{\lor} (g \land s) \leq (h \underset{―}{\lor} g) \land (h \underset{―}{\lor} s)$ ,
(4)	for all $h, g, s \in R$ ; $h, g \leq s \Rightarrow h \underset{―}{\lor} g \leq s$ .

Definition 2.5.

(Aglianó et al., Citation2007) By a join hoop, we mean a hoop $R$ satisfying one and hence all other conditions of Lemma 2.4.

3. Some new results on filters

Filters in hoops extend the notion of filters in Boolean algebras and serve as a powerful tool for characterizing subsets that exhibit certain desirable properties.

Definition 3.1.

(Aglianó et al., Citation2007). A filter in a hoop $R$ is a subset U of $R$ such that:

(F1)	$1 \in U$
(F2)	$h, h \to g \in U \Rightarrow g \in U$ for all $h, g \in R$

The collection of all filters in $R$ will be denoted by $F (R)$ .

Theorem 3.2.

A set U contained in $R$ is a filter if and only if the following conditions are satisfied:

(F1)	$1 \in U$
(F2)	$h, g \in U \Rightarrow h \circ g \in U$
(F3)	$h \in U$ and $h \leq g$ together imply $g \in U$

for all $h, g \in R$ .

Proof.

Assume that U is a filter. We show that U satisfies $(F 3)$ and $(F 4)$ . Let $h, g \in U$ . Then $h \to (g \to (h \circ g)) = 1 \in U$ . This implies that $g \to (h \circ g) \in U$ and so $h \circ g \in U .$ Therefore $(F 3)$ holds. Again $h \in U$ and $h \leq g$ imply $h, h \to g = 1 \in U$ . Thus $g \in U$ and hence $(F 4)$ holds.

Conversely, suppose that U satisfies the conditions $(F 1)$ , $(F 3)$ and $(F 4)$ . We claim that U is a filter. It is enough to show that U satisfies $(F 2)$ . Let $h, h \to g \in U .$ Then, by $(F 3)$ , $h \circ (h \to g) \in U .$ Since $h \circ (h \to g) = g \circ (g \to h) \leq g,$ it follows from $(F 4)$ that $g \in U$ and hence U is a filter.

Lemma 3.3.

Let $h, g, z \in R$ and $U \in F (R)$ . Then, we have:

(1)	$h \to g \in U \Rightarrow (h \circ s) \to (g \circ s) \in U$ for all $s \in R$ ;
(2)	$h \to g \in U \Rightarrow (s \to h) \to (s \to g) \in U$ for all $s \in R$ ;
(3)	$h \to g \in U \Rightarrow (g \to s) \to (h \to s) \in U$ for all $s \in R$ ;
(4)	$h \to g \in U, g \to z \in U \Rightarrow h \to z \in U$ .

Lemma 3.4.

Let V be a filter in $R$ . If U is a filter of V (considering V as a hoop with the restricted operations of $R$ ), then U is a filter in $R$ .

Theorem 3.5.

A nonempty set U contained in $R$ is a filter if and only if $D (h, g) \subseteq U$ for all $h, g \in U$ ; where

\begin{array}{l} D (h, g) = {z \in R : h \to (g \to z) = 1} \end{array}

Proof.

$(\Rightarrow)$ Let $h, g \in U$ and $z \in D (h, g) .$ Then $h \to (g \to z) = 1 \in U,$ and so by assumption it is clear that $z \in U$ . $(\Leftarrow)$ Since U is nonempty, we can choose $h, g \in U$ . As $h \to (g \to 1) = 1$ , it follows from our assumption that $1 \in U .$ Let $h, h \to g \in U .$ Then, by assumption, $D (h, h \to g) \subseteq F .$ Since, $h \to ((h \to g) \to g) = 1,$ $g \in D (h, h \to g) \subseteq F .$

It is routine to verify that the class $F (R)$ of all filters of $R$ is closed under arbitrary intersection. It follows then that for any set $X \subseteq R$ always we can find the smallest filter of $R$ containing X. Usually, it is called the filter of $R$ generated by X, and it is denoted by Fg(X). If X is empty, then it is trivial to check that $Fg (X) = {1}$ . The following theorem gives an equational description for filters generated by a non-empty set.

Theorem 3.6.

Let X be a non-empty set contained in $R$ . Then,

\begin{array}{l} Fg (X) = {k \in R : h_{1} \circ h_{2} \circ \dots \circ h_{n} \leq k for some \\ h_{1}, \dots, h_{n} \in X} \end{array}

Proof.

Put $W = {k \in R : h_{1} \circ h_{2} \circ \dots \circ h_{n} \leq k for some h_{1}, \dots, h_{n} \in X} .$ We show that W is the least filter in $R$ containing X. Clearly $X \subseteq W$ and $1 \in W .$ Let $h, h \to g \in W .$ Then $h_{1} \circ h_{2} \circ \dots \circ h_{n} \leq h$ and $g_{1} \circ g_{2} \circ \dots \circ g_{m} \leq h \to g for some h_{1}, \dots, h_{n}, g_{1}, \dots, g_{m} \in X$ and so

\begin{array}{l} h_{1} \circ h_{2} \circ \dots \circ h_{n} \circ g_{1} \circ g_{2} \circ \dots \circ g_{m} \leq h \circ (h \to g) \leq g . \end{array}

Therefore W is a filter. Suppose that Z is any filter containing X and let $t \in W$ . Then, we can choose $h_{1}, \dots, h_{n} \in X$ with $h_{1} \circ h_{2} \circ \dots \circ h_{n} \leq t .$ Then it is clear that $t \in Z .$ This proves that $W = Fg (X) .$

In order to get another description for Fg(X), let us define sets ${X_{n}}_{n \geq 1}$ inductively as follows:

\begin{array}{l} X_{1} = X \cup {1} and\ for n \geq 1, \\ X_{n + 1} = {g \in R : h \in X_{n} and h \to g \in X_{n}} \end{array}

Then we have the following theorem.

Theorem 3.7.

For any $X \subseteq R$ :

\begin{array}{l} Fg (X) = \cup {X_{n} : n \in N} \end{array}

where $N$ denotes the set of positive integers.

Proof.

Put $T = \cup {X_{n} : n \in N} .$ Then, it can be easily shown that $1 \in X_{n}$ and $X_{n} \subseteq X_{n + 1} . ∀n \in N$ . Then $1 \in T$ and $X \subseteq T .$ Let $h, h \to g \in T$ . Then $∃n, m \in N$ such that $h \in X_{n}$ and $h \to g \in X_{m} .$ Let $p = max {n, m} .$ Then $h, h \to g \in X_{p}$ and so $g \in X_{p + 1} \subseteq ⋃ {X_{n} : n \in N} = T .$ Therefore T is a filter. Let K be any filter such that $X \subseteq K .$ Now we show $X_{n} \subseteq K$ for all $n \in N,$ by induction. Since $X \subseteq K$ and $1 \in K,$ we have $X_{1} = X \cup {1} \subseteq K .$ Thus the statement is true for n = 1. Let $n \geq 1$ and assume that $X_{n} \subseteq K .$ Now $g \in X_{n + 1}$ implies that $∃h \in X_{n}$ and $h \to g \in X_{n}$ and thus $g \in K .$ Therefore $X_{n + 1} \subseteq K$ and so by mathematical induction, $X_{n} \subseteq K$ for all $n \in N$ and hence $T = ⋃ {X_{n} : n \in N} \subseteq K .$ This proves that $Fg (X) = ⋃ {X_{n} : n \in N}$

For any $X \subseteq R$ define sets ${e_{n} (X)}_{n = 1}^{\infty}$ inductively as follows:

\begin{array}{l} e_{1} (X) = X \cup {1} and\ for n \geq 1, \\ e_{n + 1} (X) = {z \in R : h \circ g \leq z for some h, g \in e_{n} (X)} \end{array}

Then we have the following theorem.

Theorem 3.8.

For any set X contained in $R$ :

\begin{array}{l} Fg (X) = ⋃ {e_{n} (X) : n \in N} \end{array}

where $N$ denotes the set of positive integers.

Proof.

Put $M = ⋃ {e_{n} (X) : n \in N} .$ Then, we can easily show that clearly $1 \in e_{n} (X)$ and $e_{n} (X) \subseteq e_{n + 1} (X), ∀n \in N .$ Thus $X \subseteq M$ and $1 \in M .$ Let $z, z \to w \in M .$ Then there exist $n, m \in N$ such that $z \in e_{n} (X)$ and $z \to w \in X_{m} .$ Let $p = max {n, m} .$ Then $z, z \to w \in e_{p} (X)$ and since $z \circ (z \to w) \leq w,$ $w \in e_{p + 1} (X) \subseteq ⋃ {e_{n} (X) : n \in N} = T .$ Therefore T is a filter. Let K be any filter such that $X \subseteq K .$ Now we show $e_{n} (X) \subseteq K$ for all $n \in N .$ We use induction on n. Since $X \subseteq K$ and $1 \in K,$ we have $e_{1} (X) = X \cup {1} \subseteq K .$ Thus the statement is true for n = 1. Let $n \geq 1$ and assume that $e_{n} (X) \subseteq K .$ Now $z \in e_{n + 1} (X)$ implies that $∃h, g \in e_{n} (x)$ such that $h \circ g \leq z .$ Thus $h, g \in e_{n} (X) \subseteq K$ implying $h \circ g \in K$ and hence $z \in K .$ Therefore $e_{n + 1} (X) \subseteq K .$ Thus, by mathematical induction, $e_{n} (X) \subseteq K$ for all $n \in N$ and hence $T = ⋃ {e_{n} (X) : n \in N} \subseteq K$ . This proves that $T = Fg (X)$ .

Lemma 3.9

For any $U, V \in F (R),$ the infimum $U \land V$ and supremum $U \lor V$ of U and V are given, respectively, by:

\begin{array}{l} U \land V & = U \cap V and U \lor V \\ = {z \in R : u \circ v \leq z for some u \in U and v \in V} . \end{array}

Proof.

Put $T = {z \in R : u \circ v \leq z for some u \in U and v \in V} .$ Now we claim that T is the least filter containing $U \cup V .$ Clearly, $U \cup V \subseteq T$ . Let $z, w \in R$ be such that $z, z \to w \in T .$ There exist $u_{1}, u_{2} \in U and v_{1}, v_{2} \in V$ such that $u_{1} \circ v_{1} \leq z$ and $u_{2} \circ v_{2} \leq z \to w .$ Thus we have

\begin{array}{l} (u_{1} \circ u_{2}) \circ (v_{1} \circ v_{2}) \leq z \circ (z \to w) \leq w \end{array}

Since $u_{1} \circ u_{2} \in U$ and $v_{1} \circ v_{2} \in V,$ we have $w \in T$ . Therefore, T is a filter. Let K be any filter such that $U \cup V \subseteq K$ . Then $z \in T$ implies that $u \circ v \leq z$ for some $u \in U$ and $v \in V .$ Thus $u, v \in K$ and $u \circ v \in K$ and so $z \in K$ . This proves that $U \lor V = T .$

Corollary 3.10.

$⟨ F (R), \land, \lor ⟩$ forms a lattice.

Theorem 3.11

The map $X \mapsto Fg (X) : P (R) \to P (R)$ is an algebraic closure operator.

Proof.

Clearly, the given map forms a closure operator on $R$ . Now we show that for any $X \subseteq R,$

\begin{array}{l} Fg (X) = ⋃_{E \subset\subset X} Fg (E), \end{array}

where $E \subset\subset X$ is to mean E is a finite subset of $X .$ Let $k \in Fg (X) .$ Then $h_{1} \circ h_{2} \circ \dots \circ h_{n} \leq k$ for some $h_{1}, \dots, h_{n} \in X$ . Put $E = {h_{1}, \dots, h_{n}} .$ Then $E \subset\subset X$ and $k \in Fg (E) \subseteq ⋃_{E \subset\subset X} Fg (E) .$ Thus, $Fg (X) \subseteq ⋃_{E \subset\subset X} Fg (E)$ , and the other inclusion is obvious. Therefore, the given map is an algebraic closure operator.

Corollary 3.12.

The pair $⟨ F (R), \subseteq ⟩$ forms an algebraic lattice.

Proof.

It is proved in the above theorem that the map $X \mapsto Fg (X)$ is an algebraic closure operator on $R$ . Then, the class $L_{C} = {U \subseteq R : Fg (U) = U}$ of closed elements of $P (R)$ with respect to this closure operator is an algebraic lattice (see Theorem 5.5 of (Burris & Sankappanavor, Citation1981)). It is also clear that $Fg (U) = U$ if and only if U is a filter in $R$ , and hence we have $L_{C} = F (R)$ . Therefore, $F (R)$ is an algebraic lattice.

4. Implicative filters

In this section, we recall the definition of implicative filters in hoops, and we present some results mainly focusing on the characterization of implicative filters generated by a set.

Definition 4.1.

A set M in $R$ is defined to be an implicative filter if:

(IF1)	$1 \in M$
(IF2)	$h \to (g \to z) \in M$ and $h \to g \in M$ imply $h \to z \in M .$

Theorem 4.2.

Every implicative filter is a filter.

Proof.

Let M be an implicative filter in $R$ and $h, g \in M .$ We first show that $h \to (h \circ g) \in M$ . Put $r = 1, s = g$ and $t = h \to (h \circ g) .$ Then, it can be shown that $r \to (s \to t) \in M$ and $r \to s = g .$ Thus by $(FI 2)$ , $r \to t \in M,$ i.e. $h \to (h \circ g) \in M .$ Now we claim that $h \circ g \in M .$ Put $r = 1, s = h$ and $t = h \circ g .$ Then $r \to (s \to t) = s \to t = h \to (h \circ g) \in M .$ Also $r \to s = 1 \to h = h \in M .$ Thus, by $(FI 2),$ $r \to s \in M$ so $h \circ g \in M .$ Next let $h, g \in R$ such that $h \leq g$ and $h \in M .$ Then $h \to g = 1 \in M .$ Put $r = 1, s = h$ and $t = g .$ Then $r \to (s \to t) = s \to t = h \to g = 1 \in M$ and $r \to s = 1 \to h = h \in M .$ Thus, by $(FI 2),$ $r \to s = s = g \in M .$ There fore M is a filter.

The following theorem gives a set of equivalent conditions for a set to be an implicative filter and the proof is ilar to that of Haveshki et al. (Citation2006).

Theorem 4.3

For any set M contained in $R$ the following are equivalent.

(1)	M is an implicative filter;
(2)	M is a filter and $g \to (g \to h) \in M \Rightarrow g \to h \in M, ∀h, g \in R$ ;
(3)	M is a filter and $z \to (g \to h) \in M \Rightarrow (z \to g) \to (z \to h) \in M, ∀h, g, z \in R$ ;
(4)	$1 \in M$ , $z \to (g \to (g \to h)) \in M$ and $z \in M$ imply $g \to h \in M \forall h, g, z \in R$ .

Theorem 4.4

Any filter U of $R$ is an implicative filter if and only if $h \to h^{2} \in U$ for all $h \in R$ ; where $h^{2} = h \circ h$ .

Given a set X in $R$ , in the following theorem we describe the least implicative filter containing X.

Theorem 4.5

For a set X contained in $R$ , let us define an indexed family ${I_{n} (X)}_{n = 1}^{\infty}$ of sets as follows: $I_{1} (X) = X \cup {1}$ and for $n \geq 1$ ;

\begin{array}{l} I_{n + 1} (X) = {h \to z : h \to (g \to z) \in I_{n} (X) and \\ h \to g \in I_{n} (X) for some g \in R} . \end{array}

Then $⋃_{n = 1}^{\infty} I_{n} (X)$ is the least implicative filter containing $X .$

Proof.

We first show that $z \in I_{n} (X)$ for all $n \in N$ . Clearly, $1 \in I_{1} (X) .$ We use induction on $n .$ Let $n \geq 2$ and assume that $1 \in I_{n} (X) .$ We claim that $1 \in I_{n + 1} (X) .$ As $1 \in I_{n} (X)$ we can write 1 as $1 \to (1 \to 1) \in I_{n} (X)$ and $1 = 1 \to 1 \in I_{n} (X) .$ This implies that $1 = 1 \to 1 \in I_{n + 1} (X) .$ Therefore $1 \in I_{n} (X), ∀n \in N .$

Next, we show that $I_{n} (X) \subseteq I_{n + 1} (X)$ for all $n \in N$ . Taking n = 1, $I_{1} (X) = X \cup {1} .$ It is observed that $1 \in I_{2} (X)$ . Thus, it remains to show that $X \subseteq I_{2} (X) .$ Let $h \in X .$ Then $h = 1 \to (1 \to h) \in X = I_{n} (X)$ and $1 = 1 \to 1 \in I_{1} (X) .$ This implies that $h = 1 \to h \in I_{2} (X) .$ Therefore $h \in I_{2} (X)$ and hence $I_{1} (X) \subseteq I_{2} (X) .$ Now let $n \geq 1$ and assume that the result holds for $n,$ i.e. $I_{n} (X) \subseteq I_{n + 1} (X) .$ Now we claim that $I_{n + 1} (X) \subseteq I_{n + 2} (X) .$ Let $s \in I_{n + 1} (X) .$ Then $s = h \to z$ for some $h, z \in R$ such that $h \to (g \to z) \in I_{n} (X)$ and $h \to g \in I_{n} (X)$ for some $g \in R .$ Since, by induction assumption, $I_{n} (X) \subseteq I_{n + 1} (X)$ we get $h \to (g \to z) \in I_{n + 1} (X)$ and $h \to g \in I_{n + 1} (X) .$ Then $h \to z \in I_{n + 2} (X) .$ Therefore $I_{n + 1} (X) \subseteq I_{n + 2} (X), ∀n \in N .$

Put $S = ⋃_{n = 1}^{\infty} I_{n} (X)$ . We claim to show that S is an implicative filter. Clearly $1 \in S .$ Let $h, g, z \in S$ such that $h \to (g \to z)$ and $h \to g \in S .$ Then $h \to (g \to z) \in I_{n} (X)$ and $h \to g \in I_{m} (X),$ for some $n, m \geq 1.$ Since ${I_{n} (X)}_{n = 1}^{\infty}$ is a chain, assume without loss of generality $I_{m} (X) \subseteq I_{n} (X)$ and hence $h \to (g \to z) \in I_{n} (X)$ and $h \to g \in I_{n} (X)$ implying that $h \to z \in I_{n + 1} (X) \subseteq S .$ Therefore S is an implicative filter. Finally, it remains to show that S is the least such a filter containing X. Let G be any implication filter such that $X \subseteq G .$ It suffices to show that $I_{n} (X) \subseteq G, ∀n \in N .$ We use induction on $n .$ If $n = 1,$ then $I_{1} (X) = X \cup {1} \subseteq G .$ Let $n \geq 2$ and assume that $I_{n} (X) \subseteq G .$ Now we claim that $I_{n + 1} (X) \subseteq G .$ Let $s \in I_{n + 1} (X) .$ Then $s = h \to z,$ where $h \to (g \to z), h \to g \in I_{n} (X) \subseteq G .$ This implies that $h \to (g \to z), h \to g \in G .$ Since G is an implicative filter $s = h \to z \in G .$ Therefore $I_{n + 1} (X) \subseteq G .$ and hence $I_{n} (X) \subseteq G, ∀n \in N .$ Therefore $S = ⋃_{n = 1}^{\infty} I_{n} (X) \subseteq G .$ Hence proved.

Definition 4.6.

For any filter U in $R$ , define a set $I_{U} (R)$ to be:

\begin{array}{l} I_{U} (R) = {h \in R : h \to h^{2} \in U} \end{array}

Observe that $I_{U} (R)$ is contained in $R$ containing U and is closed with respect to $\circ$ . Moreover, U is an implicative filter if and only if $I_{U} (R) = R .$ Furthermore, $R$ is an idempotent hoop if and only if $I_{{1}} (R) = R .$

Notation: For $a \in R$ let us denote by $\overset{―}{a}$ the element $a \to a \circ a$ . It is clear that $a \leq \overset{―}{a}$ for all $a \in R$ , and $\overset{―}{a} = 1$ for all $a \in R$ if and only if $R$ is an idempotent hoop.

Definition 4.7.

Define a set $I_{0} (R)$ in $R$ by:

\begin{array}{l} I_{0} (R) = {x \in R : \overset{―}{a_{1}} \circ \overset{―}{a_{n}} \circ \dots \circ \overset{―}{a_{n}} \leq x for some a_{1}, \\ \dots, a_{n} \in R} \end{array}

We call $I_{0} (R)$ the implicative center $R$ .

Theorem 4.8

$I_{0} (R)$ is the least implicative filter in $R$ . Moreover, $R$ is idempotent hoop if and only if $I_{0} (R) = {1}$ .

Proof.

One can easily observe that $I_{0} (R)$ is the filter generated by the set

\begin{array}{l} {a \to a^{2} : a \in R} \end{array}

and hence it is an implicative filter. Moreover, if J is any other implicative filter in $R$ , then it follows from Theorem 4.4 that $I_{0} (R) \subseteq J$ and hence $I_{0} (R)$ is the least among all implicative filters.

Corollary 4.9.

For any filter U in $R$ , the smallest implicative filter $I_{g} (U)$ of $R$ containing U can be described as follows:

\begin{array}{l} I_{g} (U) = U \lor I_{0} (R) \end{array}

where the supremum being computed over all filters of $R$ .

Lemma 4.10

Let $IF (R)$ be the collection of all implicative filters in $R .$ Then the following are true.

(1)	$IF (R)$ is a complete lattice;
(2)	$IF (R)$ is a filter of $F (R) .$

Proof.

(1) It is clear that $R$ itself is an implicative filter. So $IF (R)$ has the largest element. Now it is enough to show that $IF (R)$ is closed under arbitrary intersection. Let ${M_{α}}_{α \in Δ}$ be a family of implicative filters in $R .$ Now we claim that $M = ⋂_{α \in Δ} M_{α} \in IF (R) .$ Clearly $1 \in M .$ Let $h, g \in R$ such that $h \to (g \to z) \in M$ and $h \to g \in M .$ Then $h \to (g \to z) \in M_{α}$ and $h \to g \in M_{α}$ for all $α \in Δ .$ Then $h \to z \in F_{α}, ∀α \in Δ .$ Therefore $h \to z \in M$ and hence $M = ⋂_{α \in Δ} M_{α}$ is an implicative filter of $R .$ Thus $IF (R)$ is a complete lattice.

(2) Let $M, N \in IF (R)$ . Then, by (1), we get $M \cap N \in IF (R)$ . Moreover let $M \in IF (R)$ and N be any filter in $R$ such that $M \subseteq N .$ Now our aim is to show that M is an implicative filter. As N is given to be a filter, it is enough to show that $h \to h^{2} \in N$ for all $h \in R$ . As M is an implicative filter taking $r = h, s = h$ and $t = h^{2}$ we get $r \to (s \to t) = h \to (h \to h^{2}) = h^{2} \to h^{2} = 1 \in M$ and $r \to s = h \to h = 1 \in M .$ Since $M \in IF (R)$ it follows that $r \to t \in M,$ i.e. $h \to h^{2} \in M \subseteq N$ Therefore $h \to h^{2} \in N, ∀h \in R$ (*). Now let $h, g, z \in R$ such that $h \to (g \to z) \in N$ and $r \to g \in N .$ This implies that $(h \circ g) \to z \in N$ and $h^{2} \to (h \circ g) \in N .$ By (*) we have $h \to h^{2} \in N .$ Thus by transitivity $h \to Z \in N .$ Therefore $N \in IF (R) .$

Note. During our investigation, we noticed that most of the results in this section can also be readily extended to those positive implicative filters.

5. Congruence relations

An equivalence relation Φ on $R$ is said to be a congruence relation on $R$ provided that it satisfies the substitution property for $\circ$ and $\to$ ; i.e. for all $h, g, h^{'}, g^{'} \in R$ :

\begin{array}{l} (h, g), (h^{'}, g^{'}) \in Φ \Rightarrow (h \circ h^{'}, g \circ g^{'}), (h \to h^{'}, g \to g^{'}) \in Φ \end{array}

The collection of all congruence relations on $R$ will be denoted by $Con (R)$ . For $h \in R$ and $Φ \in Con (R)$ let

\begin{array}{l} [h]^{Φ} = {g \in R : (h, g) \in Φ} \end{array}

In other words, $[h]^{Φ}$ is an equivalence class of Φ to which h belongs. Put

\begin{array}{l} R / Φ = {[h]^{Φ} : h \in R} \end{array}

Let us define the binary operations $\circ$ and $\to$ on $R / Φ$ as follows. For $[h]^{Φ}, [g]^{Φ} \in R / Φ$ :

\begin{array}{l} [h]^{Φ} \circ [g]^{Φ} = [h \circ g]^{Φ} and [h]^{Φ} \to [g]^{Φ} = [h \to g]^{Φ} \end{array}

For all $[h]^{Φ}, [g]^{Φ} \in R / Φ$ . Then, it is routine to check that the binary operations are well defined and $⟨ R / Φ, \circ, \to, [1]^{Φ} ⟩$ is a hoop called the quotient of $R$ modulo Φ.

Lemma 5.1.

For any $Φ \in Con (R)$ , $[1]^{Φ}$ is a filter of $R$ .

Proof.

$[1]^{Φ} = {h \in R : (1, h) \in Φ} .$ Since $(1, 1) \in Φ,$ $1 \in [1]^{Φ} .$ Let $h, h \to g \in [1]^{Φ} .$ Then

\begin{array}{l} (1, h), (1, h \to g) \in Φ & \Rightarrow & (1, h \circ (h \to g)) \in Φ \\ \Rightarrow & < 1 \to g, (h \circ (h \to g)) \\ \to g >\in Φ \\ \Rightarrow & (g, 1) \in Φ \\ \Rightarrow & g \in [1]^{Φ} . \end{array}

Therefore $[1]^{Φ}$ is a filter in $R$ .

It is evident that the class $Con (R)$ of all congruence relations on $R$ contains $▽_{R} = R \times R$ and is closed under arbitrary intersection so that it is a complete lattice under the usual inclusion order. Moreover, for any $Φ, Ψ \in Con (R)$ :

\begin{array}{l} Φ \land Ψ = Φ \cap Ψ and Φ \lor Ψ = ⟨ Φ \cup Ψ ⟩ \end{array}

where $⟨ Φ \cup Ψ ⟩$ is the congruence relation on $R$ generated by $Φ \cup Ψ$ . It is described in (Burris & Sankappanavor, Citation1981) for general universal algebras in the following way:

\begin{array}{l} ⟨ Φ \cup Ψ ⟩ = ⋃_{n = 1}^{\infty} Π_{Φ, Ψ}^{n} \end{array}

where $Π_{Φ, Ψ}^{n}$ is defined as follows:

\begin{array}{l} Π_{Φ, Ψ}^{1} & = & Φ, and for n \geq 1 \\ Π_{Φ, Ψ}^{n + 1} & = & Π_{Φ, Ψ}^{n} * Φ * Ψ, \end{array}

where $*$ is the relational product; i.e. $(h, g) \in Φ * Ψ$ if and only if there is $z \in R$ such that $(h, z) \in Ψ$ and $(z, g) \in Φ$ . We say that two congruences Φ and Ψ permute if $Φ \circ Ψ = Ψ \circ Φ$ , and an algebra A is called permutable, provided that all congruences of A permute each other; i.e. $Φ \circ Ψ = Ψ \circ Φ$ for all $Φ, Ψ \in Con (A)$ . Moreover, a class $K$ of algebras of a fixed type is permutable if each of its member is permutable. It was proved by Mal’cev in (Mal’cev, Citation1954) that a variety $K$ of algebras is permutable if and only if it has a 3-ary term $p (h, g, z)$ satisfying the identities

\begin{array}{l} g \approx p (h, h, g) and h \approx p (h, g, g) \end{array}

The term p is called the Ma’cev term. In this regard, the class of hoops is congruence permutable, where its Mal’cev term is:

\begin{array}{l} p (h, g, z) = ((h \to g) \to z) \land ((z \to g) \to h) \end{array}

Therefore every hoop $R$ is congruence permutable.

Corollary 5.2.

For any $Φ, Ψ \in Con (R)$ : $Φ \lor Ψ = Φ \circ Ψ$ .

Lemma 5.3.

For any $Φ, Ψ \in Con (R)$ :

\begin{array}{l} Φ \subseteq Ψ if and only if [1]^{Φ} \subseteq [1]^{Ψ} . \end{array}

Proof.

Suppose that $Φ \subseteq Ψ .$ Then we have the following:

\begin{array}{l} h \in [1]^{Φ} & \Rightarrow & (1, h) \in Φ \subseteq Ψ \\ \Rightarrow & (1, h) \in Ψ \\ \Rightarrow & h \in [1]^{Ψ} . \end{array}

Therefore $[1]^{Φ} \subseteq [1]^{Ψ}$ . Conversely, suppose that $[1]^{Φ} \subseteq [1]^{Ψ} .$

\begin{array}{l} (h, g) \in Φ & \Rightarrow & (h \to g, 1)) \in Φ \\ \Rightarrow & h \to g \in [1]^{Φ} \subseteq [1]^{Ψ} \\ \Rightarrow & (h \to g, 1) \in Ψ \\ \Rightarrow & (h \circ (h \to g), h) \in Ψ \end{array}

Similarly $(g \circ (g \to h), g) \in Ψ .$ Since $(h \circ (h \to g) = (g \circ (g \to h)$ , by transitivity we get $(h, g) \in Ψ .$ Therefore $Φ \subseteq Ψ .$

Remark. It can be deduced from the above lemma that $Φ = Ψ if and only if [1]^{Φ} = [1]^{Ψ}$ for all $Φ, Ψ \in Con (R)$ , and hence every hoop is 1-regular.

Lemma 5.4.

For any $Φ, Ψ \in Con (R)$ :

(1)	$[1]^{Φ \cap Ψ} = [1]^{Φ} \cap [1]^{Ψ}$
(2)	$[1]^{Φ \lor Ψ} = [1]^{Φ} \lor [1]^{Ψ}$

Proof.

(1) It is obvious.

(2) Since $Φ, Ψ \subseteq Φ \lor Ψ$ it follows from Lemma 5.3 that $[1]^{Φ}, [1]^{Ψ} \subseteq [1]^{Φ \lor Ψ}$ . Thus $[1]^{Φ} \lor [1]^{Ψ} \subseteq [1]^{Φ \lor Ψ} .$ To prove the other inclusion let $h \in [1]^{Φ \lor Ψ} .$ Then

\begin{array}{l} (1, h) \in Φ \lor Ψ = Φ \circ Ψ & \Rightarrow & \exists g \in R such that \\ (1, g) \in Ψ and (g, h) \in Φ \\ \Rightarrow & (1, g) \in Ψ and (1, g \to h) \in Φ \\ \Rightarrow & g \in [1]^{Ψ} \subseteq [1]^{Φ} \lor [1]^{Ψ} and \\ g \to h \in [1]^{Φ} \subseteq [1]^{Φ} \lor [1]^{Ψ} \\ \Rightarrow & h \in [1]^{Φ} \lor [1]^{Ψ} . \end{array}

Therefore $[1]^{Φ \lor Ψ} \subseteq [1]^{Φ} \lor [1]^{Ψ} .$ Hence (2) holds.

Definition 5.5.

For a given filter U in $R$ consider the binary relation on $R$ denoted by $Θ_{U}$ and given by:

\begin{array}{l} (h, g) \in Θ_{U} if and only if h \to g \in U and g \to h \in U . \end{array}

Theorem 5.6

$Θ_{U} \in Con (R)$ for all $U \in F (R)$ . And conversely if $Θ_{U}$ is a congruence relation on $R$ , then U is a filter.

Proof.

Let $U \in F (R)$ . Clearly $Θ_{U}$ is reflexive and symmetric. Now we proceed to show that $Θ_{U}$ is transitive. Let $(h, g) \in Θ_{U}$ and $(g, s) \in Θ_{U} .$ Then $h \to g,$ $g \to h \in U$ and $g \to s,$ $s \to g \in U .$ Then, by Lemma 3.4 (2), we have $h \to s \in U$ and $s \to h \in U$ and hence $< h, s >\in Θ_{U} .$

Next we prove the substitution property for $\circ .$ Let $(h, g) \in Θ_{U}$ and $(h^{'}, g^{'}) \in Θ_{U} .$ Then $h \to g,$ $g \to h \in U$ and $h^{'} \to g^{'},$ $g^{'} \to h^{'} \in U .$ Then, by Lemma 3.4(1), we have $(h \circ h^{'}) \to (g \circ h^{'}) \in U$ and $(g \circ h^{'}) \to (g \circ g^{'}) \in U$ and hence by Lemma 3.4(2), we have $(h \circ h^{'}) \to (g \circ g^{'}) \in Θ_{U} .$ In a ilar fashion it can be shown that $(g \circ g^{'}) \to (h \circ h^{'}) \in U$ . Thus, $(h \circ h^{'}, g \circ g^{'}) \in Θ_{F}$ . Finally, we prove the substitution property for $^{'} \to^{'} .$ Let $(h, g) \in Θ_{U}$ and $(h^{'}, g^{'}) \in Θ_{U} .$ Then $h \to g,$ $g \to h \in U$ and $h^{'} \to g^{'},$ $g^{'} \to h^{'} \in U .$ Then, by Lemma 3.4(1), we have $(h \to h^{'}) \to (g \to h^{'}) \in U$ and $(g \to h^{'}) \to (g \to g^{'}) \in U$ and hence by Lemma 3.4(2), we have $(h \to h^{'}) \to (g \to g^{'}) \in Θ_{U} .$ In a ilar fashion, it can be shown that $(g \to g^{'}) \to (h \to h^{'}) \in U$ . Thus $(h \to h^{'}, g \to g^{'}) \in Θ_{U}$ . Therefore $Θ_{U} \in Con (R) .$

Conversely suppose that $Θ_{U}$ is a congruence relation on $R .$ Then, by Lemma 4.1 $[1]^{Θ_{U}}$ is a filter of $R .$ Now we claim that $[1]^{Θ_{U}} = U .$ Now let $h \in [1]^{Θ_{U}} .$ Then $(1, h) \in Θ_{U} .$ That is, $h \in U$ and hence $[1]^{Θ_{U}} \subseteq U$ . Again, to show the other inclusion, let $h \in U$ . Then $1 \to h = h \in U$ and $h \to 1 = 1 \in U .$ Therefore $(1, h) \in Θ_{U}$ and hence $h \in [1]^{Θ_{U}}$ . Thus $U \subseteq [1]^{Θ_{U}}$ . This proves the result.

Lemma 5.7.

For any $U \in F (R)$ and $Φ \in Con (R)$ we have:

(1)	$[1]^{Θ_{U}} = U$
(2)	$Θ_{[1]^{Φ}} = Φ$

Proof.

(1) It follows from the converse part of the above theorem.

(2) Let $(h, g) \in Θ_{[1]^{Φ}} .$ Then $h \to g,$ $g \to h \in [1]^{Φ}$ and so $(1, h \to g) \in Φ$ and $(1, g \to h) \in Φ .$ This implies that $(h, h \circ (h \to g)) \in Φ$ and $(g, g \circ (g \to h)) \in Φ$ . Since Φ is symmetric and $h \circ (h \to g) = g \circ (g \to h),$ by transitivity, we get $(h, g) \in Φ$ and so $Θ_{[1]^{Φ}} \subseteq Φ .$ On the other hand let $(h, g) \in Φ$ . Then $(1, h \to g) \in Φ$ and $(g \to h, 1) \in Φ .$ This implies that $(h \to g), (g \to h) \in [1]^{Φ}$ and so $(h, g) \in Θ_{[1]^{Φ}} .$ Therefore $Φ \subseteq Θ_{[1]^{Φ}} .$

Theorem 5.8.

There is a lattice isomorphism between $Con (R)$ and $F (R)$ .

Proof.

Define a map $π : Con (R) ⟶ F (R)$ by $π (Φ) = [1]^{Φ},$ for all $Φ \in Con (R) .$ If $π^{'} : F (R) ⟶ Con (R)$ is a map defined by $π^{'} (U) = Θ_{U}$ , then $π (π^{'} (U)) = π (Θ_{U}) = [1]^{Θ_{U}} = U$ and $π^{'} (π (Φ)) = π ([1]^{Φ}) = Θ_{[1]^{Φ}} = Φ$ . Therefore, π is invertible, and hence, it is bijective. Now, it remains to show that π is a lattice homomorphism.

Let $Θ_{1}, Θ_{2} \in Con (R)$ . Then

\begin{array}{l} π (Θ_{1} \cap Θ_{2}) = [1]^{Θ_{1} \cap Θ_{2}} = [1]^{Θ_{1}} \cap [1]^{Θ_{2}} = π (Θ_{1}) \cap π (Θ_{2}) \end{array}

and

\begin{array}{l} π (Θ_{1} \lor Θ_{2}) = [1]^{Θ_{1} \lor Θ_{2}} = [1]^{Θ_{1}} \lor [1]^{Θ_{2}} = π (Θ_{1}) \lor π (Θ_{2}) . \end{array}

Thus π is a lattice homomorphism and hence it is a lattice isomorphism. Therefore $Con (R) ≅ F (R)$ .

Corollary 5.9.

The map $U \mapsto Θ_{U}$ is a lattice isomorphism of $F (R)$ onto $Con (R)$ .

Proof.

Since the map $U \mapsto Θ_{U}$ is the inverse of π and π is a lattice isomorphism of $Con (R)$ onto $F (R)$ , the given map is a lattice isomorphism.

Corollary 5.10

The following holds for all filters U and V in $R$ :

(1)	$Θ_{U \cap V} = Θ_{U} \cap Θ_{V}$
(2)	$Θ_{U \lor V} = Θ_{U} \lor Θ_{V}$
(3)	$U \subseteq V$ if and only if $Θ_{U} \subseteq Θ_{V}$

Proof.

It is a direct consequence of Corollary 5.10.

Next, we see distributivity

Theorem 5.11.

Every hoop $R$ is congruence distributive.

Proof.

Applying Theorem 4.8 it suffices to show that the lattice $(F (R), \subseteq)$ of filters of $R$ is distributive. Let $U, V$ and W be any filters in $R$ . Then $U = [1]^{Φ}, V = [1]^{Ψ}$ and $W = [1]^{Υ}$ for some $Φ, Ψ, Υ \in Con (R) .$ It is sufficient to verify that

\begin{array}{l} U \cap (V \lor W) \subseteq (U \cap V) \lor (U \cap W) . \end{array}

Now

\begin{array}{l} x \in U \cap (V \lor W) & \Rightarrow & x \in [1]^{Φ} \cap ([1]^{Ψ} \lor [1]^{Υ}) \\ = [1]^{Φ \cap (Ψ \lor Υ)} \\ \Rightarrow & (x, 1) \in Φ \cap (Ψ \lor Υ) \\ \Rightarrow & (x, 1) \in Φ and (x, 1) \in Ψ \lor Υ \\ = Ψ \circ Υ \\ \Rightarrow & (x, 1) \in Φ and \exists z \in R \\ such that (x, z) \in Υ and \\ (z, 1) \in Ψ \\ \Rightarrow & (x, u) \in Φ \cap Ψ and \\ (1, u) \in Φ \cap Υ, \\ where u = z \to x \\ \Rightarrow & (1, x) \in (Φ \cap Ψ) \circ (Φ \cap Υ) \\ = (Φ \cap Ψ) \lor (Φ \cap Υ) \\ \Rightarrow & x \in [1]^{(Φ \cap Ψ) \lor (Φ \cap Υ)} \\ \Rightarrow & x \in ([1]^{Φ} \cap [1]^{Ψ}) \\ \lor ([1]^{Φ} \cap [1]^{Υ}) = (U \cap V) \\ \lor (U \cap W) \end{array}

Therefore $(F (R), \subseteq)$ is distributive.

Recall from (Chajda et al., Citation2003) that a permutable and regular variety of algebras is called arithmetic. Therefore, it is evident from the above theorem that the variety of hoops is arithmetic.

Notation. Given a filter U of $R$ and $a \in R$ ; we write $[a]^{U}$ to denote the congruence class of $Θ_{U}$ containing a, and by $R / U$ we mean the quotient $R / Θ_{U}$ .

6. Hoop Homomorphisms

In this section, we study homomorphism and isomorphism Theorems in hoops.

Definition 6.1.

A map $f : R ⟶ R^{'}$ is said to be a hoop homomorphism if it satisfies the following conditions: for all $h, g \in R$

(1)	$f (h \circ g) = f (h) \circ f (g);$
(2)	$f (h \to g) = f (h) \to f (g)$

Note that if $f : R ⟶ R^{'}$ is a hoop homomorphism, then one can easily check that $f (1) = 1.$ A hoop homomorphism $f : R ⟶ R^{'}$ is called a hoop monomorphism (respectively, a hoop epimorphism) if it is injective (respectively, surjective), and it is called a hoop isomorphism if it is both hoop monomorphism and epimorphism. The following result follows from the definition of hoop homomorphism.

Theorem 6.2.

Let, f be a hoop homomorphism from $R$ to $R^{'}$ . Then

(1)	If V is a filter in $R^{'}$ , then $f^{- 1} (V)$ is a filter in $R .$
(2)	If f is onto and U is a filter in $R$ , then f(U) is a filter in $R^{'}$ .

Definition 6.3.

Given a hoop homomorphism $f : R \to R^{'}$ , its kernel is defined to be the set:

\begin{array}{l} \ker (f) = {h \in R : f (h) = 1} \end{array}

Lemma 6.4.

For any hoop homomorphism $f : R \to R^{'}$ , its kernel ker(f) is a filter in $R$ .

Proof.

One can easily observe that $\ker (f) = f^{- 1} (1)$ , and hence it follows from (1) of Theorem 6.2 that ker(f) is a filter in $R$ .

Theorem 6.5.

A hoop homomorphism $f : R ⟶ R^{'}$ is a monomorphism if and only if $\ker f = {1}$ .

Proof.

Suppose that f is a monomorphism. Then, f is an injection and therefore $h \in \ker f \Rightarrow f (h) = 1$

Theorem 6.6

[The first isomorphism Theorem] Let $f : R ⟶ R^{'}$ be a hoop homomorphism. Then

\begin{array}{l} R / \ker f ≅ f (R) \end{array}

Proof.

Put $U = \ker f .$ Define $\overset{―}{f} : R / U ⟶ f (R)$ by: $\overset{―}{f} ([h]^{U}) = f (h), \forall [h]^{U} \in R / U .$ Now we show that $\overset{―}{f}$ is a hoop isomorphism. Let $[h]^{U}, [g]^{U} \in R / U$ such that $[h]^{U} = [g]^{U} .$ Then $h \to g \in U$ and $g \to h \in U$ and so $f (h) = f (g) .$ This shows that $\overset{―}{f}$ is well defined. Now for any $[h]^{U}, [g]^{U} \in R / U$ , we have $\overset{―}{f} ([h]^{U} \to [g]^{U}) = \overset{―}{f} ([h \to g]^{U}) = f (h \to g) = f (h) \to f (g) = \overset{―}{f} (h) \to \overset{―}{f} (g)$ and $\overset{―}{f} ([h]^{U} \circ [g]^{U}) = \overset{―}{f} ([h \circ g]^{U}) = f (h \circ g) = f (h) \circ f (g) = \overset{―}{f} (h) \circ \overset{―}{f} (g)$ and so $\overset{―}{f}$ is a hoop homomorphism. Since any element of $f (R)$ is of the form $f (h) = \overset{―}{f} ([h]^{U})$ for some $h \in R$ and $[h]^{U} \in R / U$ , $\overset{―}{f}$ is a surjection. Also, for any $[h]^{U}, [g]^{U} \in R / U,$

\begin{array}{l} \overset{―}{f} ([h]^{U}) = \overset{―}{f} ([g]^{U}) & \Rightarrow & f (h) = f (g) \\ \Rightarrow & f (h \to g) = f (h) \to f (g) = 1 \\ and f (g \to h) = f (g) \to f (h) = 1 \\ \Rightarrow & h \to g and g \to h \in U \\ \Rightarrow & h \leq g and g \leq h \\ \Rightarrow & h = g \end{array}

Therefore $\overset{―}{f}$ is an injection. Thus, $\overset{―}{f}$ is a hoop isomorphism and so $R / \ker f ≅ f (R) .$

Note that, in particular, if f is surjective, then $R / \ker f ≅ R^{'} .$

Theorem 6.7

[The second Isomorphism Theorem] Let U and V be filters in. Then $U / (U \cap V) ≅ (U \lor V) / V .$

Proof.

Define a map $f : U ⟶ (U \lor V) / V$ by $f (h) = [h]^{V}$ , for all $h \in R .$ Now for any $h, g \in U$ , $f (h \to g) = [h \to g]^{V} = [h]^{V} \to [g]^{V} = f (h) \to f (g)$ and similarly we can show $f (h \circ g) = f (h) \circ f (g)$ and hence f is a hoop homomorphism. Let $[z]^{V} \in (U \lor V) / V .$ Then $z \in U \lor V$ and so there exists $h \in U$ and $g \in V$ such that $h \circ g \leq z .$ Now $h \circ g \leq z$ implies that $h \leq g \to z .$ Thus $g \to z \in R .$ Put $j = g \to z$ and we claim that $[j]^{V} = [z]^{V} .$ Now

\begin{array}{l} g \to (j \to z) & = & g \to ((g \to z) \to z) \\ = & (g \circ (g \to z)) \to z \\ = & (z \circ (z \to g)) \to z \\ = & (z \to g) \to (g \to g) = 1 \end{array}

Therefore $g \leq j \to z .$ Since $g \in V,$ we have $j \to z \in V$ and

\begin{array}{l} z \to j & = & z \to (g \to z) \\ = & (z \circ g) \to z \\ = & g \to (z \to z) \\ = & g \to 1 = 1 \in V \end{array}

Therefore $f (j) = [j]^{V} = [z]^{V}$ and hence f is onto. Therefore, by first isomorphism theorem, we have $U / \ker f ≅ (U \lor V) / V .$ But

\begin{array}{l} \ker f & = {h \in U : f (h) = [1]^{V}} = {h \in U : [h]^{V} = [1]^{V}} \\ = {h \in U : h \in V} = U \cap V \end{array}

This proves the result.

Theorem 6.8

[The third Isomorphism Theorem] Let U and V be filters in $R$ such that $U \subseteq V .$ Then

\begin{array}{l} (R / U) / (V / U) ≅ R / V \end{array}

Proof.

Define a map $f : R / U ⟶ R / V$ by:

\begin{array}{l} f ([h]^{U}) = [h]^{V} . \end{array}

Let $[h]^{U}, [g]^{U} \in R / V$ such that $[h]^{U} = [g]^{V} .$ Then $h \to g \in U \subseteq V$ and $g \to h \in U \subseteq V .$ This implies that $[h]^{V} = [g]^{V}$ and hence $f ([h]^{U}) = f ([g]^{U}) h$ . Therefore, $\overset{―}{f}$ is well defined. It is also clear that f is a hoop epimorphism. Thus, by first isomorphism theorem, $(R / U) / \ker f ≅ R / V .$ But

\begin{array}{l} \ker f & = {[h]^{U} : f ([h]^{U}) = [1]^{V}} = {[h]^{U} : [h]^{V} = [1]^{V}} \\ = {[h]^{U} : h \in V} = V / U . \end{array}

Hence proved.

Theorem 6.9

(Correspondence Theorem) Let U be a filter in $R .$ Then there is a one-to-one correspondence between the set $[U, R]$ of all filters of $R$ containing U and the set $F [R / U]$ of all filters of $R / U$ .

Proof.

Note that every filter of $R / U$ is of the form $W / U$ where W is a filter of $R$ which contains U. Define a map $g : [U, R] ⟶ F [R / U]$ by:

\begin{array}{l} g (W) = W / U, for all W \in [U, R] \end{array}

Let $W_{1}, W_{2} \in [U, R]$ such that $g (W_{1}) = g (W_{2}) .$ Then $W_{1} / U = W_{2} / U$ . Now

\begin{array}{l} h \in W_{1} & \Rightarrow & [h]^{U} \in W_{1} / U = W_{2} / U \\ \Rightarrow & [h]^{U} = [g]^{U} for some g \in W_{2} \\ \Rightarrow & h \to g, g \to h \in U \subseteq W_{2} \\ \Rightarrow & h \in W_{2} . \end{array}

Therefore $W_{1} \subseteq W_{2} .$ Similarly, we have $W_{2} \subseteq W_{1} .$ Therefore $W_{1} = W_{2}$ and hence g is injective. Let $P \in F [R / U] .$ Put $W = {h \in R : [h]^{U} \in P}$ . Now we claim that W is a filter which contains $U .$ Now let $h, h \to g \in W .$ Then $[h]^{U}, [h \to g]^{U} = [h]^{U} \to [g]^{U} \in P .$ Since P is a filter of $R / U$ , we have $[g]^{U} \in P .$ This implies that $g \in W .$ therefore W is a filter of $R .$ Let $h \in U$ . Then $h \to 1 \in U$ and $1 \to h \in U$ and hence $[h]^{U} = [1]^{U} \in P .$ This implies that $h \in W .$ Therefore $U \subseteq W$ and $g (W) = W / U = P .$ Therefor g is bijective, i.e. there is a one-to-one correspondence between $[U, R]$ and $F [R / U] .$

7. Conclusions

This manuscript has presented several new and significant results on filters in hoops. We have derived conditions for sets in a hoop to be filters and have provided multiple characterizations for filters generated in different ways. Furthermore, we have demonstrated that the class of filters in hoops forms an algebraic lattice. Additionally, our investigation into congruence relations on hoops has revealed an interesting correspondence between the lattice of filters and the lattice of congruences. This finding confirms that the variety of hoops is indeed ideally determined. Moreover, our exploration of hoop homomorphisms in connection with congruences and quotient structures has enabled us to establish various homomorphism and correspondence theorems. Through these theorems, we have developed a deeper understanding of the relationships between different structures within hoops. Overall, this manuscript contributes to the existing body of knowledge on hoops by providing new insights into the properties of filters, congruences, and hoop homomorphisms. We believe that these results have the potential to impact future research in this field, and we hope that they will inspire further investigations and applications in the study of hoops.

Authors contribution

All listed authors have made a significant scientific contribution to the findings of this manuscript.

Human and animal rights

This article does not contain any studies with human participants or animals performed by any of the authors.

Disclosure statement

No potential conflict of interest was reported by the author(s).

Funding

The work was supported by the University of Gondar.

Data availability

No data were used to support the results of this study.

Additional information

Funding

The work was supported by the University of Gondar .

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Some Results on Filters and Congruence Relations of Hoops

ABSTRACT

1. Introduction

2. Preliminaries

3. Some new results on filters

4. Implicative filters

5. Congruence relations

6. Hoop Homomorphisms

7. Conclusions

Authors contribution

Human and animal rights

Disclosure statement

Funding

Data availability

References

Information for

Open access

Opportunities

Help and information

Some Results on Filters and Congruence Relations of Hoops

ABSTRACT

1. Introduction

2. Preliminaries

3. Some new results on filters

4. Implicative filters

5. Congruence relations

6. Hoop Homomorphisms

7. Conclusions

Authors contribution

Human and animal rights

Disclosure statement

Funding

Data availability

Additional information

Funding

References

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