Full article: Structure of an Autometrized Algebra

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ABSTRACT

This paper introduces the concept of subalgebra and discusses some properties of subalgebra. We also examine different types of ideals and how they behave. We also describe some properties of a homomorphism. Finally, we present quotient algebra and discuss isomorphism theorems.

KEYWORDS:

1. Introduction

Swamy (Citation1964b) introduced the notion of an autometrized algebra to obtain a unified theory of the then-known autometrized algebras: Boolean algebras (Blumenthal, Citation1952; Ellis, Citation1951), Brouwerian algebras (Nordhaus & Lapidus, Citation1954), Newman algebras (Roy, Citation1960), autometrized lattices (Nordhaus & Lapidus, Citation1954) and commutative lattice ordered groups or l-groups (Swamy, Citation1964a). Swamy &and Rao (Citation1977), Rachŭnek (Citation1987, Citation1989, Citation1990, Citation1998), Hansen (Citation1994), Kovář (Citation2000), and Chajda & Rachunek (Citation2001) developed the theory of autometrized algebra.

In addition, Subba Rao and Yedlapalli (Citation2018) and Rao et al. (Citation2019, Citation2021) studied the concept of representable autometrized algebra.

In Section 1.1., we recall some definitions and terms (Swamy & Rao, Citation1977; Swamy, Citation1964b). Section 2. introduces the concept of subalgebra and discusses the characteristics of subalgebra. Section 3. discusses different types of ideals and how they behave. We also discuss some properties of a homomorphism. In Section 4., we introduce quotient algebra and discuss isomorphism theorems.

1.1. Preliminaries

In this section, we look at some essential concepts, definitions, and terms that are important in other sections.

Definition 1.1 ([1])

A system A= $(A, +, 0, \leq, *)$ is called an autometrized algebra if

$(A, +, 0)$ is a commutative monoid.
$(A, \leq)$ is a partial ordered set, and $\leq$ is translation invariant, that is, $∀a, b, c \in A; a \leq b \Rightarrow + c \leq b + c$ .
$* : A \times A \to A$ is autometric on $A$ , that is, $*$ satisfies metric operation axioms:

(M₁) $∀a, b \in A; a * b \geq 0$ and, $a * b = 0 \Leftrightarrow a = b$ ,

(M_{2}) ∀a, b \in A; a * b = b * a,

(M₃) $∀a, b, c \in A$ ; $a * c \leq a * b + b * c$ .

Definition 1.2 ([7])

An autometrized algebraA = $(A, +, 0, \leq, *)$ is called normal if and only if

(i) a \leq a * 0 ∀a \in A .

(ii) (a + c) * (b + d) \leq (a * b) + (c * d) ∀a, b, c, \break d \in A .

(iii) (a * c) * (b * d) \leq (a * b) + (c * d) ∀a, b, c, \break d \in A .

(iv) For any $a$ and $b$ in A, $a \leq b \Rightarrow ∃x \geq 0$ such that $a + x = b$ .

Definition 1.3 (Citation[7]) Let $A = (A, +, 0, \leq, *)$ be a system. Then, $A$ is said to be a lattice ordered autometrized algebra (or) autometrized l-algebra if

$(A, +, 0)$ is a commutative semigroup with 0.
$(A, \leq)$ is a lattice, and $\leq$ is translation invariant, that is, $∀a, b, c \in A$ ;
$a + (b \lor c) = (a + b) \lor (a + c)$
$a + (b \land c) = (a + b) \land (a + c)$

(iii) $* : A \times A \to A$ is autometric on $A$ , that is, $*$ satisfies metric operation axioms: $M_{1}$ , $M_{2}$ and $M_{3}$ .

Definition 1.4 ([McKenzie, Citation2018) Let $C$ be a closure operator on the set $A$ . $C$ is said to be algebraic if and only if $C (X) = \cup {C (Z) | Z \subseteq X and Z is finite}$ , for all $X \subseteq A$ .

Theorem 1.5 ([18])

Let $A$ be a set. Let $C : P (A) \to P (A)$ be an algebraic closure operator on a set $A$ . Then, the lattice of all $C$ -closed elements ( $L_{C}$ ) is an algebraic lattice, and the compact elements of $L_{C}$ are exactly the finitely generated closed sets $C (X)$ , where $X$ is a finite subset of $A$ . Conversely, every algebraic lattice is isomorphic to the lattice of closed sets for some algebraic closure operator.

Definition 1.6 ([7])

A nonempty subset $I$ of a normal autometrized algebra A = $(A, +, 0, \leq, *)$ is called an ideal if and only if

$a, b \in I$ imply $a + b \in I$ .
$a \in I, b \in A and b * 0 \leq a * 0$ imply $b \in I$ .

Definition 1.7 (Citation[10])Let $A$ be an autometrized algebra. Let $\MathI (A)$ be the set of all ideals of $A$ . Let $I \in \MathI (A)$ . Then, $I$ is called a regular ideal in A if $I = ⋂_{α \in Γ} J_{α}$ , where $J_{α} \in \MathI (A)$ for each $α \in Γ$ implies $I = J_{β}$ for some $β \in Γ$ . Further regular ideal if and only if prime ideal.

Definition 1.8 (Citation[7]) Let A = $(A, +, 0, \leq, *)$ and B = $(B, +, 0, \leq, *)$ be normal autometrized algebras. Let $f : A \to B$ be a map. Then, $f$ is said to be a homomorphism from $A$ to $B$ if and only if

$f (a + b) = f (a) + f (b) ∀a, b \in A,$
$f (a * b) = f (a) * f (b) ∀a, b \in A$ and
$a \leq b \Rightarrow f (a) \leq f (b) ∀a, b \in A .$

Definition 1.9 ([7])

Let A = $(A, +, 0, \leq, *)$ and B = $(B, +, 0, \leq, *)$ be normal autometrized algebras. Let $f : A \to B$ be a homomorphism. Then, $ker f = {x \in A | f (x) = \overline{0}}$ where $\overline{0}$ is the zero element of $B$ .

Definition 1.10 ([7])

Let $A$ be an autometrized algebra. Then, $A$ is said to be semiregular if for any $a \in A$ , $a \geq 0 \Rightarrow a * 0 = a$ .

2. Subalgebras

This section introduces the definition of subalgebra and discusses some properties of subalgebra.

Definition 2.1

Let A = $(A, +, 0, \leq, *)$ be autometrized algebra. Let $B \subseteq A$ . Then, $B$ is said to be a subalgebra of $A$ if;

$(B, +, 0)$ is a commutative monoid.
$(B, \leq)$ is a subposet, and $\leq$ is translation invariant, that is, $a \leq b \Rightarrow a + c \leq b + c$ for any $a, b, c \in B$ .
$* | B : B \times B \to B$ is metric.

Remark 2.2 Any subalgebra of an autometrized algebra is an autometrized algebra.

Definition 2.3

Let $A$ be an autometrized algebra. Then, the set of all subalgebras of $A$ is denoted by $Sub (A)$ .

Definition 2.4

Let $A$ be an autometrized algebra. Then, ${0}$ is a subalgebra of $A$ called trivial subalgebra of $A$ and $A$ is itself a subalgebra of $A$ called improper subalgebra. Any other subalgebras of $A$ are called non-trivial proper subalgebras.

Lemma 2.5

In an autometrized algebra $A$ , the intersection of any collection of subalgebras in $A$ is again a subalgebra.

Proof.

Obvious.

Definition 2.6

Let $A$ be an autometrized algebra. Let $X \subseteq A$ . Then, smallest subalgebra of $A$ containing $X$ is called the subalgebra generated by $X$ and is denoted by $Sg (X)$ . That is; $Sg (X) = ⋂ {B \in Sub (A) ∣ X \subseteq B}$ .

If ${S_{i}}_{i \in I}$ be a collection of subalgebra of $A$ , then $\cup_{i \in I} S_{i}$ is not a subalgebra of $A$ .The subalgebra $⟨ \cup_{i \in I} S_{i} ⟩$ generated by the set collection of subalgebra $\cup_{i \in I} S_{i}$ is called the subalgebra generated by the collection of subalgebra ${S_{i}}_{i \in I}$ . It is easy to show that $Sub (A)$ is a lattice. Since if $S_{1}, S_{2} \in Sub (A)$ , then $S_{1} \land S_{2}$ , $S_{1} \lor S_{2} \in Sub (A)$ . Generally, if ${S_{i}}_{i \in I}$ be a family of subalgebra in $Sub (A)$ , then

\underset{i \in I}{\land} S_{i} = \cap_{i \in I} S_{i} and \underset{i \in I}{\lor} S_{i} = Sg (\cup_{i \in I} S_{i}) .

Hence, $(Sub (A), \lor, \land)$ is a complete lattice.

Example 2.7

Let $A = {0, a, b, c}$ with $0 \leq a, b \leq c$ and elements $a, b$ are incomparable. Define $*$ and $+$ by the following tables.

Table

Display Table

Table

Display Table

So $A$ is an autometrized algebra. Let the set $S_{1} = {0, a}$ and $S_{2} = {0, b}$ . Clearly $S_{1}, S_{2}$ are subalgebras of $A$ . Since $a + b = c$ and $c \notin S_{1} \cup S_{2}$ ; and hence, $S_{1} \cup S_{2} = {0, a, b}$ is not a subalgebra of $A$ . But $⟨ S_{1} \cup S_{2} ⟩ = {0, a, b, c}$ is a subalgebra of $A$ .

Definition 2.8

Let $A$ be an autometrized algebra. Let $\MathS = {S_{i}}_{i \in I}$ be a family of subalgebras of $A$ . Then, ${S_{i}}_{i \in I}$ is said to be a directed family of subalgebras, if for any $S_{i}, S_{j} \in \MathS$ there exists $S_{k} \in \MathS$ such that $S_{i}, S_{j}$ are subalgebras of $S_{k}$ .

Theorem 2.9

Let $A$ be an autometrized algebra. The union of any directed family of subalgebras of $A$ is again a subalgebra of $A$ .

Proof.

Let $\MathS = {S_{i}}_{i \in I}$ be a directed family of subalgebras of $A$ . Let $S = \cup_{i \in I} S_{i}$ .

To show that

S

is a subalgebra of

A

To show that $(S, +, 0)$ is a commutative monoid. Since $0 \in S_{i} ∀i \in I$ ; imply that $0 \in S$ .

Let $a, b \in S$ . Suppose $a \in S_{i}$ and $b \in S_{j}$ for some $i, j \in I$ . Since $\MathS$ is directed; there exists $S_{k} \in \MathS$ such that $S_{i}, S_{j}$ are subalgebras of $S_{k}$ . Then, $a, b \in S_{k}$ . Since each $S_{k}$ are subalgebra; $a + b \in S_{k}$ . Since $S_{k} \subseteq S$ ; therefore, $a + b \in S$ .

(ii) Since $S \subseteq A$ ; implies $(S, \leq)$ is a subposet and $\leq$ is translation invariant.
(iii) Let $a, b \in S$ . Suppose $a \in S_{i}$ and $b \in S_{j}$ for some $i, j \in I$ . Since $\MathS$ is directed; there exists $S_{k} \in \MathS$ such that $S_{i}, S_{j}$ are subalgebras of $S_{k}$ . Then, $a, b \in S_{k}$ . Since each $S_{k}$ are subalgebra; $a * b \in S_{k}$ . Since $S_{k} \subseteq S$ ; therefore, $a * b \in S$ . Hence, $S$ is a subalgebra.

Remark 2.10 Let $A$ be an autometrized algebra. Let ${S_{i}}_{i \in I}$ be a chain of subalgebras. Then, $\cup_{i \in I} S_{i}$ is a subalgebra in $A$ .

Remark 2.11 Let $A$ be an autometrized algebra. Then, $Sub (A)$ is closed under directed unions.

Theorem 2.12

Let $A$ be an autometrized algebra. Suppose $A = Sg (X)$ for $X \subseteq A$ . For any finite subset $Y$ of $X$ , define $Sg (Y)$ = subalgebra of $A$ generated by $Y$ . Then, $\MathS = {Sg (Y) ∣ Y is finite subset of X}$ is a directed family of subalgebras of $A$ and $A = \cup {Sg (Y) ∣ Y is finite subset of X}$ .

Proof.

To show that $\MathS = {Sg (Y) ∣ Y is finite \breaksubset of X}$ is a directed family of subalgebras of $A$ .

Let $Sg (Y_{1}), Sg (Y_{2}) \in \MathS$ where $Y_{1}, Y_{2}$ are finite subsets of $X$ . Let $Y_{3} = Y_{1} \cup Y_{2}$ . This implies that $Y_{3}$ is a finite subset of $X$ . Therefore, $Sg (Y_{3}) \in \MathS$ .

Since $Y_{1}, Y_{2} \subseteq Y_{3}$ ; $Sg (Y_{1}), Sg (Y_{2})$ are subalgebras of $Sg (Y_{3})$ . Hence, $\MathS$ is a directed family of subalgebras of $A$ . By Theorem 2.9; $\cup {Sg (Y) ∣ Y is finite subset of X}$ is a subalgebra of $A$ .

To show that $A = \cup {Sg (Y) ∣ Y is finite subset of X}$ .

Since each $Sg (Y) \subseteq A$ ; therefore,

(1)

\cup {Sg (Y) ∣ Y is finite subset of X} \subseteq A .

(1)

Let $a \in X$ . Let $Y = {a}$ . Then, $Y$ is a finite subset of $X$ . This implies that; $a \in Sg (Y)$ and $Sg (Y) \in \MathS$ . Therefore, $a \in \cup {Sg (Y) | Y is finite subset of X}$ . Thus, $X \subseteq \cup {Sg \break (Y) | Y is finite subset of X}$ . So, $Sg (X) \subseteq \cup {Sg (Y) | Y is \break finite subset of X}$ . Then,

(2)

A \subseteq \cup {Sg (Y) | Y is finite subset of X} .

(2)

By Equationequations (1)(1) $\cup {Sg (Y) ∣ Y is finite subset of X} \subseteq A .$ (1) and (Equation2(2) $A \subseteq \cup {Sg (Y) | Y is finite subset of X} .$ (2) ); $A = \cup {Sg (Y) | Y is finite \break subset of X}$ .

Corollary 2.13

Every autometrized algebra can be written as a union of its directed family of subalgebras.

Proof.

From the above theorem; take $X = A$ to conclude that every autometrized algebra is the directed union of its finitely generated subalgebras.

Theorem 2.14

Let $A$ be an autometrized algebra. For any $X, Y \subseteq A$ , the following properties hold:

If $X \subseteq Y$ , then $Sg (X) \subseteq Sg (Y)$ .
$X \subseteq Sg (X) .$
$S g^{2} (X) = Sg (X) .$
$Sg (X) = \cup^{{} S (Y) | Yisfinite ∖breaksubset ofX}$

Proof.

Suppose $X \subseteq Y$ . Since $Y \subseteq Sg (Y)$ and $X \subseteq Sg (Y)$ ; implies that $Sg (X) \subseteq Sg (Y)$ .
Clearly $X \subseteq Sg (X)$ .
$S g^{2} (X) = Sg (Sg (X)) = Sg (X) .$
By Theorem 2.12, $Sg (X) = \cup {S (Y) | Y is finite \breaksubset of X}$ .

Theorem 2.15

Let $A$ be an autometrized algebra. Let $P (A)$ = Power set of $A$ and $X \subseteq A$ . Define $Sg : P (A) \to P (A)$ by $Sg (X)$ = subalgebra generated by $X$ . Then, $Sg$ is an algebraic closure operator on $A$ .

Proof.

By (i), (ii) and (iii) of Theorem 2.14, $Sg$ is a closure operator on $A$ . And again by (iv) of Theorem 2.14; $Sg$ is a closure operator on $A$ .

Remark 2.16 The set of all $Sg$ -closed elements is given by

L_{Sg} = {X \in P (A) ∣ Sg (X) = X} = {Sg (X) ∣ X \in P (A)} = Sub (A) .

Theorem 2.17

Let $A$ be an autometrized algebra. Let $P (A)$ = Power set of $A$ and $X \subseteq A$ . Define $Sg : P (A) \to P (A)$ by $Sg (X)$ = subalgebra generated by $X$ is a closure operator on $A$ . Then, $L_{Sg} = Sub (A)$ is an algebraic lattice.

Proof.

Clearly, $Sg$ is an algebraic closure operator on $A$ . And $Sg (Y)$ where $Y$ is a finite subset of $A$ is a compact element. By Theorem 2.14, $Sg (X) = \cup {S (Y) ∣ Y is finite \break subset of X}$ implies that $A$ is compactly generated. Therefore, $L_{Sg} = Sub (A)$ is an algebraic lattice.

Definition 2.18

Let $A$ be an autometrized algebra. Then, $A$ is an algebraically closed set system if and only if $A$ is closed under arbitrary intersections and unions of chains.

Theorem 2.19

Let $A$ be an autometrized algebra. Then, $Sub (A)$ is an algebraically closed set system.

Proof.

Since $Sub (A)$ is closed under arbitrary intersection by lemma (2.5) and unions of chains by remark (2.11). Hence, $Sub (A)$ is an algebraically closed set system.

3. Ideals and Homomorphism Theorems

In this section, we will discuss the different types of ideals and their behavior. We also discuss some properties of a homomorphism.

Definition 3.1

A nonempty subset $I$ of an autometrized algebra $A$ is called an ideal if and only if

$a, b \in I$ imply $a + b \in I$ .
$a \in I, b \in A and b * 0 \leq a * 0$ imply $b \in I$ .

Definition 3.2

Let $A$ be an autometrized algebra. Let $M$ be an ideal of $A$ . $M$ is called maximal if and only if whenever $J$ is an ideal such that $M \subseteq J \subseteq A$ , then either $M = J$ or $J = A$ .

Theorem 3.3

Let $A$ be an autometrized algebra. Let $M$ be an ideal of $A$ . If $M$ is a maximal ideal, then $M$ is a prime ideal.

Proof.

To show that $M$ is a prime ideal. Suppose $M$ is a maximal ideal. We know that regular ideal if and only if prime ideal.

To show that $M$ is regular ideal.

Suppose $M = ⋂_{α \in Γ} J_{α}$ . Therefore, $M \subseteq J_{α} \forall α \in Γ$ . Since $M$ is maximal ideal; $M = J_{α} \forall α \in Γ$ . Thus, $M$ is regular. Hence, $M$ is prime.

Example 3.4

Let $A = {0, a, b, c}$ with $0 \leq a \leq b \leq c$ . Define $*$ and $+$ by the following tables.

Table

Display Table

Table

Display Table

It is clear to show that $A$ is an autometrized algebra. Here ${0, a}$ and ${0, a, b}$ are ideals of $A$ . Clearly ${0, a, b}$ is a maximal ideal of $A$ . But ${0, a}$ is prime but not maximal.

Definition 3.5

Let $A$ be an autometrized algebra. Then, radical of $A$ is the set $Rad (A) = ⋂ {M | M is a ideal of A}$ .

Theorem 3.6

Let $A$ be an autometrized algebra. Let $P$ be a prime ideal of $A$ . Let $M_{1}, \dots, M_{k}$ are maximal ideals. Assume that $⋂_{i = 1}^{k} M_{i} = {0}$ . Then, $P = M_{i}$ for some $i$ .

Proof.

We know that $⋂_{i = 1}^{k} M_{i} = {0} \subseteq P$ . So, $M_{1} \subseteq P$ or $⋂_{i = 2}^{k} M_{i} = {0} \subseteq P$ . Therefore, $M_{i} \subseteq P$ for some $i$ by induction on $k$ . Since $M_{i}$ is maximal; and hence, $P = M_{i}$ for some $i$ .

Definition 3.7

Let $A$ be an autometrized algebra. An ideal $I$ of $A$ is called a strong ideal if

$a \in I \Leftrightarrow a * I = I$ and
$a * I = b * I \Leftrightarrow a * b \in I$ for $a, b \in A$ .

Remark 3.8 Here $a \in I \Leftrightarrow a * I = I$ and $a * I = b * I \Leftrightarrow a * b \in I$ are not true for any ideal of $A$ . Consider the following examples.

Example 3.9

Let $A = {0, a, b, c, d, e}$ with $0 \leq a \leq b \leq c \leq d \leq e$ . Define $+$ and $*$ by the following tables.

Table

Display Table

Table

Display Table

Clearly, $A$ is an autometrized algebra and $I = {0, a, b}$ is an ideal of $A$ . So, $0 * I = I$ ; $a * I = {a, 0}$ and $b * I = I$ . We see that $a \in I$ but $a * I \neq I$ . Thus, $I$ is not a strong ideal.

Example 3.10

Let $A = {0, a, b, c}$ with $0 \leq a, b \leq c$ and elements $a, b$ are incomparable. Define $*$ and $+$ by the following tables.

Table

Display Table

Table

Display Table

Clearly, $A$ is an autometrized algebra. Here ${0, a}$ is an ideal of $A$ , which is a strong ideal.

Definition 3.11

Let $A$ be an autometrized algebra. Let $\MathI (A)$ be the set of all ideals of $A$ . Let $I, J \in ∖MathI (A)$ . Then,

(i) the sum of $I$ and $J$ , denoted $I + J$ , is the set

I + J = {x + y | x \in I and y \in J} .

(ii) the distance of $I$ and $J$ , denoted $I * J$ , is the set

I * J = {x * y | x \in I and y \in J} .

Theorem 3.12

Let $A$ be an autometrized algebra and $I, J \in \MathI (A)$ .

If $I + J$ is an ideal, then $I + J = (I \cup J)$ .
If $A$ is semiregular and $I * J$ is an ideal, then $I * J = (I \cup J)$ .

Proof.

Suppose that $I + J$ is an ideal. We know that $I, J \subseteq I + J$ . So, $I \cup J \subseteq I + J$ . Therefore, $(I \cup J) \subseteq I + J$ . Conversely, we know that $I, J \subseteq (I \cup J)$ . Therefore, $I + J \subseteq (I \cup J)$ . Hence, $I + J = (I \cup J)$
Suppose that $I * J$ is an ideal. Let $a \in I$ . Then, $a * 0 \in I * J$ . Since $A$ is semiregular $a \in I * J$ . This implies that $I \subseteq I * J$ . Similarly, $J \subseteq I * J$ . Therefore, $I, J \subseteq I * J$ . So, $I \cup J \subseteq I * J$ . Therefore, $(I \cup J) \subseteq I * J$ . Conversely, we know that $I, J \subseteq (I \cup J)$ . Therefore, $I * J \subseteq (I \cup J)$ . Hence, $I * J = (I \cup J)$ .

Remark 3.13 Let A be a semiregular autometrized algebra. By Theorem 3.12; if $I + J$ and $I * J$ are ideals, then $I + J = I * J$ .

Definition 3.14

An autometrized algebra $(A, +, 0, \leq, *)$ is called monoid if and only if

$a * (b * c) = (a * b) * c ∀a, b, c \in A$ .[Associative]
$a * 0 = a ∀a \in A$ .[Identity]

Then, we say that $A$ is a monoid autometrized algebra.

Example 3.15

In Example 2.7, it is easily to show that $*$ is associative and $0$ is identity with respect to $*$ . Therefore, $A$ is a monoid autometrized algebra.

Theorem 3.16

Let $A$ be a monoid autometrized algebra. Then, every ideal of $A$ is strong.

Proof.(i) Let $I$ is an ideal of $A$ . To show that $a \in I \Leftrightarrow a * I = I$

Suppose $a \in I$ . To show that $a * I = I$ . Let $a * x \in a * I$ where $x \in I$ . Since $a, x \in I$ and $A$ is monoid; we get $a * x = (a * x) * 0 \leq (a * 0 + x * 0) = (a * 0 + x * 0) * 0.$ Therefore, $a * x \in I$ . Thus, $a * I \subseteq I$ .

Conversely, let $x \in I$ . So, $a * x \in I$ . We know that $a, a * x \in I$ . Clearly, $a * (a * x) \in a * I$ . Since $A$ is associative; $a * (a * x) = (a * a) * x = 0 * x$ . Again $A$ is monod; implies $a * (a * x) = x$ . As a result, $x \in a * I$ . Therefore, $I \subseteq a * I$ . Hence, $a * I = I$ .

Now to show $a * I = I \Leftrightarrow a \in I$ . Suppose $a * I = I$ . To show that $a \in I$ . Let $x \in I$ . This implies that $x = a * y$ for some $y \in I$ . Therefore, $y, a * y \in I$ . Clearly, $(a * y) * y \in I$ . Since $A$ is monoid; $(a * y) * y = (y * y) * a = 0 * a = a$ . Hence, $a \in I$ .

(ii) To show that $a * I = b * I \Leftrightarrow a * b \in I$ .

Let $a * b \in I$ . Clearly, $(a * b) * I = I$ by $(i)$ . So, $(a * b) * x = y$ for some $x, y \in I$ . Therefore,

a * [(a * b) * x] = a * y .

[a * (a * b)] * x = a * y . [Since * associative]

[(a * a) * b] * x = a * y .

(0 * b) * x = a * y . [Since A is monoid]

b * x = a * y .

Hence, $a * I = b * I$ .

Conversely, suppose $a * I = b * I$ . Therefore, $a * x = b * y$ for some $x, y \in I$ . Therefore,

a * (a * x) = a * (b * y) .

(a * a) * x = (a * b) * y . [Since * associative]

x = 0 * x = (a * b) * y . [Since A is monoid]

Therefore, $(a * b) * I = I$ . Hence, $a * b \in I$ by $(i)$ .

Definition 3.17

Let $A$ be an autometrized algebra. A strong ideal $I$ of $A$ is called a distant ideal if there exists a strong ideal $J$ of $A$ such that $I * J = A$ and $I \cap J = {0}$ . The set of all distant ideals of $A$ is denoted by $Dis (A)$ .

If $A$ is a monoid autometrized algebra, then ${0}, \breakA \in Dis (A)$ .

Theorem 3.18

Let $A$ be an autometrized l-algebra. Let $P, I \in ∖MathI (A)$ and $P \subseteq I$ . If $P$ is prime, then $I$ is prime.

Proof.

Let $a, b \in A$ . Suppose $a \land b = 0$ . To show that either $a \in I$ or $b \in I$ . Since $0 \in P; a \land b \in P$ . Since $P$ is prime; either $a \in P$ or $b \in P$ . Since $P \subseteq I$ ; either $a \in I$ or $b \in I$ .

Definition 3.19

Let $A$ and $B$ be autometrized algebras. Let $f : A \to B$ be a map. Then, $f$ is said to be a homomorphism from $A$ to $B$ if and only if

(i) $f (a + b) = f (a) + f (b) ∀a, b \in A,$

(ii) $f (a * b) = f (a) * f (b) ∀a, b \in A and$

(iii) $a \leq b \Rightarrow f (a) \leq f (b) ∀a, b \in A .$

A homomorphism $f : A \to B$ is called

an epimorphism if and only if $f$ is onto.
a monomorphism (embedding) if and only if $f$ is one-to-one.
an isomorphism if and only if $f$ is a bijection.

Definition 3.20

Let $A$ and $B$ be autometrized algebras. Let $f : A \to B$ be a map. If $a \leq b \Leftrightarrow f (a) \leq f (b), b \in A$ , then $f$ is said to be an order-embedding of $A$ into $B$ . That is; f is both order-preserving and order-reversing.

Remark 3.21 If $f$ is an order-embedding of $A$ into $B$ , then $f$ is necessary injective. Since $f (a) = f (b)$ implies $a \leq b$ and $b \leq a$ and in turn $a = b$ according to antisymmetry of $\leq$ .

Definition 3.22

Let $A$ and $B$ be autometrized algebras. Let $f : A \to B$ be a homomorphism. Then, $ker f = {x \in A | f (x) = \overset{ˉ}{0}}$ where $\overset{ˉ}{0}$ is the zero element of $B$ .

Clearly, $f$ is one-to-one if and only if $ker f = {0}$ .

Theorem 3.23

Let $A, B$ be autometrized algebras. Let $f : A \to B$ be a homomorphism. Let $J$ be an ideal of B. Then, $N = f^{- 1} (J) = {a \in A | f (a) \in J}$ is an ideal of $A$ .

Proof.

To show that $N = f^{- 1} (J) = {a \in A | f (a) \in J}$ is an ideal of A.

Let $a, b \in N = f^{- 1} (J)$ . Therefore, $f (a), f (b) \in J$ . Since $J$ is an ideal; $f (a) + f (b) \in J$ . Then $f (a + b) \in J$ . And thus, $a + b \in N = f^{- 1} (J)$ .

Let $a \in N$ and $x \in A$ . Therefore, $f (a) \in J$ . Suppose $x * 0 \leq a * 0$ . Since $f$ is a homomorphism; $f (x * 0) \leq (a * 0)$ and implies $f (x) * f (0) \leq f (a) * f (0)$ . Clearly, $f (x) * 0 \leq f (a) * 0$ . Since $J$ is ideal; implies that $f (x) \in J$ . Therefore, $x \in N$ . Hence, $N = f^{- 1} (J) = \break {a \in A | f (a) \in J}$ is an ideal of $A$ .

Theorem 3.24

Let $A, B$ be autometrized algebras. Let $f : A \to B$ be an epimorphism and order-reversing. Let $I$ be an ideal of $A$ . Then, $L = f (I) = {f (a) \in B | a \in I}$ is an ideal of B.

Proof.

To show that $L = f (I) = {f (a) \in B | a \in I}$ is an ideal of A.

Let $x, y \in L$ . Therefore, there exists $a, b \in A$ such that $x = f (a), y = f (b)$ . Clearly, $a, b \in I$ . Since $I$ is ideal; $a + b \in I$ . Now consider $x + y = f (a) + f (b) = (a + b)$ . Then, $f (a + b) \in L$ . And thus, $x + y \in L$ .

Let $x \in L$ and $z \in B$ . Since $f$ is an onto; there exists $a, c \in A$ such that $x = f (a), z = f (c)$ . Clearly, $a \in I$ . Suppose $z * 0 \leq x * 0$ . Therefore, $f (c) * 0 \leq f (a) * 0$ . Since $f$ is homomorphism; $f (c * 0) \leq f (a * 0)$ . Since $f$ is an order-reversing; $c * 0 \leq a * 0$ . Since $I$ is ideal; implies that $c \in I$ . Therefore, $f (c) = z \in L$ . Hence, $L = f (I) = {f (a) \in B | a \in I}$ is an ideal of $A$ .

Theorem 3.25

Let $A$ and $B$ be autometrized algebras. Let $X \subseteq A$ and $A = Sg (X)$ . If $f : A \to B$ and $g : A \to B$ are homomorphisms and $f (x) = g (x) ∀x \in X$ . Then, $f = g$ .

Proof.

Let $C = {x \in A ∣ f (x) = g (x)}$ . Clearly, $X \subseteq C \subseteq A$ .

To show that $C$ is a subalgebra of $A$ .

Let $a, b \in C$ . Therefore, $f (a) = g (a)$ and $f (b) = g (b)$ . Consider

\begin{aligned} f (a + b) = f (a) + f (b) . [Since f is a homomorphism] \\ = g (a) + g (b) . \\ = g (a + b) . [Since g is a homomorphism] \end{aligned}

Therefore, $a + b \in C$ .

Since $C \subseteq A$ ; clearly $(C, \leq)$ is a subposet and $\leq$ is translation invariant.

Let $a, b \in C$ . Therefore, $f (a) = g (a)$ and $f (b) = g (b)$ . Consider

\begin{aligned} f (a * b) = f (a) * f (b) . [Since f is a homomorphism] \\ = g (a) * g (b) . \\ = g (a * b) . [Since g is a homomorphism] \end{aligned}

Therefore, $a * b \in C$ .

Therefore, $C$ is a subalgebra of $A$ . Since $X \subseteq C$ ; $Sg (X) \subseteq C$ . So, $A \subseteq C$ . Therefore, $A = C$ . As result, $f (x) = g (x) ∀x \in A$ . Hence, $f = g$ .

Theorem 3.26

Let A, B and C be autometrized algebras. Let f : A → B and g : B → C be homomorphisms. Then, g ◦ f : A → C is also a homomorphism.

Proof.

Let , $a, b \in A$ .

(i) Consider

g $\circ$ f (a + b) = g(f (a + b)).

= g(f (a) + f (b)).[Since f is a homomorphism]

= g(f (a)) + g(f (b)).[Since g is a homomorphism]

= g $\circ$ f (a) + g $\circ$ f (b).

(ii) Consider

g $\circ$ f (a $*$ b) = g(f (a $*$ b)).

= g(f (a) $*$ f (b)). [Since $f$ is a homomorphism]; = g(f (a)) $*$ g(f (b)). [Since $g$ is a homomorphism]; $= g o f (a) * g o f (b)$ .

(iii) Suppose $a \leq b .$ Then,

$f (a) \leq f (b) . [Since f is a homomorphism]$

$g (f (a)) \leq g (f (b)) . [Since g is a homomorphism]$

$g \circ f (a) \leq g \circ f (b)$

Hence, $a \leq b \Rightarrow g \circ f (a) \leq g \circ f (b)$ .Therefore, g $\circ$ f is a homomorphism.

Theorem 3.27

Let $A$ and $B$ be autometrized algebras and $f : A \to B$ be a homomorphism. Then, $S$ is a subalgebra of $A \Rightarrow f (S)$ is a subalgebra of $B$ .

Proof.

Suppose $S$ is a subalgebra of $A$ . To show that $f (S)$ is a subalgebra of B. Let $a, b \in f (S)$ . Therefore, $a = f (s)$ and $b = f (t)$ for $s, t \in S$ .

(i) Since f is a homomorphism; a + b = f (s) + f (t) = f (s + t). Since S is a subalgebra; $s, t \in S \Rightarrow s + t \in S .$ Thus, $a + b \in f (S) .$ (ii) Since $f (S) \subseteq B$ ; implies $(f (S), \leq)$ is a subposet and $\leq$ is translation invariant.

(iii) Since $f$ is a homomorphism; $a * b = f (s) * f (t) \break = f (s * t)$ . Since $S$ is a subalgebra; $s, t \in S \Rightarrow s * t \in S$ . Thus, $a * b \in f (S)$ .

Hence, $f (S)$ is a subalgebra of $B$ .

Theorem 3.28

Let $A$ and $B$ be autometrized algebras and $f : A \to B$ be a homomorphism. Then, $S$ is a subalgebra of $B \Rightarrow f^{- 1} (S)$ is a subalgebra of $A$ .

Proof.

Suppose $S$ is a subalgebra of $B$ . To show that $f^{- 1} (S)$ is a subalgebra of $A$ . Let $a, b \in f^{- 1} (S)$ . Therefore, $f (a), f (b) \in S$ .

Since $S$ is a subalgebra of $B$ ; $f (a) + f (b) \break = f (a + b) \in S$ . Therefore, $a + b \in f^{- 1} (S)$ .
Since $(f^{- 1} (S) \subseteq A$ ; implies $(f^{- 1} (S), \leq) is a$ subposet and $\leq$ is translation invariant.
Since $S$ is a subalgebra of $B$ ; $f (a) * f (b) \break = f (a * b) \in S$ . Therefore, $a * b \in f^{- 1} (S)$ .

Hence, $f^{- 1} (S)$ is a subalgebra of $A$ .

Theorem 3.29 Let $A$ and $B$ be autometrized algebras. Let $f : A \to B$ be a homomorphism. Then, $f (Sg (X)) = \breakSg (f (X))$ for any $X \subseteq A$ .

Proof.

Since $X \subseteq Sg (X)$ ; implies that $f (X) \subseteq f (Sg (X))$ . Again since $Sg (X)$ is a subalgebra of A and by Theorem 3.27 $f (Sg (X))$ is a subalgebra; implies $Sg (f (X)) \subseteq f (Sg (X))$ .

Conversely, let $a \in X$ . This implies that $f (a) \in f (X)$ . Clearly, $a \in Sg (X)$ and $f (a) \in Sg (f (X))$ . Therefore, $f (a) \in f (Sg (X))$ implies that $f (a) \in Sg (f (X))$ . Hence, $f (Sg (X)) \subseteq Sg (f (X))$ . Thus, $f (Sg (X)) = Sg (f (X))$ .

Definition 3.30

Let $A, B, C$ be autometrized algebras. Let $f : B \to A$ and $g : C \to A$ be homomorphisms. Then, $f$ is said to be an extension homomorphism of g(denoted by $f / C = g$ ) if

$C$ is subalgebra of $B$ .
$f (x) = g (x) ∀x \in C .$

Definition 3.31

Let $A, {B_{i}}_{i \in I}$ be autometrized algebras. Let $f_{i} : B_{i} \to A$ be homomorphisms. Then, ${f_{i}}_{i \in I}$ is said to be a directed family of homomorphism if for any $f_{i}, f_{j}$ there exists $f_{k}$ which is a common extension of $f_{i}, f_{j}$ . That is, $f_{k} / B_{i} = f_{i}$ and $f_{k} / B_{j} = f_{j}$ .

Theorem 3.32

Let $A$ be an autometrized algebra. Any directed family of homomorphisms that maps into $A$ ( $A$ - valued homomorphisms) has a common extension.

Proof.

Let $f_{i} : B_{i} \to A$ be homomorphism. Suppose ${f_{i}}_{i \in I}$ is a directed family.

To show that $\MathB = {B_{i}}_{i \in I}$ is a directed family of autometrized algebras.

Let $B_{i}, B_{j} \in \MathB$ . We have $f_{i} : B_{i} \to A$ , $f_{j} : B_{j} \to A$ are homomorphisms. Therefore, there exists $f_{k} : B_{k} \to A$ such that $f_{k} / B_{i} = f_{i}$ and $f_{k} / B_{j} = f_{j}$ . So, $B_{i}, B_{j}$ are subalgebras of $B_{k}$ and $B_{k} \in \MathB$ . Whence, $\MathB$ is a directed family. By Theorem 2.9; $B = \cup_{i \in I} {B_{i}}$ is an autometrized algebra.

Define a map $f : B \to A$ by $f / B_{i} = f_{i} ∀i \in I$ . Clearly, $f$ is well defined.

To show that $f$ is a homomorphism.

Let $a, b \in B = \cup_{i \in I} {B_{i}}$ . Say $a \in B_{i}$ and $b \in B_{j}$ . Then, there exists $B_{k}$ such that $B_{i}, B_{j} \subseteq B_{k}$ . Therefore, $a, b \in B_{k}$ .

(i) Consider

\begin{aligned} f (a + b) = f_{k} (a + b) . [Since a + b \in B_{k}] \\ = f_{k} (a) + f_{k} (b) . [Since f_{k} is a homomorphism] \\ = f (a) + f (b) . [Since a, b \in B_{k}] \end{aligned}

(ii) Consider

\begin{aligned} f (a * b) = f_{k} (a * b) . [Since a * b \in B_{k}] \\ = f_{k} (a) * f_{k} (b) . [Since f_{k} is a homomorphism] \\ = f (a) * f (b) . [Since a, b \in B_{k}] \end{aligned}

(iii) Suppose $a \leq b$ . Since $f_{k}$ is a homomorphism; $f_{k} (a) \leq f_{k} (b)$ . Since $a, b \in B_{k}$ ; implies that $f (a) \break \leq f (b)$ .

Hence, $f$ is a homomorphism. Since $f / B_{i} = f_{i} ∀i \in I$ ; $f$ is a common extension homomorphism of ${f_{i}}_{i \in I}$ .

Definition 3.33

Let $A$ and $B$ be autometrized l-algebras. Let $f : A \to B$ be a map. Then, $f$ is said to be a homomorphism from $A$ to $B$ if and only if

$f (a + b) = f (a) + f (b) ∀a, b \in A,$
$f (a \lor b) = f (a) \lor f (b) ∀a, b \in A,$
$f (a \land b) = f (a) \land f (b) ∀a, b \in A$ , and
$f (a * b) = f (a) * f (b) ∀a, b \in A .$

Theorem 3.34

Let $A, B$ be autometrized l-algebras. Let $f : A \to B$ be a homomorphism. Let $J$ be a prime ideal of $B$ . Then, $N = f^{- 1} (J) = {a \in A | f (a) \in J}$ is a prime ideal of $A$ .

Proof.

By Theorem 3.23; $N$ is an ideal. To show that $N$ is prime. Let $a, b \in A$ . Suppose $a \land b = 0$ .

To show that either $a \in N$ or $b \in N$ .

Since $a \land b = 0$ and $f$ is a homomorphism; $f (a \land b) = 0$ and implies that $f (a) \land f (b) = 0$ . Since $J$ is an ideal; $f (a) \land f (b) = 0 \in J$ . Also $J$ is a prime ideal; either $f (a) \in J$ or $f (b) \in J$ . Therefore, either $a \in N$ or $b \in N$ . Hence, $N = f^{- 1} (J) = {a \in A | f (a) \in J}$ is a prime ideal of $A$ .

Theorem 3.35

Let $A, B$ be autometrized l-algebras. Let $f : A \to B$ be an epimorphism and order-reversing. Let $I$ be a prime ideal of $A$ . Then, $L = f (I) = {f (a) \in B | a \in I}$ is a prime ideal of $B$ .

Proof.

By Theorem 3.24; $L$ is an ideal. To show that $L$ is prime.

Let $x, y \in B$ . Since $f$ is an onto; there exists $a, b \in A$ such that $x = f (a), y = f (b)$ . Suppose $x \land y = 0$ . Therefore, $f (a) \land f (b) = 0$ .

To show that either $f (a) \in L$ or $f (b) \in L$ .

Since $f (a) \land f (b) = 0$ and $f$ is a homomorphism; $f (a \land b) = 0$ . Since $L$ is an ideal; implies that $f (a \land b) = 0 \in L$ . Therefore, $a \land b \in I$ . Since $I$ is prime; implies that either $a \in I$ or $b \in I$ . Thus, either $f (a) \in L$ or $f (b) \in L$ . Hence, $L = f (I) = {f (a) \in B | a \in I}$ is a prime ideal of $A$ .

4. Quotient Algebra

In this section, we introduce quotient algebra and discuss isomorphism theorems.

Theorem 4.1

Let $A$ be an autometrized algebra. Let $M$ be an ideal of $A$ . Let $A / M = {a * M | a \in A}$ . For any $a * M, b * M \in A / M$ , define the operations:

(a * M) + (b * M) = (a + b) * M .

(a * M) * (b * M) = (a * b) * M .

a * M \leq b * M \Leftrightarrow a \leq b .

Then, $(A / M, +, \leq, *)$ is an autometrized algebra is called the quotient algebra of $A$ by ideal $M$ .

Proof.

To claim that $(A / M, +, \leq, *)$ is an autometrized algebra.

(a) To show that $(A / M, +)$ is a commutative semigroup with additive identity $0 * M$ .

Let $a * M, b * M, c * M \in A / M$ . Therefore, $a, b, \breakc \in A$ .

$Closure$ : $(a * M) + (b * M) = (a + b) * M$ . Since $a + b \in A$ ; implies that $(a + b) * M \in A / M$ .

Associative : [(a * M) + (b * M)] + (c * M) = [(a + b) * M] + (c * M) .

= [(a + b) + c] * M .

= [a + (b + c)] * M .

= (a * M) + [(b + c) * M)] .

= (a * M) + [(b * M) + (c * M)] .

Commutative:

$(a * M) + (b * M) = (a + b) * M = (b + a) * M \break = (b * M) + (a * M)$ .

$Identity$ : $(a * M) + (0 * M) = (a + 0) * M = a * M$ .

Therefore, $0 * M$ is identity for $+$ .

(b) To show that $(A / M, \leq)$ is a poset and translation invariant.

$Reflexive$ : Let $a * M \in A / M$ . Clearly $a \in A$ . Since $a \leq a$ ; implies $a * M \leq a * M$ .

$Antisymmetric$ : Let $a * M, b * M \in A / M$ . Therefore, $a, b \in A$ . Suppose $a * M \leq b * M$ and $b * M \leq a * M$ . Therefore, $a \leq b$ and $b \leq a$ . Since $A$ is an autometrized algebra; $a = b$ . Hence, $a * M = b * M$ .

$Transitive$ : Let $a * M, b * M, c * M \in A / M$ . There-fore, $a, b, c \in A$ .

Suppose $a * M \leq b * M$ and $b * M \leq c * M$ . Therefore, $a \leq b$ and $b \leq c$ . Since $A$ is an autometrized algebra; $a \leq c$ . It follows that $a * M \leq c * M$ . Therefore, $(A / M, \leq)$ is a poset.

Suppose $a * M \leq b * M$ . Then,

a * M \leq b * M \Leftrightarrow a \leq b .

\Rightarrow a + r \leq b + r . [Since a, b \in A for any r \in A] .

\Leftrightarrow (a + r) * M \leq (b + r) * M .

\Rightarrow (a * M) + (r * M) \leq (b * M) + (r * M) .

Therefore, $(A / M, \leq)$ is translation invariant.

Hence, $(A / M, \leq)$ is a partially ordered set such that the semigroup translations are invariant under inclusion.

(M₁) Let $a * M, b * M \in A / M$ . Therefore, $a, b \in A$ . Consider $(a * M) * (b * M) = (a * b) * M$ . We know that; $0 \leq a * b$ . Therefore, $0 * M$ $\leq (a * b) * M = (a * M) * (b * M) . Hence,$ $(a * M) * (b * M) = (a * b) * M \geq 0 * M$ .

Suppose $a * M = b * M$ . So, $(a * M) * (b * M) \break = (a * M) * (a * M) = (a * a) * M = 0 * M$ . Therefore, $(a * M) * (b * M) = 0 * M$ .

Conversely, suppose

(a * M) * (b * M) = 0 * M .

\Rightarrow (a * b) * M = 0 * M .

\Rightarrow a * b = 0.

\Rightarrow a = b .

\Rightarrow a * M = b * M .

We proved that $(a * M) * (b * M) \geq 0 * M$ and $(a * M) * (b * M) = 0 * M \Leftrightarrow a * M = b * M$ .

$(M_{2}) (a * M) * (b * M) = (a * b) * M \break = (b * a) * M ∖break = (b * M) * (a * M) .$

(M₃) Let $a * M, b * M, c * M \in A / M$ . Therefore, $a, b, c \in A$ . Consider

(a * M) * (b * M) = (a * b) * M \leq [(a * c) + (c * b)] * M . [Since A is metric]

\leq [(a * c) * M] + [(c * b) * M] .

\leq (a * M) * (c * M) + (c * M) * (b * M) .

Therefore, $*$ is metric on $A / M$ . And thus $(A / M, +, \leq, *)$ is an autometrized algebra.

Example 4.2

We saw in Example 2.7 that $A$ is an autometrized algebra. Now consider an ideal $M = {0, a}$ in $A$ . Clearly, $M = 0 * M = a * M = {0, a}$ and $b * M = c * M = {b, c}$ . We can easily verify that $A / M = {0 * M, a * M, b * M, c * M}$ satisfies $(a), (b), (c)$ of Theorem 4.1. Hence, $A / M$ is an autometrized algebra.

Theorem 4.3

Let $A$ be an autometrized algebra. Let $M$ be a strong ideal of $A$ . Define a map $ϕ : A \to A / M$ by $ϕ (a) = a * M$ . Then, $ϕ$ is an epimorphism and $ker ϕ = M$ .

Proof.

Clearly, $ϕ$ is an onto map.

To show that $ϕ$ is a homomorphism.

Let $a, b \in A$ . Consider

(i)

ϕ (a + b) = (a + b) * M .

= (a * M) + (b * M) .

= ϕ (a) + ϕ (b) .

(ii)

ϕ (a * b) = (a * b) * M .

= (a * M) * (b * M) .

= ϕ (a) * ϕ (b) .

(iii) Suppose $a \leq b$ . Therefore, $a * M \leq b * M$ . Hence, $ϕ (a) \leq ϕ (b)$ .

Thus, $ϕ$ is a homomorphism.

Now, we shall prove that $ker ϕ = M$ .

ker ϕ = {a \in A | ϕ (a) = 0 * M = M} .

= {a \in A | a * M = M} .

= {a \in A | a \in M} .

= M .

Therefore, $ϕ$ is an epimorphism and $ker ϕ = M$ .

Theorem 4.4 (First Isomorphism Theorem)

Let $A$ be a monoid autometrized algebra. Let $B$ be an autometrized algebra. Let $f : A \to B$ be a homomorphism. Then, $A / ker f ≅ Imf$ .

In particular, if $f$ is onto; then, $A / ker f ≅ B$ .

Proof.

Let $M = ker f$ . By a known theorem, $M$ is a strong ideal of $A$ . Therefore, $A / M$ is an autometrized algebra.

Define a map $ϕ : A / M \to Imf$ by $ϕ (a * M) = f (a)$ .

Now we need to show if $ϕ$ is well-defined. Let $a * M, b * M \in A / M$ . Suppose $a * M = b * M$ .

\Rightarrow a * b \in M .

\Rightarrow f (a * b) = 0.

\Rightarrow f (a) * f (b) = 0.

\Rightarrow f (a) = f (b) .

\Rightarrow ϕ (a * M) = ϕ (b * M) .

Therefore, $ϕ$ is well defined.

Let $b \in Imf$ . Clearly, there exists $a \in A$ such that $f (a) = b$ . Now $a \in A$ implies that $a * M \in A / M$ . Therefore, $ϕ (a * M) = f (a) = b$ . Hence, $ϕ$ is an onto map.

To show that $ϕ$ is a homomorphism.

Let $a * M, b * M \in A / M$ . Consider

(i)

ϕ ((a * M) + (b * M)) = ϕ ((a + b) * M) .

= f (a + b) .

= f (a) + f (b) .

= ϕ (a * M) + ϕ (b * M) .

(ii)

ϕ ((a * M) * (b * M)) = ϕ ((a * b) * M) .

= f (a * b) .

= f (a) * f (b) .

= ϕ (a * M) * ϕ (b * M) .

(iii) Suppose $a \leq b$ . Since $f$ is homomorphism; $f (a) \leq f (b)$ . Therefore, $ϕ (a * M) \leq ϕ (b * M)$ .

Thus, $ϕ$ is a homomorphism.

Now, we shall prove that $ker ϕ = M$ .

ker ϕ = {a * M \in A / M | ϕ (a * M) = 0 in B} .

= {a * M \in A / M | f (a) = 0} .

= {a * M \in A / M | a \in ker f = M * 0 = M} .

= M . [Since a \in M \Rightarrow a * M = M * 0 = M]

Therefore, $ϕ$ is one-to-one. Hence, $A / M ≅ Imf$ . If $f$ is onto; then $A / M ≅ B$ .

Theorem 4.5

Let $A$ be a semiregular autometrized algebra. Let $N$ be an ideal of $A$ . Let $B$ be a subalgebra of $A$ . Define $B * N = {a \in A | a = b * n$ for $b \in B$ and $n \in N}$ such that for any $a * N, b * N \in B * N$ ; $(a * N) + (b * N) = (a + b) * N$ and $(a * N) * (b * N) \break = (a * b) * N$ . Then, $B * N$ is a subalgebra of $A$ .

Proof.

To show that $B * N$ is a subalgebra of A.

(i) To show that $(B * N, +, 0)$ is a commutative monoid. Since $0 \in B$ and $0 \in N$ ; imply that $0 \in B * N$ .

Let $a * m, b * n \in B * N$ where $a, b \in B$ and $m, n \in N$ . Therefore, $a + b \in B$ and $m * n \in N$ . Then, $(a * m) + (b * n) = (a + b) * (m * n) \in B * N$ by definition.

(ii) Clearly $(B * N, \leq)$ is a subposet, and $\leq$ is translation invariant.
(iii) Let $a * m, b * n \in B * N$ where $a, b \in B$ and $m, n \in N$ . Therefore, $(a * m) * (b * n) = (a * b) \break * (m * n) \in B * N$ by definition. Since $a * b \in B$ and $m * n \in N$ . Hence, $B * N$ is a subalgebra of $A$ .

Theorem 4.6

Let $A$ be autometrized algebra. Let $N$ be an ideal of A. Let $B$ be a subalgebra of $A$ . Then, $B \cap N$ is ideal of $B$ .

Proof.

To show that $B \cap N$ is an ideal of $B$ .

Let $a, b \in B \cap N$ . Therefore, $a, b \in B$ and $a, b \in N$ . Clearly, $a + b \in B$ and $a + b \in N$ . Hence, $a + b \in B \cap N$ .
Let $a \in B \cap N$ and $b \in B$ . Suppose $b * 0 \leq a * 0$ . So, $a \in B$ and $a \in N$ . Since $N$ is an ideal; implies that $b \in N$ . Therefore, $b \in B \cap N$ . Hence, $B \cap N$ is an ideal.

Remark 4.7 Let $A$ be a semiregular autometrized algebra. Let $N$ be an ideal of A. Let $B$ be a subalgebra of $A$ . Clearly, $N$ is an ideal of $B * N$ .

Theorem 4.8 (Second Isomorphism Theorem)

Let $A$ be a monoid autometrized algebra. Let $N$ be an ideal of $A$ . Let $B$ be a subalgebra of $A$ . Then, $\frac{B * N}{N} ≅ \frac{B}{B \cap N}$ .

Proof.

Clearly, $N$ and $B \cap N$ are strong ideals.

Define a map $ψ : B \to B * N / N$ by $ψ (a) = a * N$ . It is obvious that; $ψ$ is well defined and onto map.

To show that $ψ$ is a homomorphism. Let $a, b \in B$ . Consider

(i)

ψ (a + b) = (a + b) * N = (a * N) + (b * N) .

= ψ (a) + ψ (b) .

(ii)

ψ (a * b) = (a * b) * N = (a * N) * (b * N) .

= ψ (a) * ψ (b) .

(iii) Suppose $a \leq b$ .

\Leftrightarrow a * N \leq b * N .

\Leftrightarrow ψ (a) \leq ψ (b) .

Thus, $ψ$ is a homomorphism.

Now, we shall prove that $ker ψ = B \cap N$ .

ker ψ = {a \in B | ψ (a) = 0 * N = N} .

= {a \in B | a * N = N} .

= {a \in B | a \in N} . [Since N is a strong ideal]

= B \cap N .

Therefore, by first isomorphism theorem: $\frac{B}{ker ψ} ≅ Imψ$ . Hence, $\frac{B}{B \cap N} ≅ \frac{B * N}{N}$ .

Theorem 4.9

Let $A$ be autometrized algebra. Let $M, N$ be ideas of $A$ . Suppose $M \subseteq N$ . Then, $N / M$ is ideal.

Proof.

To show that $N / M$ is ideal.

Let $a * M, b * M \in N / M$ . Therefore, $a, b \in N$ . Therefore, $a + b \in N$ . Hence, $(a + b) * M \break \in N * M$ .
Let $a * M \in N / M$ and $x * M \in A / M$ . Therefore, $a \in N$ .

Suppose that $(x * M) * (0 * M) \leq (a * M) * (0 * M)$ . Therefore, $(x * 0) * M \leq (a * 0) * M$ . This implies that $x * 0 \leq a * 0$ . Since $N$ is an ideal and $a \in N$ ; imply that $x \in N$ . Hence, $x * M \in N / M$ . Thus, $N / M$ is ideal.

Theorem 4.10 (Third Isomorphism Theorem)

Let $A$ be a monoid autometrized algebra. Let $M, N$ be ideas of $A$ . Suppose $M \subseteq N$ . Then, $A / N ≅ \frac{A / M}{N / M}$ .

Proof.

Clearly, $M, N$ and $N / M$ are strong ideals. Define a map $ψ : A / M \to A / N$ by $ψ (a * M) = a * N$ .

$ψ$ is well defined because $∀a, b \in A$ ;

a * M = b * M .

\Rightarrow a * b \in M .

\Rightarrow a * b \in N . [Since M \subseteq N]

\Rightarrow a * N = b * N .

\Rightarrow ψ (a * M) = ψ (b * M) .

Therefore, $ψ$ is well defined.

To show that $ψ$ is onto. It is clear that $a * N \in A / N$ for any $a \in A$ . This implies that $a * M \in A * M$ such that $ψ (a * M) = a * N$ . Hence, $ψ$ is onto map.

To show that $ψ$ is a homomorphism.

Let $a * M, b * M \in A / M$ . Therefore, $a, b \in A$ . Consider

(i)

ψ ((a * M) + (b * M)) = ψ ((a + b) * M) .

= (a + b) * N .

= (a * N) + (b * N) .

= ψ (a * M) + ψ (b * M) .

(ii)

ψ ((a * M) * (b * M)) = ψ ((a * b) * M) .

= (a * b) * N .

= (a * N) * (b * N) .

= ψ (a * M) * ψ (b * M) .

(iii) Suppose $a \leq b$ . Therefore, $a * N \leq b * N$ . As a result $ψ (a * M) \leq ψ (b * M)$ . Thus, $ψ$ is a homomorphism.

Now, we shall prove that $ker ψ = N / M$ .

ker ψ = {a * M \in A / M | ψ (a * M) = 0 * N = N} .

= {a * M \in A / M | a * N = N} .

= {a * M \in A / M | a \in N} .

= N / M .

Therefore, by first isomorphism theorem: $\frac{A / M}{ker ψ} ≅ Imψ$ . Hence, $\frac{A / M}{N / M} ≅ A / N$ .

Theorem 4.11 (Correspondence Theorem)

Let $A$ be autometrized algebra. Let $M$ be an idea of $A$ .

Let $\MathI_{M} (A)$ = set of all ideals of $A$ containing $M$ .

Let $\MathI (A / M)$ = set of all ideals of $A / M$ .

Define $ψ : \MathI_{M} (A) \to \MathI (A / M)$ by $ψ (N) = N / M$ . Then, $ψ$ is one-to-one and onto. That is; $\MathI_{M} (A)$ and $\MathI (A / M)$ are in one-to-one correspondence.

Proof.

To show that $ψ$ is well defined.

Let $N_{1}, N_{2} \in \MathI_{M} (A)$ . That is; $N_{1}, N_{2}$ are ideals of A and $M \subseteq N_{1}, N_{2}$ .

Suppose $N_{1} = N_{2}$ . Therefore, $N_{1} / M = N_{2} / M$ . Thus, $ψ (N_{1}) = ψ (N_{2})$ . Hence, $ψ$ is well defined.

To show that $ψ$ is one-to-one.

Let $N_{1}, N_{2} \in \MathI_{M} (A)$ . Suppose $ψ (N_{1}) = ψ (N_{2})$ . Therefore, $N_{1} / M = N_{2} / M$ . Let $a \in N_{1}$ . Therefore, $a * M \in N_{1} / M = N_{2} / M$ . This implies that $a * M = b * M$ for some $b \in N_{2}$ . Since $a \leq b$ and $b \leq a$ ; imply that $a = b$ . Since $b \in N_{2}$ ; $a \in N_{2}$ . Therefore, $N_{1} \subseteq N_{2}$ . Similarly, $N_{2} \subseteq N_{1}$ . Therefore, $N_{1} = N_{2}$ . Hence, $ψ$ is one-to-one.

To show that $ψ$ is onto.

Let $J \in \MathI (A / M)$ . Therefore, $J$ is an ideal of $A / M$ . Consider the natural homomorphism: $f : A \to A / M$ defined by $f (p) = p * M$ . Therefore, by Theorem 3.23, $N = f^{- 1} (J) = {a \in A | f (a) \in J}$ is an ideal of $A$ containing $M$ . Thus, $N \in \MathI_{M} (A)$ and $ψ (N) = N / M$ .

Let $a * M \in N / M$ . This implies that $a \in N$ . So, $f (a) \in J$ . Therefore, $a * M \in J$ . Hence, $N / M \subseteq J$ . Conversely, let $a * M \in J$ . So, $f (a) \in J$ . This implies that $a \in N$ . Therefore, $a * M \in N / M$ . Thus, $J \subseteq N / M$ . Therefore, $ψ (N) = N / M = J$ . This shows that $ψ$ is onto.

Hence, $\MathI_{M} (A)$ and $\MathI (A / M)$ are in one-to-one correspondence.

Example 4.12

We saw in Example 3.4 that $A$ is an autometrized algebra. Also, $M = {0, a}$ and $N = {0, a, b}$ are ideals of $A$ . Clearly, $M$ is contained in the ideals $N$ and $A$ . Therefore, $\MathI_{M} (A) = {M, N, A}$ and $\MathI (A / M) = {{0 * M, a * M}, {0 * M, a * M, b * M}, A / M} .$ We also see that $ψ (M) = M / M = \{0 * M, a * M\}$ , $ψ (N) = N / M = \{0 * M, a * M, b * M\}$ and $ψ (A) = A / M$ . Thus, $∖Math I_{M} (A)$ and $∖MathI (A / M)$ are in one-to-one correspondence.

Remark 4.13 Let $A$ be an autometrized l-algebra. Let $M$ be an ideal of $A$ . Let $A / M = \{a * M ∣ a \in A\}$ . For any $a * M, b * M \in A / M$ , define the operations:

\begin{matrix} (a * M) + (b * M) = (a + b) * M \\ (a * M) + (b * M) = (a * b) * M \\ (a * M) \land (b * M) = (a \land b) * M \\ (a * M) \lor (b * M) = (a \lor b) * M \end{matrix}

Then, $(A / M, +, \land, \lor, *)$ is an autometrized l-algebra is called the quotient algebra of $A$ by ideal $M$ .

5. Conclusion

This paper introduced the concept of subalgebra and discussed some properties of subalgebra. We also examined different types of ideals and how they behaved. We also described some properties of a homomorphism. Finally, we presented quotient algebra and discussed isomorphism theorems.

Acknowledgements

The authors are very grateful to the anonymous reviewers for their helpful comments and suggestions that helped for improving the paper’s quality.

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No potential conflict of interest was reported by the authors.

Additional information

Funding

The authors received no direct funding from public or non-public organizations for this research work.

Notes on contributors

Gebrie Yeshiwas Tilahun

Gebrie Yeshiwas Tilahun is a PhD scholar at the Department of Mathematics, College of Natural Sciences, Arba Minch University, Ethiopia. His research interest includes the development of an autometrized algebra.

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Structure of an Autometrized Algebra

ABSTRACT

1. Introduction

1.1. Preliminaries

2. Subalgebras

3. Ideals and Homomorphism Theorems

4. Quotient Algebra

5. Conclusion

Acknowledgements

Disclosure statement

Notes on contributors

Gebrie Yeshiwas Tilahun

References

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Structure of an Autometrized Algebra

ABSTRACT

1. Introduction

1.1. Preliminaries

2. Subalgebras

3. Ideals and Homomorphism Theorems

4. Quotient Algebra

5. Conclusion

Acknowledgements

Disclosure statement

Additional information

Funding

Notes on contributors

Gebrie Yeshiwas Tilahun

References

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