Congruency, Homomorphism  and Isomorphism on Autometrized Algebras

Definition 2.1

(K. N. Swamy 1964) A system A= $(A,+,0,\le ,\ast )$ is called an autometrized algebra if

Definition 2.2

(K. Swamy & Rao 1977) An autometrized algebra A = $(A,+,0,\le ,\ast )$ is called normal if and only if

Definition 2.3

(K. Swamy & Rao 1977) Let $A=(A,+,0,\le ,\ast )$ be a system. Then $A$ is said to be a lattice ordered autometrized algebra (or) autometrized l-algebra if

Definition 2.4

(Tilahun et al. 2023d) Let A = $(A,+,0,\le ,\ast )$ be autometrized algebra. Let $B\subseteq A$ . Then $B$ is said to be a subalgebra of $A$ if;

Definition 2.5

(Tilahun et al. 2023d) A nonempty subset $I$ of an autometrized algebra A = $(A,+,0,\le ,\ast )$ is called an ideal if and only if

Definition 2.6

(K. Swamy & Rao 1977) Let $A$ be an autometrized algebra. Then $A$ is said to be semiregular if for any $a\in A$ , $a\ge 0\Rightarrow a\ast 0=a$ .

Definition 2.7

(K. Swamy & Rao 1977) Let $A$ be a normal autometrized algebra. An equivalence relation $\mathrm{\Theta }$ on $A$ is called a congruence relation if

Definition 2.8

(K. Swamy & Rao 1977) Let $A$ be a normal autometrized algebra and $\mathrm{\Theta }$ be a congruence on $A$ . Then, $A/\mathrm{\Theta }$ = the set of all equivalence classes of $\mathrm{\Theta }$ in $A=\{\overline{x}=\mathrm{\Theta }\left(x\right)|x\in A\}$ .

Theorem 2.9

(K. Swamy & Rao 1977) Let $A$ be a normal autometrized algebra. Let $\mathrm{\Theta }$ be a congruence relation on $A$ . For any $\overline{a},\overline{b}\in A/\mathrm{\Theta }$ , Define:

Then $(A/\mathrm{\Theta },+,\le ,\ast )$ is a normal autometrized algebra if and only if $\mathrm{\Theta }$ has the following property: For any $a,b\in A$ and $z_1\ge 0,z_2\ge 0$ ;

Theorem 2.10

(K. Swamy & Rao 1977) Let $A$ be normal autometrized algebra. Let $\mathrm{ℐ}\left(A\right)$ = set of all ideals of $A$ . Let $\mathit{Con}\left(A\right)$ = set of all congruences on $A$ . Then $\mathrm{ℐ}\left(A\right)$ and $\mathit{Con}\left(A\right)$ are in one-to-one correspondence.

Definition 2.11

(Tilahun et al. 2023d) Let A = $(A,+,0,\le ,\ast )$ and B = $(B,+,0,\le ,\ast )$ 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

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

Definition 2.12

(Tilahun et al. 2023d) Let $A,B$ be normal autometrized algebras. Let $f:A\to B$ be a homomorphism. Then $kerf=\{x\in A|f\left(x\right)=\overline{0}\}$ where $\overline{0}$ is the zero element of $B$ .

Clearly, $f$ is one-to-one if and only if $kerf=\left\{0\right\}$ .

Definition 2.13

Let $A$ and $B$ be autometrized algebras. Let $f:A\to B$ be a map. If $a\le b\iff f\left(a\right)\le f\left(b\right),\forall a,b\in A$ , then $f$ is said to be an order-embedding of $A$ in to $B$ . That is, $f$ is both order-preserving and order-reversing.

Remark 2.14

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

Definition 3.1

Let $A$ be an autometrized algebra. Let $\mathrm{\Theta }$ be a relation on $A$ . Then $\mathrm{\Theta }$ is said to be satisfied compatibility property, if

Definition 3.2

Let $A$ be an autometrized algebra. The set of all equivalence relations on $A$ is denoted by $\mathit{Eq}\left(A\right)$ .

Definition 3.3

Let $A$ be an autometrized algebra. Let $\mathrm{\Theta }\in \mathit{Eq}\left(A\right)$ . Then $\mathrm{\Theta }$ is said to be a congruence on $A$ if it satisfies compatibility property.

Definition 3.4

Let $A$ be an autometrized algebra. The set of all congruence relations on $A$ is denoted by $\mathit{Con}\left(A\right)$ .

Definition 3.5

Let $A$ be an autometrized algebra. Then $\mathrm{\Delta }=\{(a,b)\in A\times A|a=b\}=\left\{\right(a,a\left)\right|a\in A\}$ is called diagonal relation on $A$ . And also $\nabla =A\times A$ is called all relation on $A$ .

Example 3.6

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

It is clear to show that $A$ is an autometrized algebra. Let us consider the equivalent relations $\Delta =\left\{\right(0,0),(a,a),(b,b),(c,c)\}$ , ${\mathrm{\Theta }}_1=\left\{\right(0,0),(a,a),(b,b),(c,c),(0,a),(a,0)\}$ , ${\mathrm{\Theta }}_2=\left\{\right(0,0),(a,a),(b,b),(c,c),(a,0),(0,a),(0,b),(b,0),(a,b),(b,a)\}$ and $\nabla =A\times A$ on $A$ . Then $\Delta$ , ${\mathrm{\Theta }}_1$ , ${\mathrm{\Theta }}_2$ and $\nabla$ are all the congruence relations on $A$ . Therefore, $\mathit{Con}\left(A\right)=\{\Delta ,{\mathrm{\Theta }}_1,{\mathrm{\Theta }}_2,\nabla \}$ .

Definition 3.7

Let $A$ be an autometrized algebra and $X\subseteq A$ . Then the smallest congruence containing $X\times X$ is called the congruence generated by $X\times X$ or $X$ and denoted by $\mathrm{\Theta }(X\times X)$ or $\mathrm{\Theta }\left(X\right)$ . Further, if $X=\{a,b\}$ , then $\mathrm{\Theta }\left(X\right)$ = $\mathrm{\Theta }(a,b)$ is called principal congruence.

Example 3.8

In Example (3.6), $A$ is an autometrized algebra. Take $X=\{a,b\}$ . Then, $X\times X=\left\{\right(a,a),(b,b),(a,b),(b,a)\}$ . Therefore, $\mathrm{\Theta }(X\times X)=\mathrm{\Theta }\left(X\right)=\left\{\right(0,0),(a,a),(b,b),(c,c),(a,0),(0,a),(0,b),(b,0),(a,b),(b,a)\}$ . Hence, $\mathrm{\Theta }\left(X\right)$ is a principal congruence.

Definition 3.9

Let ${\mathrm{\Theta }}_1$ , ${\mathrm{\Theta }}_2$ be relations on an autometrized algebra $A$ . Then their product is denoted by ${\mathrm{\Theta }}_1◦{\mathrm{\Theta }}_2$ , defined as:

That is;

Example 3.10

It is clear that in Example (3.6), $A$ is an autometrized algebra. And we know that ${\mathrm{\Theta }}_1=\left\{\right(0,0),(a,a),(b,b),(c,c),(0,a),(a,0)\}$ , ${\mathrm{\Theta }}_2=\left\{\right(0,0),(a,a),(b,b),(c,c),(a,0),(0,a),(0,b),(b,0),(a,b),(b,a)\}$ . Therefore, ${\mathrm{\Theta }}_1◦{\mathrm{\Theta }}_2=\left\{\right(0,0),(a,a),(b,b),(c,c),(a,0),(0,a),(0,b),(b,0),(a,b),(b,a)\}$ .

Remark 3.11

If ${\mathrm{\Theta }}_1,{\mathrm{\Theta }}_2\in \mathit{Con}\left(A\right)$ , then ${\mathrm{\Theta }}_1◦{\mathrm{\Theta }}_2$ need not be a congruence relation on $A$ .

Theorem 3.12

If ${\mathrm{\Theta }}_1$ , ${\mathrm{\Theta }}_2$ are two congruence relations on a normal autometrized algebra $A$ satisfying $a\le b\Rightarrow a\ast c\le b\ast c\forall a,b,c\in A$ and $b\ne c$ , then ${\mathrm{\Theta }}_1◦{\mathrm{\Theta }}_2$ satisfies compatibility property.

Proof.

Now to show that ${\mathrm{\Theta }}_1◦{\mathrm{\Theta }}_2$ satisfies compatibility property.

Theorem 3.13

Let Θ₁, Θ₂ be two congruence relations on a normal autometrized algebra A satisfying a ≤ b ⇒ a ∗ c ≤ b ∗ c ∀ a, b, c ∈ A and b ≠ c. Suppose that Θ₁ $◦$ Θ₂ = Θ₂ $◦$ Θ₁. Then Θ₁ $◦$ Θ₂ is a congruence relation on A.

Proof.

To show that the product Θ₁ $◦$ Θ₂ is an equivalence relation.

By theorem (3.12); we know that Θ₁ $◦$ Θ₂ satisfies compatibility property. Hence, Θ₁ $◦$ Θ₂ is a congruence relation on A.

Example 3.14

It is clear that in Example (3.6), A is a normal autometrized algebra satisfying a ≤ b ⇒ a ∗ c ≤ b ∗ c ∀ a, b, c ∈ A and b ≠ c. And we know that Θ₁ = {(0, 0), (a, a), (b, b), (c, c), (0, a), (a, 0)}, Θ₂ = {(0, 0), (a, a), (b, b), (c, c), (a, 0), (0, a), (0, b), (b, 0), (a, b), (b, a)}. Therefore, Θ₁ $◦$ Θ₂ = Θ₂ $◦$ Θ₁ = Θ₂. Hence, Θ₁ $◦$ Θ₂ is a congruence relation on A.

Theorem 3.15

Let A be a normal autometrized algebra A satisfying a ≤ b ⇒ a ∗ c ≤ b ∗ c ∀ a, b, c ∈ A and b ≠ c. Then (Con(A), ⊆) is a complete sublattice of (Eq(A), ⊆).

Proof.

Let {Θ_i} _i∈I ⊆ Con(A). Therefore, {Θ_i} _i∈I ⊆ Eq(A). We know that

Clearly, ${\cap }_{i\in I}{\mathrm{\Theta }}_i\in \mathit{Con}\left(A\right)$ . So, ${\mathit{Inf}}_{\mathit{Eq}\left(A\right)}{\left\{{\mathrm{\Theta }}_i\right\}}_{i\in I}={\wedge }_{i\in I}{\mathrm{\Theta }}_i\in \mathit{Con}\left(A\right)$ .

To show that ${\mathit{Sup}}_{\mathit{Eq}\left(A\right)}{\left\{{\mathrm{\Theta }}_i\right\}}_{i\in I}={\vee }_{i\in I}{\mathrm{\Theta }}_i\in \mathit{Con}\left(A\right)$ . Let $\mathrm{\Psi }=\cup \{{\mathrm{\Theta }}_0◦{\mathrm{\Theta }}_1◦\cdots ◦{\mathrm{\Theta }}_k|0,1,\cdots k\in I\}$ . To show that $\mathrm{\Psi }\in \mathit{Con}\left(A\right)$ . Clearly, $\mathrm{\Psi }\in \mathit{Eq}\left(A\right)$ .

To show that $\mathrm{\Psi }$ satisfies compatibility property. Let $(a,b)\in \mathrm{\Psi }$ and $x\ast y\le a\ast b$ . Therefore, $(a,b)\in {\mathrm{\Theta }}_0◦{\mathrm{\Theta }}_1◦\cdots ◦{\mathrm{\Theta }}_i$ for $i=0,\cdots ,k$ . By theorem (3.12); $(x,y)\in {\mathrm{\Theta }}_0◦{\mathrm{\Theta }}_1◦\cdots ◦{\mathrm{\Theta }}_i$ . It is clear that $(x,y)\in \mathrm{\Psi }$ . Therefore, ${\mathit{Sup}}_{\mathit{Eq}\left(A\right)}{\left\{{\mathrm{\Theta }}_i\right\}}_{i\in I}={\vee }_{i\in I}{\mathrm{\Theta }}_i\in \mathit{Con}\left(A\right)$ . Hence, $\mathit{Con}\left(A\right)$ is a complete sublattice of $\mathit{Eq}\left(A\right)$ .

Definition 3.16

Let A be an autometrized algebra and θ, φ ∈ Con(A). Then θ, φ are said to be permutable congruences if θ $◦$ φ = φ $◦$ θ.

Example 3.17

It is clear that in Example (3.6); ${\mathrm{\Theta }}_1,{\mathrm{\Theta }}_2$ are permutable congruences. Since ${\mathrm{\Theta }}_1◦{\mathrm{\Theta }}_2={\mathrm{\Theta }}_2◦{\mathrm{\Theta }}_1={\mathrm{\Theta }}_2$ .

Definition 3.18

Let $A$ be an autometrized algebra. If every pair of elements in $\mathit{Con}\left(A\right)$ are permutable, then $A$ is said to be a congruence-permutable autometrized algebra.

Example 3.19

It is clear that in Example (3.6); $\mathit{Con}\left(A\right)=\{\Delta ,{\mathrm{\Theta }}_1,{\mathrm{\Theta }}_2,\nabla \}$ . Further, $\mathrm{\Delta }◦{\mathrm{\Theta }}_1={\mathrm{\Theta }}_1◦\Delta ,\mathrm{\Delta }◦{\mathrm{\Theta }}_2={\mathrm{\Theta }}_2◦\Delta ,\mathrm{\Delta }◦\nabla =\nabla ◦\Delta ,{\mathrm{\Theta }}_1◦{\mathrm{\Theta }}_2={\mathrm{\Theta }}_2◦{\mathrm{\Theta }}_1,{\mathrm{\Theta }}_1◦\Delta =\Delta ◦{\mathrm{\Theta }}_1$ , and ${\mathrm{\Theta }}_2◦\Delta =\Delta ◦{\mathrm{\Theta }}_2$ . Hence, $A$ is a congruence-permutable autometrized algebra.

Definition 3.20

Let $A$ be an autometrized algebra. Then $A$ is said to be a congruence-modular algebra if $\mathit{Con}\left(A\right)$ is a modular lattice.

Theorem 3.21

(Burris 1981) Let $A$ be an autometrized algebra and $\theta ,\varphi \in \mathit{Eq}\left(A\right)$ . Then the following are equivalent.

Theorem 3.22

Let $A$ be a congruence-permutable autometrized algebra. Then $A$ is congruence-modular.

Proof.

Suppose $A$ is a congruence-permutable. That is every pair of congruences on $A$ are permutable.

Now to show that $A$ is congruence-modular. To show that $\mathit{Con}\left(A\right)$ is a modular.

Let ${\mathrm{\Theta }}_1,{\mathrm{\Theta }}_2,{\mathrm{\Theta }}_3\in \mathit{Con}\left(A\right)$ and ${\mathrm{\Theta }}_1\subseteq {\mathrm{\Theta }}_3$ . Clearly, ${\mathrm{\Theta }}_1\vee ({\mathrm{\Theta }}_2\wedge {\mathrm{\Theta }}_3)\le ({\mathrm{\Theta }}_1\vee {\mathrm{\Theta }}_2)\wedge {\mathrm{\Theta }}_3$ .

To show that $({\mathrm{\Theta }}_1\vee {\mathrm{\Theta }}_2)\wedge {\mathrm{\Theta }}_3\le {\mathrm{\Theta }}_1\vee ({\mathrm{\Theta }}_2\wedge {\mathrm{\Theta }}_3)$ .

Let $(a,b)\in ({\mathrm{\Theta }}_1\vee {\mathrm{\Theta }}_2)\wedge {\mathrm{\Theta }}_3$ . Therefore, $(a,b)\in {\mathrm{\Theta }}_1\vee {\mathrm{\Theta }}_2$ and $(a,b)\in {\mathrm{\Theta }}_3$ . Since ${\mathrm{\Theta }}_1,{\mathrm{\Theta }}_2$ are permutable; ${\mathrm{\Theta }}_1\vee {\mathrm{\Theta }}_2={\mathrm{\Theta }}_1◦{\mathrm{\Theta }}_2$ . As a result, $(a,b)\in {\mathrm{\Theta }}_1◦{\mathrm{\Theta }}_2$ . There exists $c\in A$ such that $(a,c)\in {\mathrm{\Theta }}_1$ and $(c,b)\in {\mathrm{\Theta }}_2$ . Since ${\mathrm{\Theta }}_1\subseteq {\mathrm{\Theta }}_3$ ; $(a,c)\in {\mathrm{\Theta }}_3$ . Since ${\mathrm{\Theta }}_3$ is a congruence relation; $(c,a)\in {\mathrm{\Theta }}_3$ . Now we have $(c,a),(a,b)\in {\mathrm{\Theta }}_3$ . Clearly, $(c,b)\in {\mathrm{\Theta }}_3$ . As a result, $(c,b)\in {\mathrm{\Theta }}_2\wedge {\mathrm{\Theta }}_3$ .

Now we have $(a,c)\in {\mathrm{\Theta }}_1$ and $(c,b)\in {\mathrm{\Theta }}_2\wedge {\mathrm{\Theta }}_3$ . Therefore, $(a,b)\in {\mathrm{\Theta }}_1◦({\mathrm{\Theta }}_2\wedge {\mathrm{\Theta }}_3)$ . Since $A$ is a congruence-permutable; $(a,b)\in {\mathrm{\Theta }}_1\vee ({\mathrm{\Theta }}_2\wedge {\mathrm{\Theta }}_3)$ . Therefore, $({\mathrm{\Theta }}_1\vee {\mathrm{\Theta }}_2)\wedge {\mathrm{\Theta }}_3\le {\mathrm{\Theta }}_1\vee ({\mathrm{\Theta }}_2\wedge {\mathrm{\Theta }}_3)$ . Consequently, ${\mathrm{\Theta }}_1\vee ({\mathrm{\Theta }}_2\wedge {\mathrm{\Theta }}_3)=({\mathrm{\Theta }}_1\vee {\mathrm{\Theta }}_2)\wedge {\mathrm{\Theta }}_3$ . Hence, $\mathit{Con}\left(A\right)$ is modular.

Theorem 4.1

Let $A,B$ be normal autometrized algebras. Let $f:A\to B$ be a homomorphism. Then the set

is called kernel of the homomorphism $f$ .

Theorem 4.2

Let $A,B$ be normal autometrized algebras. Let $f:A\to B$ be a homomorphism. Then, $kerf$ is a congruence on $A$ . That is $kerf\in \mathit{Con}\left(A\right)$ .

Proof.

It is clear that $kerf$ is an equivalence relation.

Thus, $kerf\in \mathit{Con}\left(A\right)$ .

Theorem 4.3

Let $A$ be a normal autometrized algebra and $\mathrm{\Theta }\in \mathit{Con}\left(A\right)$ . Define a map $f:A\to A/\mathrm{\Theta }$ by $f\left(a\right)=a/\mathrm{\Theta }=\overline{a}$ . Then $f$ is an epimorphism and $kerf=\mathrm{\Theta }$ .

Proof.

Clearly $\mathrm{\Theta }$ satisfies the condition. Hence $A/\mathrm{\Theta }$ is a normal automertrized algebra. Clearly $f$ is onto.

To show that $f$ is homomorphism. Let $a,b\in A$ .

Let $a\le b$ . Then $\exists x\ge 0$ such that $a+x=b$ . Therefore $(a+x,b)\in \mathrm{\Theta }$ . Which implies that $\overline{a}\le \overline{b}.$ That is; $f\left(a\right)\le f\left(b\right)$ . Hence $f$ is a homomorphism. So that $f$ is an epimorphism from $A$ to $A/\mathrm{\Theta }$ . Consider;

Theorem 4.4

(Homomorphism theorem) Let A, B be normal autometrized algebras. Let f: A → B be a homomorphism. Then A/ ker f $\cong$ Imf. In particular if f is onto, then A/ ker f $\cong$ B.

Proof.

Let ker f = Θ. By the above theorem (4.3), the map

$g:A\to A/\mathrm{\Theta }$ is an epimorphism and $kerg=\mathrm{\Theta }$ . Define a map $h:A/\mathrm{\Theta }\to \mathit{Imf}$ by $h\left(\overline{a}\right)=f\left(a\right)$ . Clearly, $h$ is onto map. To show that $h$ is well-defined and one to one. Suppose $\overline{a}=\overline{b}$ , then

Therefore, $h$ is well-defined and one to one.

Now to show that $h$ is homomorphism. Let $\overline{a},\overline{b}\in A/\mathrm{\Theta }$ . Consider;

Again consider;

Suppose let $\overline{a}\le \overline{b}$ . Therefore $\exists x\ge 0$ such that $(a+x,b)\in \mathrm{\Theta }$ . Hence,

Hence $f\left(a\right)+f\left(x\right)=f\left(b\right)$ . Since $x\ge 0\Rightarrow f\left(x\right)\ge \overline{0}$ . Then $f\left(a\right)\le f\left(b\right)$ . Therefore $h\left(\overline{a}\right)\le h\left(\overline{b}\right)$ .

And thus $h$ is a homomorphism. Therefore $h$ is an isomorphism from $A/\mathrm{\Theta }$ to $\mathit{Imf}$ . That is $A/\mathrm{\Theta }\cong \mathit{Imf}$ . If $f$ is onto map, then $\mathit{Imf}=B$ . Hence $A/\mathrm{\Theta }\cong B$ .

Definition 4.5

Let $A$ and $B$ be autometrized algebras and $\theta ,\varphi \in \mathit{Con}\left(A\right)$ . Suppose $\theta \subseteq \varphi$ .

Theorem 4.6

Let $A$ be a normal autometrized algebra and $\theta ,\varphi \in \mathit{Con}\left(A\right)$ . Suppose $\theta \subseteq \varphi$ . Then,

$\varphi /\theta \in \mathit{Con}(A/\theta ).$

Proof.

First let us show that $\varphi /\theta$ is an equivalence relation on A.

To show that $\varphi /\theta$ is congruence on A. Let $(a/\theta ,b/\theta ),(c/\theta ,d/\theta )\in \varphi /\theta$ . Then, $(a,b),(c,d)\in \varphi$ .

Therefore, $\varphi /\theta \in \mathit{Con}\left(A\right)$ .

Theorem 4.7

(Second Isomorphism Theorem) Let A be a normal autometrized algebras. Let θ,φ ∈ Con (A) and θ ⊆ φ. Then

Proof.

Define a map $g:A/\theta \to A/\varphi$ by $g(a/\theta )=a/\varphi$ .

To show that $g$ is well-defined.

Suppose; $\overline{a}=\overline{b}$ . That is;

Therefore, $g(a/\theta )=g(b/\theta )$ . Hence, $g$ is well-defined and one to one. Clearly $g$ is onto map.

To show that $g$ is a homomorphism.

Let $a/\theta ,b/\theta \in A/\theta$ .

Again consider;