Some Examples of BL-Algebras Using Commutative Rings

Abstract

BL-algebras are algebraic structures corresponding to Hajek’s basic fuzzy logic. The aim of this paper is to analyze the structure of BL-algebras using commutative rings. Due to computational considerations, we are interested in the finite case. We present new ways to generate finite BL-algebras using commutative rings and provide summarizing statistics. Furthermore, we investigated BL-rings, i.e., commutative rings whose the lattice of ideals can be equipped with a structure of BL-algebra. A new characterization for these rings and their connections to other classes of rings is established. Furthermore, we give examples of finite BL-rings for which the lattice of ideals is not an MV-algebra and, using these rings, we construct BL-algebras with $2^r+1$ elements, $r\ge 2,$ and BL-chains with k elements, $k\ge 4.$ In addition, we provide an explicit construction of isomorphism classes of BL-algebras of small n size ($2\le n\le 5$).

1. Introduction

The origin of residuated lattices is in mathematical logic. They were introduced by Dilworth and Ward, through the papers [1,2]. The study of residuated lattices originated in 1930 in the context of the theory of rings, with the study of ring ideals. It is known that the lattice of ideals of a commutative ring is a residuated lattice; see [3]. Several researchers ([3,4,5,6], etc.) have been interested in this construction.

Two important subvarieties of residuated lattices are BL-algebras (corresponding to Hajek’s logic; see [7]) and MV-algebras (corresponding to Łukasiewicz’s many-valued logic; see [8,9]). For instance, rings for which the lattice of ideals is a BL-algebra are called BL-rings and were introduced in [5].

In this paper, we obtain a description for BL-rings using a new characterization of BL-algebras, given in Theorem 1, i.e., residuated lattices L in which $[x\odot (x\to y\left)\right]\to z=(x\to z)\vee (y\to z)$ for every $x,y,z\in L.$ Then, BL-rings are unitary and commutative rings A with the property that $K:$$[I\otimes (J:I\left)\right]=(K:I)+(K:J),$ for every $I,J,K\in Id\left(A\right);$ see Corollary 1.

Additionally, we show that the class of BL-rings contains other known classes of commutative rings: rings that are principal ideal domains and some types of finite unitary commutative rings; see Theorem 2, Corollaries 2 and 3.

One recent application of BCK algebras is in coding theory. In fact, MV-algebras are commutative BCK-algebras, see [10].

Due to computational considerations, in this paper, we are interested in finding ways to generate finite BL-algebras using finite commutative rings, since a solution that is computationally tractable is to consider algebras with a small number of elements. First, we give examples of finite BL-rings whose lattice of ideals is not an MV-algebra. Using these rings, we construct BL-algebras with $2^r+1$ elements, $r\ge 2$ (see Theorem 3) and BL-chains with $k\ge 4$ elements (see Theorem 4).

In [11], isomorphism classes of BL-algebras of size $n\le 12$ were only counted, not constructed, using computer algorithms. Up to an isomorphism, there is 1 BL-algebra of size 2, 2 BL-algebras of size 3, 5 BL-algebras of size 4, 9 BL-algebras of size 5, 20 BL-algebras of size 6, 38 BL-algebras of size 7, 81 BL-algebras of size 8, 160 BL-algebras of size 9, 326 BL-algebras of size 10, 643 BL-algebra of size 11 and 1314 BL-algebras of size 12. In Theorem 6, we present a way to generate (up to an isomorphism) finite BL-algebras with $2\le n\le 5$ elements by using the ordinal product of residuated lattices, and we present summarizing statistics. The described method can be used to construct finite BL-algebras of larger size, the inconvenience being the large number of BL-algebras that must be generated.

2. Preliminaries

Definition 1 

([1,2]). A (commutative) residuated lattice is an algebra $(L,\wedge ,\vee ,\odot ,\to ,0,1)$ such that:

(LR1) $(L,\wedge ,\vee ,0,1)$ is a bounded lattice;

(LR2) $(L,\odot ,1)$ is a commutative ordered monoid;

(LR3) $z\le x\to y$ iff $x\odot z\le y,$ for all $x,y,z\in L.$

The property (LR3) is called residuation, where ≤ is the partial order of the lattice $(L,\wedge ,\vee ,0,1).$

In a residuated lattice, an additional operation is defined; for $x\in L,$ we denote $x^{*}=x\to 0.$

Example 1 

([12]). Let $(\mathcal{B},\wedge ,\vee {,}^{\prime },0,1)$ be a Boolean algebra. If we define for every $x,y\in \mathcal{B},x\odot y=x\wedge y$ and $x\to y=x^{\prime }\vee y,$ then $(\mathcal{B},\wedge ,\vee ,\odot ,\to ,0,1)$ becomes a residuated lattice.

Example 2. 

It is known that, for a commutative unitary ring $A,$ if we denote by $Id\left(A\right)$ the set of all ideals, then for $I,J\in Id\left(A\right)$, the following sets

$$I+J=<I\cup J>=\{i+j,i\in I,j\in J\},$$

$$I\otimes J=\{\underset{k=1}{\sum ^n}{i}_k{j}_k,i_k\in I,j_k\in J\},$$

$$\left(I:J\right)=\{x\in A,x·J\subseteq I\},$$

$$Ann\left(I\right)=\left(\mathbf{0}:I\right),\mathit{where}\mathbf{0}=<0>,$$

are also ideals of $A,$ called sum, product, quotient and annihilator; see [13]. If we preserve these notations, $\left(Id\right(A),\cap ,+,\otimes \to ,0=\{0\},1=A)$ is a residuated lattice in which the order relation is ⊆ and $I\to J=(J:I),$ for every $I,J\in Id\left(A\right)$; see [6].

In a residuated lattice $(L,\wedge ,\vee ,\odot ,\to ,0,1)$, we consider the following identities:

$$\left(prel\right)(x\to y)\vee (y\to x)=1(\mathit{prelinearity});$$

$$\left(div\right)x\odot (x\to y)=x\wedge y(\mathit{divisibility}).$$

Definition 2 

([10,12,14]). A residuated lattice L is called a BL-algebra if L verifies $\left(prel\right)+\left(div\right)$ conditions.

A BL-chain is a totally ordered BL-algebra, i.e., a BL-algebra such that its lattice order is total.

Definition 3 

([8,9]). An MV-algebra is an algebra $\left(L,\oplus {,}^{*},0\right)$ satisfying the following axioms:

(MV1) $\left(L,\oplus ,0\right)$ is an abelian monoid;

(MV2) ${\left(x^{*}\right)}^{*}=x;$

(MV3) $x\oplus 0^{*}=0^{*};$

(MV4) ${\left(x^{*}\oplus y\right)}^{*}\oplus y=$${\left(y^{*}\oplus x\right)}^{*}\oplus x$, for all $x,y\in L.$

In fact, a residuated lattice L is an MV-algebra iff it satisfies the additional condition:

$$(x\to y)\to y=(y\to x)\to x,$$

for every $x,y\in L;$ see [12].

Remark 1 

([12]). If, in a BL- algebra L, $x^{**}=x,$ for every $x\in L$, and for $x,y\in L$ we denote

$$x\oplus y={(x^{*}\odot y^{*})}^{*},$$

then we obtain an MV-algebra $(L,\oplus {,}^{*},0).$ Conversely, if $(L,\oplus {,}^{*},0)$ is an MV-algebra, then $(L,\wedge ,\vee ,\odot ,\to ,0,1)$ becomes a BL-algebra, in which for $x,y\in L:$

$$x\odot y={(x^{*}\oplus y^{*})}^{*},$$

$$x\to y=x^{*}\oplus y,1=0^{*},$$

$$x\vee y=(x\to y)\to y=(y\to x)\to x\mathit{and}x\wedge y={(x^{*}\vee y^{*})}^{*}.$$

In fact, MV-algebras are exactly involutive BL-algebras.

Example 3 

([10]). We give an example of a finite BL-algebra which is not an MV-algebra. Let $L=\{0,a,b,c,1\}$; we define the following operations on L:

$$\begin{array}{cccccc}\to & 0& c& a& b& 1\\ 0& 1& 1& 1& 1& 1\\ c& 0& 1& 1& 1& 1\\ a& 0& b& 1& b& 1\\ b& 0& a& a& 1& 1\\ 1& 0& c& a& b& 1\end{array},\begin{array}{cccccc}\odot & 0& c& a& b& 1\\ 0& 0& 0& 0& 0& 0\\ c& 0& c& c& c& c\\ a& 0& c& a& c& a\\ b& 0& c& c& b& b\\ 1& 0& c& a& b& 1\end{array}.$$

We have, $0\le c\le a,b\le 1,$ but $a,b$ are incomparable; hence, L is a BL-algebra that is not a chain. We remark that $x^{**}=1$ for every $x\in L,x\ne 0.$

3. BL-Rings

Definition 4 

([5]). A commutative ring whose lattice of ideals is a BL-algebra is called a BL-ring.

In particular, we can call a commutative ring whose lattice of ideals is an MV-algebra an MV-ring.

We recall that, in [15], we showed that a commutative unitary ring A is an MV-ring iff it has the Chang property, i.e.,

$$I+J=(J:(J:I\left)\right),$$

for every $I,J\in Id\left(A\right).$ Obviously, every MV-ring is also a BL-ring.

BL-rings are closed under finite direct products, arbitrary direct sums and homomorphic images; see [5].

In the following, using the connections between BL-algebras and BL-rings, we give new characterizations for commutative and unitary rings for which the lattice of ideals is a BL-algebra.

Proposition 1 

([10]). Let $(L,\vee ,\wedge ,\odot ,\to ,0,1)$ be a residuated lattice. Then, we have the equivalences:

(i) $L$ satisfies $\left(prel\right)$ condition;

(ii) $(x\wedge y)\to z=(x\to z)\vee (y\to z),$ for every $x,y,z\in L.$

Lemma 1. 

Let $(L,\vee ,\wedge ,\odot ,\to ,0,1)$ be a residuated lattice. The following assertions are equivalent:

(i) L satisfies $\left(prel\right)$ condition;

(ii) $\mathit{For\; every}$$x,y,z\in L,$ if $x\wedge y\le z,$ then $(x\to z)\vee (y\to z)=1.$

Proof. 

$\left(i\right)\Rightarrow \left(ii\right).$ Following Proposition 1.

$\left(ii\right)\Rightarrow \left(i\right).$ Using (ii), for $z=$$x\wedge y$ we deduce that $1=(x\to (x\wedge y\left)\right)\vee (y\to (x\wedge y\left)\right)=\left[\right(x\to x)\wedge (x\to y\left)\right]\vee \left[\right(y\to x)\wedge (y\to y\left)\right]$$=(x\to y)\vee (y\to x),$ so L satisfies $\left(prel\right)$condition. □

Lemma 2. 

Let $(L,\vee ,\wedge ,\odot ,\to ,0,1)$ be a residuated lattice. The following assertions are equivalent:

(i) L satisfies $\left(div\right)$ condition;

(ii) $\mathit{For\; every}$$x,y,z\in L,$ if $x\odot (x\to y)\le z,$ then $x\wedge y\le z.$

Proof. 

$\left(i\right)\Rightarrow \left(ii\right),$ evidently.

$\left(ii\right)\Rightarrow \left(i\right).$ Using (ii), for $z=$$x\odot (x\to y)$ we can deduce that $x\wedge y\le x\odot (x\to y).$ Since in a residuated lattice, $x\odot (x\to y)\le x\wedge y$, we deduce that L satisfies $\left(div\right)$condition. □

Using Lemmas 1 and 2 we deduce Proposition 2.

Proposition 2. 

Let $(L,\vee ,\wedge ,\odot ,\to ,0,1)$ be a residuated lattice. The following assertions are equivalent:

(i) L is a BL-algebra;

(ii) For every $x,y,z\in L,$ if $x\odot (x\to y)\le z,$ then $(x\to z)\vee (y\to z)=1;$

(iii) $[x\odot (x\to y\left)\right]\to z=(x\to z)\vee (y\to z),$ for every $x,y,z\in L.$

Proof. 

$\left(i\right)\Rightarrow \left(ii\right).$ Let $x,y,z\in L$ such that $x\odot (x\to y)\le z.$ Since every BL-algebra satisfies $\left(div\right)$ condition, by Lemma 2, we can deduce that $x\wedge y\le z.$ Since every BL-algebra satisfies $\left(prel\right)$ condition, following Lemma 1, we can deduce that $1=(x\to z)\vee (y\to z).$

$\left(ii\right)\Rightarrow \left(i\right).$ First, we prove that L satisfies condition (ii) from Lemma 1. Therefore, let $x,y,z\in L$ such that $x\wedge y\le z.$ Thus, $(x\wedge y)\to z=1$. Since $x\odot (x\to y)\le x\wedge y,$ we deduce that $1=(x\wedge y)\to z\le (x\odot (x\to y\left)\right)\to z.$ Then, $x\odot (x\to y)\le z.$ By hypothesis, $(x\to z)\vee (y\to z)=1.$

To prove that L verifies condition $\left(ii\right)$ from Lemma 2, let $x,y,z\in L$ such that $x\odot (x\to y)\le z.$ By hypothesis, we deduce that, $(x\to z)\vee (y\to z)=1.$ Since $(x\to z)\vee (y\to z)\le (x\wedge y)\to z$, we obtain $(x\wedge y)\to z=1,$ that is, $x\wedge y\le z.$

$\left(iii\right)\Rightarrow \left(ii\right),$ evidently.

$\left(ii\right)\Rightarrow \left(iii\right).$ If we denote $t=[x\odot (x\to y\left)\right]\to z,$ we have $1=t\to t=t\to \left[\right(x\odot (x\to y))\to z]=[x\odot (x\to y\left)\right]\to (t\to z)$; hence, $x\odot (x\to y)\le t\to z.$

By hypothesis, we deduce that, $(x\to (t\to z\left)\right)\vee (y\to (t\to z\left)\right)=1.$

Then, $1=(t\to (x\to z\left)\right)\vee (t\to (y\to z\left)\right)\le t\to \left[\right(x\to z)\vee (y\to z\left)\right].$ Thus, $t\le (x\to z)\vee (y\to z).$

However, $(x\to z)\vee (y\to z)\le (x\wedge y)\to z\le [x\odot (x\to y\left)\right]\to z=t.$

We conclude that $t=(x\to z)\vee (y\to z),$ that is, $[x\odot (x\to y\left)\right]\to z=(x\to z)\vee (y\to z),$ for every $x,y,z\in L.$ □

Using Proposition 2 we obtain a new characterization for BL-algebras:

Theorem 1. 

A residuated lattice L is a BL-algebra if and only if for every $x,y,z\in L,$

$$[x\odot (x\to y\left)\right]\to z=(x\to z)\vee (y\to z).$$

Using this result, we can give a new description for BL-rings:

Corollary 1. 

Let A be a commutative and unitary ring. The following assertions are equivalent:

(i) $A$ is a BL-ring;

(ii) $K:$$[I\otimes (J:I\left)\right]=(K:I)+(K:J),$ for every $I,J,K\in Id\left(A\right).$

Theorem 2. 

Let A be a commutative ring that is a principal ideal domain. Then, A is a BL-ring.

Proof. 

Since A is a principal ideal domain, let $I=<a>$, $J=<b>$ be the principal non-zero ideals generated by $a,b\in A$$\setminus \left\{0\right\}$.

If $d=$gcd$\{a,b\}$, then $d=a·\alpha +b·\beta ,$$a,b\in A$, $a=a_{1}d$ and $b=b_{1}d$, with $1=$gcd$\{a_1,b_1\}$. Thus, $I+J=<d>,$$I\cap J=<ab/d>,$$I\otimes J=<ab>$ and $\left(I:J\right)=$$<a_1>.$

The conditions $\left(prel\right)$ are satisfied, $(I:J)+(J:I)=<a_1>+<b_1>=<1>=A$ and $\left(div\right)$ is also satisfied: $J\otimes (I:J)=<b>\otimes <a_1>=<ab/d>=I\cap J$.

If $I=\left\{0\right\}$, since A is an integral domain, we have that $\left(\mathbf{0}:J\right)+(J:\mathbf{0})=Ann\left(J\right)+A=A$ and $J\otimes (\mathbf{0}:J)=J\otimes Ann\left(J\right)=\mathbf{0}=\mathbf{0}\cap J=$$\mathbf{0}\otimes (J:\mathbf{0})$ for every $J\in Id\left(A\right)\setminus \left\{0\right\}.$

Moreover, we remark that $\left(I:\left(I:J\right)\right)=\left(J:\left(J:I\right)\right)=I+J$ for every non-zero ideal $I,J\in Id\left(A\right).$ Additionally, since A is an integral domain, we obtain $Ann\left(Ann\right(I\left)\right)=A,$ for every $I\in Id\left(A\right)\setminus \left\{0\right\}.$ We conclude that $Id\left(A\right)$ is a BL-algebra that is not an MV-algebra. □

Corollary 2. 

A ring factor of a principal ideal domain is a BL-ring.

Proof. 

We use Theorem 2 since BL-rings are closed under homomorphic images; see [5]. Moreover, we remark that a ring factor of a principal ideal domain is, in particular, an MV-ring, see [15]. □

Corollary 3. 

A finite commutative unitary ring of the form $A={\mathbb{Z}}_{k_1}\times {\mathbb{Z}}_{k_2}\times ...\times {\mathbb{Z}}_{k_r}$ (direct product of rings, equipped with componentwise operations) where $k_i=p_i^{{\alpha }_i}$, with $p_i$ a prime number, is a BL-ring.

Proof. 

We apply Corollary 2 using the fact that BL-rings are closed under finite direct products; see [5].

Moreover, we remark that if $A$ is a finite commutative unitary ring of the above form, then $Id\left(A\right)=$$Id\left({\mathbb{Z}}_{k_1}\right)\times Id\left({\mathbb{Z}}_{k_2}\right)\times ...\times Id\left({\mathbb{Z}}_{k_r}\right)$ is an MV-algebra $(Id\left(A\right),\oplus {,}^{*},0=\left\{0\right\})$ in which

$$I\oplus J=Ann(Ann\left(I\right)\otimes Ann\left(J\right))\mathrm{and}{I}^{*}=Ann\left(I\right)$$

for every $I,J\in Id\left(A\right)$ since, $Ann\left(Ann\right(I\left)\right)=I;$ see [15]. □

Example 4. 

(1) Following Theorem 2, the ring of integers $(\mathbb{Z},+,·)$ is a BL-ring in which $\left(Id\right(\mathbb{Z}),$$\cap ,+,\otimes \to ,0=\left\{0\right\},1=A)$ is not an MV-algebra. Indeed, since $\mathbb{Z}$ is the principal ideal domain, we have $Ann\left(Ann\left(I\right)\right)=\mathbb{Z}$, for every $I\in Id\left(\mathbb{Z}\right)\setminus \left\{0\right\}.$

(2) Let K be a field and $K\left[X\right]$ be the polynomial ring. For $f\in K\left[X\right]$, the quotient ring $A=K\left[X\right]/\left(f\right)$ is a BL-ring. Indeed, the lattice of ideals of this ring is an MV-algebra; see [15].

4. Examples of BL-Algebras Using Commutative Rings

In this section, we present ways to generate finite BL-algebras using finite commutative rings.

First, we give examples of finite BL-rings whose lattice of ideals is not an MV-algebra. Using these rings we construct BL-algebras with $2^r+1$ elements, $r\ge 2$ (see Theorem 3) and BL-chains with $k\ge 4$ elements (see Theorem 4).

We recall that, in [15], we proved the following proposition.

Proposition 3 

([15]). If A is a finite commutative unitary ring of the form ${\mathbb{Z}}_{k_1}\times {\mathbb{Z}}_{k_2}\times ...\times {\mathbb{Z}}_{k_r}$ (direct product of rings, equipped with componentwise operations), where $k_i=p_i^{{\alpha }_i}$, with $p_i$ a prime number, for all $i\in \{1,2...,r\}$ and $Id\left(A\right)$ denotes the set of all ideals of the ring A, then $\left(Id\left(A\right),\vee ,\wedge ,\odot ,\to ,0,1\right)$ is an MV-algebra, where the order relation is $\subseteq , I\odot J=I\otimes J$, $I^{*}=Ann\left(I\right)$, $I\to J=\left(J:I\right)$, $I\vee J=I+J$, $I\wedge J=I\cap J$, $0=\left\{0\right\}$ and $1=A$. The set $Id\left(A\right)$ has ${\mathcal{N}}_A=\stackrel{r}{\prod _{i=1}}\left({\alpha }_i+1\right)$ elements.

In the following, we give examples of finite BL-rings whose lattice of ideals is not an MV-algebra.

Definition 5 

([13]). Let R be a commutative unitary ring. The ideal M of the ring R is maximal if it is maximal with respect of the set inclusion, amongst all proper ideals of the ring R. That means there are no other ideals different from R contained in M. The ideal J of the ring R is a minimal ideal if it is a nonzero ideal that contains no other nonzero ideals. A commutative local ring R is a ring with a unique maximal ideal.

Example 5. 

(i) A field F is a local ring, with $\left\{0\right\}$ being the maximal ideal in this ring.

(ii) In $\left({\mathbb{Z}}_8,+,·\right)$, the ideal $J=\{\widehat{0},\widehat{4}\}$ is a minimal ideal and the ideal $M=\{\widehat{0},\widehat{2},\widehat{4},\widehat{6}\}$ is the maximal ideal.

Remark 2. 

Let R be a local ring with M its maximal ideal. Then, the quotient ring $R\left[X\right]/\left(X^n\right)$ with n being a positive integer is local. Indeed, the unique maximal ideal of the ring $R\left[X\right]/\left(X^n\right)$ is $\overrightarrow{M}=\{\overrightarrow{f}\in R\left[X\right]/\left(X^n\right)/f\in R\left[X\right],f=a_0+a_{1}X+...+a_{n-1}{X}^{n-1}$, with $a_0\in M\}$. For other details, the reader is referred to [16].

In the following, we consider the ring $({\mathbb{Z}}_n,+,·)$ with $n=p_1{p}_2...p_r,p_1,p_2,...,p_r$ being distinct prime numbers, $r\ge 2$ and the factor ring $R={\mathbb{Z}}_n\left[X\right]/\left(X^2\right).$

Remark 3. 

(i)With the above notations, in the ring $({\mathbb{Z}}_n,+,·)$, the ideals generated by ${\widehat{p}}_i,M_{p_i}=\left({\widehat{p}}_i\right),$ are maximals. The ideals of ${\mathbb{Z}}_n$ are of the form $I_d=\left(\widehat{d}\right)$, where d is a divisor of n.

(ii) Each element from ${\mathbb{Z}}_n-\{M_{p_1}\cup M_{p_2}\cup ...\cup M_{p_r}\}$ is an invertible element. Indeed, if $\widehat{x}\in {\mathbb{Z}}_n-\{M_{p_1}\cup M_{p_2}\cup ...\cup M_{p_r}\}$, we have gcd $\{x,n\}=1$; therefore, x is an invertible element.

Proposition 4. 

(i) With the above notations, the factor ring $R={\mathbb{Z}}_n\left[X\right]/\left(X^2\right)$ has $2^r+1$ ideals including $\left\{0\right\}$ and R.

(ii) For $\widehat{\gamma }\in {\mathbb{Z}}_n-\{M_{p_1}\cup M_{p_2}\cup ...\cup M_{p_r}\}$, the element $X+\widehat{\gamma }$ is an invertible element in R.

Proof. 

(i) Indeed, the ideals are: $J_{p_i}=\left(\widehat{\alpha }X+{\widehat{\alpha }}_i\right),{\widehat{\alpha }}_i\in M_{p{}_i},i\in \{1,2,...,r\},$ which are maximal, $J_d=\left(\widehat{\beta }X+{\widehat{\beta }}_d\right),{\widehat{\beta }}_d\in I_d,I_d$ is not maximal, $\widehat{\alpha },\widehat{\beta }\in R,d\ne n$, where d is a proper divisor of n, the ideals $\left(X\right)$, for $d=n$ and $\left(0\right)$. Therefore, we have ${\complement }_n^0$ ideals for ideal $\left(X\right)$, ${\complement }_n^1$ ideals for ideals $J_{p_i}$, ${\complement }_n^2$ ideals for ideals $J_{p_i{p}_j}$, $p_i\ne p_j$,...,${\complement }_n^n$ ideals for ideal R, for $d=1$, resulting in a total of $2^r+1,$ if we add ideal $\left(0\right)$. Here, ${\complement }_n^k=\left(\begin{array}{c}k\\ n\end{array}\right)$ are combinations.

(ii) Since $\widehat{\gamma }\in {\mathbb{Z}}_n-\{M_{p_1}\cup M_{p_2}\cup ...\cup M_{p_r}\}$, we have that $\widehat{\gamma }$ is invertible, with $\widehat{\delta }$ being its inverse. Therefore, $\left(X+\widehat{\gamma }\right)[-{\widehat{\delta }}^{-2}\left(X-\widehat{\gamma }\right)]=1$. As a result, $X+\widehat{\gamma }$ is invertible; therefore, $\left(X+\widehat{\gamma }\right)=R$. □

Since, for any commutative unitary ring, the lattice of ideals is a residuated lattice (see [6]), in particular, for the unitary and commutative ring $A={\mathbb{Z}}_n\left[X\right]/\left(X^2\right)$, we have that $(Id({\mathbb{Z}}_n/\left(X^2\right)),\cap ,+,\otimes \to ,0=\left\{0\right\},1=A)$ is a residuated lattice with $2^r+1$ elements.

Remark 4. 

As we remarked above, the ideals in the ring $R={\mathbb{Z}}_n\left[X\right]/\left(X^2\right)$ are:

(i) $\left(0\right);$

(ii) of the form $J_d=\left(\widehat{\alpha }X+{\widehat{\beta }}_d\right),\widehat{\alpha }\in R,{\widehat{\beta }}_d\in I_d,$ where d is a proper divisor of $n=p_1{p}_2...p_r,p_1,p_2,...,p_r$ being distinct prime numbers, $r\ge 2,$ by using the notations from Remark 3. If $I_d=\left(\widehat{p_i}\right),$ then $J_d$ is denoted $J_{p_i}$ and is a maximal ideal in $R={\mathbb{Z}}_n\left[X\right]/\left(X^2\right)$;

(iii) The ring $R,$ if $d=1$;

(iv) $\left(X\right),$ if $d=n$.

Remark 5. 

We remark that for all nonzero ideals I of the above ring R, we have $\left(X\right)\subseteq I$ and the ideal $\left(X\right)$ is the only minimal ideal of ${\mathbb{Z}}_n\left[X\right]/\left(X^2\right)$.

Remark 6. 

Let $D_d=\{p/$$p\in \{p_1,p_2,...,p_r\}$ such that $d=\prod p\},d\ne 1$.

(1) We have $J_{d_1}\cap J_{d_2}=J_{d_1}\otimes J_{d_2}=J_{d_3}$, where $D_{d_3}=\{p\in D_{d_1}\cup D_{d_2},d_3=\prod p\}$ for $d_1,d_2$ proper divisors.

If $d_1=1,$ we have $R\otimes J_{d_2}=J_{d_2}=R\cap J_{d_2}$.

If $d_1=n$, $d_2\ne n$, we have $\left(X\right)\otimes J_{d_2}=\left(X\right)\cap J_{d_2}=\left(X\right)$. If $d_2=n,$ we have $\left(X\right)\otimes \left(X\right)=\left(0\right)$.

(2) We have $(J_{d_1}:J_{d_2})=J_{d_3}$, with $D_{d_3}=D_{d_1}-D_{d_2}$. Indeed, $(J_{d_1}:J_{d_2})=\{y\in R,y·J_{d_2}\subseteq J_{d_1}\}=J_{d_3}$, for $d_1,d_2$ proper divisors.

If $J_{d_1}=\left(0\right),$ we have $(0:J_{d_2})=\left(0\right)$. Indeed, if $(0:J_{d_2})=J\ne \left(0\right)$, $J\otimes J_{d_2}=\left(0\right)$. However, from the above, $J\otimes J_{d_2}=$$J\cap J_{d_2}$$\ne \left(0\right)$, which is false

If $J_{d_2}=\left(0\right)$, we have $(J_{d_1}:0)=R$.

If $d_1=1$, we have $\left(R:J_{d_2}\right)=R$ and $\left(J_{d_2}:R\right)=J_{d_2}.$

If $d_1=n$, $d_2\ne n$, we have $J_{d_1}=\left(X\right)$; therefore, $(J_{d_1}:J_{d_2})=J_{d_1}=\left(X\right)$. If $d_1\ne n$, $d_2=n$, we have $J_{d_2}=\left(X\right)$; therefore, $(J_{d_1}:J_{d_2})=R$. If $d_1=d_2=n$, we have $J_{d_1}=J_{d_2}=\left(X\right)$ and $(J_{d_1}:J_{d_2})=R$.

Theorem 3. 

(i) For $n\ge 2$,with the above notations, the residuated lattice $(Id({\mathbb{Z}}_n\left[X\right]/\left(X^2\right)),\cap ,$$+,\otimes \to ,0=\left\{0\right\},1=R),R={\mathbb{Z}}_n\left[X\right]/\left(X^2\right)$ is a BL-algebra with $2^r+1$ elements.

(ii) By using notations from Remark 4, we have that $(I{d}_{p_i}({\mathbb{Z}}_n\left[X\right]/\left(X^2\right)),\cap ,+,\otimes \to ,0=\left\{0\right\},1=R),$ where $I{d}_{p_i}({\mathbb{Z}}_n\left[X\right]/\left(X^2\right))=\{\left(0\right),J_{p_i},R\}$ is a BL-sublattice of the lattice $Id({\mathbb{Z}}_n\left[X\right]/\left(X^2\right))$ with 3 elements.

Proof. 

(i) First, we will prove the $\left(prel\right)$ condition:

$$\left(I\to J\right)\vee \left(J\to I\right)=(J:I)\vee (I:J)={\mathbb{Z}}_n\left[X\right]/\left(X^2\right),$$

for every $I,J\in Id({\mathbb{Z}}_n\left[X\right]/\left(X^2\right))$.

Case 1. If $d_1$ and $d_2$ are proper divisors of n, we have $\left(J_{d_1}\to J_{d_2}\right)\vee \left(J_{d_2}\to J_{d_1}\right)=(J_{d_2}:J_{d_1})\vee (J_{d_1}:J_{d_2})=J_{d_4}\vee J_{d_5},$ where $D_{d_5}=D_{d_1}-D_{d_2}$ and $D_{d_4}=D_{d_2}-D_{d_1}$. We remark that $D_{d_4}\cap D_{d_5}=\varnothing ;$ then, gcd $\{d_4,d_5\}=1$. From here, there are the integers a and b such that $a{d}_4+b{d}_5=1$. We obtain that $J_{d_4}\vee J_{d_5}=<J_{d_4}\cup J_{d_5}>=R$ from Proposition 4, (ii).

Case 2. If $d_1$ is a proper divisor of n and $d_2=n$, we have $J_{d_2}=\left(X\right)$. Therefore, $\left(J_{d_1}\to J_{d_2}\right)\vee \left(J_{d_2}\to J_{d_1}\right)=(J_{d_2}:J_{d_1})\vee (J_{d_1}:J_{d_2})=J_{d_2}\vee R=R$ using Remark 6.

Case 3. If $d_1$ is a proper divisor of n and $J_{d_2}=\left(0\right)$, we have $\left(J_{d_1}\to J_{d_2}\right)\vee \left(J_{d_2}\to J_{d_1}\right)=(0:J_{d_1})\vee (J_{d_1}:0)=0\vee R=R$ using Remark 6.

Case 4. If $d_1$ is a proper divisor of n and $J_{d_2}=R$, it is clear. From here, the condition $\left(prel\right)$ is satisfied.

Now, we prove condition $\left(div\right):$

$$I\otimes \left(I\to J\right)=I\otimes \left(J:I\right)=I\cap J,$$

for every $I,J\in Id({\mathbb{Z}}_n\left[X\right]/\left(X^2\right))$.

Case 1. If $d_1$ and $d_2$ are proper divisors of n, we have $J_{d_1}\otimes \left(J_{d_2}:J_{d_1}\right)=J_{d_1}\otimes J_{d_3}=J_{d_4}=J_{d_1}\cap J_{d_2},$ since $D_{d_3}=D_{d_2}-D_{d_1}$ and $D_{d_4}=\{p\in D_{d_1}\cup D_{d_3},d_4=\prod p\}=\{p\in D_{d_1}\cup D_{d_2},d_4=\prod p\}$.

Case 2. If $d_1$ is a proper divisor of n and $d_2=n$, we have $J_{d_2}=\left(X\right)$. We obtain $J_{d_1}\otimes \left(J_{d_2}:J_{d_1}\right)=J_{d_1}\otimes \left(\left(X\right):J_{d_1}\right)=J_{d_1}\otimes \left(X\right)=J_{d_1}\cap \left(X\right)$ since $\left(X\right)\subset J_{d_1}$.

Case 3. If $d_1=n$ and $d_2$ is a proper divisor of n, we have $J_{d_1}=\left(X\right)$. We obtain $J_{d_1}\otimes \left(J_{d_2}:J_{d_1}\right)=\left(X\right)\otimes \left(J_{d_2}:\left(X\right)\right)=\left(X\right)\otimes R=\left(X\right)=J_{d_2}\cap \left(X\right)$ since $\left(X\right)\subset J_{d_2}$.

Case 4. If $d_1$ is a proper divisor of n and $J_{d_2}=\left(0\right)$, we have $J_{d_1}\otimes \left(J_{d_2}:J_{d_1}\right)=J_{d_1}\otimes \left(0:J_{d_1}\right)=J_{d_1}\otimes \left(0\right)=\left(0\right)=J_{d_1}\cap \left(0\right)$ from Remark 6.

Case 5. If $J_{d_1}=\left(0\right)$ and $d_2$ is a proper divisor of n, we have $J_{d_1}\otimes \left(J_{d_2}:J_{d_1}\right)=0\otimes \left(J_{d_2}:0\right)=0.$

Case 6. If $d_1$ is a proper divisor of n and $J_{d_2}=R$, we have $J_{d_1}\otimes \left(J_{d_2}:J_{d_1}\right)=J_{d_1}\otimes \left(R:J_{d_1}\right)=J_{d_1}\otimes R=J_{d_1}$. If $J_{d_1}=R$ and $d_2$ is a proper divisor of n, we have $J_{d_1}\otimes \left(J_{d_2}:J_{d_1}\right)=R\otimes \left(J_{d_2}:R\right)=R\otimes J_{d_2}=J_{d_2}$. From here, the condition $\left(div\right)$ is satisfied and the proposition is proven.

(ii) It is clear that $J_{p_i}\odot J_{p_i}=J_{p_i}\otimes J_{p_i}=J_{p_i}$; we obtain the following tables:

$$\begin{array}{cccc}\hfill \to \hfill & O& \hfill J_{p_i}\hfill & R\\ \hfill O\hfill & R& \hfill R\hfill & R\\ \hfill J_{p_i}\hfill & O& \hfill R\hfill & R\\ \hfill R\hfill & O& \hfill J_{p_i}\hfill & R\end{array}\begin{array}{cccc}\hfill \odot \hfill & O& \hfill J_{p_i}\hfill & R\\ \hfill O\hfill & O& \hfill O\hfill & O\\ \hfill J_{p_i}\hfill & O& \hfill J_{p_i}\hfill & J_{p_i}\\ \hfill R\hfill & O& \hfill J_{p_i}\hfill & R\end{array},$$

therefore showing a BL-algebra of order $3.$ □

Theorem 4. 

Let $n=p^r$ with p a prime number, $p\ge 2,$ r a positive integer, $r\ge 2$. We consider the ring $R={\mathbb{Z}}_n\left[X\right]/\left(X^2\right)$. The set $(Id({\mathbb{Z}}_n\left[X\right]/\left(X^2\right)),\cap ,+,\otimes \to ,0=\left\{0\right\},1=R)$ is a BL-chain with $r+2$ elements. In this way, for a given positive integer $k\ge 4$, we can construct BL-chains with k elements.

Proof. 

The ideals in ${\mathbb{Z}}_n$ are of the form: $\left(0\right)\subseteq \left(p^{r-1}\right)\subseteq \left(p^{r-2}\right)\subseteq ...\subseteq \left(p\right)\subseteq {\mathbb{Z}}_n$. The ideal $\left(p^{r-1}\right)$ and the ideal $\left(p\right)$ are the only maximal ideals of ${\mathbb{Z}}_n$. The ideals in the ring R are $\left(0\right)\subseteq \left(X\right)\subseteq \left({\alpha }_{r-1}X+{\beta }_{r-1}\right)\subseteq \left({\alpha }_{r-2}X+{\beta }_{r-2}\right)\subseteq ...\subseteq \left({\alpha }_{1}X+{\beta }_1\right)\subseteq R,$ where ${\alpha }_i\in {\mathbb{Z}}_n,i\in \{1,...,r-1\},$${\beta }_{r-1}\in \left(p^{r-1}\right),{\beta }_{r-2}\in \left(p^{r-2}\right),...,{\beta }_1\in \left(p\right)$, meaning $r+2$ ideals. We denote these ideals with $\left(0\right),\left(X\right),$$I_{p^{r-1}},I_{p^{r-2}},...I_p,R$, with $I_p$ being the only maximal ideal in R.

First, we prove the $\left(prel\right)$ condition:

$$\left(I\to J\right)\vee \left(J\to I\right)=(J:I)\vee (I:J)={\mathbb{Z}}_n\left[X\right]/\left(X^2\right),$$

for every $I,J\in Id({\mathbb{Z}}_n\left[X\right]/\left(X^2\right))$.

Case 1. We suppose that I and J are proper ideals and $I\subseteq J$. We have $\left(I\to J\right)\vee \left(J\to I\right)=(J:I)\vee (I:J)=R\vee (I:J)=R$.

Case 2. $I=\left(0\right)$ and J are a proper ideal, we have $\left(I\to J\right)\vee \left(J\to I\right)=(J:\left(0\right))\vee (\left(0\right):J)=R$. Therefore, the condition $\left(prel\right)$ is satisfied.

Now, we prove the $\left(div\right)$ condition:

$$I\otimes \left(I\to J\right)=I\otimes \left(J:I\right)=I\cap J,$$

for every $I,J\in Id({\mathbb{Z}}_n\left[X\right]/\left(X^2\right))$.

Case 1. We suppose that I and J are proper ideals and $I\subseteq J$. We have $I\otimes \left(I\to J\right)=I\otimes \left(J:I\right)=I\otimes R=I=I\cap J$. If $J\subseteq I$, we have $I\otimes \left(I\to J\right)=I\otimes \left(J:I\right)=J=I\cap J$.

Case 2. $I=\left(0\right)$ and J is a proper ideal. We have $\left(0\right)\otimes \left(\left(0\right)\to J\right)=\left(0\right)\otimes \left(J:\left(0\right)\right)=\left(0\right)=I\cap J$. If $I\ne \left(X\right)$ is a proper ideal and $J=\left(0\right)$, we have $I\otimes \left(I\to J\right)=I\otimes \left(\left(0\right):I\right)=I\otimes \left(0\right)=\left(0\right)$. If $I=\left(X\right)$ and $J=\left(0\right)$, we have $I\otimes \left(I\to J\right)=\left(X\right)\otimes \left(\left(0\right):\left(X\right)\right)=\left(X\right)\otimes \left(X\right)=\left(0\right)$ and $\left(0\right)\cap \left(X\right)=\left(0\right)$.

From here, the condition $\left(div\right)$ is satisfied and the theorem is proven. □

Example 6. 

In Theorem 3, we take $n=2·3$; therefore, the ideals of ${\mathbb{Z}}_6$ are $\left(0\right),\left(2\right),\left(3\right),{\mathbb{Z}}_6$, with $\left(2\right)$ and $\left(3\right)$ maximal ideals. The ring ${\mathbb{Z}}_6\left[X\right]/\left(X^2\right)$ has five ideals: $O=\left(0\right)\subset A=\left(X\right),B=\left(\alpha X+\beta \right),C=\left(\gamma X+\delta \right),E={\mathbb{Z}}_6\left[X\right]/\left(X^2\right)$, with $\alpha ,\gamma \in$${\mathbb{Z}}_4,\beta \in \left(2\right)$ and $\delta \in \left(3\right)$. From the following tables, we have a BL-structure on $Id({\mathbb{Z}}_6\left[X\right]/\left(X^2\right))$:

$$\begin{array}{cccccc}\hfill \to \hfill & O& \hfill A\hfill & B& C& E\\ \hfill O\hfill & E& \hfill E\hfill & E& E& E\\ \hfill A\hfill & A& \hfill E\hfill & E& E& E\\ \hfill B\hfill & O& \hfill C\hfill & E& C& E\\ \hfill C\hfill & O& \hfill B\hfill & B& E& E\\ \hfill E\hfill & O& \hfill A\hfill & B& C& E\end{array}\begin{array}{cccccc}\hfill \odot \hfill & O& \hfill A\hfill & B& C& E\\ \hfill O\hfill & O& \hfill O\hfill & O& O& O\\ \hfill A\hfill & O& \hfill O\hfill & A& A& A\\ \hfill B\hfill & O& \hfill A\hfill & B& A& B\\ \hfill C\hfill & O& \hfill A\hfill & A& C& C\\ \hfill E\hfill & O& \hfill A\hfill & B& C& E\end{array}.$$

From Theorem 3, if we consider $J_{p_i} = B$, we have the following BL-algebra of order 3:

$$\begin{array}{cccc}\hfill \to \hfill & O& \hfill B\hfill & E\\ \hfill O\hfill & E& \hfill E\hfill & E\\ \hfill B\hfill & O& \hfill E\hfill & E\\ \hfill E\hfill & O& \hfill B\hfill & E\end{array}\begin{array}{cccc}\hfill \odot \hfill & O& \hfill B\hfill & E\\ \hfill O\hfill & O& \hfill O\hfill & O\\ \hfill B\hfill & O& \hfill B\hfill & B\\ \hfill E\hfill & O& \hfill B\hfill & E\end{array}.$$

Example 7. 

In Theorem 3, we take $n=2·3·5$; therefore, the ideals of the ring ${\mathbb{Z}}_{30}$ are $\left(0\right),\left(2\right),\left(3\right),$$\left(5\right),\left(6\right),\left(10\right),\left(15\right),{\mathbb{Z}}_{30}$, with $\left(2\right),\left(3\right)$ and $\left(5\right)$ being maximal ideals. The ring ${\mathbb{Z}}_{30}\left[X\right]/\left(X^2\right)$ has nine ideals: $O=\left(0\right)\subset A=\left(X\right),B=\left({\alpha }_{1}X+{\beta }_1\right),C=\left({\alpha }_{2}X+{\beta }_2\right),$ $D=\left({\alpha }_{3}X+{\beta }_3\right),E=\left({\alpha }_{4}X+{\beta }_4\right),F=\left({\alpha }_{5}X+{\beta }_5\right),G=\left({\alpha }_{6}X+{\beta }_6\right),R={\mathbb{Z}}_{30}\left[X\right]/\left(X^2\right)$, with ${\alpha }_i\in$${\mathbb{Z}}_{30},i\in \{1,2,3,4,5,6\},$${\beta }_1\in \left(6\right)$, ${\beta }_2\in \left(10\right),{\beta }_3\in \left(15\right),{\beta }_4\in \left(2\right),{\beta }_5\in \left(3\right)$ and ${\beta }_6\in \left(5\right)$. The ideals $E,F$ and G are maximal. From the following tables, we have a BL-structure on $Id({\mathbb{Z}}_{30}\left[X\right]/\left(X^2\right))$:

$$\begin{array}{cccccccccc}\hfill \to \hfill & O& \hfill A\hfill & B& C& D& E& F& G& R\\ \hfill O\hfill & R& \hfill R\hfill & R& R& R& R& R& R& R\\ \hfill A\hfill & A& \hfill R\hfill & R& R& R& R& R& R& R\\ \hfill B\hfill & O& \hfill G\hfill & R& G& G& R& R& G& R\\ \hfill C\hfill & O& \hfill F\hfill & F& R& F& R& F& R& R\\ \hfill D\hfill & O& \hfill E\hfill & E& E& R& E& R& R& R\\ \hfill E\hfill & O& \hfill D\hfill & F& G& D& R& F& G& R\\ \hfill F\hfill & O& \hfill C\hfill & E& C& G& E& R& G& R\\ \hfill G\hfill & O& \hfill B\hfill & B& E& F& E& F& R& R\\ \hfill R\hfill & O& \hfill A\hfill & B& C& D& E& F& G& R\end{array}\begin{array}{cccccccccc}\hfill \odot \hfill & O& \hfill A\hfill & B& C& D& E& F& G& R\\ \hfill O\hfill & O& \hfill O\hfill & O& O& O& O& O& O& O\\ \hfill A\hfill & O& \hfill O\hfill & A& A& A& A& A& A& A\\ \hfill B\hfill & O& \hfill A\hfill & B& A& A& B& B& A& B\\ \hfill C\hfill & O& \hfill A\hfill & A& C& A& C& C& A& C\\ \hfill D\hfill & O& \hfill A\hfill & A& A& D& A& D& D& D\\ \hfill E\hfill & O& \hfill A\hfill & B& C& A& E& A& A& E\\ \hfill F\hfill & O& \hfill A\hfill & B& A& D& A& F& A& F\\ \hfill G\hfill & O& \hfill A\hfill & A& C& D& A& A& G& G\\ \hfill R\hfill & O& \hfill A\hfill & B& C& D& E& F& G& R\end{array}.$$

Example 8. 

In Theorem 4, we consider $p=2,r=2$. The ideals in $\left({\mathbb{Z}}_4,+,·\right)$ are $\left(0\right)\subset \left(2\right)\subset {\mathbb{Z}}_4$ and ${\mathbb{Z}}_4$ is a local ring. The ring ${\mathbb{Z}}_4\left[X\right]/\left(X^2\right)$ has four ideals: $O=\left(0\right)\subset A=\left(X\right)\subset B=\left(\alpha X+\beta \right)\subset E={\mathbb{Z}}_4\left[X\right]/\left(X^2\right)$, with $\alpha \in$${\mathbb{Z}}_4,\beta \in \left(2\right)$. From the following tables, we have a BL-structure for $Id({\mathbb{Z}}_4\left[X\right]/\left(X^2\right))$:

$$\begin{array}{ccccc}\hfill \to \hfill & O& \hfill A\hfill & B& E\\ \hfill O\hfill & E& \hfill E\hfill & E& E\\ \hfill A\hfill & A& \hfill E\hfill & E& E\\ \hfill B\hfill & O& \hfill B\hfill & E& E\\ \hfill E\hfill & O& \hfill A\hfill & B& E\end{array}\begin{array}{ccccc}\hfill \odot \hfill & O& \hfill A\hfill & B& E\\ \hfill O\hfill & O& \hfill O\hfill & O& O\\ \hfill A\hfill & O& \hfill O\hfill & A& A\\ \hfill B\hfill & O& \hfill A\hfill & A& B\\ \hfill E\hfill & O& \hfill A\hfill & B& E\end{array}.$$

Example 9. 

In Theorem 4, we consider $p=2,r=3$. The ideals in $\left({\mathbb{Z}}_8,+,·\right)$ are $\left(0\right)\subset \left(4\right)\subset \left(2\right)\subset {\mathbb{Z}}_8$. The ring ${\mathbb{Z}}_8\left[X\right]/\left(X^2\right)$ has five ideals: $O=\left(0\right)\subset A=\left(X\right)\subset B=\left(\alpha X+\beta \right)\subset C=\left(\gamma X+\delta \right)\subset E={\mathbb{Z}}_8\left[X\right]/\left(X^2\right)$, with $\alpha ,\gamma \in$${\mathbb{Z}}_8,\beta \in \left(4\right)$ and $\delta \in \left(2\right)$. From the following tables, we have a BL-structure for $Id({\mathbb{Z}}_8\left[X\right]/\left(X^2\right))$:

$$\begin{array}{cccccc}\hfill \to \hfill & O& \hfill A\hfill & B& C& E\\ \hfill O\hfill & E& \hfill E\hfill & E& E& E\\ \hfill A\hfill & A& \hfill E\hfill & E& E& E\\ \hfill B\hfill & O& \hfill B\hfill & E& E& E\\ \hfill C\hfill & O& \hfill B\hfill & B& E& E\\ \hfill E\hfill & O& \hfill A\hfill & B& C& E\end{array}\begin{array}{cccccc}\hfill \odot \hfill & O& \hfill A\hfill & B& C& E\\ \hfill O\hfill & O& \hfill O\hfill & O& O& O\\ \hfill A\hfill & O& \hfill O\hfill & A& A& A\\ \hfill B\hfill & O& \hfill A\hfill & A& A& B\\ \hfill C\hfill & O& \hfill A\hfill & A& B& C\\ \hfill E\hfill & O& \hfill A\hfill & B& C& E\end{array}.$$

In the following, we present a way to generate finite BL-algebras using the ordinal product of residuated lattices.

We recall that, in [10], Iorgulescu studied the influence of the conditions $\left(prel\right)$ and $\left(div\right)$ on the ordinal product of two residuated lattices.

It is known that if ${\mathcal{L}}_1=(L_1,{\wedge }_1,{\vee }_1,{\odot }_1,{\to }_1,0_1,1_1)$ and ${\mathcal{L}}_2=(L_2,{\wedge }_2,{\vee }_2,{\odot }_2,{\to }_2,$$0_2,1_2)$ are two residuated lattices such that $1_1=0_2$ and $(L_1\setminus \left\{1_1\right\})\cap (L_2\setminus \left\{0_2\right\})=\oslash ,$ then the ordinal product of ${\mathcal{L}}_1$ and ${\mathcal{L}}_2$ is the residuated lattice ${\mathcal{L}}_1⊠{\mathcal{L}}_2=(L_1\cup L_2,\wedge ,\vee ,\odot ,\to ,$$0,1)$ where

$$0=0_1\mathrm{and}1=1_2,$$

$$x\le y\mathrm{if}(x,y\in L_1\mathrm{and}x{\le }_{1}y)\mathrm{or}(x,y\in L_2\mathrm{and}x{\le }_{2}y)\mathrm{or}(x\in L_1\mathrm{and}y\in L_2),$$

$$x\to y=\left\{\begin{array}{c}1,\mathrm{if}x\le y,\\ x{\to }_{i}y,\mathrm{if}x\nleqq y,x,y\in L_i,i=1,2,\\ y,\mathrm{if}x\nleqq y,x\in L_2,y\in L_1\setminus \left\{1_1\right\}.\end{array}\right.$$

$$x\odot y=\left\{\begin{array}{c}x{\odot }_{1}y,\mathrm{if}x,y\in L_1,\\ x{\odot }_{2}y,\mathrm{if}x,y\in L_2,\\ x,\mathrm{if}x\in L_1\setminus \left\{1_1\right\}\mathrm{and}y\in L_2.\end{array}\right.$$

The ordinal product is associative, but is not commutative; see [10].

Proposition 5 

([10] (Corollary 3.5.10)). Let ${\mathcal{L}}_1$ and ${\mathcal{L}}_2$ be BL-algebras.

(i) If ${\mathcal{L}}_1$ is a chain, then the ordinal product ${\mathcal{L}}_1⊠{\mathcal{L}}_2$ is a BL-algebra;

(ii) $\mathit{If}$ ${\mathcal{L}}_1$ is not a chain, then the ordinal product ${\mathcal{L}}_1⊠{\mathcal{L}}_2$ is only a residuated lattice satisfying (div) condition.

Remark 7. 

(i) An ordinal product of two BL-chains is a BL-chain. Indeed, using the definition of implication in an ordinal product for every $x,y$ we have $x\to y=1$ or $y\to x=1;$

(ii) An ordinal product of two BL-algebras is a BL-algebra that is not an MV-algebra. Indeed, if ${\mathcal{L}}_1$ and ${\mathcal{L}}_2$ are two BL-algebras (the first being a chain), using Proposition 5, the residuated lattice ${\mathcal{L}}_1⊠{\mathcal{L}}_2$ is a BL-algebra in which we have ${\left(1_1\right)}^{**}={(1_1\to 0_1)}^{*}={\left(0_1\right)}^{*}=0_1\to 0_1=1=1_2\ne 1_1.$ Thus, ${\mathcal{L}}_1⊠{\mathcal{L}}_2$ is not an MV-algebra.

For a natural number $n\ge 2$, we consider the decomposition (which is not unique) of n in factors greater than $1.$ We only count the decompositions one time with the same terms, but with other orders of terms in the product. We denote by $\pi \left(n\right)$ the number of all such decompositions. Obviously, if n is prime, then $\pi \left(n\right)=0.$

We recall that an MV-algebra is finite iff it is isomorphic to a finite product of MV-chains; see [17]. Furthermore, for two MV-algebras $L_1$ and $L_2,$ the algebras $L_1\times L_2$ and $L_2\times L_1$ are isomorphic; see [18]. Using these results, in [15], we showed that for every natural number $n\ge 2$, there are $\pi \left(n\right)+1$ non-isomorphic MV-algebras with n elements of which only one is a chain.

Example 10. 

For $n=6,$ we have $6=2·3=3·2$; thus, $\pi \left(6\right)=1.$ Therefore, there are $\pi \left(6\right)+1=2$ types (up to an isomorphism) of MV-algebras with six elements.

In

Table 1. Rings that Generate MV-algebras of order n, 2 ≤ n ≤ 8.

$\left|\mathit{M}\right|=\mathit{n}$	No. of MVs	Rings that Generate MV
$n=2$	1	$Id\left({\mathbb{Z}}_2\right)$ (chain)
$n=3$	1	$Id\left({\mathbb{Z}}_4\right)$ (chain)
$n=4$	2	$Id\left({\mathbb{Z}}_8\right)$ (chain) and $Id({\mathbb{Z}}_2\times {\mathbb{Z}}_2)$
$n=5$	1	$Id\left({\mathbb{Z}}_{16}\right)$ (chain)
$n=6$	2	$Id\left({\mathbb{Z}}_{32}\right)$ (chain) and $Id({\mathbb{Z}}_2\times {\mathbb{Z}}_4)$
$n=7$	1	$Id\left({\mathbb{Z}}_{64}\right)$ (chain)
$n=8$	3	$Id\left({\mathbb{Z}}_{128}\right)$ (chain) and $Id({\mathbb{Z}}_2\times {\mathbb{Z}}_8)$ and $Id({\mathbb{Z}}_2\times {\mathbb{Z}}_2\times {\mathbb{Z}}_2)$

, we briefly describe a way of generating finite MV-algebras M with $2\le n\le 8$ elements using commutative rings; see [15].

Using the construction of the ordinal product, Proposition 5 and Remark 7, we can generate BL-algebras (which are not MV-algebras) using commutative rings.

Example 11. 

In [15] we show that there is one MV-algebra with three elements (up to an isomorphism); see Table 1. This MV-algebra is isomorphic to $Id\left({\mathbb{Z}}_4\right)$ and is a chain. To generate a BL-chain with three elements (which is not an MV-algebra) using the ordinal product, we must consider only the MV-algebra with two elements (which is, in fact, a Boolean algebra). In the commutative ring $({\mathbb{Z}}_2,+,·)$, the ideals are $Id\left({\mathbb{Z}}_2\right)=\{\left\{\widehat{0}\right\},$${\mathbb{Z}}_2\}$. Obviously, $(Id\left({\mathbb{Z}}_2\right),\cap ,+,\otimes \to ,0=\left\{0\right\},1={\mathbb{Z}}_2)$ is an MV-chain. Now we consider two MV-algebras isomorphic with $Id\left({\mathbb{Z}}_2\right)$ denoted ${\mathcal{L}}_1=(L_1=\{0,a\},\wedge ,\vee ,\odot ,\to ,0,a)$ and ${\mathcal{L}}_2=(L_2=\{a,1\},\wedge ,\vee ,\odot ,\to ,a,1).$ Using Proposition 5, we can construct the BL-algebra ${\mathcal{L}}_1⊠{\mathcal{L}}_2=(L_1\cup L_2=\{0,a,1\},\wedge ,\vee ,\odot ,\to ,0,1)$ with $0\le a\le 1$ and the following operations:

$$\begin{array}{cccc}\hfill \to \hfill & 0& \hfill a\hfill & 1\\ \hfill 0\hfill & 1& \hfill 1\hfill & 1\\ \hfill a\hfill & 0& \hfill 1\hfill & 1\\ \hfill 1\hfill & 0& \hfill a\hfill & 1\end{array}\mathit{and}\begin{array}{cccc}\hfill \odot \hfill & 0& \hfill a\hfill & 1\\ \hfill 0\hfill & 0& \hfill 0\hfill & 0\\ \hfill a\hfill & 0& \hfill a\hfill & a\\ \hfill 1\hfill & 0& \hfill a\hfill & 1\end{array},$$

obtaining the same BL-algebra of order 3 as in Example 6.

Obviously, ${\mathcal{L}}_1⊠{\mathcal{L}}_2$ is a BL-chain that is not an MV-chain, since, for example, $a^{**}=1\ne a.$

Example 12. 

To generate the non-linearly ordered BL-algebra with five elements from Example 3, we consider the commutative rings $({\mathbb{Z}}_2,+,·)$ and $({\mathbb{Z}}_2\times {\mathbb{Z}}_2,+,·).$ For ${\mathbb{Z}}_2\times {\mathbb{Z}}_2=\{\left(\widehat{0},\widehat{0}\right),\left(\widehat{0},\widehat{1}\right),$$\left(\widehat{1},\widehat{0}\right),\left(\widehat{1},\widehat{1}\right)\}$, we obtain the lattice $Id\left({\mathbb{Z}}_2\times {\mathbb{Z}}_2\right)=\{\left(\widehat{0},\widehat{0}\right),\{\left(\widehat{0},\widehat{0}\right),\left(\widehat{0},\widehat{1}\right)\},\{\left(\widehat{0},\widehat{0}\right),\left(\widehat{1},\widehat{0}\right)\},$${\mathbb{Z}}_2\times {\mathbb{Z}}_2\}=\{O,R,B,E\}$, which is an MV-algebra $(Id\left({\mathbb{Z}}_2\times {\mathbb{Z}}_2\right),\cap ,+,\otimes \to ,0=\left\{\left(\widehat{0},\widehat{0}\right)\right\},1={\mathbb{Z}}_2\times {\mathbb{Z}}_2)$. In $Id\left({\mathbb{Z}}_2\times {\mathbb{Z}}_2\right)$, we have the following operations:

$$\begin{array}{ccccc}\hfill \to \hfill & O& \hfill R\hfill & B& E\\ \hfill O\hfill & E& \hfill E\hfill & E& E\\ \hfill R\hfill & B& \hfill E\hfill & B& E\\ \hfill B\hfill & R& \hfill R\hfill & E& E\\ \hfill E\hfill & O& \hfill R\hfill & B& E\end{array},\begin{array}{ccccc}\otimes =\cap \hfill & O& \hfill R\hfill & B& E\\ O\hfill & O& \hfill O\hfill & O& O\\ R\hfill & O& \hfill R\hfill & O& R\\ B\hfill & O& \hfill O\hfill & B& B\\ E\hfill & O& \hfill R\hfill & B& E\end{array}\mathit{and}\begin{array}{ccccc}\hfill +\hfill & O& \hfill R\hfill & B& E\\ \hfill O\hfill & O& \hfill R\hfill & B& E\\ \hfill R\hfill & R& \hfill R\hfill & E& E\\ \hfill B\hfill & B& \hfill E\hfill & B& E\\ \hfill E\hfill & E& \hfill E\hfill & E& E\end{array}.$$

If we consider two MV-algebras isomorphic with $(Id\left({\mathbb{Z}}_2\right),\cap ,+,\otimes \to ,0=\left\{0\right\},1={\mathbb{Z}}_2)$ and $(Id\left({\mathbb{Z}}_2\times {\mathbb{Z}}_2\right),\cap ,+,\otimes \to ,0=\left\{\left(\widehat{0},\widehat{0}\right)\right\},1={\mathbb{Z}}_2\times {\mathbb{Z}}_2),$ denoted by ${\mathcal{L}}_1=(L_1=\{0,c\},{\wedge }_1,{\vee }_1,$${\odot }_1,{\to }_1,0,c)$ and ${\mathcal{L}}_2=(L_2=\{c,a,b,1\},{\wedge }_2,{\vee }_2,{\odot }_2,{\to }_2,c,1),$ then, using Proposition 5, we generate the BL-algebra ${\mathcal{L}}_1⊠{\mathcal{L}}_2=(L_1\cup L_2=\{0,c,a,b,1\},\wedge ,\vee ,\odot ,\to ,0,1)$ from Example 3.

Remark 8. 

Using the model from Examples 11 and 12 for two BL-algebras ${\mathcal{L}}_1$ and ${\mathcal{L}}_2$ we can use these algebras to obtain two BL-algebras ${\mathcal{L}}_1^{\prime }$ and ${\mathcal{L}}_2^{\prime }$, isomorphic with ${\mathcal{L}}_1$ and ${\mathcal{L}}_2$, respectively, that satisfy the conditions imposed by the ordinal product.

We denote by ${\mathcal{L}}_1⊡{\mathcal{L}}_2$ the ordinal product ${\mathcal{L}}_1^{\prime }⊠{\mathcal{L}}_2^{\prime }.$

From Proposition 5 and Remark 7, we deduce the following.

Theorem 5. 

(i) To generate a BL-algebra with $n\ge 3$ elements as an ordinal product ${\mathcal{L}}_1⊠{\mathcal{L}}_2$ of two BL-algebras ${\mathcal{L}}_1$ and ${\mathcal{L}}_2$ we have the following possibilities:

$${\mathcal{L}}_1\mathit{is}a\mathit{BL}\text{-}\mathit{chain}\mathit{with}i\mathit{elements}\mathit{and}{\mathcal{L}}_2\mathit{is}a\mathit{BL}\text{-}\mathit{algebra}\mathit{with}j\mathit{elements}$$

and

$${\mathcal{L}}_1\mathit{is}a\mathit{BL}\text{-}\mathit{chain}\mathit{with}j\mathit{elements}\mathit{and}{\mathcal{L}}_2\mathit{is}a\mathit{BL}\text{-}\mathit{algebra}\mathit{with}i\mathit{elements}$$

or

$${\mathcal{L}}_1\mathit{is}a\mathit{BL}\text{-}\mathit{chain}\mathit{with}k\mathit{elements}\mathit{and}{\mathcal{L}}_2\mathit{is}a\mathit{BL}\text{-}\mathit{algebra}\mathit{with}k\mathit{elements}$$

for $i,j\ge 2,i+j=n+1,$$i<j$ and $k\ge 2,$$k=\frac{n+1}{2}$$\in N,$

(ii) To generate a BL-chain with $n\ge 3$ elements as the ordinal product ${\mathcal{L}}_1⊠{\mathcal{L}}_2$ of two BL-algebras ${\mathcal{L}}_1$ and ${\mathcal{L}}_2$, we have the following possibilities:

$${\mathcal{L}}_1\mathit{is}a\mathit{BL}\text{-}\mathit{chain}\mathit{with}i\mathit{elements}\mathit{and}{\mathcal{L}}_2\mathit{is}a\mathit{BL}\text{-}\mathit{chain}\mathit{with}j\mathit{elements}$$

and

$${\mathcal{L}}_1\mathit{is}a\mathit{BL}\text{-}\mathit{chain}\mathit{with}j\mathit{elements}\mathit{and}{\mathcal{L}}_2\mathit{is}a\mathit{BL}\text{-}\mathit{chain}\mathit{with}i\mathit{elements}$$

or

$${\mathcal{L}}_1\mathit{is}a\mathit{BL}\text{-}\mathit{chain}\mathit{with}k\mathit{elements}\mathit{and}{\mathcal{L}}_2\mathit{is}a\mathit{BL}\text{-}\mathit{chain}\mathit{with}k\mathit{elements}$$

for $i,j\ge 2,i+j=n+1,$$i<j$ and $k\ge 2,$$k=\frac{n+1}{2}$$\in N.$

We make the following notations:

$${\mathcal{BL}}_n=\mathrm{the}\mathrm{set}\mathrm{of}\mathrm{BL}\text{-}\mathrm{algebras}\mathrm{with}n\mathrm{elements};$$

$${\mathcal{BL}}_n\left(c\right)=\mathrm{the}\mathrm{set}\mathrm{of}\mathrm{BL}\text{-}\mathrm{chains}\mathrm{with}n\mathrm{elements};$$

$${\mathcal{MV}}_n=\mathrm{the}\mathrm{set}\mathrm{of}\mathrm{MV}\text{-}\mathrm{algebras}\mathrm{with}n\mathrm{elements};$$

$${\mathcal{MV}}_n\left(c\right)=\mathrm{the}\mathrm{set}\mathrm{of}\mathrm{MV}\text{-}\mathrm{chains}\mathrm{with}n\mathrm{elements}.$$

Theorem 6. 

(i) Finite BL-algebras (up to an isomorphism) that are not MV-algebras with $3\le n\le 5$ elements can be generated using the ordinal product of BL-algebras.

(ii) The number of non-isomorphic BL-algebras with n elements (with $2\le n\le 5$) is

$$\left|{\mathcal{BL}}_2\right|=\left|{\mathcal{MV}}_2\right|=\pi \left(2\right)+1,$$

$$\left|{\mathcal{BL}}_3\right|=\left|{\mathcal{MV}}_3\right|+\left|{\mathcal{BL}}_2\right|=\pi \left(3\right)+\pi \left(2\right)+2,$$

$$\left|{\mathcal{BL}}_4\right|=\left|{\mathcal{MV}}_4\right|+\left|{\mathcal{BL}}_3\right|+\left|{\mathcal{BL}}_2\right|=\pi \left(4\right)+\pi \left(3\right)+3·\pi \left(2\right)+4,$$

$$\left|{\mathcal{BL}}_5\right|=\left|{\mathcal{MV}}_5\right|+\left|{\mathcal{BL}}_4\right|+\left|{\mathcal{BL}}_3\right|+\left|{\mathcal{BL}}_2\right|=$$

$$=\pi \left(5\right)+\pi \left(4\right)+2·\pi \left(3\right)+5·\pi \left(2\right)+8.$$

Proof. 

From Proposition 5 and Remark 7, we remark that using the ordinal product of two BL-algebras, we can generate only BL-algebras that are not MV-algebras.

We generate all BL-algebras with n elements ($2\le n\le 5)$ that are not MV-algebras.

Case$n=2.$

We obviously only have a BL-algebra (up to an isomorphism) isomorphic with

$$(Id\left({\mathbb{Z}}_2\right),\cap ,+,\otimes \to ,0=\left\{0\right\},1={\mathbb{Z}}_2).$$

In fact, this residuated lattice is a BL-chain and is the only MV-algebra with 2 elements. We deduce that

$$\left|{\mathcal{MV}}_2\right|=\left|{\mathcal{BL}}_2\right|=\pi \left(2\right)+1=1$$

$$\left|{\mathcal{MV}}_2\left(c\right)\right|=\left|{\mathcal{BL}}_2\left(c\right)\right|=1.$$

Case$n=3.$

Using Theorem 5, to generate a BL-algebra with 3 elements as an ordinal product ${\mathcal{L}}_1⊠{\mathcal{L}}_2$ of two BL-algebras ${\mathcal{L}}_1$ and ${\mathcal{L}}_2$, we must consider:

$${\mathcal{L}}_1\mathrm{is}\mathrm{a}\mathrm{BL}\text{-}\mathrm{chain}\mathrm{with}\mathrm{two}\mathrm{elements}\mathrm{and}{\mathcal{L}}_2\mathrm{is}\mathrm{a}\mathrm{BL}\text{-}\mathrm{algebra}\mathrm{with}\mathrm{two}\mathrm{elements}.$$

Since there is only one BL-algebra (up to an isomorphism) with two elements and it is a chain, we obtain the BL-algebra

$$Id\left({\mathbb{Z}}_2\right)⊡Id\left({\mathbb{Z}}_2\right),$$

which is a chain.

We deduce that

$$\left|{\mathcal{MV}}_3\right|=\pi \left(3\right)+1\mathrm{and}\left|{\mathcal{BL}}_3\right|=\left|{\mathcal{MV}}_3\right|+1·\left|{\mathcal{BL}}_2\right|=\pi \left(3\right)+\pi \left(2\right)+2=2$$

$$\left|{\mathcal{MV}}_3\left(c\right)\right|=1\mathrm{and}\left|{\mathcal{BL}}_3\left(c\right)\right|=\left|{\mathcal{MV}}_3\left(c\right)\right|+1=1+1=2.$$

We remark that $\left|{\mathcal{BL}}_3\right|=\left|{\mathcal{MV}}_3\right|+\left|{\mathcal{BL}}_2\right|.$

Case$n=4.$

Using Theorem 5, to generate a BL-algebra with four elements as the ordinal product ${\mathcal{L}}_1⊠{\mathcal{L}}_2$ of two BL-algebras ${\mathcal{L}}_1$ and ${\mathcal{L}}_2$, we must consider:

$${\mathcal{L}}_1\mathrm{is}\mathrm{a}\mathrm{BL}\text{-}\mathrm{chain}\mathrm{with}\mathrm{two}\mathrm{elements}\mathrm{and}{\mathcal{L}}_2\mathrm{is}\mathrm{a}\mathrm{BL}\text{-}\mathrm{algebra}\mathrm{with}\mathrm{three}\mathrm{elements};$$

$${\mathcal{L}}_1\mathrm{is}\mathrm{a}\mathrm{BL}\text{-}\mathrm{chain}\mathrm{with}\mathrm{three}\mathrm{elements}\mathrm{and}{\mathcal{L}}_2\mathrm{is}\mathrm{a}\mathrm{BL}\text{-}\mathrm{algebra}\mathrm{with}\mathrm{two}\mathrm{elements}.$$

We obtain the following BL-algebras:

$$Id\left({\mathbb{Z}}_2\right)⊡Id\left({\mathbb{Z}}_4\right)\mathrm{and}Id\left({\mathbb{Z}}_2\right)⊡(Id\left({\mathbb{Z}}_2\right)⊡Id\left({\mathbb{Z}}_2\right))$$

and

$$Id\left({\mathbb{Z}}_4\right)⊡Id\left({\mathbb{Z}}_2\right)\mathrm{and}(Id\left({\mathbb{Z}}_2\right)⊡\left(Id\left({\mathbb{Z}}_2\right)\right)⊡Id\left({\mathbb{Z}}_2\right).$$

Since ⊠ is associative, we obtain three BL-algebras (up to an isomorphism) that are chains with Remark 7.

We deduce that

$$\left|{\mathcal{MV}}_4\right|=\pi \left(4\right)+1$$

$$\left|{\mathcal{BL}}_4\right|=\left|{\mathcal{MV}}_4\right|+1·\left|{\mathcal{BL}}_3\right|+2·\left|{\mathcal{BL}}_2\right|-1=\pi \left(4\right)+\pi \left(3\right)+3·\pi \left(2\right)+4=5$$

$$\left|{\mathcal{MV}}_4\left(c\right)\right|=1\mathrm{and}\left|{\mathcal{BL}}_4\left(c\right)\right|=\left|{\mathcal{MV}}_3\left(c\right)\right|+3=1+3=4.$$

We remark that $\left|{\mathcal{BL}}_4\right|=\left|{\mathcal{MV}}_4\right|+\left|{\mathcal{BL}}_3\right|+\left|{\mathcal{BL}}_2\right|.$

Case$n=5.$

To generate a BL-algebra with five elements as the ordinal product ${\mathcal{L}}_1⊠{\mathcal{L}}_2$ of two BL-algebras ${\mathcal{L}}_1$ and ${\mathcal{L}}_2$, we must consider:

$${\mathcal{L}}_1\mathrm{is}\mathrm{a}\mathrm{BL}\text{-}\mathrm{chain}\mathrm{with}\mathrm{two}\mathrm{elements}\mathrm{and}{\mathcal{L}}_2\mathrm{is}\mathrm{a}\mathrm{BL}\text{-}\mathrm{algebra}\mathrm{with}\mathrm{four}\mathrm{elements};$$

$${\mathcal{L}}_1\mathrm{is}\mathrm{a}\mathrm{BL}\text{-}\mathrm{chain}\mathrm{with}\mathrm{four}\mathrm{elements}\mathrm{and}{\mathcal{L}}_2\mathrm{is}\mathrm{a}\mathrm{BL}\text{-}\mathrm{algebra}\mathrm{with}\mathrm{two}\mathrm{elements};$$

$${\mathcal{L}}_1\mathrm{is}\mathrm{a}\mathrm{BL}\text{-}\mathrm{chain}\mathrm{with}\mathrm{three}\mathrm{elements}\mathrm{and}{\mathcal{L}}_2\mathrm{is}\mathrm{a}\mathrm{BL}\text{-}\mathrm{algebra}\mathrm{with}\mathrm{three}\mathrm{elements}.$$

We obtain the following BL-algebras:

$$\begin{array}{ccc}& & Id\left({\mathbb{Z}}_2\right)⊡Id\left({\mathbb{Z}}_8\right),Id\left({\mathbb{Z}}_2\right)⊡Id({\mathbb{Z}}_2\times {\mathbb{Z}}_2),Id\left({\mathbb{Z}}_2\right)⊡[Id\left({\mathbb{Z}}_2\right)⊡Id\left({\mathbb{Z}}_4\right)],\hfill \\ & & Id\left({\mathbb{Z}}_2\right)⊡[Id\left({\mathbb{Z}}_4\right)⊡Id\left({\mathbb{Z}}_2\right)]\mathrm{and}Id\left({\mathbb{Z}}_2\right)⊡[Id\left({\mathbb{Z}}_2\right)⊡(Id\left({\mathbb{Z}}_2\right)⊡Id\left({\mathbb{Z}}_2\right))]\hfill \end{array}$$

and

$$\begin{array}{ccc}& & Id\left({\mathbb{Z}}_8\right)⊡Id\left({\mathbb{Z}}_2\right),[Id\left({\mathbb{Z}}_2\right)⊡Id\left({\mathbb{Z}}_4\right)]⊡Id\left({\mathbb{Z}}_2\right),\hfill \\ & & [Id\left({\mathbb{Z}}_4\right)⊡Id\left({\mathbb{Z}}_2\right)]⊡Id\left({\mathbb{Z}}_2\right)\mathrm{and}[Id\left({\mathbb{Z}}_2\right)⊡(Id\left({\mathbb{Z}}_2\right)⊡Id\left({\mathbb{Z}}_2\right))]⊡Id\left({\mathbb{Z}}_2\right)\hfill \end{array}$$

and

$$\begin{array}{ccc}& & Id\left({\mathbb{Z}}_4\right)⊡Id\left({\mathbb{Z}}_4\right),[Id\left({\mathbb{Z}}_2\right)⊡Id\left({\mathbb{Z}}_2\right)]⊡\left[Id\left({\mathbb{Z}}_2\right)\right)⊡Id\left({\mathbb{Z}}_2\right)]\hfill \\ & & Id\left({\mathbb{Z}}_4\right)⊡[Id\left({\mathbb{Z}}_2\right)⊡Id\left({\mathbb{Z}}_2\right)]\mathrm{and}[Id\left({\mathbb{Z}}_2\right)⊡Id\left({\mathbb{Z}}_2\right)]⊡Id\left({\mathbb{Z}}_4\right)\hfill \end{array}$$

Since ⊠ is associative, $Id\left({\mathbb{Z}}_2\right)⊡[Id\left({\mathbb{Z}}_4\right)⊡Id\left({\mathbb{Z}}_2\right)]=[Id\left({\mathbb{Z}}_2\right)⊡Id\left({\mathbb{Z}}_4\right)]⊡Id\left({\mathbb{Z}}_2\right),$$Id\left({\mathbb{Z}}_2\right)⊡[Id\left({\mathbb{Z}}_2\right)⊡(Id\left({\mathbb{Z}}_2\right)⊡Id\left({\mathbb{Z}}_2\right))]$=$[Id\left({\mathbb{Z}}_2\right)⊡(Id\left({\mathbb{Z}}_2\right)⊡Id\left({\mathbb{Z}}_2\right))]⊡Id\left({\mathbb{Z}}_2\right)$$=[Id\left({\mathbb{Z}}_2\right)⊡Id\left({\mathbb{Z}}_2\right)]⊡\left[Id\left({\mathbb{Z}}_2\right)\right)⊡Id\left({\mathbb{Z}}_2\right)],$$[Id\left({\mathbb{Z}}_4\right)⊡Id\left({\mathbb{Z}}_2\right)]⊡Id\left({\mathbb{Z}}_2\right)=Id\left({\mathbb{Z}}_4\right)⊡[Id\left({\mathbb{Z}}_2\right)$$⊡Id\left({\mathbb{Z}}_2\right)]$ and $Id\left({\mathbb{Z}}_2\right)⊡(Id\left({\mathbb{Z}}_2\right)$$⊡Id\left({\mathbb{Z}}_4\right))=[Id\left({\mathbb{Z}}_2\right)⊡Id\left({\mathbb{Z}}_2\right)]⊡Id\left({\mathbb{Z}}_4\right).$

We obtain eight BL-algebras of which seven are chains from Remark 7.

We deduce that

$$\left|{\mathcal{MV}}_5\right|=\pi \left(5\right)+1=1\mathrm{and}\left|{\mathcal{BL}}_5\right|=9=\left|{\mathcal{MV}}_5\right|+\left|{\mathcal{BL}}_4\right|+\left|{\mathcal{BL}}_3\right|+\left|{\mathcal{BL}}_2\right|$$

$$\left|{\mathcal{MV}}_5\left(c\right)\right|=1\mathrm{and}\left|{\mathcal{BL}}_5\left(c\right)\right|=8.$$

□

Table 2. The structure of BL-algebras of order n, 2 ≤ n ≤ 5.

$\left|\mathit{L}\right|=\mathit{n}$	No. of BL-alg	Structure
$n=2$	1	$\left\{Id\left({\mathbb{Z}}_2\right)(\mathrm{chain},\mathrm{MV})\right.$
$n=3$	2	$\left\{\begin{array}{c}Id\left({\mathbb{Z}}_4\right)(\mathrm{chain},\mathrm{MV})\\ Id\left({\mathbb{Z}}_2\right)⊡Id\left({\mathbb{Z}}_2\right)\left(\mathrm{chain}\right)\end{array}\right.$
$n=4$	5	$\left\{\begin{array}{c}Id\left({\mathbb{Z}}_8\right)(\mathrm{chain},\mathrm{MV})\\ Id({\mathbb{Z}}_2\times {\mathbb{Z}}_2)\left(\mathrm{MV}\right)\\ Id\left({\mathbb{Z}}_2\right)⊡Id\left({\mathbb{Z}}_4\right)\left(\mathrm{chain}\right)\\ Id\left({\mathbb{Z}}_4\right)⊡Id\left({\mathbb{Z}}_2\right)\left(\mathrm{chain}\right)\\ Id\left({\mathbb{Z}}_2\right)⊡(Id\left({\mathbb{Z}}_2\right)⊡Id\left({\mathbb{Z}}_2\right))\left(\mathrm{chain}\right)\end{array}\right.$
$n=5$	9	$\left\{\begin{array}{c}Id\left({\mathbb{Z}}_{16}\right)(\mathrm{chain},\mathrm{MV})\\ Id\left({\mathbb{Z}}_2\right)⊡Id\left({\mathbb{Z}}_8\right)\left(\mathrm{chain}\right)\\ Id\left({\mathbb{Z}}_2\right)⊡Id({\mathbb{Z}}_2\times {\mathbb{Z}}_2)\\ Id\left({\mathbb{Z}}_2\right)⊡(Id\left({\mathbb{Z}}_2\right)⊡Id\left({\mathbb{Z}}_4\right))\left(\mathrm{chain}\right)\\ Id\left({\mathbb{Z}}_2\right)⊡(Id\left({\mathbb{Z}}_4\right)⊡Id\left({\mathbb{Z}}_2\right))\left(\mathrm{chain}\right)\\ Id\left({\mathbb{Z}}_2\right)⊡(Id\left({\mathbb{Z}}_2\right)⊡(Id\left({\mathbb{Z}}_2\right)⊡Id\left({\mathbb{Z}}_2\right)))\left(\mathrm{chain}\right)\\ Id\left({\mathbb{Z}}_8\right)⊡Id\left({\mathbb{Z}}_2\right)\left(\mathrm{chain}\right)\\ (Id\left({\mathbb{Z}}_4\right)⊡Id\left({\mathbb{Z}}_2\right))⊡Id\left({\mathbb{Z}}_2\right)\left(\mathrm{chain}\right)\\ Id\left({\mathbb{Z}}_4\right)⊡Id\left({\mathbb{Z}}_4\right)\left(\mathrm{chain}\right)\end{array}\right.$

presents a basic summary of the structure of BL-algebras L with $2\le n\le 5$ elements:

Finally,

Table 3. A summary of the number of the obtained BL-algebras.

	$\mathit{n}=2$	$\mathit{n}=3$	$\mathit{n}=4$	$\mathit{n}=5$
MV-algebras	1	1	2	1
MV-chains	1	1	1	1
BL-algebras	1	2	5	9
BL-chains	1	2	4	8

present a summary of the number of MV-algebras, MV-chains, BL-algebras and BL-chains with $n\le 5$ elements obtained used commutative rings:

5. Conclusions

It is known that BL-algebras are a particular kind of residuated lattices.

In this paper, we studied rings whose ideals have a BL-algebra structure and we used some commutative rings to build certain finite BL-algebras by passing to the ideal lattice.

Using the results obtained in this paper, in further research, we will try to describe a recursive algorithm to construct all isomorphism classes of finite BL-algebras of a given size. Furthermore, we hope to obtain important results about BL-rings by studying the binary block codes associated with a BL-algebra in further research.