On the Monoid of Unital Endomorphisms of a Boolean Ring

Abstract

Let X be a nonempty set and $\mathcal{P}\left(X\right)$ the power set of X. The aim of this paper is to provide an explicit description of the monoid ${\mathrm{End}}_{1_{\mathcal{P}\left(X\right)}}\left(\mathcal{P}\left(X\right)\right)$ of unital ring endomorphisms of the Boolean ring $\mathcal{P}\left(X\right)$ and the automorphism group $\mathrm{Aut}\left(\mathcal{P}\right(X\left)\right)$ when X is finite. Among other facts, it is shown that if X has cardinality $n\ge 1$, then ${\mathrm{End}}_{1_{\mathcal{P}\left(X\right)}}\left(\mathcal{P}\left(X\right)\right)\cong T_n^{op}$, where $T_n$ is the full transformation monoid on the set X and $\mathrm{Aut}\left(\mathcal{P}\left(X\right)\right)\cong S_n$.

1. Introduction

In mathematics, endomorphisms, and especially automorphisms of algebraic structures, play an important role. In fact, many fundamental results have been demonstrated by analyzing the automorphism group of a given structure. For instance, Galois identified solvable (univariate) polynomials f in the field of rational numbers via the structure of the automorphism group of the splitting field of f. We mention also that automorphisms have played an important role in computer science, especially in the understanding of the complexity of several algebraic problems. Perhaps the most important structure in computer science is that of finite rings (see [1,2]).

Considerable attention has been paid over the years to the study of the endomorphism semigroup and the automorphism group of some mathematical structures. For example, in [3] (Theorem 3.1), Maxson has proved that if $B={\prod }_{i=1}^n{B}_i$, $B_i=K$, K being a rigid ring (that is, $K^2\ne \left(0\right)$ and $\mathrm{End}\left(K\right)=\{0,{\mathrm{id}}_{\mathrm{K}}\}$), then the semigroup of endomorphisms of B is isomorphic to the semigroup of $n\times n$ column monomial $(0,1)$-matrices and the semigroup of unital endomorphisms of B is isomorphic to the semigroup of $n\times n$ strictly column monomial $(0,1)$-matrices. This result applies, of course, easily to finite Boolean rings since any finite Boolean ring is isomorphic to ${\prod }_{i=1}^n{\mathbb{Z}}_2$ for some positive integer n, where ${\mathbb{Z}}_2$ is the ring of integers modulo 2. Schreier [4] and Mal’cev [5] have described all automorphisms of $\mathrm{End}\left(X\right)$, where X is a set, and Gluskin [6] has described the automorphisms of $\mathrm{End}\left(V\right)$, where V is a vector space. More examples are provided by, among others, Formanek [7], Levi [8,9], Liber [10], Magill [11], Mashevitzky and Schein [12], Schein [13], Sullivan [14], and Sutov [15]. Recently, the subject has attracted renewed attention owing to its links with universal algebraic geometry (see, for instance, [16]). Note also that in [17], Araujo and Konieczny have provided a theoretical description of the automorphism group of any semigroup.

The aim of this paper is to provide an explicit description of the semigroup of unital endomorphisms and the group of automorphisms of a finite Boolean ring. It is well known from the work of M. H. Stone [18] that $B\cong \mathcal{P}(\Omega )$ for an appropriate set $\Omega$. For this reason, we focus on the power set ring. Our work is motivated on one hand by [3,6,7,8,11,17,19,20,21,22,23,24], where the authors have studied some questions related to the semigroup of endomorphisms of some rings, especially Boolean rings, and on the other hand by the importance of finding automorphisms of some semigroups and some (finite) rings, as explained above.

For any unital ring ℜ, we let $0_{\Re }$ and $1_{\Re }$, denote, respectively, its additive identity and its multiplicative identity. If $\Omega$ is a set, we let $T_{\Omega }$ denote the full transformation monoid of the set $\Omega$ and $\mathrm{Sym}(\Omega )$ the symmetric group of $\Omega$ (that is, the group of all bijections from $\Omega$ to itself). We let $|\Omega |$ denote the cardinal number of the set $\Omega$. If $|\Omega |=n$, then we write $T_{\Omega }=T_n$ and $\mathrm{Sym}(\Omega )=S_n$. We let $T_{\Omega }^{*}=\{f^{*}\mid f\in T_{\Omega }\}$ and ${\mathrm{Sym}}^{*}(\Omega )=\{f^{*}\mid f\in \mathrm{Sym}(\Omega )\}$, where $f^{*}:\mathcal{P}(\Omega )⟶\mathcal{P}(\Omega )$ is the mapping defined by $f^{*}\left(A\right)=f^{-1}\left(A\right)$ for any $A\in \mathcal{P}(\Omega )$. If G is a group, we let $\mathrm{End}\left(G\right)$ denote the monoid (under composition) of endomorphisms of the group G. Let $\mathrm{End}(\Re )$ denote the monoid (under composition) of endomorphisms of a (unital) ring ℜ and ${\mathrm{End}}_{1_{\Re }}(\Re )$ the submonoid of unital ring endomorphisms. More precisely, ${\mathrm{End}}_{1_{\Re }}(\Re ):=\{\phi \in \mathrm{End}(\Re )\mid \phi \left(1_{\Re }\right)=1_{\Re }\}$. Note that if ℜ is a nonzero unital ring, then the trivial endomorphism is not in ${\mathrm{End}}_{1_{\Re }}(\Re )$.

The titular result of this paper is Theorem 1, which states that X is a finite nonempty set if and only if ${\mathrm{End}}_{1_{\mathcal{P}\left(X\right)}}\left(\mathcal{P}\left(X\right)\right)=T_X^{*}$. As a consequence, we show in Corollary 1 that $\mathrm{Aut}\left(\mathcal{P}\left(X\right)\right)={\mathrm{Sym}}^{*}\left(X\right)$ in the event that X is a finite nonempty set. Corollary 3 recovers [22] (Theorem C). In fact, we prove that if X is a finite nonempty set, then the monoids ${\mathrm{End}}_{1_{\mathcal{P}\left(X\right)}}\left(\mathcal{P}\left(X\right)\right)$ and ${T_X}^{op}$ (resp., the groups $\mathrm{Aut}\left(\mathcal{P}\right(X\left)\right)$ and $\mathrm{Sym}\left(X\right)$) are isomorphic. In particular, if $\left|X\right|=n\ge 1$, then ${\mathrm{End}}_{1_{\mathcal{P}\left(X\right)}}\left(\mathcal{P}\left(X\right)\right)\cong T_n^{op}$ and $\mathrm{Aut}\left(\mathcal{P}\left(X\right)\right)\cong S_n$. We derive from [3] (Theorem 4.1) that every finite semigroup S is isomorphic to $T_{\Omega }^{*}$ for some (finite) set $\Omega$ (see Corollary 4). We emphasize that the proofs presented here are elementary. Moreover, it is the hope of the authors that the proof techniques used here can be applied to other classes of rings.

All rings considered in this paper are assumed to be commutative and unital. Any undefined terminology is standard as in [25,26].

2. Main Results

To avoid unnecessary repetition, let us define the notation for this paper. The data consist of a nonempty set X and the power set $\mathcal{P}\left(X\right)$ of X. Note that $\mathcal{P}\left(X\right)$ is the typical example of a Boolean ring, with the symmetric difference playing the role of addition and the intersection playing the role of multiplication. Thus, X is the multiplication identity of $\mathcal{P}\left(X\right)$ and ∅ is the addition identity. Recall that, by a Boolean ring, we mean a ring in which every element is idempotent. It is worth noting that every finite Boolean ring is isomorphic to a power set ring for some set $\Omega$ (cf. [18]).

Henceforth, we let G denote the additive group $\left(\mathcal{P}\right(X),▵)$ and we let R denote the ring $\left(\mathcal{P}\right(X),▵,\cap )$. So $1_R=X$ and $0_R=\varnothing$. For any $f\in T_X$, we define a mapping $f^{*}\in T_R$ by $f^{*}\left(Y\right)=f^{-1}\left(Y\right)$ for any $Y\in R$ and we let $T_X^{*}:=\{f^{*}\mid f\in T_X\}$, and ${\mathrm{Sym}}^{*}\left(X\right)=\{f^{*}\mid f\in \mathrm{Sym}\left(X\right)\}$.

We start our investigation with the following result. However, first, one should recall that, given a semigroup $(S,·)$, one usually defines the opposite semigroup as ${(S,·)}^{op}=(S,•)$, where $x•y=y·x$ for all $x,y\in S$.

Proposition 1.

The following statements hold true:

(1) $T_X^{*}\subseteq {\mathrm{End}}_{1_R}\left(R\right)\subseteq \mathrm{End}\left(R\right)\subseteq \mathrm{End}\left(G\right)$.

(2) The mapping $\varrho :T_X⟶T_X^{*}$ defined by $\varrho \left(f\right)=f^{*}$ is bijective.

(3) The semigroups $T_X^{op}$ and $T_X^{*}$ are isomorphic.

(4) ${\mathrm{Sym}}^{*}\left(X\right)$ is a group isomorphic to $\mathrm{Sym}\left(X\right)$.

(5) If $\phi \in {\mathrm{End}}_{1_R}\left(R\right)$ and $A\in R$, then $\phi (X\backslash A)=X\backslash \phi \left(A\right)$.

Proof. 

(1) Let $f^{*}\in T_X^{*}$ and let $A,B\in R$. Then, we have clearly $f^{*}(A+B)=f^{*}\left(A\right)+f^{*}\left(B\right)$, $f^{*}\left(AB\right)=f^{*}\left(A\right)f^{*}\left(B\right)$, and $f^{*}\left(1_R\right)=f^{-1}\left(1_R\right)=1_R$. Thus, $f^{*}\in {\mathrm{End}}_{1_R}\left(R\right)$. This proves the inclusion relation $T_X^{*}\subseteq {\mathrm{End}}_{1_R}\left(R\right)$. The inclusion relations ${\mathrm{End}}_{1_R}\left(R\right)\subseteq \mathrm{End}\left(R\right)\subseteq \mathrm{End}\left(G\right)$ are trivial.

(2) Let $f,g\in T_X$ such that $f^{*}=g^{*}$. For any $x\in X$, we have $f^{*}\left(\left\{f\left(x\right)\right\}\right)=g^{*}\left(\left\{f\left(x\right)\right\}\right)$. Thus, $f^{-1}\left(\left\{f\left(x\right)\right\}\right)=g^{-1}\left(\left\{f\left(x\right)\right\}\right)$. As $x\in f^{-1}\left(\left\{f\left(x\right)\right\}\right)$, then $x\in g^{-1}\left(\left\{f\left(x\right)\right\}\right)$. Hence, $g\left(x\right)=f\left(x\right)$. This proves that $f=g$ and so $\varrho$ is injective. It is obvious that $\varrho$ is onto by definition of $T_X^{*}$.

(3) It follows from assertion (2) that the mapping ${\varrho }^{\prime }:T_X^{op}⟶T_X^{*}$ defined by ${\varrho }^{\prime }\left(f\right)=f^{*}$ is bijective. We claim that ${\varrho }^{\prime }$ is a semigroup homomorphism. Indeed, the binary operation defined on $T_X^{op}$ is • such that $f•g=g\circ f$ for any $f,g\in T_X$. Thus, if $f,g\in T_X^{op}$, then ${\varrho }^{\prime }(f•g)={(f•g)}^{*}={(g\circ f)}^{*}=f^{*}\circ g^{*}={\varrho }^{\prime }\left(f\right)\circ {\varrho }^{\prime }\left(g\right)$. This proves our claim. Hence, ${\varrho }^{\prime }$ is an isomorphism of semigroups.

(4) Firstly, we verify that ${\mathrm{Sym}}^{*}\left(X\right)$ is a group. More precisely, we show that ${\mathrm{Sym}}^{*}\left(X\right)$ is a subgroup of $\mathrm{Sym}\left(R\right)$. Note that $i{d}_X^{*}=i{d}_{\mathcal{P}\left(X\right)}\in {\mathrm{Sym}}^{*}\left(X\right)$. Moreover, one can easily check that for any $f,g\in T_X$, we have ${(f\circ g)}^{*}=g^{*}\circ f^{*}$. Thus, if $f\in \mathrm{Sym}\left(X\right)$, then $f^{*}\in \mathrm{Sym}\left(R\right)$ and ${\left(f^{*}\right)}^{-1}={\left(f^{-1}\right)}^{*}$. Hence, ${\mathrm{Sym}}^{*}\left(X\right)\subseteq \mathrm{Sym}\left(R\right)$. Remark that if $f^{*},g^{*}\in {\mathrm{Sym}}^{*}\left(X\right)$, then $f^{*}\circ {\left(g^{*}\right)}^{-1}={(g^{-1}\circ f)}^{*}\in {\mathrm{Sym}}^{*}\left(X\right)$. Therefore, ${\mathrm{Sym}}^{*}\left(X\right)$ is a subgroup of $\mathrm{Sym}\left(R\right)$ and so ${\mathrm{Sym}}^{*}\left(X\right)$ is a group. Now, let us consider the following mapping:

$$\begin{array}{cc}\hfill \rho :& \mathrm{Sym}\left(X\right)⟶{\mathrm{Sym}}^{*}\left(X\right)\hfill \\ \hfill & f⟼{\left(f^{-1}\right)}^{*}.\hfill \end{array}$$

We claim that $\rho$ is a group isomorphism. Indeed, $\varrho$ is onto by definition of ${\mathrm{Sym}}^{*}\left(X\right)$. Moreover, $\rho$ is injective since $\rho ={\varrho }_{\mid \mathrm{Sym}\left(X\right)}$ and $\varrho$ is injective by (2). Thus, $\rho$ is a bijection. Now, let $f,g\in \mathrm{Sym}\left(X\right)$; then, $\rho (f\circ g)={\left({(f\circ g)}^{-1}\right)}^{*}={(g^{-1}\circ f^{-1})}^{*}={\left(f^{-1}\right)}^{*}\circ {\left(g^{-1}\right)}^{*}=\rho \left(f\right)\circ \rho \left(g\right)$. Thus, $\rho$ is a group homomorphism and so $\rho$ is a group isomorphism.

(5) Let $\phi \in {\mathrm{End}}_{1_R}\left(R\right)$ and let $A\in R$. It is obvious that

$$\begin{array}{cc}\hfill \varnothing & =\phi (\varnothing )\hfill \\ \hfill & =\phi \left(A\right(X\backslash A\left)\right)\hfill \\ \hfill & =\phi \left(A\right)\phi (X\backslash A)\hfill \end{array}$$

and

$$\begin{array}{cc}\hfill X& =\phi \left(X\right)\hfill \\ \hfill & =\phi (A\cup (X\backslash A\left)\right)\hfill \\ \hfill & =\phi \left(A\right)\cup \phi (X\backslash A).\hfill \end{array}$$

Thus, $\phi (X\backslash A)=X\backslash \phi \left(A\right)$. This completes the proof. □

Now, we are in a position to establish the main result of this paper. However, first, one should recall from [25,27] that if R is a ring, then an ideal M of R is said to be maximal if M is a proper ideal of R (that is, $M\ne R$) and there are no other ideals contained between M and R. An ideal P of a commutative ring R is said to be prime if $P\ne R$ and for any $a,b\in R$ such that $ab\in P$, then either $a\in P$ or $b\in P$. It is well known that in a commutative ring with identity, any maximal ideal is prime. However, the converse is false. For instance, $\left(0\right)$ is a prime ideal of the ring of integers $\mathbb{Z}$, but not a maximal ideal. Another important fact that will be used in the proof of our next result is Krull’s Theorem, which states that in a commutative ring with identity, every proper ideal is contained in a maximal ideal (see [28]).

Theorem 1.

The following statements are equivalent:

(1) $T_X^{*}={\mathrm{End}}_{1_R}\left(R\right)$.

(2) X is finite.

(3) $\mathrm{End}\left(R\right)$ is finite.

(4) ${\mathrm{End}}_{1_R}\left(R\right)$ is finite.

In particular, if $\left|X\right|=n$, then $|{\mathrm{End}}_{1_R}\left(R\right)|=n^n$.

Proof. 

(1)⟹(2) Assume by way of contradiction that X is an infinite set. Then, $\mathrm{Fin}\left(X\right):=\{A\in \mathcal{P}(X)\mid |A|<\infty \}$ would be a proper ideal of R. Thus, there exists a maximal ideal $\mathcal{M}$ of R containing $\mathrm{Fin}\left(X\right)$ by virtue of Krull’s Theorem. Let us consider the following mapping:

$$\begin{array}{cc}\hfill {\theta }_{\mathcal{M}}:& R⟶R\hfill \\ \hfill & A⟼\left\{\begin{array}{cc}{0}_R\hfill & \mathrm{if}A\in \mathcal{M}\hfill \\ 1_R\hfill & \mathrm{if}A\notin \mathcal{M}.\hfill \end{array}\right.\hfill \end{array}$$

First claim: ${\theta }_{\mathcal{M}}\in {\mathrm{End}}_{1_R}\left(R\right)$.

For, let $A,B\in R$ and consider the following cases:

Case 1: $A\in \mathcal{M}$ or $B\in \mathcal{M}$.

In this case, $AB\in \mathcal{M}$ since $\mathcal{M}$ is an ideal of R. Thus, ${\theta }_{\mathcal{M}}\left(AB\right)=0_R$ and either ${\theta }_{\mathcal{M}}\left(A\right)=0_R$ or ${\theta }_{\mathcal{M}}\left(B\right)=0_R$. Hence, ${\theta }_{\mathcal{M}}\left(AB\right)={\theta }_{\mathcal{M}}\left(A\right){\theta }_{\mathcal{M}}\left(B\right)$. On the other hand, it is not difficult to check that ${\theta }_{\mathcal{M}}(A+B)={\theta }_{\mathcal{M}}\left(A\right)+{\theta }_{\mathcal{M}}\left(B\right)$.

Case 2: $A\notin \mathcal{M}$ and $B\notin \mathcal{M}$.

In this case, $AB\notin \mathcal{M}$ since $\mathcal{M}$ is a prime ideal of R. Thus, ${\theta }_{\mathcal{M}}\left(AB\right)=1_R$ and ${\theta }_{\mathcal{M}}\left(A\right){\theta }_{\mathcal{M}}\left(B\right)=1_R$. Consequently, ${\theta }_{\mathcal{M}}\left(AB\right)={\theta }_{\mathcal{M}}\left(A\right){\theta }_{\mathcal{M}}\left(B\right)$. Now, we claim that $A+B\in \mathcal{M}$. Indeed, assume the contrary. Then, as $\mathcal{M}$ is a prime ideal, it follows that $A(A+B)\notin \mathcal{M}$ and $B(A+B)\notin \mathcal{M}$. Hence, $A+AB\notin \mathcal{M}$ and $AB+B\notin \mathcal{M}$. Again, as $\mathcal{M}$ is a prime ideal, then $(A+AB)(AB+B)\notin \mathcal{M}$. However, $(A+AB)(AB+B)=4AB=0_R$ (this follows from the fact that $x^2=x$ and $2x=0$ for any element x in R), a contradiction proving our claim. Thus, ${\theta }_{\mathcal{M}}(A+B)=0_R$, ${\theta }_{\mathcal{M}}\left(A\right)+{\theta }_{\mathcal{M}}\left(B\right)=1_R+1_R=21_R=0_R$. Therefore, ${\theta }_{\mathcal{M}}(A+B)={\theta }_{\mathcal{M}}\left(A\right)+{\theta }_{\mathcal{M}}\left(B\right)$.

It follows from the above discussion that ${\theta }_{\mathcal{M}}\in \mathrm{End}\left(R\right)$. Finally, note that ${\theta }_{\mathcal{M}}\left(1_R\right)=1_R$ since $1_R=X\notin \mathcal{M}$. Therefore, ${\theta }_{\mathcal{M}}\in {\mathrm{End}}_{1_R}\left(R\right)$. This proves our first claim.

Second claim: ${\theta }_{\mathcal{M}}\notin T_X^{*}$.

Assume the contrary and let $f\in T_X$ such that ${\theta }_{\mathcal{M}}=f^{*}$. Pick an element $x\in X$ (such an element exists because $X\ne \varnothing$). As $\left\{f\right(x\left)\right\}\in \mathrm{Fin}\left(X\right)\subseteq \mathcal{M}$, then ${\theta }_{\mathcal{M}}\left(\left\{f\left(x\right)\right\}\right)=0_R$. Thus, $f^{-1}\left(\left\{f\left(x\right)\right\}\right)=0_R=\varnothing$, which is impossible since $x\in f^{-1}\left(\left\{f\left(x\right)\right\}\right)$.

(2)⟹(3) As X is finite, then so is $T_R$ because $|T_R{|=|R|}^{\left|R\right|}=2^{\left|X\right|2^{\left|X\right|}}$. It follows that $\mathrm{End}\left(R\right)$ is also finite since $\mathrm{End}\left(R\right)\subseteq T_R$.

(3)⟹(4) Trivial.

(4)⟹(1) Assume (4). Then, it follows from Proposition 1 (1) and (3) that $T_X$ is finite. Therefore, X is finite. In order to prove assertion (1), it is enough according to Proposition 1 (1) to show that ${\mathrm{End}}_{1_R}\left(R\right)\subseteq T_X^{*}$. To this end, let $\phi \in {\mathrm{End}}_{1_R}\left(R\right)$. For any $x\in X$, it is well known (and easy to show) that ${\mathcal{M}}_x=\mathcal{P}(X\backslash \left\{x\right\})$ is a maximal ideal of R (in fact, $R/{\mathcal{M}}_x\cong \mathcal{P}\left(\left\{x\right\}\right)\cong {\mathbb{Z}}_2$). In particular, ${\mathcal{M}}_x$ is a prime ideal of R. Thus, ${\phi }^{-1}\left({\mathcal{M}}_x\right)$ is a prime ideal of R as the inverse image of a prime ideal by a ring homomorphism. However, as, in any Boolean ring, each prime ideal is maximal, it follows that ${\phi }^{-1}\left({\mathcal{M}}_x\right)$ is a maximal ideal of R. We claim that there is a unique $x^{*}\in X$ such that ${\phi }^{-1}\left({\mathcal{M}}_x\right)={\mathcal{M}}_{x^{*}}$. Indeed, we have ${\displaystyle \bigcap _{y\in X}}{\mathcal{M}}_y=\left(0_R\right)$. Thus, ${\displaystyle \bigcap _{y\in X}}{\mathcal{M}}_y\subseteq {\phi }^{-1}\left({\mathcal{M}}_x\right)$. As X is finite, the last inclusion relation yields that ${\mathcal{M}}_y\subseteq {\phi }^{-1}\left({\mathcal{M}}_x\right)$ for some $y\in X$ because ${\phi }^{-1}\left({\mathcal{M}}_x\right)$ is a prime ideal of R. As ${\mathcal{M}}_y$ is a maximal ideal of R and ${\phi }^{-1}\left({\mathcal{M}}_x\right)\ne R$, then the inclusion relation ${\mathcal{M}}_y\subseteq {\phi }^{-1}\left({\mathcal{M}}_x\right)$ ensures that ${\mathcal{M}}_y={\phi }^{-1}\left({\mathcal{M}}_x\right)$. Let $x^{*}=y$. It remains to show the uniqueness of the element $y=x^{*}$. Indeed, suppose that ${\mathcal{M}}_{x^{*}}={\mathcal{M}}_z$ for some $z\ne x^{*}$. Then, $\left\{x^{*}\right\}\in {\mathcal{M}}_z={\mathcal{M}}_{x^{*}}=\mathcal{P}(X\backslash \left\{x^{*}\right\})$, the desired contradiction showing the uniqueness of the element $x^{*}$. Thus, we have constructed a mapping $f:X\to X$ defined by $f\left(x\right)=x^{*}$, where $x^{*}$ is the unique element in X satisfying ${\phi }^{-1}\left({\mathcal{M}}_x\right)={\mathcal{M}}_{x^{*}}$. We claim that $\phi =f^{*}$. To this end, let $A\in R$. Then, we have:

$$\begin{array}{cc}\hfill x\in f^{*}\left(A\right)& \iff f\left(x\right)\in A\hfill \\ \hfill & \iff X\backslash A\subseteq X\backslash \left\{f\right(x\left)\right\}\hfill \\ \hfill & \iff X\backslash A\in \mathcal{P}(X\backslash \left\{f\left(x\right)\right\})=\mathcal{P}(X\backslash \left\{x^{*}\right\})={\mathcal{M}}_{x^{*}}\hfill \\ \hfill & \iff X\backslash A\in {\phi }^{-1}\left({\mathcal{M}}_x\right)\hfill \\ \hfill & \iff \phi (X\backslash A)\in {\mathcal{M}}_x=\mathcal{P}(X\backslash \left\{x\right\})\hfill \\ \hfill & \iff X\backslash \phi \left(A\right)\subseteq X\backslash \left\{x\right\}\left(\mathrm{see}\mathrm{Proposition}\right(5\left)\right)\hfill \\ \hfill & \iff \left\{x\right\}\subseteq \phi \left(A\right)\hfill \\ \hfill & \iff x\in \phi \left(A\right).\hfill \end{array}$$

It follows that $f^{*}\left(A\right)=\phi \left(A\right)$ for any $A\in R$. Hence, $\phi =f^{*}$ as claimed. Finally, if $\left|X\right|=n$, then $|{\mathrm{End}}_{1_R}\left(R\right)|=|T_X^{*}|=|T_X|$ (the last equality follows from Proposition 1 (2)). As $|T_X|=n^n$, we readily obtain $|{\mathrm{End}}_{1_R}\left(R\right)|=n^n$. The proof is complete. □

Corollary 1.

If X is finite, then $\mathrm{Aut}\left(R\right)={\mathrm{Sym}}^{*}\left(X\right)$.

Proof. 

Note that $\mathrm{Aut}\left(R\right)={\mathrm{End}}_{1_R}\left(R\right)\cap \mathrm{Sym}\left(R\right)$. However, ${\mathrm{End}}_{1_R}\left(R\right)=T_X^{*}$ by virtue of Theorem 1. Thus, $\mathrm{Aut}\left(R\right)=T_X^{*}\cap \mathrm{Sym}\left(R\right)={\mathrm{Sym}}^{*}\left(X\right)$, the desired conclusion completing the proof. □

Combining Proposition 1 and Theorem 1, we derive the following.

Corollary 2.

If either $T_X^{*}=\mathrm{End}\left(R\right)$ or $T_X^{*}=\mathrm{End}\left(G\right)$, then X is finite.

Remark 1.

The converse of Corollary 2 does not hold in general. Indeed, let X be a (nonempty) finite set and let A be a proper subset of X. Consider the mapping $\phi :\mathcal{P}\left(X\right)⟶\mathcal{P}\left(X\right)$, defined by $\phi \left(B\right)=BA=B\cap A$ for any $B\in R$. Clearly, $\phi \in \mathrm{End}\left(R\right)$ (in particular, $\phi \in \mathrm{End}\left(G\right)$) and $\phi (X\backslash A)=\varnothing$. However, $\phi \notin T_X^{*}$. Indeed, assume the contrary. Then, there exists $f\in T_X$ such that $\phi =f^{*}$. Thus, $\varnothing =\phi (X\backslash A)=f^{-1}(X\backslash A)=X\backslash f^{-1}\left(A\right)=X\backslash \phi \left(A\right)=X\backslash A$, which is a contradiction.

Assertion (1) of the next corollary recovers [22] (Theorem C).

Corollary 3.

If X is finite, then the following statements hold true:

(1) The semigroups ${\mathrm{End}}_{1_R}\left(R\right)$ and ${T_X}^{op}$ are isomorphic.

(2) The groups $\mathrm{Aut}\left(R\right)$ and $\mathrm{Sym}\left(X\right)$ are isomorphic.

In particular, if $\left|X\right|=n$, then ${\mathrm{End}}_{1_R}\left(R\right)\cong T_n^{op}$ and $\mathrm{Aut}\left(R\right)\cong S_n$.

Proof. 

(1) According to Theorem 1, we have ${\mathrm{End}}_{1_R}\left(R\right)=T_X^{*}$. Moreover, $T_X^{*}\cong {T_X}^{op}$ by Proposition 1. Thus, ${\mathrm{End}}_{1_R}\left(R\right)\cong {T_X}^{op}$.

(2) By virtue of Corollary 1, we have $\mathrm{Aut}\left(R\right)={\mathrm{Sym}}^{*}\left(X\right)$. On the other hand, Proposition 1 ensures that ${\mathrm{Sym}}^{*}\left(X\right)\cong \mathrm{Sym}\left(X\right)$. Hence, $\mathrm{Aut}\left(R\right)\cong \mathrm{Sym}\left(X\right)$. This completes the proof. □

We derive from [3] (Theorem 4.1) the following result.

Corollary 4.

Every finite semigroup S is isomorphic to $T_{\Omega }^{*}$ for some (finite) set Ω.

Proof. 

According to [3] (Theorem 4.1), $S\cong {\mathrm{End}}_{1_B}\left(B\right)$ for some finite Boolean ring B. However, as is well known from the work of M. H. Stone [18], $B\cong \mathcal{P}(\Omega )$ for an appropriate set $\Omega$. Thus, ${\mathrm{End}}_{1_B}\left(B\right)\cong {\mathrm{End}}_{1_{\mathcal{P}(\Omega )}}\left(\mathcal{P}(\Omega )\right)$. By using Theorem 1, we obtain $S\cong T_{\Omega }^{*}$. This completes the proof. □

3. Conclusions

In this article, we have provided an explicit description of the monoid of unital endomorphisms of the finite Boolean ring $R=\left(\mathcal{P}\right(X),▵,\cap )$. When X is finite, we have derived that $\mathrm{Aut}\left(R\right)={\mathrm{Sym}}^{*}\left(X\right)$. Moreover, among other facts, we have shown that ${\mathrm{End}}_{1_R}\left(R\right)\cong T_n^{op}$ and $\mathrm{Aut}\left(R\right)\cong S_n$ whenever X is finite with cardinality equal to n. Concerning the future plans for our work, we will consider the semigroups $S_1=(\mathcal{P}\left(X\right),\cap )$, $S_2=(\mathcal{P}\left(X\right),\cup )$ and the group $G=\left(\mathcal{P}\right(X),▵)$. As one of the main tasks, we will perform further investigation of $\mathrm{End}\left(S_1\right)$, $\mathrm{End}\left(S_2\right)$, $\mathrm{End}\left(G\right)$, $\mathrm{Aut}\left(S_1\right)$, $\mathrm{Aut}\left(S_2\right)$, and $\mathrm{Aut}\left(G\right)$, where X is a finite set.