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
,
,
K being a rigid ring (that is,
and
), then the semigroup of endomorphisms of
B is isomorphic to the semigroup of
column monomial
-matrices and the semigroup of unital endomorphisms of
B is isomorphic to the semigroup of
strictly column monomial
-matrices. This result applies, of course, easily to finite Boolean rings since any finite Boolean ring is isomorphic to
for some positive integer
n, where
is the ring of integers modulo 2. Schreier [
4] and Mal’cev [
5] have described all automorphisms of
, where
X is a set, and Gluskin [
6] has described the automorphisms of
, 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
for an appropriate set
. 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 and , denote, respectively, its additive identity and its multiplicative identity. If is a set, we let denote the full transformation monoid of the set and the symmetric group of (that is, the group of all bijections from to itself). We let denote the cardinal number of the set . If , then we write and . We let and , where is the mapping defined by for any . If G is a group, we let denote the monoid (under composition) of endomorphisms of the group G. Let denote the monoid (under composition) of endomorphisms of a (unital) ring ℜ and the submonoid of unital ring endomorphisms. More precisely, . Note that if ℜ is a nonzero unital ring, then the trivial endomorphism is not in .
The titular result of this paper is Theorem 1, which states that
X is a finite nonempty set if and only if
. As a consequence, we show in Corollary 1 that
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
and
(resp., the groups
and
) are isomorphic. In particular, if
, then
and
. We derive from [
3] (Theorem 4.1) that every finite semigroup
S is isomorphic to
for some (finite) set
(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
of
X. Note that
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
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
(cf. [
18]).
Henceforth, we let G denote the additive group and we let R denote the ring . So and . For any , we define a mapping by for any and we let , and .
We start our investigation with the following result. However, first, one should recall that, given a semigroup , one usually defines the opposite semigroup as , where for all .
Proposition 1. The following statements hold true:
(1) .
(2) The mapping defined by is bijective.
(3) The semigroups and are isomorphic.
(4) is a group isomorphic to .
(5) If and , then .
Proof. (1) Let and let . Then, we have clearly , , and . Thus, . This proves the inclusion relation . The inclusion relations are trivial.
(2) Let such that . For any , we have . Thus, . As , then . Hence, . This proves that and so is injective. It is obvious that is onto by definition of .
(3) It follows from assertion (2) that the mapping defined by is bijective. We claim that is a semigroup homomorphism. Indeed, the binary operation defined on is • such that for any . Thus, if , then . This proves our claim. Hence, is an isomorphism of semigroups.
(4) Firstly, we verify that
is a group. More precisely, we show that
is a subgroup of
. Note that
. Moreover, one can easily check that for any
, we have
. Thus, if
, then
and
. Hence,
. Remark that if
, then
. Therefore,
is a subgroup of
and so
is a group. Now, let us consider the following mapping:
We claim that is a group isomorphism. Indeed, is onto by definition of . Moreover, is injective since and is injective by (2). Thus, is a bijection. Now, let ; then, . Thus, is a group homomorphism and so is a group isomorphism.
(5) Let
and let
. It is obvious that
and
Thus, . 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,
) 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
and for any
such that
, then either
or
. It is well known that in a commutative ring with identity, any maximal ideal is prime. However, the converse is false. For instance,
is a prime ideal of the ring of integers
, 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) .
(2) X is finite.
(3) is finite.
(4) is finite.
In particular, if , then .
Proof. (1)⟹(2) Assume by way of contradiction that
X is an infinite set. Then,
would be a proper ideal of
R. Thus, there exists a maximal ideal
of
R containing
by virtue of Krull’s Theorem. Let us consider the following mapping:
First claim: .
For, let and consider the following cases:
Case 1: or .
In this case, since is an ideal of R. Thus, and either or . Hence, . On the other hand, it is not difficult to check that .
Case 2: and .
In this case, since is a prime ideal of R. Thus, and . Consequently, . Now, we claim that . Indeed, assume the contrary. Then, as is a prime ideal, it follows that and . Hence, and . Again, as is a prime ideal, then . However, (this follows from the fact that and for any element x in R), a contradiction proving our claim. Thus, , . Therefore, .
It follows from the above discussion that . Finally, note that since . Therefore, . This proves our first claim.
Second claim: .
Assume the contrary and let such that . Pick an element (such an element exists because ). As , then . Thus, , which is impossible since .
(2)⟹(3) As X is finite, then so is because . It follows that is also finite since .
(3)⟹(4) Trivial.
(4)⟹(1) Assume (4). Then, it follows from Proposition 1 (1) and (3) that
is finite. Therefore,
X is finite. In order to prove assertion (1), it is enough according to Proposition 1 (1) to show that
. To this end, let
. For any
, it is well known (and easy to show) that
is a maximal ideal of
R (in fact,
). In particular,
is a prime ideal of
R. Thus,
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
is a maximal ideal of
R. We claim that there is a unique
such that
. Indeed, we have
. Thus,
. As
X is finite, the last inclusion relation yields that
for some
because
is a prime ideal of
R. As
is a maximal ideal of
R and
, then the inclusion relation
ensures that
. Let
. It remains to show the uniqueness of the element
. Indeed, suppose that
for some
. Then,
, the desired contradiction showing the uniqueness of the element
. Thus, we have constructed a mapping
defined by
, where
is the unique element in
X satisfying
. We claim that
. To this end, let
. Then, we have:
It follows that for any . Hence, as claimed. Finally, if , then (the last equality follows from Proposition 1 (2)). As , we readily obtain . The proof is complete. □
Corollary 1. If X is finite, then .
Proof. Note that . However, by virtue of Theorem 1. Thus, , the desired conclusion completing the proof. □
Combining Proposition 1 and Theorem 1, we derive the following.
Corollary 2. If either or , 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 , defined by for any . Clearly, (in particular, ) and . However, . Indeed, assume the contrary. Then, there exists such that . Thus, , 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 and are isomorphic.
(2) The groups and are isomorphic.
In particular, if , then and .
Proof. (1) According to Theorem 1, we have . Moreover, by Proposition 1. Thus, .
(2) By virtue of Corollary 1, we have . On the other hand, Proposition 1 ensures that . Hence, . This completes the proof. □
We derive from [
3] (Theorem 4.1) the following result.
Corollary 4. Every finite semigroup S is isomorphic to for some (finite) set Ω.
Proof. According to [
3] (Theorem 4.1),
for some finite Boolean ring
B. However, as is well known from the work of M. H. Stone [
18],
for an appropriate set
. Thus,
. By using Theorem 1, we obtain
. This completes the proof. □