# A Rickart-Like Theorem for the Additivity of Multiplicative Maps on Rings

Authors: Bana Al Subaiei; Noômen Jarboui
Journal: Journal of Mathematics
 (2022)

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## Abstract

Rickart’s theorem states that every bijective multiplicative mapping of a Boolean ringR onto an arbitrary ring S is necessarily additive. We prove a version of Rickart’s theorem for non-bijective mappings. This enables us to partially answer a question that was left open (Al Subaiei, B., Jarboui, N. On the Monoid of Unital Endomorphisms of a Boolean Ring. Axioms 2021, 10, 305).


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## Body

## 1. Introduction

LetR and S be arbitrary associative rings (not necessarily with identity elements). A mapping φ of R into S such that φx+y=φx+φy (resp., φxy=φxφy) for all x,y∈R will be called an additive (resp., a multiplicative) mapping ofRintoS. The question of when a bijective multiplicative mapping is additive has been considered by Rickart [1] in 1948. It is a surprising result of Rickart (1, Theorem I) that every multiplicative bijective map of a Boolean ring R onto an arbitrary ring S is necessarily additive. Thus, the multiplicative structure determines the ring structure for some rings. Rickart’s result has attracted the attention of many mathematicians. Some generalizations have been established by Johnson [2] and Martindale [3] and have been utilized by Sˇ emrk in [4] to characterize the semigroup isomorphisms of standard operator algebras on Banach spaces. Some other results on the additivity of multiplicative maps between operator algebras can be found in [5–7].Our aim in this study is twofold. First, we establish an analogous result to that of Rickart, but for not necessarily bijective mappings. Second, we pause to partially answer a question that was left open in [8]. We prove that any multiplicative mapping φ of a Boolean ring R with identity into an arbitrary ring S with characteristic 2, satisfying φx+φ1R−x=φ1R for any x∈R, is necessarily additive (see Theorem 1). We recover Rickart’s result but for Boolean rings with identity. Examples 1 and 2 show the limits of our main result. As an application of Theorem 1, we provide, in Corollary 1, an explicit description of unital endomorphisms of the monoid S1=PX,∩ satisfying φA+φX∖A=X for any A∈PX, where X is a finite set and PX is its power set. This partially answers a question which was left open in [8]. Any undefined terminology is standard as in [9, 10].


## 2. Main Results

By a Boolean ring [11], we mean a ring R each of whose elements satisfies the condition x2=x. It is not difficult to show that a Boolean ring R is necessarily commutative and each of its elements also satisfies the condition 2x=x+x=0R. A Boolean ring may, but need not, contain an identity element for multiplication. In 1948, Rickart has proved (Theorem I of [1]) that any bijective multiplicative mapping of a Boolean ring R onto an arbitrary ring S is necessarily additive. This seems not to be true for non-bijective mappings. Consider, for instance, the mapping φ:R⟶R defined by φx=1R for any x∈R, where R is a nonzero Boolean ring. We are going to establish a result which is analogous to Rickart’s theorem.Theorem 1.
LetR be a Boolean ring with identity and let S be an arbitrary ring of characteristic 2. Assume that φ:R⟶S is a multiplicative mapping and consider the following four conditions:(i)
φ is bijective(ii)
φ is injective, and for any x∈R, there exists ax∈R such that φx+φ1R−x=φax(iii)
φx+φ1R−x=φ1R, for any x∈R(iv)
φ is additive
Then, (i)⇒ (ii) ⇒ (iii) ⇔ (iv).Proof.
(i)⇒ (ii) As φ is surjective, then it follows that, for any x∈R, there exists ax∈R such that φx+φ1R−x=φax.
(ii)⇒ (iii) Let x∈R. By assumption, there exists an element ax∈R such that(1)φx+φ1R−x=φax.
Multiplying both sides of (1) by φx and using the fact that φ is multiplicative, x2=x, and x1R−x=0R, we obtain(2)φx+φ0R=φxax.
Similarly, multiplying both sides of (1) by φ1R−x, we obtain(3)φ0R+φ1R−x=φ1R−xax.
Now, we claim thatφ0R=0S. Indeed, set α=φ0R. Notice that α is idempotent since α2=φ0R2=φ0R2=φ0R=α. According to (1), we have φ0R+φ1R=φa0R. Thus, α+φ1R=φa0R. This implies when multiplying both sides of the latter equation by α that α2+α=α. Hence, α2=0S, and consequently, α=0S as claimed. It follows from (2) and (3) that φx=φxax and φ1R−x=φ1R−xax. However, as φ is injective, we obtain x=xax and 1R−x=1R−xax. Therefore, 1R=1R−x+x=1R−xax+xax=ax. Thus, equation (1) becomes φx+φ1R−x=φ1R, as desired.
The implication (iv)⇒ (iii) is trivial. Let us prove that (iii) implies (iv). Assume (iii) and let x,y∈R. Then, by assumption, we have(4)φx+y=φ1R−φ1R−x−y.
Note that(5)1R−x−y=1R−x1R−y1R−y1R−x.
Equation (5) follows from the fact that 2z=0R and z1R−z=0R for any z∈R since R is a Boolean ring. Therefore, by using the fact that φ is multiplicative, we get immediately, from (5), the following equation:(6)φ1R−x−y=φ1R−x1R−yφ1R−y1R−x.
However, hypothetically, we have(7)φ1R−x1R−y=φ1R−φx1R−y,(8)φ1R−y1R−x=φ1R−φy1R−x.
Combining (6)–(8) and using the fact that φxφy=φxy=φyx=φyφx, we obtain(9)φ1R−x−y=φ1R−φx1R−yφ1R−φy1R−x=φ1R−φxφ1R−yφ1R−φyφ1R−x=φ1R−φxφ1R−φyφ1R−φyφ1R−φx=φ1R−φx+φxφyφ1R−φy+φyφx=φ1R−φx−φy+2φxφy.
AscharS=2, we have 2φxφy=0. Hence, we obtain(10)φ1R−x−y=φ1R−φx−φy.
Combining (4) and (10), we infer that(11)φx+y=φx+φy.
Note that the implication (i)⇒ (iv) recovers Rickart’s theorem but in the case where the domain of φ is (a Boolean ring) with identity.Example 1.
This example provides a mappingφ satisfying the hypotheses of Theorem 1 and shows that the implication (iii) ⇒ (ii) does not hold. To this end, let R be a nonzero Boolean ring with identity and let φ:R⟶R be defined by φx=0R, for any x∈R. Clearly, φ is multiplicative and additive. Thus, condition (iii) is satisfied. However, φ is not injective.
In Example2, we will construct a multiplicative mapping φ of a Boolean ring R into an arbitrary ring S of characteristic 2, satisfying condition (ii) but not condition (i).Proposition 1.
LetR be a nonzero Boolean ring and let M be a prime ideal of R. Let S be an arbitrary ring and α,β∈S. Then, the following mapping,(12)θM:R⟶Sx↦α,ifx∈M,β,ifx∉M,satisfies the following conditions:(1)
θM is multiplicative if and only if α2=α, β2=β, and αβ=α(2)
IfR has an identity, then θMx+θM1R−x=θM1R, for any x∈R if and only if α=0S(3)
θM is additive if and only if α=0S and 2β=0S(4)
θM is surjective if and only if S=α,β(5)
θM is injective if and only if R=2 and α≠βTo prove this proposition, we need the following straightforward lemma. We include proof for the sake of completeness. However, first, recall from [9, 10] that an ideal P of a commutative ring R is said to be prime if P≠R and for any a,b∈R such that ab∈P; then, either a∈P or b∈P.Lemma 1.
LetR be a nonzero Boolean ring with identity and let M be a prime ideal of R. Then, for x∈R, we have(13)x∈M⇔1R−x∉M.Proof.
“⇒” Assume by the way of contradiction that 1R−x∈M. Therefore, 1R=x+1R−x∈M, which is a contradiction.
”⇐” As 0R=x1R−x∈M and M is a prime ideal; then, either x∈M or 1R−x∈M. However, by assumption, 1R−x∉M. Thus, necessarily, x∈M. This completes the proof.Proof of Proposition1.
(1)
For the “Only if” part, pick an elementx∈M and an element y∈R∖M. As θM is multiplicative, we get θMx2=(θMx)2 and θMy2=(θMy)2. However, x2=x and y2=y, so α=α2 and β=β2. Now, since xy∈M, we obtain α=θMxy=θMxθMy=αβ. For the “If” part, let x,y∈R. If x∈M or y∈M, then xy∈M since M is an ideal of R. Thus, θMxy=α. On the other hand, as θMx=α or θMy=α, then θMxθMy=α2 or αβ. Hence, θMxy=θMxθMy. Assume now that x∉M and y∉M. In this case, xy∉M since M is a prime ideal of R. It follows that θMxy=β and θMx=θMy=β. Therefore, θMxy=θMxθMy. We derive that θM is a multiplicative mapping.(2)
Assume thatα=0S and let x∈R. If x∈M, then 1R−x∉M by Lemma 1. So, θMx=α and θM1R−x=β. Thus, θMx+θM1R−x=α+β=0S+β=β=θM1R. Similarly, if x∉M, then 1R−x∈M by using Lemma 1. Thus, θMx=β and θM1R−x=α=0S. It follows that θMx+θM1R−x=θM1R. Conversely, assume that θMx+θM1R−x=θM1R for any x∈R. Pick an element x∈R∖M. We have θMx+θM1R−x=β+α and θM1R=β. Thus, α=0S.(3)
Suppose thatθM is additive. It follows that θM0R=0S and so α=0S. We need to show that 2β=0S. Pick x∈R∖M. We have θMx+x=θMx+θMx. Hence, θM0R=2θMx. This yields 0S=2β. Conversely, assume that α=2β=0S and let x,y∈R. If x∈M and y∈M, then x+y∈M. Thus, θMx+y=α=0S and θMx+θMy=α+α=0S. So, θMx+y=θMx+θMy. Now, if x∈M and y∉M, then a fortiori, x+y∉M. Thus, θMx+y=β and θMx+θMy=α+β=β. Hence, θMx+y=θMx+θMy. The case x∉M and y∈M can be treated similarly. Suppose now that x∉M and y∉M. We claim that x+y∈M. Indeed, it follows from Lemma 1 that 1R−x,1R−y∈M. Hence, x+y=1R−x+1R−y∈M as claimed. We deduce that θMx+y=α=0S and θMx+θMy=2β=0S. This yields θMx+y=θMx+θMy.(4)
This is trivial.(5)
For the “Only if” part, pick an elementx∈R∖M. As x≠0R and θM is injective, then θM0R≠θMx. Thus, α≠β. It remains to show that R=2. Assume the contrary. It follows that R≥4 and so M=R/2≥2. Choose x≠y in M. The equality θMx=θMy contradicts the injectivity of θM. For the “If” part, assume that R=2 and α≠β. Then, R=0R,x, where x∉M. Hence, θM0R=α≠β=θMx, which implies that θM is injective.Example 2.
LetR:=ℤ2 be the ring of integers modulo 2. Let S be an arbitrary ring (not necessarily commutative) containing an idempotent element β≠0S and such that charS=2 and S≥3. Then, according to Proposition 1, M=0R and the following mapping,(14)θM:R⟶Sx↦0S,ifx=0R,β,ifx=1R,satisfies the following conditions:(1)
θM is multiplicative(2)
θMx+θM1R−x=θM1R, for any x∈R(3)
θM is additive(4)
θM is injective but not surjectiveRemark 1.
We can replace the assumption “is injective” in Theorem1 (ii) with the following condition:(15)∀x,y∈R,φxy=φx⇒xy=x.
However, we can show easily that, for a multiplicative mappingφ, we have(16)φis injective⇔∀x,y∈R,φxy=φx⇒xy=x.
Indeed, the implication”⇒” is obvious. Let us prove the inverse implication. To this end, let x,y∈R such that φx=φy∗. Multiplying both sides of the equation ∗ by φx, we obtain(17)φx=(φx)2=φxφy=φxy.
Hence,x=xy. On the contrary, multiplying both sides of the equation ∗ by φy, we obtain(18)φxy=φxφy=φy2=φy.
Thus,xy=y. It follows that x=y, as desired.
In what follows, we will retain the same notation and terminology as in our paper [8]. Data will consist of a nonempty set X and the power set PX of X. The set PX is equipped with symmetric difference as addition and intersection as multiplication so that it is a typical example of a Boolean ring. We let R denote the ring (PX,+,·). Clearly, 1R=X and 0R=∅. For any f∈TX (where TX denotes the monoid of full transformations of X), we define a mapping f∗∈TR by f∗Y=f−1Y for any Y∈R and we let TX∗:=f∗|f∈TX. Perhaps the most important result in [8] is Theorem 1, which states that X is finite if and only if TX∗=End1RR. We are unable to give an explicit description of the monoid End1S1S1 of unital endomorphisms of S1:=PX,∩. However, we can provide an explicit description of its submonoid:(19)End1S1+S1≔φ∈End1S1S1|φa+φ1S1−a=1S1for anya∈S1.Corollary 1.
The following statements hold true:(1)
End1S1+S1=End1RR(2)
X is finite if and only if TX∗=End1S1+S1Proof.
(1)
The equalityEnd1S1+S1=End1RR is an immediate consequence of Theorem 1.(2)
Combine assertion (1) and Theorem 1 of [8].



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