# The EA-Dimension of a Commutative Ring

Authors: Mosbah Eljeri
Journal: ISRN Algebra
 (2013)

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## Abstract

An elementary annihilator of a ringA is an annihilator that has the form (0:a)A; a∈R∖(0). We define the elementary annihilator dimension of the ring A, denoted by EAdim(A), to be the upper bound of the set of all integers n such that there is a chain (0:a0)⊂⋯⊂(0:an) of annihilators of A. We use this dimension to characterize some zero-divisors graphs.


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## Body

## 1. Introduction

In this paper, all rings are considered to be commutative and unitary.LetA be a ring and S be a nonempty subset of A. We call the annihilator of S in A denoted by (0:S)A or (0:S) the set {a∈A/aS=(0)}. If S={a} is a singleton then (0:S) will be denoted by (0:a). If a≠0 then (0:a) is called an elementary annihilator. An annihilator is said to be maximal if it is maximal in the set of all proper annihilators of A. It is well known that all maximal annihilators are elementary. For n∈ℕ an elementary annihilator chain (0:a1)⊂(0:a2)⊂⋯⊂(0:an+1) is said to be a chain of elementary annihilators with length n ending in (0:an+1). The upper bound of the set of all lengths of elementary annihilator chains ending in (0:a) is called the elementary annihilator height of an+1 (or (0:an+1)). In this paper, we introduce a dimension of a ring R using elementary annihilator chains called elementary annihilator dimension, denoted by EAdim(R). The EAdim(R) is the upper bound of the set of elementary annihilator heights. We use this dimension to study zero-divisor graphs.We introduce a class of rings calledisometric maximal elementary annihilator rings, in short IMEA-rings. That is the class of rings with finite EAdimension whose all maximal annihilators have the same height.


## 2. Elementary Annihilator Dimension of a Ring

Definition 1.
(1) Letn∈ℕ and (0:a1)⊂(0:a2)⊂⋯⊂(0:an+1) be chain of elementary annihilators in the ring A. One says that this chain is an elementary annihilator chain of length n ending in (0:an+1).
(2) Leta be a nonzero element of A. One defines the elementary annihilator height of a, denoted by EAht(a), as the upper bound of the set of all lengths of elementary annihilator chains ending in (0:a).
(3) One calls elementary annihilator dimension ofA, denoted by EAdim(A), the upper bound of the set {EAht(a);a∈A∖{0}}.Example 2.
(1)EAdim(ℤ/4ℤ)=1. Indeed, (0:1)⊂(0:2) is the longest chain of elementary annihilators in ℤ/4ℤ.
(2)EAht(1)=0.
(3) All nonzero zero-divisorsa satisfy EAht(a)≥1. Indeed, (0:1)⊂(0:a) is a chain of length one.It is easy to check the following results.Remark 3.
(1) Leta∈A∖{0}, EAht(a)=0 if and only if a is regular.
(2)EAdim(A)=0 if and only if A is a domain.
(3) For an idealI of A, EAdim(A/I)=0 if and only if I is prime.
(4) Ifa is a nonzero noninvertible element EAdim(A/(a))=0 if and only if a is prime.We denote bynil(A) the set of all nilpotent elements of A. A is said to be reduced if it has no nilpotents other then zero.Theorem 4.
Leta∈nil(A)∖{0} and n(a) be its index of nilpotency; one has: EAht(a)+ n(a)≤ EAdim(A)+2.Proof.
IfEAht(a)+n(a) or EAdim(A) is infinite the result is obvious. Otherwise, there exists a chain whose length is EAht(a) and it ends in (0:a). Let (0:1)⊂(0:a1)⊂⋯⊂(0:ar-1)⊂(0:a) be this chain. Moreover, we have (0:a)⊂⋯⊂(0:an(a)-1). So we obtain the following chain: (0:1)⊂(0:a1) ⊂⋯⊂(0:ar-1)⊂ (0:a)⊂(0:a2) ⊂⋯⊂(0:an(a)-1) whose length is EAht(a)+n(a)-2. Consequently, EAht(a)+n(a)-2≤EAdim(A).Corollary 5.
Ifa∈nil(A)∖{0} satisfies EAht(a)=EAdim(A) is finite then n(a)=2. In particular, if EAdim(A)=1 then for all a∈nil(A)∖{0}, n(a)=2.Theorem 6.
LetA1 and A2 be two rings; then;(1)
EAdim(A1×A2) is finite if and only if EAdim(A1) and EAdim(A2) are finite.(2)
EAdim(A1×A2)=EAdim(A1)+EAdim(A2)+1.Proof.
Let(a,b)∈A1×A2 be a nonzero zero-divisor. If a and b are nonzero then (0:(a,b))A1×A2=(0:a)A1×(0:b)A2.
If one of them is zero, for example,a=0 then (0:(a,b))A1×A2=A1×(0:b)A2.
(1) “⇒” Let EAdim(A1×A2)=n, suppose that EAdim(A1) or EAdim(A2) is infinite; for example, EAdim(A1) is infinite. Then there exists r≥n+1 and (0:1)⊂(0:a1)⊂⋯⊂(0:ar) a chain in A1; then (0:(1,0))⊂(0:(a1,0)⊂⋯⊂(0:(ar,0)) is a chain of elementary annihilators in A1×A2 whose length is r>n, contradiction.
“⇐” If we assume that n=EAdim(A1), then there is a chain of length n in A1; let (0:1)A1⊂(0:a1)A1⊂⋯⊂(0:an)A1 be this chain. In the same way we put m=EAdim(A2) and we take (0:1)A2⊂(0:b1)A2⊂⋯⊂(0:bm)A2 as a chain of length m. Then (0:(1,1))A1×A2⊂ (0:(a1,1))A1×A2⊂⋯⊂ (0:(an,1))A1×A2 ⊂(0:(an,b1))A1×A2⊂⋯⊂(0:(an, bm))A1×A2 ⊂(0:(0,bm))A1×A2 is an elementary annihilator chain of A1×A2 whose length is n+m+1 that is maximal, because of the inclusion (0:(a,b))A1×A2⊂(0:(c,d))A1×A2⇔(0:a)A1⊂(0:c)A1 or (0:b)A2⊂(0:d)A2. Then EAdim(A1×A2) is finite and EAdim(A1×A2)=n+m+1.
(2) IfEAdim(A1×A2) is infinite (that is EAdim(A1) or EAdim(A2) is infinite, by (1)) then the result is obvious. The finite case is shown in the proof of (1) “⇐”.By induction, we have the following result.Corollary 7.
(1) LetA1,…,Ar be some rings, one has EAdim(A1×⋯×Ar)=∑i=1r‍EAdim(Ai)+r-1.
(2) IfA is a domain and n∈ℕ* then EAdim(An)=n-1.


## 3. The EAdimension and the Zero-Divisor Graph

LetA be a ring. The zero-divisor graph of A is defined to be the graph whose vertices are the nonzero zero-divisors of A and its edges are the pairs {a,b} satisfying ab=0. We denote this graph by Γ(A). For the simplicity of writing we still denote by Γ(A) the set of nonzero zero-divisors of A.Γ
(
A
) is said to be connected if for every two different vertices a and b of Γ(A) there is a sequence a1,…,an∈Γ(A) such as a=a1, b=an and {ai,ai+1} is an edge, ∀1≤i≤n-1. This sequence is called a path connecting a and b with length n-1. Γ(A) is said to be complete if each two distinct vertices form an edge. We call the distance between a and b the least length of a path connecting them, denoted by dA(a,b) or d(a,b). We call the diameter of Γ(A), denoted diam(Γ(A)), the supremum of the set {d(a,b);a,b∈Γ(A)}. In [1], Anderson and Livingston showed that Γ(A) is connected and diam(Γ(A))∈{0,1,2,3}.For an integerr≥2, Anderson and Livingston defined Γ(A) to be r-partite complete if Γ(R)=Γ1∪⋯∪Γr, where the Γis are nonempty disjoined sets and for all x≠y in Γ(A) satisfy xy≠0 if and only if there exists 1≤i≤r such that x,y∈Γi. In this paper we extend the definition of r-partite complete graph to the case when r is infinite.Lemma 8 (see [2], Theorem 6).
(1) If(0:a) is an elementary annihilator that is maximal (in the set of proper annihilators of A) then it is prime.
(2) Leta,b∈A∖{0}; if (0:a) is maximal and (0:b)⊈(0:a) then b∈(0:a).Proposition 9.
LetA be a reduced ring that is not a domain and a≠b be two nonzero zero-divisors such that (0:a) and (0:b) are maximal; then ab≠0 if and only if (0:a)=(0:b).Proof.
“⇒” Immediately, by Lemma 8 “⇐” If (0:a)=(0:b), suppose that ba=0. ba=0⇒ b∈(0:a) =(0:b)⇒b2=0, contradiction. Then ab≠0.Theorem 10.
IfA is a nonreduced ring then EAdim(A)=1 if and only if Γ(A) is complete.Proof.
“⇐” if Γ(A) is complete then, according to [1, Theorem 2.8], we have for all x,y∈Z(A), xy=0. Then Z(A)=(0:c), ∀c∈Z(A)∖{0}. So all nonzero elementary annihilators are equal, and then EAdim(A)=1.
“⇒” Let a∈nil(A)∖{0}; then a2=0, according to Corollary 5. If Z(A)={0,a} then Γ(A) is complete. Otherwise, for all b∈Z(A)∖{0,a} we have (0:a) and (0:b) are maximal. Suppose that (0:a)≠(0:b), according to Lemma 8(2), ab=0 and there, for example, y∈(0:a)∖(0:b); then (0:b)⊂(0:yb), contradiction to the maximality of (0:b). Then (0:a)=(0:b) and ab=0. Consequently, Z(A)=nil(A) and all nonzero zero-divisors b satisfy b2=0. It follows that for all a,b∈Z(A), ab=0. According to [1, Theorem 2.8], Γ(A) is complete.Theorem 11.
IfΓ(A)=⋃i∈IΓi is r-partite complete graph with r=card(I)∈ℕ∪{∞}, r≥2 then r=2 under one of the following conditions:  (1)
two of theΓi's contain, each one, more than one element;(2)
A is reduced.Proof.
Suppose thatr≥3.
First case: If two among the
Γ
i
's contain, each one, more then one element. Assume that Γ(A)=Γ1∪Γ2∪Γ3∪⋯, where Γ1 and Γ2 have, each one, at least two elements. Let ai∈Γi, 1≤i≤3 and ai≠bi∈Γi, 1≤i≤2. We have (a1+a2)a3=0; then a1+a2 is a divisor of zero in A. Suppose that a1+a2=0; then b1(a1+a2)=0⇒b1a1=0, contradiction. Then a1+a2∈Γ(A). Suppose that a1+a2∈Γ3∪⋯; then b1(a1+a2)=0⇒b1a1=0, contradiction; then a1+a2∈Γ1∪Γ2. If a1+a2∈Γ1 then b2(a1+a2)=0⇒b2a2=0, contradiction. Then a1+a2∈Γ2 and b1(a1+a2)= 0⇒b1a1 =0, contradiction.
Second Case: If
A
Is Reduced. Assume that Γ(A)=Γ1∪Γ2∪Γ3∪⋯. Let ai∈Γi, 1≤i≤3, we have (a1+a2)a3=0 then a1+a2 is a divisor of zero in A. Suppose that a1+a2=0, then a1(a1+a2)=0⇒a12=0, contradiction. Then a1+a2∈Γ(A). Suppose that a1+a2∈Γ3∪⋯ then a1(a1+ a2)=0⇒a12=0, contradiction. Then a1+a2∈Γ1∪Γ2. If a1+a2 ∈Γ1 then a2(a1+ a2)=0⇒a22=0, contradiction. Then a1+a2 ∈Γ2 and then a1(a1+ a2)=0⇒ a12=0, contradiction. We conclude that r=2.Theorem 12.
IfA is reduced then EAdim(A)=1 if and only if Γ(A) is bipartite complete.Proof.
“⇒” In Γ(A), we define the relation “~” by the following: x~y if (0:x)=(0:y). ~ is a relation of equivalence. For all x∈Γ(A), we denote by Γx its equivalence class. The different classes Γx form a partition of Γ(A) and we write Γ(A)=⋃i∈IΓxi. Since A is reduced and not a domain, then there exist nonzero elements b≠a satisfying ba=0. Now, EAdim(A)=1; then (0:a) and (0:b) are maximal. According to Proposition 9, (0:a)≠(0:b); then Γa≠Γb; then r=card(I)≥2.
Ify≠z∈Γx then (0:z)=(0:y) then yz≠0, by Proposition 9.
LetΓx≠Γy; then (0:x)≠(0:y); then, by Lemma 8, xy=0. And we conclude that Γ(A) is r-partite complete. According to Theorem 11, Γ(A) is bipartite complete.
“⇐” Assume that Γ(A)=Γ1∪Γ2 is bipartite complete. If a∈Γi then (0:a)=Γj∪{0}, for i≠j∈{1,2}. Let a,b∈Γ(A), if a,b∈Γi then (0:a)=(0:b)=Γj∪{0}. Otherwise (0:a) and (0:b) are incomparable. Then for all a∈Γ(A), EAht(a)=1. It follows that EAdim(A)=1.Theorem 13.
LetR be a ring.(1)
IfEAdim(R)=1 then diam(R)≤2.(2)
Ifdiam(R)≤1 or diam(R)=2 and R is reduced then EAdim(R)=1.(3)
Ifdiam(R)=2 and R is not reduced or diam(R)=3 then EAdim(R)≥2.Proof.
(1) IfR is reduced then, by Theorem 12, Γ(R) is bipartite complete; then diam(R)≤2.
IfR is not reduced then, by Theorem 10, Γ(R) is complete then, by Theorem 2.8 of [1], for all x,y∈Z(R), xy=0. Then, by Theorem 2.6 of [3], diam(R)≤1.
(2) Ifdiam(R)=2 and R is reduced then, by Theorem 2.6 of [3], R is reduced with exactly two minimal primes and at least three nonzero zero-divisors. Then Z(R)=P1∪P2, where P1, P2 are the two minimal primes of R; they satisfy P1∩P2=(0). Then for all p1∈P1 and p2∈P2, p1,p2=0 and for x≠y∈Pi∖{0}, xy≠0. Consequently, Γ(R)=[P1∖{0}]∪[P2∖{0}] is bipartite complete graph. According to Theorem 12, EAdim(R)=1.
Ifdiam(R)≤1: if diam(R)=0, by Theorem 2.6 of [3], R is isomorphic to either ℤ4 or ℤ2[y]/(y2) and in both cases R has a unique nonzero elementary annihilator then EAdim(R)=1.
Now, ifdiam(R)=1, using Theorem 2.6 of [3], xy=0 for each distinct pair of zero-divisors and R has at least two nonzero zero-divisors. According to Theorem 2.8 of [1], Γ(R) is complete and R is not reduced. By Theorem 10, EAdim(R)=1.
(3) Ifdiam(R)=3 then, by (1), EAdim(R)≠1. Since diam(R)=3 then R is not a domain then EAdim(R)≠0. Consequently, EAdim(R)≥2. If diam(R)=2 and R is not reduced: diam(R)=3 then R is not a domain then EAdim(R)≠0. Suppose that EAdim(R)=1; then by Theorem 10, Γ(R) is complete then diam(R)=1, contradiction. Then EAdim(R)≥2.Lemma 14.
LetA be a ring, and F={P1,…,Pr}, r>1 a set of distinct prime ideals which are incomparable. For a an element of R we denote E(a)={P∈F;a∉P}. Let R=A/(P1∩⋯∩Pr).(1)
For all subsetsE of F there is a∈A such that E(a)=E.(2)
Leta∈A, a¯∈Z(R)⇔E(a)≠F. In particular, a¯=0¯⇔E(a)=∅.(3)
LetE1 and E2 be two nonempty subsets of F. If ⋂P∈E1‍P⊆⋂P∈E2‍P then E2⊆E1.(4)
Leta,b∈A. b¯∈(0¯:a¯)⇔b∈⋂P∈E(a)‍P.(5)
The mapφ:{(0¯:a¯),a¯≠0¯}→𝒫(F)∖{∅},(0¯:a¯)⟼E(a) is a decreasing bijection, here 𝒫(F) denotes the set of F subsets.Proof.
(1) IfE=∅, take a=0 then a∈P, for all P∈F then E(a)=∅ and the result is true in this case.
IfE=F, take a=1 then a∉P, for all P∈F then E(a)=F.
Now ifE∉{F,∅}: E(a)=E⇔a∈P, ∀P∉E and a∉P, ∀P∈E⇔a∈(⋂P∈F∖E‍P)∖(⋃P∈E‍P). Suppose that (⋂P∈F∖E‍P)∖(⋃P∈E‍P)=∅⇒⋂P∈F∖E‍P⊆⋃P∈E‍P; then there exists Pi0∈E such that ⋂P∈F∖E‍P⊆Pi0; then there exists Pi1∈F∖E such that Pi1⊆Pi0. Since the P’s in F are incomparable under inclusion then Pi1=Pi0, then Pi0∈(F∖E)∩E=∅, contradiction. Consequently, there exists a∈A such that E(a)=E.
(2) Leta∈A, a¯∈Z(R)⇔∃b¯≠0¯/a¯b¯ =0¯⇔ ∃b∈A/b ∉⋂P∈F‍P and ba∈⋂P∈F‍P⇔∃b∈A/E(b)≠∅ and ba∈P, ∀P∈E(a)⇔∃b∈A/E(b)≠∅ and b∈P, ∀P∈E(a)⇔∃b∈A/E(b)≠∅ and E(a)⊆F∖E(b)⇔E(a)≠F. In the last equivalence the indirect sense “⇐” is obtained by (1).
(3) Suppose thatE1={Q1,…,Qs} and E2={R1,…,Rt}, where the Qi’s (resp., Pi’s) are pairwise different. Q1∩⋯∩Qs⊆ R1∩⋯∩Rt ⇒Q1∩⋯∩Qs⊆R1 then one of the Qi's is contained in R1; for example, Q1⊆R1. Since the elements of F are incomparable under inclusion then R1=Q1.
Q
1
∩
⋯
∩
Q
s
⊆
R
1 ∩⋯∩Rt⇒Q1∩⋯∩ Qs⊆R2 then one of the Qi’s is contained in R2. Suppose that Q1⊆R2; then Q1=R2 then R1=R2, contradiction. Then there exists i≥2 such that Qi⊆R2; for example, Q2⊆R2; then R2=Q2.
We repeat this process until reaching the stage numbern=min(s,t).
Suppose thats<t; then Q1∩⋯∩Qs=R1∩⋯∩Rs⊆Rs+1; then there exists k≤s such that Rk⊆Rs+1 that is, Rk=Rs+1, contradiction. Consequently, s≥t and we get E1={R1,…,Rt, Qt+1,…,Qs}; then E2⊆E1.
(4) Leta,b∈A, b¯∈(0¯:a¯)⇔ba∈⋂P∈F‍P⇔b∈⋂P∈E(a)‍P.
(5) We check thatφ is well defined: put (0¯:a¯)=(0¯:b¯). According to (4), x¯∈(0¯:a¯)⇔x∈⋂P∈E(a)‍P. Then x∈⋂P∈E(a)‍P⇔x∈⋂P∈E(b)‍P; then ⋂P∈E(b)‍P=⋂P∈E(a)‍P. According to (3), E(b)=E(a).
Fora¯≠0¯,  E(a)≠∅; then E(a)∈𝒫(F)∖{∅}. Then φ is well defined.
φ is injective, by (4).
We show thatφ is surjective: let E∈𝒫(F)∖{∅}. According to (1), there exists a∈A such that E(a)=E. Since E≠∅ then a¯≠0¯ and φ(0¯:a¯)=E.
(
0
¯
:
a
¯
)
⊆
(
0
¯
:
b
¯
)
⇒
⋂
P
∈
E
(
a
)
‍
P
⊆
⋂
P
∈
E
(
b
)
‍
P. Then, according to (3), E(b)⊆E(a). Thus φ is a decreasing bijection.Theorem 15.
LetA be a ring and P1,…,Pr,r≥1 be different incomparable prime ideals of A. Then EAdim(A/(P1∩⋯∩Pr))=r-1.Proof .
Ifr=1 then A/P1 is a domain and the result is checked.
Now letr≥2. {P1}⊂{P1,P2}⊂⋯⊂{P1,…,Pr-1}⊂{P1,…,Pr} is decreasing sequence in 𝒫(F)∖{∅} with maximal length. Using the bijection φ defined in Lemma 14, φ-1({P1,…,Pr}) ⊂φ-1({P1,…,Pr-1}) ⊂⋯⊂φ-1({P1,P2})⊂φ-1({P1}) is a chain of elementary annihilators in A/(P1∩⋯∩Pr) that has a maximal length. Then EAdim(A/(P1∩⋯∩Pr))=r-1.Example 16.
A ring is said to be semilocal ring if it has a finite number of maximal ideals. LetA be a semilocal ring with n maximal ideal; then EAdim(A/J)=n-1, where J is the Jacobson radical of A. We obtain this result by using the previous theorem.A ring is called a noetherian spectrum ring if it satisfies the ascending chain condition (acc) on radical ideals; equivalently each radical ideal is a radical of finitely generated ideal. The set of prime ideals of a ringA which are minimal over an ideal I, denoted by minI(A), is finite in the case when A is a noetherian spectrum ring. If I=(0), we denote by min(A) instead of minI(A). For more information about noetherian spectrum rings see [4, Chapter 2].Proposition 17.
LetA be a noetherian spectrum ring, for all ideals I, EAdim(A/I)=|minI(A)|-1.Proof.
SinceA is a noetherian spectrum ring then minI(A) is finite ([4], Chapter 2, Corollary 2.1.10). Assume that minI(A)={P1,…,Pr}, r=|minI(A)|∈ℕ*. The Pi’s are incomparable, then we get the result by using Theorem 15.Definition 18 (according to [5]).
LetR be a ring.(1)
One calls the chromatic number ofR the minimal number of colors used to color the elements of R such that each two adjacent elements (with zero product) have different colors, denoted by χ(R).(2)
One says that the ringR is a coloring if its chromatic number is finite.Theorem 19.
IfR is a reduced coloring then EAdim(R)=χ(R)-2.Proof.
IfR is a reduced coloring, according to [5, Theorem 3.8], min(R) is finite. And if |min(R)|=n then χ(R)=n+1. Let min(R)={P1,…,Pn}; then R=R/(P1∩⋯∩Pn). According to Theorem 15, EAdim(R)=n-1=χ(R)-2.
LetA be a ring such that EAdim(A)=n≥1. For all vertices a in Γ(A), we denote i(a)=EAht(a) and j(a)=max{EAht(b);(0:a)⊆(0:b)}. For all i≤j≤n, we denote by Γi,j(A) the subgraph of Γ(A) whose vertices form the following set {a;i(a)=i et j(a)=j}.
Theoretically we can writeΓ(A)=⋃1≤i≤j≤n‍Γi,j(A).Remark 20.
LetA be a ring such that EAdim(A)=n≥2. If Γ1,1(A)≠∅ then diam(Γ(A))=2. Indeed, consider a∈Γ1,1(A) and b∈Γ(A)∖Γ1,1(A). There exists i≥1 and j≥2 such that b∈Γi,j(A). Suppose that ab≠0 then b∉(0:a). According to Lemma 8(2) (0:b)⊆(0:a), this contradicts the fact that a∈Γ1,1(A). Then ab=0 and {a,b} is an edge. Now take x≠y∈Γ(A), three cases are possible. If x,y∈Γ1,1(A) then take b∈Γ(A)∖Γ1,1(A) and the chain x-b-y is of length 2 then d(x,y)≤2. If only x is in Γ1,1(A) then x-y is an edge. If x,y∈Γ(A)∖Γ1,1(A) then take a∈Γ1,1(A) and x-a-y is a chain of length 2. Then, in all cases, d(x,y)≤2 that is diam(Γ(A))≤2. Now EAdim(A)=n≥2, then there exists a,b such that (0:1)⊂(0:a)⊂(0:b). Let x∈(0:b)∖(0:a) then xa≠0 then d(x,a)≥2. Consequently, diam(Γ(A))=2.


## 4. Isometric Maximal Elementary Annihilator Rings

Definition 21.
LetR be a ring with finite EA dimension; one says that R is an isometric maximal elementary annihilator ring, in short an IMEA-ring if its all maximal elementary annihilators have the same height.Example 22.
The ringℤ/8ℤ is an IMEA-ring. Indeed, the elementary annihilators of ℤ/8ℤ are (0¯:1¯), (0¯:2¯), (0¯:4¯) and (0¯:6¯). They satisfy (0¯:1¯)⊂(0¯:2¯)⊂(0¯:4¯) and (0¯:1¯)⊂(0¯:2¯)⊂(0¯:6¯).Theorem 23.
LetR1 and R2 be two rings; then R1×R2 is an IMEA-ring if and only if R1 and R2 are two IMEA-rings.Proof.
By Theorem6(1), EAdim(R1×R2) is finite if and only if EAdim(R1) and EAdim(R2) are finite.
(
0
:
(
a
,
b
)
) is a maximal elementary annihilator in R1×R2 if and only if (0:(a,b))=R1×(0:b), and (0:b) is a maximal elementary annihilator in R2 or (0:(a,b))=(0:a)×R2, and (0:a) is a maximal elementary annihilator in R1, then all maximal elementary annihilators of R1×R2 have the same hight if and only if all maximal elementary annihilators of R1 have the same height and also for the maximal elementary annihilators of R2.We get the following result, inductively.Corollary 24.
LetR1,…,Rn be some rings. We have R1×R2×⋯×Rn is an IMEA-ring if and only if each Ri is an IMEA-ring.LetR be a domain, we say that R is atomic if each nonzero nonunit element of R decomposes into a finite product of irreducibles, according to [6]. An atomic domain is called a half factorial domain, in short a HFD if x1⋯xn=y1⋯ym are two decompositions into irreducibles then n=m. This concept was introduced by Zaks in [7]. A HFD is called a unique factorization domain, in short a UFD if x1⋯xn=y1⋯yn are two decompositions into irreducibles then the xi’s and the yi’s are associates after reordering them. It is well known that a UFD is an atomic domain in which each irreducible is primed by [8, Theorem 1].Proposition 25.
IfR is a UFD, then for all nonzero nonunit a of R we have R/(a) is an IMEA-ring. Moreover if a=p1m1⋯prmr is the decomposition of a into prime elements then EAdim(R/(a))=m1+⋯+mr-1.Proof.
Leta=p1m1⋯prmr∈R. Suppose that x∈R; then x¯∈Z(R/(a))∖{0¯}(1)⟺{x¯≠0¯∃y¯≠0¯x¯y¯=0¯⟺{x=p1α1⋯prαrx1;αi∈ℕ,x1∉(pi),hhhhhhhhhhhhhhhhll∀i,∃iαi<mi,∃y=p1β1⋯prβry1;βi∈ℕ,y1∉(pi),hhhhhhhhhhhhhhhhll∀i,∃iβi<mi,αi+βi≥mi,∀i.
Then
(2)Z(R(a))={⋃i(pi)p1α1⋯prαrx1¯;(α1,…,αr)∈ℕr∖{(0,…,0)},x1∈R∖⋃i(pi)p1α1⋯prαrx1¯}.
And for x¯=p1α1⋯prαrx1¯≠0¯ (one among the αi’s is <mi and x1∈R∖∪i(pi)), we have the following:
(3)(0¯:x¯)={⋃i(pi)p1β1⋯prβry1¯;(β1,…βr)∈ℕr∖{(0,…,0)}/βi≥mi-αi,∀i,y1∈R∖⋃i(pi)p1β1⋯prβry1¯}.
It is easy to check that the set of all elementary annihilators ofR/(a) is
(4){(0¯:p1α1⋯prαr¯);(α1,…,αr)∈[[0,m1]]×⋯×[[0,mr]]∖(m1,…,mr)(0¯:p1α1⋯prαr¯)}.
The maximal ones among them are
(5)(0¯:p1m1-1p2m2⋯prmr¯),(0¯:p1m1p2m2-1p3m3⋯prmr¯),…,(0¯:p1m1p2m2⋯prmr-1¯).
A longest chain ending in one of them, for example, (0¯:1¯)⊂(0¯:p1¯) ⊂⋯⊂ (0¯:p1m1-1¯)⊂(0¯:p1m1-1p2¯) ⊂⋯⊂ (0¯:p1m1-1p2m2¯)⊂ ⋯⊂(0¯:p1m1p2m2⋯pr-1mr-1pr¯) ⊂⋯⊂ (0¯: p1m1-1p2m2⋯prmr¯), has the length m1+⋯+mr-1. Thus all maximal elementary annihilators have the same height m1+⋯+mr-1 then R/(a) is an IMEA-ring and EAdim(R/(a))=m1+⋯+mr-1.Proposition 26.
LetR be a HFD. R is a UFD if and only if EAdim(R/(a))=λ(a)-1, for all nonzero nonunit a. Where λ(a) is the number of factors in a decomposition of a into irreducibles (counted with multiplicities).Proof.
“⇒” is due to the previous proposition.
“⇐” Let a be an irreducible of R then λ(a)=1; then EAdim(R/(a))=λ(a)-1=0. According to Remark 3, R/(a) is a domain; then (a) is prime. According to [8, Theorem 1], R is a UFD.Question 1.
Are all finite EAdimensional rings IMEA-rings?Question 2.
Are all finite rings IMEA-rings?



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