Throughout this paper, all rings are associative rings with identity and all near-rings are abelian left near-rings with unity. Recall that a non-empty set S with two binary operations "+" and "⋅" is an abelian left near-ring if (S,+) forms an abelian group, (S,⋅) forms a semi-group, and a⋅(b+c)=a⋅b+a⋅c for each a,b,c∈S. Clearly, every ring is a near-ring. The zero-symmetric part of a near-ring S is the set of all elements a ∈ S such that 0⋅a=0 and it is denoted by S₀. Moreover, a near-ring N is called zero-symmetric if S=S0. Let S be a near-ring and A⊆S. Then annS(A)=l.annS(A)∪r.annS(A), where

l.annS(A)={s∈S∣sa=0 for each a∈A}

and r.annS(A)={s∈S∣as=0 for each a∈A}. Also, we write Zl(S), Zr(S) and Z(S) for the set of all left zero-divisors ofS, the set of all right zero-divisors and the set Zl(S)∪Zr(S), respectively. Moreover, we use 〈A〉 to denote the ideal generated by A. For basic definitions and comprehensive discussion on near-rings, we refer the reader to [21].

Let G be a graph. Recall that G is connected if there is a path between any two distinct vertices of G. Also, the diameter of G is

diam(G)=sup{d(a,b)|a,b are vertices of G},

where d(a,b) is the length of the shortest path from a to b. Moreover, the girth of G, gr(G), is the length of the shortest cycle of the graph, and gr(G)=∞ if G has no cycles.

The concept of a zero-divisor graph of a commutative ring R was introduced by Beck in [5]. However, he let all elements of R be vertices of the graph and was mainly interested in coloring. Inspired by his study, Anderson and Livingston [3], redefined and studied the (undirected) zero-divisor graph Γ(R), whose vertices are the non-zero zero-divisors of a ring such that distinct vertices x and y are adjacent if and only if xy=0. According to [3, Theorems 2.3 and 2.4], Γ(R) is connected with diam(Γ(R))≤3, and gr(Γ(R))≤4 if Γ(R) contains a cycle. Redmond [22] extended the concept of the zero-divisor graph to noncommutative rings. Several papers are devoted to studying the relationship between the zero-divisor graph and algebraic properties of rings (cf. [3, 15, 17, 18, 20, 22]).

In [8], the authors generalized this concept to a zero-symmetric near-ring S. They defined an undirected graph Γ(S) with vertices in the set Z*(S)=Z(S)∖{0} and such that for distinct vertices a and b there is an edge connecting them if and only if ab = 0 or ba = 0. Following [8, Theorem 2.2], the zero-divisor graph of zero-symmetric near-ring S is connected and diam(Γ(S))≤3.

For a ring or near-ring S, define a~bif and only if annS(a)=annS(b). As in [20], one can see that ∼ is an equivalence relation on S. For any a∈S, let [a]S={b∈S | a~b} (for short we can use [a] instead of [a]S).

For instance, it is clear that [0]S={0} and [1]S=S∖Z(S), and that [a]S⊆Z(S)∖{0} for each a∈S∖([0]S∪[1]S).

As in [23], ΓE(S) will denote the (undirected) graph, called the compressed zero-divisor graph of S, whose vertices are the elements of SE∖{[0]S,[1]S} such that distinct vertices [a]S and [b]S are adjacent if and only if ab=0 or ba=0. Note that if a and b are distinct adjacent vertices in Γ(S), then [a]_S and [b]_S are adjacent in ΓE(S) if and only if [a]S≠[b]S. Clearly, diam(ΓE(S))≤diam(Γ(S)). For a commutative ring R, Anderson and LaGrange [2], showed that gr(ΓE(R))=3if ΓE(R) contains a cycle, and determined the structure of ΓE(R) when it is a cyclic and the monoid R_E when ΓE(R) is a star graph.

Let R be a ring. Since R[x] is an abelian near-ring under addition and substitution, it is natural to investigate the near-ring of polynomials (R[x],+,°). The binary operation of substitution, denoted by "°", of one polynomial into another is both natural and important in the theory of polynomials. We adopt the convention that for polynomials f=∑ i=0maixi and g∈R[x],

g∘f=∑ i=0maigi.

For example, (a0+a1x)°x2=(a0+a1x)2=a02+(a0a1+a1a0)x+a12x2.

However, the operation °, left distributes but does not right distribute over addition. Thus (R[x],+,°) forms a left near-ring but not a ring. We use R[x] to denote the left near-ring (R[x],+,°)with coefficients from R and

R0[x]={f∈R[x]∣f has zero constant term}

is the zero-symmetric left near-ring of polynomials with coefficients in R. Also, for each f=∑ i=0maixiand g=∑ j=0nbjxj∈R[x], we write

fg=∑ k=0 n+m(∑i+j=kaibj)xk

The aim of this paper is the study of the compressed zero-divisor graphs of zero-symmetric near-ring of polynomials R0[x] and near-ring of formal power series R0[[x]] over a commutative ring R. For a reduced ring R, we prove that diam(ΓE(R0[x]))=i if and only if diam(ΓE(R[x]))=i for each i=1,2,3. Moreover, we show that diam(ΓE(R0[x]))=1 if and only if |ΓE(R)|≤2, Nil(R)2=0, Z(R)=annR(a) for some a ∈ R, and annR(c)=Nil(R) for each c∈Z(R)∖Nil(R). Also, it is proved that diam(Γ(R0[x]))=3 if and only if diam(ΓE(R0[x]))=3. In addition, we are interested in characterizing the diameter of graph ΓE(R0[[x]]). In fact, The diameter of the graphs ΓE(R[[x]]) and ΓE(R0[[x]]) are the same when R is a reduced ring. Also, we try to relate diam(ΓE(R)) to ΓE(R0[[x]]). As a corollary, it is shown that

diam(ΓE(R))≤diam(ΓE(R0[x]))≤diam(ΓE(R0[[x]])),

where R is reduced. Moreover, we give a complete characterization for the possible diameters of ΓE(R0[[x]]), where R is a non-reduced Noetherian ring.

Let R be a commutative ring. Following [1, Theorem 2.7], we have

2≤diam(Γ(R0[x]))≤3

Hence diam(ΓE(R0[x]))≤3, since diam(ΓE(R0[x]))≤diam(Γ(R0[x])).

Proposition 2.1. Let R be a commutative ring with Z(R)≠0. Then diam(ΓE(R0[x]))≥1.

Proof. First suppose that R is a reduced ring and 0≠a∈Z(R). Thus ab=0 for some non-zero b≠a of R. If [ax]=[bx], then ax∈annR0[x](ax), and so a2=0, which is a contradiction. Hence diam(ΓE(R0[x]))≥1. Now assume R is a non-reduced ring. Then there exists 0≠a∈R such that a2=0. Thus ax,ax+x2∈Z(R0[x]). Also, x2∈annR0[x](ax) but x2∉annR0[x](ax+x2), which implies that [ax]≠[ax+x2], and so diam(ΓE(R0[x]))≥1.

For any f∈R[x], we denote by Cf the set of all coefficients of f. Also, the set of all non-zero coefficients of f is denoted by Cf*=Cf∖{0}.

To characterize the diameter of ΓE(R0[x]), where R is a reduced ring, we need the following lemma.

Lemma 2.2. Let R be a reduced ring. Then

• (1) [4, Lemma 1] For each f,g ∈ R[x] , fg=0 if and only if aibj=0 for each a_i∈ C_f and b_j ∈ C_g .
  

• (2) [7, Lemma 3.4] For each f,g ∈ R₀[x] , f ◦ g=0 if and only if aibj=0 for each a_i∈ C_f and b_j ∈ C_g .
  

Let R be a reduced ring and f,g be elements of the ring R[x]. Then fg=0 if and only if aibj=0 for each ai∈Cf and bj∈Cg, by Lemma 2.2. Hence fx∘gx=0, by Lemma 2.2. On the other hand, Z(R0[x])⊆Z(R[x]), by Lemma 2.2. Thus d([f],[g])=t in ΓE(R[x]), if and only if d([fx],[gx])=t in ΓE(R0[x]). Therefore we can conclude the next result.

Proposition 2.3. Let R be a reduced ring. Then

• (1) diam(ΓE(R[x]))=1 if and only if diam(ΓE(R0[x]))=1.

• (2) diam(ΓE(R[x]))=2 if and only if diam(ΓE(R0[x]))=2.

• (3) diam(ΓE(R[x]))=3 if and only if diam(ΓE(R0[x]))=3.

Corollary 2.4. Let R be a reduced commutative ring. Then diam(Γ(R0[x]))=3 if and only if diam(ΓE(R0[x]))=3.

Proof. (⇒) Since diam(Γ(R0[x]))=3, then we have diam(Γ(R[x]))=3, by [1, Proposition 2.10]. Thus diam(ΓE(R[x]))=3, by [12, Theorem 3.3]. Hence diam(ΓE(R0[x]))=3, by Proposition 2.3.

(⇐) It is clear, since diam(ΓE(R0[x]))≤diam(Γ(R0[x]))≤3.

Now, we investigate the diameter of ΓE(R0[x]), when R is not reduced. For this purpose, we bring the following lemmas which are used extensively in the sequel.

Lemma 2.5. ([1, Lemma 2.4]) Let R be a commutative ring and f=∑ i=1naixi, g=∑ j=1mbjxj be non-zero elements of R0[x] with f∘g=0. Then

• (1) rf=0 for some non-zero r ∈ R.

• (2) f is nilpotent or sg=0 for some non-zero s ∈ R.

Lemma 2.6. ([1, Proposition 2.5]) Let R be a non-reduced commutative ring. Then

Zr(R0[x])=Zl(R0[x])∪{∑i=1naixi∈R0[x]∣annR(a1)∩Nil(R)≠0 and ai∈R for each  i≥2},

where Zl(R0[x])={f∈R0[x]∣rf=0, for some 0≠r∈R}.

Lemma 2.7. Let R be a non-reduced commutative ring and for each a,b∈Z(R), annR({a,b})∩Nil(R)≠0. Then diam(ΓE(R0[x]))≤2. Also, if there exists c∈Nil(R) such that ck=0≠ck−1 for some k≥3, then diam(ΓE(R0[x]))=2.

Proof. By [1, Theorem 2.9], we have diam(ΓE(R0[x]))≤diam(Γ(R0[x]))=2.

Now assume that ck=0 but ck−1≠0 for some c∈Nil(R) and k≥3. Since c2x∘xk−1=0, then xk−1∈Z(R0[x]). Also, cx∘xk−1≠0≠xk−1∘cx. Since xk∈annR0[x](cx) but xk∉annR0[x](xk−1), then [cx]≠[xk−1]. It follows that d(cx,xk−1)≥2, and thus diam(ΓE(R0[x]))=2.

Following [14], a ring R is called semicommutative if ab=0 implies aRb=0 for each a,b ∈ R.

Remark 2.8. Let R be a commutative ring. Then R is a semicommutative ring, and so Nil(R[x])=Nil(R)[x], by [16]. On the other hand, Nil(R0[x])=Nil(R)0[x], by [11, Corollary 2]. Therefore Nil(R0[x])=Nil(R[x])x. We use this fact freely in the sequel.

For any f∈R0[x], we use deg(f) to denote the degree of f.

Theorem 2.9. Let R be a non-reduced commutative ring. Then diam(ΓE(R0[x]))=1 if and only if |ΓE(R)|≤2, Nil(R)2=0, Z(R)=annR(a) for some a∈R, and annR(c)=Nil(R) for each c∈Z(R)∖Nil(R).

Proof. (⇒) Let diam(ΓE(R0[x]))=1. Since R is a non-reduced ring, there exists 0≠a∈R such that a2=0. Let b∈Z(R). If [ax]=[bx], then ax∈annR0[x](bx), since a2=0. Thus ax∘bx=0, and so ab=0. Also, if [ax]≠[bx], then ax∘bx=0, by hypothesis. Hence ab=0. Therefore Z(R)=annR(a). It follows that for each b∈Nil(R), b2=0, by Lemma 2.7. Now assume that b,c are distinct elements of Nil(R). If [bx]=[cx], then cx∈annR0[x](bx), and so bc=0. If [bx]≠[cx], then 0=bcx=bx∘cx, by assumption. Hence Nil(R)2=0.

Now suppose that c∈Z(R)∖Nil(R) and d∈annR(c). Thus [x2]=[cx], since diam(ΓE(R0[x]))=1. Hence dx∈annR0[x](cx)=annR0[x](x2), which implies that d2=0, and so annR(c)⊆Nil(R). Also, by a similar way as used above, we have Z(R)=annR(b) for each b∈Nil(R), since b2=0. Hence Nil(R)⊆annR(c). Therefore Nil(R)=annR(c).

Let c∈Z(R). If c is nilpotent, then annR(c)=Z(R), and if c∉Nil(R), then annR(c)=Nil(R). Hence there exist at most two different vertices [a]R and [b]R in ΓE(R), where a∈Nil(R) and b∉Nil(R). This shows that |ΓE(R)|≤2.

(⇐) We claim that for each c∈Nil(R), annR0[x](cx)=Z(R0[x]) and annR(c)=Z(R). Since Nil(R)2=0 and c∈Nil(R), then Nil(R)⊆annR(c). Now assume d∈Z(R)∖Nil(R). Hence annR(d)=Nil(R), and thus cd=0. It means that annR(c)=Z(R). Now suppose that g=∑ j=1mbjxj∈Z(R0[x]). Thus cx∘g=0, since Nil(R)2=0 and b1∈Z(R), by Lemma 2.6. Hence annR0[x](cx)=Z(R0[x]). On the other hand, since R is non-reduced, x2∈Z(R0[x]). Also, x2∈annR0[x](cx) but x2∉annR0[x](x2). Hence we have at least two vertices [cx] and [x2] in ΓE(R0[x]). Clearly, r.annR0[x](x2)=0. On the other hand, if g∈l.annR0[x](x2), then g2=0, and so g∈Nil(R)0[x]. Since Nil(R)2=0, then Nil(R)0[x]⊆l.annR0[x](x2), and thus

annR0[x](x2)=Nil(R)0[x]=Nil(R0[x])

Now let f be a non-zero element of Z(R0[x]). We can write f=f1+f2+f3 such that Cf1*⊆Nil(R), Cf2*⊆Z(R)∖Nil(R), and Cf3*⊆R∖Z(R). We consider the following cases:

Case 1. Let f=f1=∑ i=1naixi and g=∑ j=1mbjxj∈Z(R0[x]). Since Cf1*⊆Nil(R), then annR(ai)=Z(R) for each 1≤i≤n. Also, by Lemma 2.6, b1∈Z(R). Hence f∘g=0, since Nil(R)2=0. Therefore annR0[x](f)=Z(R0[x]), and so [f]=[f1]=[cx].

Case 2. Let f=f2=∑ i=1naixi. Then annR(ai)=Nil(R) for each 1≤i≤n. Suppose that g=∑ j=1mbjxj∈r.annR0[x](f). It means that

f∘g=b1f+b2f2+⋯+bmfm=0.

Thus bmanm=0, since it is the leading coefficient of f∘g=0. Also, from an∉Nil(R) yields anm∉Nil(R), and so bm∈annR0[x](anm)=Nil(R). Hence bm∈Nil(R), which implies that bmf=0, since annR(ai)=Nil(R). Thus f∘g=b1f+b2f2+⋯+bm−1fm−1=0. Continuing this process, we see that bj∈Nil(R) for each 1≤j≤m−1. Hence g is a nilpotent element of R0[x], and so r.annR0[x](f)⊆Nil(R)0[x]. Now assume that g∈l.annR0[x](f). Thus g∘f=a1g+a2g2+⋯+angn=0, and so anbmn=0. This shows that bmn∈annR(an)=Nil(R), which implies that bm∈Nil(R). Hence

g∘f=a1g1+a2g12+⋯+ang1n=0,

where g1=∑ j=1 m−1bjxj. By repeating this argument, we can conclude that bj∈Nil(R) for each 1≤j≤m−1. Therefore l.annR0[x](f)⊆Nil(R)0[x]. Since annR(ai)=Nil(R) for each 1≤i≤n, then g∘f=0=f∘g for each g∈Nil(R)0[x]. Hence annR0[x](f)=Nil(R)0[x]=Nil(R0[x]). Therefore [f]=[f2]=[x2].

Case 3. Let f=f3=∑ i=1naixi. Then a1=0, by Lemma 2.6. Since rf≠0 for each 0≠r∈R, then r.annR0[x](f)=0, by Lemma 2.5. Also, if g∈l.annR0[x](f), then g is nilpotent, by Lemma 2.5. Since Nil(R)2=0, then

h∘f=a2h2+⋯+anhn=0

for each h∈Nil(R)0[x]. Therefore

annR0[x](f)=l.annR0[x](f)=Nil(R)0[x]=Nil(R0[x]).

Hence [f]=[f3]=[x2].

Case 4. Let f=f1+f2, where 0≠f1=∑ i=1naixi and 0≠f2=∑ s=1tcsxs. Suppose that g∈l.annR0[x](f). Then rg=0 for some 0≠r∈R, by Lemma 2.5. Thus Cg*⊆Z(R). Since annR(ai)=Z(R) for each ai∈Cf1*, we have g∘f=c1g+c2g2+⋯+ctgt=g∘f2=0, which implies that g∈l.annR0[x](f2). Thus l.annR0[x](f)⊆Nil(R0[x]), by Case 2. Now, assume

g=∑ j=1mbjxj∈r.annR0[x](f).

Since f is not nilpotent, then Cg*⊆Z(R), by Lemma 2.5. Hence

0=f∘g=b1f+b2f2+⋯+bmfm=b1f2+b2f22+⋯+bmf2m=f2∘g,

which implies that g∈r.annR0[x](f2), and so g∈Nil(R0[x]), by Case 2. Since annR(ai)=Z(R) for each ai∈Cf1* and annR(cs)=Nil(R) for each cs∈Cf2*, then l.annR0[x](f)=r.annR0[x](f)=Nil(R0[x]). Hence annR0[x](f)=Nil(R0[x]). Therefore [f]=[f1+f2]=[x2].

Case 5. Let f=f1+f3, where 0≠f1=∑ i=1naixi and 0≠f3=∑ s=1tcsxs. Then a1+c1 is the coefficient of x in f. By Lemma 2.6, we have a1+c1∈Z(R). Thus c1=0, since a1∈Z(R) and Z(R)=annR(a) for some a ∈ R . Hence deg(f3)≥2. Similar to Case 3, we can conclude that r.annR0[x](f)=0. On the other hand, if g∘f=0 for some g∈R0[x], then g is nilpotent, by Lemma 2.5. Hence l.annR0[x](f)⊆Nil(R0[x]). Since Nil(R)2=0 and annR(ai)=Z(R) for each ai∈Cf1*, then g∘f=0 for each g∈Nil(R0[x]). Therefore we have annR0[x](f)=l.annR0[x](f)=Nil(R0[x]). Thus [f]=[f1+f3]=[x2].

Case 6. Let f=f2+f3, where fi≠0 for each i∈{2,3}. Since deg(f3)≥2 and annR(ai)=Nil(R) for each ai∈Cf2*, then by a similar way as used in Case 5 one can show that annR0[x](f)=l.annR0[x](f)=Nil(R0[x]). Hence [f]=[x2].

Case 7. Let f=f1+f2+f3, where fi≠0 for each i∈{1,2,3}. Since Cf3*⊆R∖Z(R), then r.annR0[x](f)=0 and l.annR0[x](f)⊆Nil(R0[x]), by Lemma 2.5. Hence annR0[x](f)=Nil(R0[x]), and so [f]=[f1+f2+f3]=[x2].

Therefore |ΓE(R0[x])|=2, and thus diam(ΓE(R0[x]))=1.

Corollary 2.10. Let R be a non-reduced commutative ring with Z(R)≠0. If Z(R)2=0, then diam(ΓE(R0[x]))=1.

From [1, Theorems 2.7 and 2.9], we immediately deduce the following result.

Proposition 2.11. Let R be a non-reduced commutative ring. Then there exist a,b∈Z(R) with annR({a,b})∩Nil(R)=0 if and only if diam(Γ(R0[x]))=3

Lemma 2.12. Let R be a commutative ring and a,b∈R. If

annR({a,b})∩Nil(R)=0,

then annR({ak,bs})∩Nil(R)=0 for each positive integer k,s with ak≠0≠bs.

Proof. Let ak≠0 for some positive integer k. On the contrary, assume that and 0≠t∈annR({ak,b})∩Nil(R). Then tak=0=tb. Hence there exists 1≤r≤k−1 such that tar≠0 but tar+1=0. Thus tar∈annR({a,b})∩Nil(R), which is a contradiction. Now suppose bs≠0 for some positive integer s. Put a′=ak≠0. Hence annR({a′,b})∩Nil(R)=0, and so by a similar way as used above, annR({a′,bs})∩Nil(R)=0, as desired.

Theorem 2.13. Let R be a non-reduced commutative ring. Then diam(Γ(R0[x]))=3 if and only if diam(ΓE(R0[x]))=3.

Proof. (⇒) Let diam(Γ(R0[x]))=3. Then there exist a,b∈Z(R), such that annR({a,b})∩Nil(R)=0, by Proposition 2.11. Notice that if a or b∈Nil(R) and ab=0, then annR({a,b})∩Nil(R)≠0, which is a contradiction. Hence we consider the following cases:

Case 1. Let a,b∉Nil(R). Since annR({a,b})∩Nil(R)=0, then either there exists c∈Nil(R) such that ca=0 but cb≠0 or for each c∈Nil(R), ca≠0≠cb.

First assume ca=0 but cb≠0 for some c∈Nil(R). There exists a positive integer k such that ck=0. Hence ax+xk,bx∈Z(R0[x]). Since

cx∈annR0[x](ax+xk)

but cx∉annR0[x](bx), then [ax+xk]≠[bx]. Also, bx∘(ax+xk)≠0≠(ax+xk)∘bx. Since for each 0≠r∈R, r(ax+xk)≠0, then

annR0[x](ax+xk)=l.annR0[x](ax+xk)⊆Nil(R0[x]),

by Lemma 2.5. Suppose that g=∑ i=sncixi∈annR0[x](ax+xk)∩annR0[x](bx) and cs≠0. Then g∘(ax+xk)=0 and either g∘bx=0 or bx∘g=0. Hence ci∈Nil(R) for each i and acs=0. If g∘bx=0, then bcs=0, which implies that cs∈annR({a,b})∩Nil(R), a contradiction. If 0=bx∘g=csbsxs+⋯+cnbnxn, then csbs=0. Since b∉Nil(R), then bs≠0. Hence cs∈annR({a,bs})∩Nil(R), which is a contradiction by Lemma 2.12. Thus bx and ax+xk have not common non-zero annihilator, and so d([ax+xk],[bx])≥3. Therefore diam(ΓE(R0[x]))=3.

Now assume for each c′∈Nil(R), c′a≠0≠c′b. Since R is not reduced, there exists c∈R such that c2=0. Thus cb≠0 and cbx+x2∈Z(R0[x]). Hence [cbx+x2]≠[ax], since cx∈annR0[x](cbx+x2)∖annR0[x](ax). Obviously, (cbx+x2)∘ax≠0≠ax∘(cbx+x2). By Lemma 2.5, we have

annR0[x](cbx+x2)=l.annR0[x](cbx+x2)⊆Nil(R0[x]).

Let g=∑ i=sncixi∈annR0[x](cbx+x2)∩annR0[x](ax) and cs≠0. Hence either g∘ax=0 or ax∘g=0. If g∘ax=0, then acs=0, which is a contradiction. If ax∘g=0, then csas=0. Since as≠0, there exists 1≤t≤s−1 such that csat≠0 but csat+1=0. Hence csat∈annR(a)∩Nil(R), which is a contradiction. Therefore d([cbx+x2],[ax])≥3, and so diam(ΓE(R0[x]))=3.

Case 2. Let a∈Nil(R), b∉Nil(R) and ab≠0. Hence there exists a positive integer k such that ak=0 but ak−1≠0. Thus ak−1x+xk,bx∈Z(R0[x]). Since ax∈annR0[x](ak−1x+xk)∖annR0[x](bx), then [bx]≠[ak−1x+xk]. Moreover, bx∘(ak−1x+xk)≠0≠(ak−1x+xk)∘bx. Let

g∈annR0[x](ak−1x+xk)∩annR0[x](bx).

Hence g=∑ i=sncixi with cs≠0 is nilpotent, since

annR0[x](ak−1x+xk)=l.annR0[x](ak−1x+xk)⊆Nil(R0[x]).

From g∘(ak−1x+xk)=0 yields ak−1cs=0. On the other hand, if g∘bx=0, then bcs=0. Therefore 0≠cs∈annR({ak−1,b})∩Nil(R), which is a contradiction by Lemma 2.12. Now assume that bx∘g=0. Then csbs=0. Since b∉Nil(R), then bs≠0. Thus 0≠cs∈annR({ak−1,bs})∩Nil(R), which is a contradiction by Lemma 2.12. Hence d([ak−1x+xk],[bx])≥3, and so the result follows.

Case 3. Let a,b∈Nil(R) and ab≠0. Then there exist positive integers t,k such that ak=bt=0 but ak−1≠0≠bt−1. Therefore

ak−1x+xk,bt−1x+xt∈Z(R0[x]).

Notice that (ak−1x+xk)∘(bt−1x+xt)≠0≠(bt−1x+xt)∘(ak−1x+xk). Moreover,

annR0[x](ak−1x+xk)=l.annR0[x](ak−1x+xk)⊆Nil(R0[x])

and annR0[x](bt−1x+xt)=l.annR0[x](bt−1x+xt). Also, if ax∈annR0[x](bt−1x+xt), then ax∘(bt−1x+xt)=0, and so a∈annR({ak−1,bt−1})∩Nil(R), which is a contradiction by Lemma 2.12. Hence

ax∈annR0[x](ak−1x+xk)∖annR0[x](bt−1x+xt),

and so [ak−1x+xk]≠[bt−1x+xt]. Let

g=∑ i=sncixi∈annR0[x](ak−1x+xk)∩annR0[x](bt−1x+xt), cs≠0.

Hence g°(ak−1x+xk)=0=g∘(bt−1x+xt). Therefore

0≠cs∈annR({ak−1,bt−1})∩Nil(R),

which is a contradiction by Lemma 2.12. Hence d([ak−1x+xk],[bt−1x+xt])≥3, as wanted.

(⇐) Let diam(ΓE(R0[x]))=3. Since diam(ΓE(R0[x]))≤diam(Γ(R0[x]))≤3, then the result follows.

By using Theorems 2.9 and 2.13, we can determine when diam(ΓE(R0[x]))=2.

Theorem 2.14. Let R be a non-reduced commutative ring with Z(R)≠0. Then diam(ΓE(R0[x]))=2 if and only if annR({a,b})∩Nil(R)≠0 for each a,b∈Z(R) and one of the following conditions holds:

• (1) ∣ΓE(R)∣≥3.
  

• (2) Z(R)≠annR(c) for each c ∈ R .

• (3) Nil(R)² ≠ 0 .

• (4) There exists 0≠c∈Z(R)∖Nil(R) such that annR(c)≠Nil(R).

Proof. (⇒) By Theorem 2.13, we have diam(Γ(R0[x]))=2. It follows that annR({a,b})∩Nil(R)≠0 for each a,b∈Z(R), by [1, Theorem 2.9], Since diam(ΓE(R0[x]))=2, then the result follows from Theorem 2.9.

(⇐) Since annR({a,b})∩Nil(R)≠0 for each a,b∈Z(R), we have diam(Γ(R0[x]))=2, by [1, Theorem 2.9]. Hence diam(ΓE(R0[x]))∈{1,2}, since diam(ΓE(R0[x]))≤diam(Γ(R0[x])). On the other hand, if one of the conditions (1)-(4) holds, then diam(ΓE(R0[x]))≠1, by Theorem 2.9, and so the result follows.