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Kyungpook Mathematical Journal 2022; 62(3): 437-453

Published online September 30, 2022 https://doi.org/10.5666/KMJ.2022.62.3.437

Copyright © Kyungpook Mathematical Journal.

An Alternative Perspective of Near-rings of Polynomials and Power series

Fatemeh Shokuhifar, Ebrahim Hashemi∗ and Abdollah Alhevaz

Faculty of Mathematical Sciences, Shahrood University of Technology, Shahrood 316-3619995161, Iran
e-mail : shokuhi.135@gmail.com, eb_hashemi@yahoo.com or eb_hashemi@shahroodut.ac.ir and a.alhevaz@gmail.com or a.alhevaz@shahroodut.ac.ir

Received: June 26, 2021; Revised: January 2, 2022; Accepted: January 13, 2022

Unlike for polynomial rings, the notion of multiplication for the near-ring of polynomials is the substitution operation. This leads to somewhat surprising results. Let S be an abelian left near-ring with identity. The relation ∼ on S defined by letting a~b if and only if annS(a)=annS(b), is an equivalence relation. The compressed zero-divisor graph ΓE(S) of S is the undirected graph whose vertices are the equivalence classes induced by ∼ on S other than [0]S and [1]S, in which two distinct vertices [a]S and [b]S are adjacent if and only if ab=0 or ba=0. In this paper, we are interested in studying the compressed zero-divisor graphs of the zero-symmetric near-ring of polynomials R0[x] and the near-ring of the power series R0[[x]] over a commutative ring R. Also, we give a complete characterization of the diameter of these two graphs. It is natural to try to find the relationship between diam(ΓE(R0[x])) and diam(ΓE(R0[[x]])). As a corollary, it is shown that for a reduced ring R, diam(ΓE(R))diam(ΓE(R0[x]))diam(ΓE(R0[[x]])).

Keywords: Near-ring of polynomials, Zero-divisor graph, Compressed zero-divisor graph, Diameter of graph, Near-ring of formal power series

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)=ab+ac for each a,b,cS. 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 0a=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 AS. Then annS(A)=l.annS(A)r.annS(A), where

l.annS(A)={sSsa=0 for each aA}

and r.annS(A)={sSas=0 for each aA}. 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 aS, let [a]S={bS | a~b} (for short we can use [a] instead of [a]S).

For instance, it is clear that [0]S={0} and [1]S=SZ(S), and that [a]SZ(S){0} for each aS([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 RE 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 gR[x],

gf= 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]={fR[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=0nbjxjR[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 cZ(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

2diam(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 0aZ(R). Thus ab=0 for some non-zero ba of R. If [ax]=[bx], then axannR0[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 0aR such that a2=0. Thus ax,ax+x2Z(R0[x]). Also, x2annR0[x](ax) but x2annR0[x](ax+x2), which implies that [ax][ax+x2], and so diam(ΓE(R0[x]))1.

For any fR[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 ai∈ Cf and bj ∈ Cg .

  • (2) [7, Lemma 3.4] For each f,g ∈ R0[x] , f ◦ g=0 if and only if aibj=0 for each ai∈ Cf and bj ∈ Cg .

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 aiCf and bjCg, by Lemma 2.2. Hence fxgx=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 fg=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=1naixiR0[x]annR(a1)Nil(R)0 and aiR for each  i2},

where Zl(R0[x])={fR0[x]rf=0, for some 0rR}.

Lemma 2.7. Let R be a non-reduced commutative ring and for each a,bZ(R), annR({a,b})Nil(R)0. Then diam(ΓE(R0[x]))2. Also, if there exists cNil(R) such that ck=0ck1 for some k3, 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 ck10 for some cNil(R) and k3. Since c2xxk1=0, then xk1Z(R0[x]). Also, cxxk10xk1cx. Since xkannR0[x](cx) but xkannR0[x](xk1), then [cx][xk1]. It follows that d(cx,xk1)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 fR0[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 aR, and annR(c)=Nil(R) for each cZ(R)Nil(R).

Proof. () Let diam(ΓE(R0[x]))=1. Since R is a non-reduced ring, there exists 0aR such that a2=0. Let bZ(R). If [ax]=[bx], then axannR0[x](bx), since a2=0. Thus axbx=0, and so ab=0. Also, if [ax][bx], then axbx=0, by hypothesis. Hence ab=0. Therefore Z(R)=annR(a). It follows that for each bNil(R), b2=0, by Lemma 2.7. Now assume that b,c are distinct elements of Nil(R). If [bx]=[cx], then cxannR0[x](bx), and so bc=0. If [bx][cx], then 0=bcx=bxcx, by assumption. Hence Nil(R)2=0.

Now suppose that cZ(R)Nil(R) and dannR(c). Thus [x2]=[cx], since diam(ΓE(R0[x]))=1. Hence dxannR0[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 bNil(R), since b2=0. Hence Nil(R)annR(c). Therefore Nil(R)=annR(c).

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

() We claim that for each cNil(R), annR0[x](cx)=Z(R0[x]) and annR(c)=Z(R). Since Nil(R)2=0 and cNil(R), then Nil(R)annR(c). Now assume dZ(R)Nil(R). Hence annR(d)=Nil(R), and thus cd=0. It means that annR(c)=Z(R). Now suppose that g= j=1mbjxjZ(R0[x]). Thus cxg=0, since Nil(R)2=0 and b1Z(R), by Lemma 2.6. Hence annR0[x](cx)=Z(R0[x]). On the other hand, since R is non-reduced, x2Z(R0[x]). Also, x2annR0[x](cx) but x2annR0[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 gl.annR0[x](x2), then g2=0, and so gNil(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*RZ(R). We consider the following cases:

Case 1. Let f=f1= i=1naixi and g= j=1mbjxjZ(R0[x]). Since Cf1*Nil(R), then annR(ai)=Z(R) for each 1in. Also, by Lemma 2.6, b1Z(R). Hence fg=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 1in. Suppose that g= j=1mbjxjr.annR0[x](f). It means that

fg=b1f+b2f2++bmfm=0.

Thus bmanm=0, since it is the leading coefficient of fg=0. Also, from anNil(R) yields anmNil(R), and so bmannR0[x](anm)=Nil(R). Hence bmNil(R), which implies that bmf=0, since annR(ai)=Nil(R). Thus fg=b1f+b2f2++bm1fm1=0. Continuing this process, we see that bjNil(R) for each 1jm1. Hence g is a nilpotent element of R0[x], and so r.annR0[x](f)Nil(R)0[x]. Now assume that gl.annR0[x](f). Thus gf=a1g+a2g2++angn=0, and so anbmn=0. This shows that bmnannR(an)=Nil(R), which implies that bmNil(R). Hence

gf=a1g1+a2g12++ang1n=0,

where g1= j=1 m1bjxj. By repeating this argument, we can conclude that bjNil(R) for each 1jm1. Therefore l.annR0[x](f)Nil(R)0[x]. Since annR(ai)=Nil(R) for each 1in, then gf=0=fg for each gNil(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 rf0 for each 0rR, then r.annR0[x](f)=0, by Lemma 2.5. Also, if gl.annR0[x](f), then g is nilpotent, by Lemma 2.5. Since Nil(R)2=0, then

hf=a2h2++anhn=0

for each hNil(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 0f1= i=1naixi and 0f2= s=1tcsxs. Suppose that gl.annR0[x](f). Then rg=0 for some 0rR, by Lemma 2.5. Thus Cg*Z(R). Since annR(ai)=Z(R) for each aiCf1*, we have gf=c1g+c2g2++ctgt=gf2=0, which implies that gl.annR0[x](f2). Thus l.annR0[x](f)Nil(R0[x]), by Case 2. Now, assume

g= j=1mbjxjr.annR0[x](f).

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

0=fg=b1f+b2f2++bmfm=b1f2+b2f22++bmf2m=f2g,

which implies that gr.annR0[x](f2), and so gNil(R0[x]), by Case 2. Since annR(ai)=Z(R) for each aiCf1* and annR(cs)=Nil(R) for each csCf2*, 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 0f1= i=1naixi and 0f3= s=1tcsxs. Then a1+c1 is the coefficient of x in f. By Lemma 2.6, we have a1+c1Z(R). Thus c1=0, since a1Z(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 gf=0 for some gR0[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 aiCf1*, then gf=0 for each gNil(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 fi0 for each i{2,3}. Since deg(f3)2 and annR(ai)=Nil(R) for each aiCf2*, 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 fi0 for each i{1,2,3}. Since Cf3*RZ(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,bZ(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,bR. If

annR({a,b})Nil(R)=0,

then annR({ak,bs})Nil(R)=0 for each positive integer k,s with ak0bs.

Proof. Let ak0 for some positive integer k. On the contrary, assume that and 0tannR({ak,b})Nil(R). Then tak=0=tb. Hence there exists 1rk1 such that tar0 but tar+1=0. Thus tarannR({a,b})Nil(R), which is a contradiction. Now suppose bs0 for some positive integer s. Put a=ak0. 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,bZ(R), such that annR({a,b})Nil(R)=0, by Proposition 2.11. Notice that if a or bNil(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,bNil(R). Since annR({a,b})Nil(R)=0, then either there exists cNil(R) such that ca=0 but cb0 or for each cNil(R), ca0cb.

First assume ca=0 but cb0 for some cNil(R). There exists a positive integer k such that ck=0. Hence ax+xk,bxZ(R0[x]). Since

cxannR0[x](ax+xk)

but cxannR0[x](bx), then [ax+xk][bx]. Also, bx(ax+xk)0(ax+xk)bx. Since for each 0rR, 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=sncixiannR0[x](ax+xk)annR0[x](bx) and cs0. Then g(ax+xk)=0 and either gbx=0 or bxg=0. Hence ciNil(R) for each i and acs=0. If gbx=0, then bcs=0, which implies that csannR({a,b})Nil(R), a contradiction. If 0=bxg=csbsxs++cnbnxn, then csbs=0. Since bNil(R), then bs0. Hence csannR({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 cNil(R), ca0cb. Since R is not reduced, there exists cR such that c2=0. Thus cb0 and cbx+x2Z(R0[x]). Hence [cbx+x2][ax], since cxannR0[x](cbx+x2)annR0[x](ax). Obviously, (cbx+x2)ax0ax(cbx+x2). By Lemma 2.5, we have

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

Let g= i=sncixiannR0[x](cbx+x2)annR0[x](ax) and cs0. Hence either gax=0 or axg=0. If gax=0, then acs=0, which is a contradiction. If axg=0, then csas=0. Since as0, there exists 1ts1 such that csat0 but csat+1=0. Hence csatannR(a)Nil(R), which is a contradiction. Therefore d([cbx+x2],[ax])3, and so diam(ΓE(R0[x]))=3.

Case 2. Let aNil(R), bNil(R) and ab0. Hence there exists a positive integer k such that ak=0 but ak10. Thus ak1x+xk,bxZ(R0[x]). Since axannR0[x](ak1x+xk)annR0[x](bx), then [bx][ak1x+xk]. Moreover, bx(ak1x+xk)0(ak1x+xk)bx. Let

gannR0[x](ak1x+xk)annR0[x](bx).

Hence g= i=sncixi with cs0 is nilpotent, since

annR0[x](ak1x+xk)=l.annR0[x](ak1x+xk)Nil(R0[x]).

From g(ak1x+xk)=0 yields ak1cs=0. On the other hand, if gbx=0, then bcs=0. Therefore 0csannR({ak1,b})Nil(R), which is a contradiction by Lemma 2.12. Now assume that bxg=0. Then csbs=0. Since bNil(R), then bs0. Thus 0csannR({ak1,bs})Nil(R), which is a contradiction by Lemma 2.12. Hence d([ak1x+xk],[bx])3, and so the result follows.

Case 3. Let a,bNil(R) and ab0. Then there exist positive integers t,k such that ak=bt=0 but ak10bt1. Therefore

ak1x+xk,bt1x+xtZ(R0[x]).

Notice that (ak1x+xk)(bt1x+xt)0(bt1x+xt)(ak1x+xk). Moreover,

annR0[x](ak1x+xk)=l.annR0[x](ak1x+xk)Nil(R0[x])

and annR0[x](bt1x+xt)=l.annR0[x](bt1x+xt). Also, if axannR0[x](bt1x+xt), then ax(bt1x+xt)=0, and so aannR({ak1,bt1})Nil(R), which is a contradiction by Lemma 2.12. Hence

axannR0[x](ak1x+xk)annR0[x](bt1x+xt),

and so [ak1x+xk][bt1x+xt]. Let

g= i=sncixiannR0[x](ak1x+xk)annR0[x](bt1x+xt), cs0.

Hence g°(ak1x+xk)=0=g(bt1x+xt). Therefore

0csannR({ak1,bt1})Nil(R),

which is a contradiction by Lemma 2.12. Hence d([ak1x+xk],[bt1x+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,bZ(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 0cZ(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,bZ(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,bZ(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.

We denote the collection of all power series with positive orders using the operations of addition and substitution by R0[[x]], unless specifically indicatedŁinebreak otherwise (i.e., R0[[x]] denotes (R0[[x]],+,)). Observe that the system (R0[[x]],+,) is a zero-symmetric left near-ring. For any fR0[[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}.

In this section, we characterize the diameter of the compressed zero-divisor graph of the near-ring R0[[x]].

Lemma 3.1. Let R be a reduced ring. Then

  • (1) [13, Proposition 2.3] For each f,gR[[x]], fg=0 if and only if aibj=0 for each ai ∈ Cf and bjCg.

  • (2) [6, Lemma 3.3] For each f,gR0[[x]] if and only if aibj=0 for each ai∈ Cf and bjCg.

By using Lemma 3.1 and a similar argument as used in the proof of Proposition 2.3, we can conclude the following nice fact.

Proposition 3.2. 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.

Let R be a commutative ring. For polynomials, McCoy's Theorem [19, Theorem 2] states that a polynomial fR[x] is a zero-divisor if and only if there is a non-zero element rR such that rf=0. Based on this theorem, a ring R is said to be McCoy ring if each finitely generated ideal contained in Z(R) has a non-zero annihilator [9].

Corollary 3.3. Let R be a reduced commutative ring. Then diam(Γ(R0[[x]]))=3 if and only if diam(ΓE(R[[x]]))=3.

Proof. () Let diam(Γ(R0[[x]]))=3. Then diam(Γ(R[[x]]))=3, by Lemma 3.1. Thus by [17, Theorem 4.9], one of the following cases occurs:

Case 1. R is a McCoy ring with Z(R) an ideal but there exist countably generated ideals I and J with non-zero annihilators such that I+J does not have a non-zero annihilator. Since Z(R) is an ideal, then R has more than two minimal primes. Therefore diam(ΓE(R[[x]]))=3, by [12, Theorem 4.3].

Case 2. Z(R) is an ideal and each two generated ideal contained in Z(R) has a non-zero annihilator but R is not a McCoy ring. Then R has more than two minimal primes and there exists K=a1,,anZ(R) with annR(K)=0, since R is not McCoy. Hence n3. Therefore one can easily show that there exist finitely generated ideals I and J with non-zero annihilators such that I+J does not have a non-zero annihilator. Hence diam(ΓE(R[[x]]))=3, by [12, Theorem 4.3].

Case 3. R has more than two minimal primes and there is a pair of zero-divisors a and b such that a+b=a,b does not have a non-zero annihilator. Then diam(ΓE(R[[x]]))=3, by [12, Theorem 4.3].

Therefore diam(ΓE(R0[[x]]))=3, by Proposition 3.2.

The backward direction is clear.

Corollary 3.4. Let R be a reduced commutative ring. If diam(ΓE(R0[x]))=3, then diam(ΓE(R0[[x]]))=3.

Proof. Let diam(ΓE(R0[x]))=3. Then diam(Γ(R0[x]))=3, by Corollary 2.4. Thus diam(Γ(R[x]))=3, by [1, Proposition 2.10], and so diam(Γ(R[[x]]))=3, by [17, Theorem 4.9]. Hence diam(Γ(R0[[x]]))=3, by Lemma 3.1. Therefore the result follows from Corollary 3.3.

Proposition 3.5. Let R be a reduced commutative ring. Then

diam(ΓE(R))diam(ΓE(R0[x]))diam(ΓE(R0[[x]])).

Proof. Clearly, if diam(ΓE(R))=0, then we have diam(ΓE(R))diam(ΓE(R0[x])). Also, diam(ΓE(R))=1 if and only if diam(ΓE(R[x]))=1 if and only if diam(ΓE(R0[x]))=1, by [12, Theorem 3.3] and Proposition 2.3. Therefore if diam(ΓE(R))=2, then diam(ΓE(R0[x]))2.

Finally, if diam(ΓE(R))=3, then diam(ΓE(R[x]))=3, by [12][Theorem 4.4]. Hence diam(ΓE(R0[x]))=3, by Proposition 2.3.

Obviously, diam(ΓE(R0[x]))diam(ΓE(R0[[x]])), if diam(ΓE(R0[x]))=1. Now assume that diam(ΓE(R0[x]))=2. Then there exist f,gZ(R0[x]) with d([f]R0[x],[g]R0[x])=2. On the contrary, suppose that diam(ΓE(R0[[x]]))=1. Since d([f]R0[x],[g]R0[x])=2, we have fg0. Therefore [f]R0[[x]]=[g]R0[[x]], which implies that [f]R0[x]=R0[x][f]R0[[x]]=R0[x][g]R0[[x]]=[g]R0[x], a contradiction. Hence diam(ΓE(R0[x]))diam(ΓE(R0[[x]])), by Corollary 3.4.

The following lemmas play an important role in proving Theorem 3.10.

Lemma 3.6. ([10, Corollary 1]) Let R be a commutative Noetherian ring. Then Nil(R[[x]])=Nil(R)[[x]].

For each fR0[x] and positive integer n, we write

f(n)= fffn.

Lemma 3.7. Let R be a commutative Noetherian ring. Then

Nil(R0[[x]])=Nil(R)0[[x]].

Proof. First, Suppose that f= r=1arxrNil(R0[[x]]). Then there exists a positive integer n such that f(n)=0. We show that for each ai1,ai2,,ainCf, we have ai1ai2ainNil(R), which implies that arNil(R) for each arCf, as wanted.

We use induction on n. Assume that n=2 and R¯=R/Nil(R). Since 0=ffNil(R)0[[x]], then f¯f¯=0¯ in R¯0[[x]]. By Lemma 3.1, we have a¯ia¯j=0¯ for each a¯i,a¯jCf¯, since R¯ is a reduced ring. Thus aiajNil(R) for each i,j. Now suppose that n>2. Let g=f(n1). Thus fgNil(R)0[[x]]. By a similar argument as used above, we have aragNil(R), where agCg and arCf. Therefore for each ai1Cf,

gai1x=f(n1)ai1x=f(n2)(fai1x)=f(n2)(ai1f)Nil(R)0[[x]].

By induction, we have ai2ai3ai1ainNil(R), where aijCf for each j and the coefficients of ai1f are ai1ain. Therefore arNil(R) for each arCf.

Now assume that fNil(R)0[[x]]. Since R is Noetherian, there exists a positive integer k such that Nil(R)k=0. It follows that Cfk=0. Since for each n1, the coefficient of xn in f(k) is a sum of such elements ai1ai2ail, where aijCf and lk, then we have f(k)=0. Hence fNil(R0[[x]]).

Lemma 3.8. Let R be a commutative ring. If f= i=1aixi is a zero-divisor of R0[[x]], then a1Z(R).

Proof. Let a10. Since fZ(R0[[x]]), then there exists g= i=1bixiR0[[x]] such that fg=0 or gf=0. Let bk be the first non-zero coefficient of g. Assume that fg=0. Then bka1k=0. Hence there exists 1tk1 such that bka1t0 but bka1t+1=0, which implies that a1Z(R). On the other hand, if gf=0, then a1bk=0, and so the result follows.

Lemma 3.9. Let R be a Noetherian commutative ring and f= i=1aixi and g= j=1bjxj be non-zero elements of the near-ring R0[[x]]. If fg=0, then

  • (1) rf=0 for some non-zero r ∈ R .

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

Proof. (1) Let bk be the first non-zero coefficient of g. Since fg=0, we have bkfk+bk+1fk+1+=0. Hence (bk+bk+1f+)fk=0. If fk=0, then there exists 1tk1 such that ft0=ft+1. Therefore rf=0 for some 0rR, by McCoy's Theorem. Thus assume that fk0. Since 0bk+bk+1f+, then the result follows by McCoy's Theorem.

(2) Notice that Cg=b1,,bn for some n1, since R is Noetherian. Suppose that f is not nilpotent. Thus there exists a=ai such that aNil(R), by Lemma 3.6. Let R¯=R/Nil(R). Since fg=0, then f¯g¯=0¯ in the near-ring R¯0[[x]]. Since R¯ is a reduced ring, it follows that a¯ib¯j=0¯, by Lemma 3.1. Since R is Noetherian, then Nil(R) is nilpotent, and so Nil(R)k=0 for some positive integer k. Thus akbjk=0 for each j1. Hence there exist integers 0tjk such that akbjtj0 but akbjtj+1=0 for each j1. Therefore there exist integers 0sjtj such that akb1s1b2s2bnsn0 but akb1s1b2s2bnsnbj=0 for each 1jn. Let s=akb1s1b2s2bnsn. Thus sg=0, since Cg=b1,,bn.

Theorem 3.10. Let R be a non-reduced commutative ring. Then

  • (1) If R is Noetherian and diam(ΓE(R0[x]))=1, then diam(ΓE(R0[[x]]))=1.

  • (2) If diam(ΓE(R0[[x]]))=1, then diam(ΓE(R0[x]))=1.

Proof. (1) Let diam(ΓE(R0[x]))=1. Then |ΓE(R)|2, Z(R)=annR(a) for some aR, Nil(R)2=0, and annR(c)=Nil(R) for each cZ(R)Nil(R), by Theorem 2.9. As shown in the proof of Theorem 2.9, for each cNil(R), annR(c)=Z(R). Assume cNil(R) and g= j=1bjxjZ(R0[[x]]). Since Nil(R)2=0, then cxg=0, by Lemma 3.8. Thus annR0[[x]](cx)=Z(R0[[x]]). It is clear that r.annR0[[x]](x2)=0. Also, we have l.annR0[[x]](x2)Nil(R)0[[x]], by Lemmas 3.7 and 3.9. Hence annR0[[x]](x2)=l.annR0[[x]](x2)=Nil(R0[[x]]), since Nil(R)2=0 and Nil(R0[[x]])=Nil(R)0[[x]]. Notice that [cx][x2], since x2annR0[[x]](cx)annR0[[x]](x2). Now suppose that 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*RZ(R).

Assume f=f1= i=1aixi and g= j=1bjxjZ(R0[[x]]). Hence we have annR(ai)=Z(R) for each aiCf*, since Cf*Nil(R). Thus fg=0, since Nil(R)2=0 and b1Z(R), by Lemma 3.8. Therefore annR0[[x]](f)=Z(R0[[x]]), which implies that [f]=[cx].

Suppose that f=f2= i=qaixi and aq0. Since Cf*Z(R)Nil(R), then annR(ai)=Nil(R) for each aiCf*. Hence for each gNil(R)0[[x]], fg=0 and gf=0. Let g= j=1bjxjr.annR0[[x]](f). Thus f°g= j=1bjfj=0. Assume that bt is the first non-zero coefficient of g. Then btannR(aqt)=Nil(R), since btaqt=0 and aqtNil(R). Hence btf=0, and so fg= j=t+1bjfj=0. By repeating this argument, one can deduce that bjNil(R) for each bjCg*. Thus gNil(R)0[[x]], and so r.annR0[[x]](f)=Nil(R)0[[x]].

Now suppose that g= j=tbjxjl.annR0[[x]](f), where bt0. Therefore

gf= i=qaigi=0

which implies that aqbtq=0. Hence btqannR(aq)=Nil(R), and so btNil(R). Then btai=0 for each aiCf*, and thus gf= i=qaig1i=0, where g1= j=t+1bjxj. Continuing this process one can show that bjNil(R) for each bjCg*, and so l.annR0[[x]](f)Nil(R)0[[x]]. Hence annR0[[x]](f)=Nil(R)0[[x]]. Therefore [f]=[f2]=[x2].

If f=f3 or f=f1+f2 (f10f2) or f=f1+f3 (f10f3) or f=f2+f3 (f20f3) or f=f1+f2+f3 (each fi be non-zero), then by using Lemmas 3.7, 3.9 and a similar argument as used in the proof of Theorem 2.9, one can show that [f]=[x2]=Nil(R)0[[x]]. Hence |ΓE(R0[[x]])|=2, and thus diam(ΓE(R0[[x]]))=1.

(2) It is clear.

Proposition 3.11. Let R be a non-reduced commutative ring. Then

  • (1) If diam(Γ(R0[[x]]))=3, then annR({a,b})Nil(R)=0 for some a,bZ(R).

  • (2) Let R be a Noetherian ring. If annR({a,b})Nil(R)=0 for some a,bZ(R), then diam(Γ(R0[[x]]))=3.

Proof. (1) Since diam(Γ(R0[[x]]))=3, then there exist f,gR0[[x]] such that d(f,g)=3. Let at and bq be the first non-zero coefficients of f and g, respectively. On the contrary, suppose that annR({a,b})Nil(R)0 for each a,bZ(R). By Lemma 3.8, we have at,bqZ(R). Hence there exists cNil(R) such that cat=bqc=0. Let cr=0cr1 for some positive integer r. Therefore fcr1xg is a path in Γ(R0[[x]]), which is a contradiction.

(2) Since R is non-reduced, there exists cR such that c2=0. It follows that x2,x3Z(R0[[x]]) and x2x30x3x2. Thus d(x2,x3)2, and so diam(Γ(R0[[x]]))2. On the contrary, suppose that diam(Γ(R0[[x]]))3. Therefore diam(Γ(R0[[x]]))=2, by [8, Theorem 2.2]. Let a,bZ(R). We show that ax+x2,bx+x2Z(R0[[x]]). If ak10=ak for some positive integer k, then ak1x(ax+x2)=0. Thus assume that aNil(R). Since ax,x2Z(R0[[x]]) and axx20x2ax, then there exists a non-zero nilpotent element f= i=rcixi with cr0 such that axfx2 is a path. If fax=0, then acr=0. By Lemma 3.6, we have crk10=crk for some positive integer k. Therefore crk1x(ax+x2)=0. If axf=0, then crar=0. Hence there exists 1tr1 such that crat0=crat+1, and so cratx(ax+x2)=0. Similarly, we have bx+x2Z(R0[[x]]). Since diam(Γ(R0[[x]]))=2 and

(ax+x2)(bx+x2)0(bx+x2)(ax+x2),

then g(ax+x2)=0=g(bx+x2) for some non-zero nilpotent element g, by Lemma 3.9. Let s be the first non-zero coefficient of g. Therefore sannR({a,b})Nil(R), which is a contradiction.

Corollary 3.12. Let R be a non-reduced commutative ring. Then

  • (1) If R is Noetherian and diam(Γ(R0[x]))=3, then diam(Γ(R0[[x]]))=3.

  • (2) If diam(Γ(R0[[x]]))=3, then diam(Γ(R0[x]))=3.

Proof. It follows from Propositions 2.11 and 3.11.

Theorem 3.13. Let R be a non-reduced commutative ring. Then

  • (1) If R is Noetherian and diam(Γ(R0[[x]]))=3, then diam(ΓE(R0[[x]]))=3.

  • (2) If diam(ΓE(R0[[x]]))=3, then diam(Γ(R0[[x]]))=3.

Proof. (1) By using Lemmas 3.7, 3.9, Proposition 3.11 and a similar argument as used in the proof of Theorem 2.13, one can prove it.

(2) It is clear.

Corollary 3.14. Let R be a non-reduced commutative ring. Then

  • (1) If R is Noetherian and diam(ΓE(R0[x]))=3, then diam(ΓE(R0[[x]]))=3.

  • (2) If diam(ΓE(R0[[x]]))=3, then diam(ΓE(R0[x]))=3.

Proof. It follows from Theorems 2.13, 3.13 and Corollary 3.12.

Proposition 3.15. Let R be a non-reduced commutative ring. Then

  • (1) If diam(ΓE(R0[x]))=2, then diam(ΓE(R0[[x]]))=2.

  • (2) If R is Noetherian and diam(ΓE(R0[[x]]))=2, then diam(ΓE(R0[x]))=2

Proof. This follows from Theorem 3.10 and Corollary 3.14.

Proposition 3.16. Let R be a non-reduced Noetherian commutative ring. If Z(R)annR(a) for each aR, then

diam(ΓE(R))diam(ΓE(R0[x]))diam(ΓE(R0[[x]])).

Proof. Clearly, diam(ΓE(R))diam(ΓE(R0[x])), if diam(ΓE(R)){0,1}. Hence suppose that diam(ΓE(R))=2. Then ΓE(R))3, which implies that diam(ΓE(R0[x]))2, by Theorem 2.9.

Now assume that diam(ΓE(R))=3. Notice that diam(ΓE(R0[x]))2, by Theorem 2.9. On the contrary, suppose that diam(ΓE(R0[x]))=2. Thus Z(R) is an ideal and each pair of zero-divisors has a non-zero annihilator, by Theorem 2.14. Since Z(R)annR(a) for every aZ(R), then diam(ΓE(R))=2, by [12, Theorem 2.3], which is a contradiction. Hence diam(ΓE(R))diam(ΓE(R0[x])).

Also, by Corollary 3.14 and Proposition 3.15, we have

diam(ΓE(R0[x]))diam(ΓE(R0[[x]])).
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