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ZERO-DIVISOR GRAPHS OF DIRECT PRODUCTS OF

COMMUTATIVE RINGS

M. AXTELL, J. STICKLES, AND J. WARFEL

Abstract. We recall several results of zero divisor graphs of commutative

rings. We then examine the preservation of diameter and girth of the zero

divisor graph of direct products of commutative rings.

1. Introduction

The concept of the graph of the zero-divisors of a ring was …rst introduced

by Beck in [5] when discussing the coloring of a commutative ring. In his work

all elements of the ring were vertices of the graph. D.D. Anderson and Naseer

used this same concept in [1]. We adopt the approach used by D.F. Anderson

and Livingston in [2] and consider only nonzero zero-divisors as vertices of

the graph. D.F. Anderson and Livingston, Mulay in [9], and DeMeyer and

Schneider in [6] examined, among other things, the diameter and girth of

the zero-divisor graph of a commutative ring. For instance, Anderson and

Livingston showed the zero-divisor graph of a commutative ring is connected

with diameter less than or equal to three, and DeMeyer and Schneider, and

Mulay showed (independently) the girth is either in…nite or less than or equal

to four. Axtell, Coykendall, and Stickles examined the preservation, or lack

thereof, of the diameter and girth of the graph of a commutative ring under

extensions to polynomial and power series rings in [3], and Axtell and Stickles

examined these same preservations in idealizations of commutative rings in

[4]. In this paper we completely characterize the girth and diameter of the

zero-divisor graph of a direct product.

For the sake of completeness, we state some de…nitions and notations used

throughout. All rings are assumed to be commutative and not necessarily with

identity. We will use Rto denote a ring and Dto denote an integral domain.

We use Z(R)to denote the set of zero-divisors of R; we use Z(R)to denote

the set of nonzero zero-divisors of R. We will use reg(R)to denote the regular

elements of R; i.e. reg(R) = RrZ(R):By the zero-divisor graph of R,

denoted (R);we mean the graph whose vertices are the nonzero zero-divisors

of R; and for distinct r; s 2Z(R), there is an edge connecting rand sif and

only if rs = 0. For two distinct vertices aand bin a graph , the distance

between aand b; denoted d(a; b);is the length of the shortest path connecting a

and b; if such a path exists; otherwise, d(a; b) = 1. The diameter of a graph

1991 Mathematics Subject Classi…cation. 13A99.

1

ZERO-DIVISOR GRAPHS OF DIRECT PRODUCTS OF COMMUTATIVE RINGS 2

is diam() = sup fd(a; b)jaand bare distinct vertices of g. We will use the

notation diam((R)) to denote the diameter of the graph of Z(R). The girth

of a graph , denoted g(), is the length of the shortest cycle in ;provided

contains a cycle; otherwise, g() = 1. We will use the notation g((R))

to denote the girth of the graph of Z(R). A graph is said to be connected if

there exists a path between any two distinct vertices, and a graph is complete

if it is connected with diameter one. We use the notation Ato refer to the

nonzero elements of A.

2. Some Preliminaries and a Brief Excursion

Before starting to classify the diameter of a direct product of rings, We will

recall and develop some tools that will be used in many results. The following

is from [2, Theorem 2.8]:

Theorem 2.1. Let Rbe a commutative ring with identity. Then (R)is

complete if and only if either R

=Z2Z2;or xy = 0 for all x; y 2Z(R).

This theorem requires the presence of a multiplicative identity. This restric-

tion will prove overly restrictive in the classi…cation of the diameters of direct

products, and so we present some results similar to those found in [2] for rings

that do not necessarily have identity.

Lemma 2.2. Let Rbe a commutative ring. If R=Z(R)and diam((R)) = 1;

then R2= 0:

Proof. Assume that a26= 0 for some a2R: If R=f0; agthen we have

R6=Z(R):Hence, there exists an element b2Rwith b6=a: Observe that

a+b6=a. Since R=Z(R)and diam((R)) = 1;we have ab = 0:So,

0 = a(a+b) = a2+ab =a2;a contradiction.

The previous lemma will be used extensively in place of Theorem 2.1 for

rings that may not have identity. The next result will also prove useful in

examining commutative rings not necessarily with identity with zero-divisor

graphs having diameter two.

Lemma 2.3. Let Rbe a commutative ring such that diam((R)) = 2:Suppose

Z(R)is a (not necessarily proper) subring of R. Then for all x; y 2Z(R),

there exists a nonzero zsuch that xz =yz = 0.

Proof. Let x; y 2Z(R):If x= 0,y= 0;or x=y, we are done. So, assume x

and yare distinct and nonzero. Since diam((R)) = 2, whenever xy 6= 0 there

exists z2Z(R)such that xz =yz = 0:Thus assume xy = 0:If x2= 0 (resp.

y2= 0);then z=x(resp. z=y) yields the desired element. So, suppose

x2; y26= 0. Let X0=fx02Z(R) : xx0= 0gand Y0=fy02Z(R) : yy0= 0g:

Observe that x2Y0and y2X0;hence X0and Y0are nonempty. If X0\Y06=;;

choose z2X0\Y0:Suppose X0\Y0=;and consider x+y: Clearly x+y6=x

and x+y6=y: Also if x+y= 0;then x2= 0. Hence x+y6= 0. Since Z(R)is

ZERO-DIVISOR GRAPHS OF DIRECT PRODUCTS OF COMMUTATIVE RINGS 3

a subring, we have x+y2Z(R). Since x2; y26= 0;we see that x+y =2X0and

x+y =2Y0:Since diam((R)) = 2;there exists w2X0such that the following

path exists: xw(x+y):Then 0 = w(x+y) = wx +wy =wy and so

w2Y0;a contradiction.

This excursion into zero-divisor graphs of diameter two leads to results that

are interesting in and of themselves, most of which relate to (R)being com-

plete bipartite, or nearly so. Recall that a graph Gis complete bipartite if it

can be partitioned into two disjoint, nonempty vertex sets Pand Qsuch that

two vertices pand qare connected by an edge if and only if, without loss of

generality, p2Pand q2Q. The following lemma considers when (R)has a

complete bipartite subgraph induced by removing only edges from (R).

Lemma 2.4. Let (R)be the zero-divisor graph of a commutative ring R. If

(R)is not complete bipartite but has a complete bipartite subgraph induced

by removing only edges from (R);then Z(R)is a subring of R.

Proof. Let a; b 2Z(R). Clearly ab 2Z(R)and a2Z(R):We need to

show a+b2Z(R). Since it is possible to form a complete bipartite graph

by removing edges from (R), there must exist nonempty sets Pand Qsuch

that P[Q=Z(R),P\Q=;, and pq = 0 for all p2P,q2Q. If a,b2P,

then for any q2Qwe have q(a+b) = qa +qb = 0 + 0 = 0. So, a+b2Z(R).

A similar argument works when a; b 2Q. Without loss of generality, assume

a2Pand b2Q. Since (R)is not complete bipartite, there must be an

additional edge that connects, without loss of generality, two vertices of P;

let it lie between p1; p22P. Then, p1(b+p2) = p1b+p1p2= 0. There are

three possibilities: b+p2= 0,b+p22Q; or b+p22P. If b+p2= 0,

then 0 = b(b+p2) = b2+bp2=b2. So, b(a+b) = 0:If b+p22Q, then

0 = a(b+p2) = ab +ap2=ap2:So, p2(a+b) = 0:If b+p22P; for any q2Q

we have q(b+p2)=0and 0 = q(b+p2) = qb +qp2=qb: Thus, q(a+b)=0.

Therefore a+b2Z(R):

Using the lemmas above, it is possible to prove the following theorem that is

nearly an analogue of Theorem 2.8 from [2] for zero-divisor graphs of diameter

two.

Theorem 2.5. Let Rbe a commutative ring. If (R)is not complete bipartite

but has a complete bipartite subgraph induced by removing only edges from

(R), then for all x; y 2Z(R)there exists z2Z(R)such that xz =yz = 0.

Proof. By Lemma 2.4, Z(R)is a subring of R. Thus Z(R)is a ring with

Z(R) = Z(Z(R)):If diam((Z(R))) = 1 then by Lemma 2.2, (Z(R))2= 0

and the result is trivial. If diam((Z(R))) = 2 then Lemma 2.3 yields the

desired result.

This result could be used to give a general description of all zero-divisor

graphs of diameter two if the following conjecture could be proven.

ZERO-DIVISOR GRAPHS OF DIRECT PRODUCTS OF COMMUTATIVE RINGS 4

Conjecture 2.6. Let Gbe a connected graph with diameter two. If Gdoes

not have a complete bipartite subgraph induced by removing only edges from

G, then G6= (R)for any commutative ring R.

3. Diameter and Direct Products

We are now ready to turn our attention toward classifying the diameters of

zero-divisor graphs of direct products of commutative rings. The reader can

quickly check that diam ( (Z2Z2)) = 1;and for domains D1and D2not

both Z2;we get diam ( (D1D2)) = 2:Thus we need examine only rings of

the form RDwhere Ris not a domain, and R1R2where R1and R2are

not domains. We begin by examining the diameter of RDwhere Ris not

a domain.

Lemma 3.1. Let R1and R2be rings, at least one of which has nonzero zero

divisors, so that reg (R1)6=;and reg (R2)6=;. Then diam ( (R1R2)) = 3.

Proof. Without loss of generality, let a2Z(R1). Then, there exists b2

Z(R1)such that ab = 0. Let r12reg (R1); r22reg (R2). Then, (r1;0)

(0; r2)(a; 0) (b; r2)is a path of length 3. Since (r1;0) is annihilated only by

an element of the form (0; x)and (b; r2)is annihilated only by an element of

the form (y; 0) with yb = 0, there is no path of length 2 from (r1;0) to (b; r2).

Hence, diam( (R1R2)) = 3.

Corollary 3.2. Let Dbe a domain and Ra ring with nonzero zero divisors.

If reg(R)6=;;then diam ( (RD)) = 3.

The next theorem accounts for the remaining RDsituation.

Theorem 3.3. Let Dbe a domain and Ra ring with R=Z(R).

(i) If diam ( (R)) 2;then diam ( (RD)) = 2:

(ii) If diam ( (R)) = 3;then diam ( (RD)) = 3:

Proof. Since R=Z(R);we have Z(RD) = RD: If a2Z(R);we see

d((0;1);(a; 1)) 2:If diam ( (R)) 2;then for (r1; d1);(r2; d2)2RDwe

have either (r1; d1)(r2; d2) = (0;0);or for some z2Rwe get (r1; d1)(z; 0) =

(0;0) and (r2; d2)(z; 0) = (0;0) using Lemma 2.3 in the diameter two case.

If diam ( (R)) = 3 with d(a1; a2)=3in (R), then for b2Dwe have

d((a1; b);(a2; b)) = 3:

For the remainder of the section, we assume that R1and R2are rings not

necessarily with identity such that Z(R1)and Z(R2)are nonempty.

Theorem 3.4. Let R1and R2be rings with diam((R1)) = 1 = diam((R2)).

Then:

(i) diam ( (R1R2)) = 1 if and only if R2

1= 0 = R2

2:

(ii) diam ( (R1R2)) = 2 if and only if (without loss of generality) R2

1=

0and R2

26= 0:

(iii) diam ( (R1R2)) = 3 if and only if R2

1; R2

26= 0:

ZERO-DIVISOR GRAPHS OF DIRECT PRODUCTS OF COMMUTATIVE RINGS 5

Proof. (i) ()) If without loss of generality R2

16= 0;then ab 6= 0 for some a; b 2

R1. Let c2Z(R2). We have (b; c);(a; 0) 2Z(R1R2)and d((a; 0);(b; c)) >

1, a contradiction.

(() Clear.

(ii) ()) Assume reg(R1);reg(R2)6=;and let a2reg(R1); b 2reg(R2); c 2

Z(R1); d 2Z(R2):Then ann((a; d)) = f(0; l)jdl = 0gand ann((c; b)) =

f(m; 0) jcm = 0g:Thus, d((a; d);(c; b)) >2;a contradiction. Without loss of

generality, assume reg(R1) = ;. Thus R1=Z(R1);and R2

1= 0 by Lemma

2.2. If R2

2= 0;then diam ( (R1R2)) = 1 by (i).

(() Assume R2

1= 0 and R2

26= 0. For any (a1; b1);(a2; b2)2Z(R1R2);

at worst we have (a1; b1)(c; 0) (a2; b2):So, diam ( (R1R2)) 2. The

result then follows from (i).

(iii) Follows from (i) and (ii).

Theorem 3.5. Let R1and R2be rings such that diam((R1)) = 1 and

diam((R2)) = 2. Then:

(i) diam((R1R2)) 6= 1.

(ii) diam((R1R2)) = 2 if and only if R1=Z(R1)or R2=Z(R2).

(iii) diam((R1R2)) = 3 if and only if R16=Z(R1)and R26=Z(R2).

Proof. (i) Since diam((R2)) = 2, there exist disinct y1; y22Z(R2)with

y1y26= 0. Then (0; y1)(0; y2) = (0; y1y2)6= (0;0):Therefore diam((R1

R2)) >1.

(ii) (() Let R1=Z(R1)and diam((R1)) = 1:Thus we have R2

1= 0 by

Lemma 2.2. Let a2R

1:Since (a; 0) (x; y) = (0;0) for all (x; y)2Z(R1R2),

we have diam((R1R2)) 2:It follows from (i) that diam((R1R2)) = 2:

()) If R16=Z(R1)and R26=Z(R2);Lemma 3.1 gives diam((R1R2)) = 3.

(iii) Follows from (i) and (ii).

Theorem 3.6. Let R1and R2be rings such that diam((R1)) = 1 and

diam((R2)) = 3. Then:

(i) diam((R1R2)) 6= 1.

(ii) diam((R1R2)) = 2 if and only if R1=Z(R1).

(iii) diam((R1R2)) = 3 if and only if R16=Z(R1).

Proof. (i) Same as Theorem 3.5 (i).

(ii) (() Same as Theorem 3.5 (ii).

()) Assume that R16=Z(R1). Let m2reg(R1). Since diam((R2)) = 3,

there exist distinct y1,y22Z(R2)with y1y26= 0 and there is no y32

Z(R2)such that y1y3=y2y3= 0. Now (m; y1);(m; y2)2Z(R1R2)

and (m; y1)(m; y2)6= 0. Since diam((R1R2)) = 2;there exists (a; b)2

Z(R1R2)such that (m; y1)(a; b)=(m; y2)(a; b)=0. Since ma = 0, we

have a= 0:Also, since y1b=y2b= 0;we have b= 0;a contradiction.

(iii) By (i) and (ii).

Theorem 3.7. Let R1and R2be rings such that diam((R1)) = diam((R2)) =

2. Then:

ZERO-DIVISOR GRAPHS OF DIRECT PRODUCTS OF COMMUTATIVE RINGS 6

(i) diam((R1R2)) 6= 1.

(ii) diam((R1R2)) = 2 if and only if R1=Z(R1)or R2=Z(R2).

(iii) diam((R1R2)) = 3 if and only if R16=Z(R1)and R26=Z(R2).

Proof. (i) Same as Theorem 3.5 (i).

(ii) (() Without loss of generality, let R1=Z(R1). Since R1=Z(R1), for

all x1; x22Z(R)there exists an x3such that x3x1=x3x2= 0 by Lemma 2.3.

So, for any (x1; y1);(x2; y2)2Z(R1R2), there exists (x3;0) 2Z(R1R2)

such that (x1; y1)(x3;0) = (x2; y2)(x3;0) = (0;0). If, without loss of generality,

(x2; y2)=(x3;0);we have (x1; y1)(x2; y2)=0:Thus, diam((R1R2)) 2.

By (i), it must be that diam((R1R2)) = 2.

()) Assume R16=Z(R1)and R26=Z(R2). Let x2Z(R1),y2Z(R2),

m2reg(R1),n2reg(R2). Then (x; n)(m; y)6= (0;0):Since diam((R1

R2)) = 2, there exists (a; b)2Z(R1R2)such that (a; b)(x; n)=(a; b)

(m; y) = (0;0). Then ma =nb = 0, so (a; b) = (0;0), a contradiction.

(iii) By (i) and (ii).

Theorem 3.8. Let R1and R2be rings such that diam((R1)) = 2 and

diam((R2)) = 3. Then:

(i) diam((R1R2)) 6= 1.

(ii) diam((R1R2)) = 2 if and only if R1=Z(R1).

(iii) diam((R1R2)) = 3 if and only if R16=Z(R1).

Proof. (i) Same as Theorem 3.5 (i).

(ii) (() Same as in proof of Theorem 3.7 (ii).

()) Assume R16=Z(R1). Let m2reg(R1). Since diam((R2)) = 3, there

exist distinct y1; y22Z(R2)with y1y26= 0 and there is no y32Z(R2)such

that y1y3=y2y3= 0. Then (m; y1)(m; y2)6= (0;0):Since diam((R1R2)) =

2, there exists (a; b)2Z(R1R2)such that (a; b)(m; y1) = (a; b)(m; y2) =

(0;0). Then ma = 0, giving a= 0. Also, we have by1=by2= 0;giving b= 0,

a contradiction.

(iii) By (i) and (ii).

Theorem 3.9. Let R1and R2be commutative rings such that diam((R1)) =

diam((R2)) = 3. Then diam((R1R2)) = 3.

Proof. Since diam((R1)) = 3, there exist distinct x1; x22Z(R1)with x1x26=

0and there is no x32Z(R1)such that x1x3=x2x3= 0. Likewise, there

exist distinct y1; y22Z(R2)with y1y26= 0 and there is no y32Z(R2)such

that y1y3=y2y3= 0. Since (x1; y1)(x2; y2)6= 0, we have diam((R1R2)) >

1. If diam((R1R2)) = 2, there exists (a; b)2Z(R1R2)such that

(a; b)(x1; y1) = (a; b)(x2; y2) = 0. Then ax1=ax2= 0, giving a= 0. Also,

by1=by2= 0, giving b= 0;a contradiction.

If we restrict the rings R1and R2to be commutative rings with identity,

then the previous six theorems yield the following result.

ZERO-DIVISOR GRAPHS OF DIRECT PRODUCTS OF COMMUTATIVE RINGS 7

Corollary 3.10. Let R1and R2rings with identity and nonzero zero divisors.

Then diam((R1R2)) = 3.

In a similar manner, if we consider a ring Rwith identity having a nontrivial

idempotent e; then we can decompose the ring as R=eR (1 e)R. If both

eR and (1 e)Rare domains, then diam((R)) = 2; if either eR or (1 e)R

contains nonzero zero-divisors, then diam((R)) = 3 by Theorem 3.1:The

following corollary summarizes this fact:

Corollary 3.11. If Ris a ring with identity having at least one non-trivial

idempotent, then diam((R)) 2.

4. Girth and Direct Products

The situation simpli…es considerably when considering girth verus diameter.

As the reader will see, a classi…cation of when a direct product obtains girth

of 3;4;or 1is arrived at relatively quickly and without appealing to the girth

of the constituent rings.

Theorem 4.1. For any nontrivial rings R1and R2,g((R1R2)) = 3 if and

only if one (or both) of the following hold:

(i) jZ(R1)j 2or jZ(R2)j 2

(ii) Both R1and R2contain a nonzero nilpotent element.

Proof. (()If (i) holds, assume, without loss of generality, that jZ(R1)j 2.

Since (R1)is connected, there must exist distinct a; b 2Z(R1)such that

ab = 0. Then, (a; 0) (b; 0) (0; c)(a; 0) is a cycle of length 3, where

c2R

2. If (ii) holds, let a2R

1and b2R

2with a2=b2= 0. Then,

(a; 0) (a; b)(0; b)(a; 0) is a cycle of length 3.

())Suppose, without loss of generality, R1has no nonzero nilpotent ele-

ments. If jZ(R1)j<2, then jZ(R1)j= 0. Let (a; b)(c; d)(e; f )(a; b)

be a cycle in (R1R2). Two of the elements a; c; and emust be zero; hence,

this cycle must have the form (a; b)(0; d)(0; f)(a; b);and dand fmust

be distinct. Thus, jZ(R2)j 2.

Theorem 4.2. For any rings R1and R2,g((R1R2)) = 4 if and only if

both of the following hold:

(i) jR1j;jR2j 3

(ii) Without loss of generality, R1is a domain, and jZ(R2)j 1.

Proof. (()Let a; b 2R

1be distinct; let c; d 2R

2be distinct. Then, (a; 0)

(0; c)(b; 0) (0; d)(a; 0) is a cycle. Hence, g((R1R2)) 4. If g((R1

R2)) = 3, Theorem 4.1 yields a contradiction.

())By Theorem 4.1 jZ(R1)j 1and jZ(R2)j 1:Without loss of

generality assume R2is not a domain, hence jZ(R2)j= 1. If R1is not

a domain then jZ(R1)j= 1 and both R1and R2contain nonzero nilpotent

ZERO-DIVISOR GRAPHS OF DIRECT PRODUCTS OF COMMUTATIVE RINGS 8

elements. Thus g((R1R2)) = 3 by Theorem 4.1, a contradiction. Therefore

R1is a domain.

Since R1is a domain, a cycle must have the form (a; b)(0; c)(d; e)

(0; f )(a; b). In this cycle, cand fmust be nonzero and distinct. Thus,

jR2j 3. If either aor dis zero, then R2has at least two distinct nonzero

zero divisors, whence g((R1R2)) = 3. If a=d, then band eare distinct.

If e= 0, then b; c; f 2Z(R2), implying b=c=f, a contradiction; if e6= 0,

then c; e; f 2Z(R2), implying c=e=f; another contradiction. We must

then have a6=d, and jR1j 3.

Corollary 4.3. For any rings R1and R2,g((R1R2)) = 1if and only if

(i) without loss of generality, R1

=Z2and jZ(R2)j 1, or

(ii) without loss of generality, R1is a domain, and jR2j= 2 with R2

2= 0.

Proof. ())By Theorem 4.1, without loss of generality, jZ(R1)j= 0 and

jZ(R2)j 1;and by Theorem 4.2, either jR1j<3or jR2j<3. If jR1j= 2;

then R1

=Z2;and (i) holds. Suppose jR2j= 2. If R2

=Z2;then (i) holds. If

not, then R2

2= 0, and (ii) holds.

(()By Theorems 4.1 and 4.2.

References

[1] Anderson, D.D. and Naseer, M., Beck’s coloring of a commutative ring, J.

Algebra 159 (1993), 500-514.

[2] Anderson, David F. and Livingston, Philip S., The zero-divisor graph of a

commutative ring;J. Algebra 217 (1999), 434-447.

[3] Axtell, M., Coykendall, J., and Stickles, J., Zero-divisor graphs of polyno-

mial and power series over commutative rings, Comm. Alg., to appear.

[4] Axtell, M., and Stickles, J., Zero-divisor graphs of idealizations, preprint.

[5] Beck, I., Coloring of commutative rings, J. Algebra 116 (1988), 208-226.

[6] DeMeyer, F.and Schneider, K., Automorphisms and zero-divisor graphs of

commutative rings, International J. of Commutative Rings 1(3) (2002),

93-106.

[7] Fields, David E., Zero divisors and nilpotent elements in power series;Proc.

Amer. Math. Soc.27(3) (1971), 427-433.

[8] McCoy, Neil H., Remarks on divisors of zero, Amer. Math. Monthly 49

(1942), 286-295.

[9] Mulay, S.B., Cycles and symmetries of zero-divisors, Comm. Alg. 30(7)

(2002), 3533-3558.

ZERO-DIVISOR GRAPHS OF DIRECT PRODUCTS OF COMMUTATIVE RINGS 9

Dept. Mathematics and Computer Science, Wabash College,

Crawfordsville IN 47933-0352

E-mail address:axtellm@wabash.edu

Dept. of Mathematics, University of Evansville, Evansville IN

47722

E-mail address:js298@evansville.edu

Wabash College, Crawfordsville, IN 47933

E-mail address:warfelj@wabash.edu