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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
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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
