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Zero-divisor graphs of direct products of commutative rings

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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.
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.
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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
... The following example shows that a ring R will not necessarily be UFR provided that γ(x) is complete for all x R*\U(R). Example 3. Consider the ring 6 . We need only consider the graphs γ(2) and γ(3) because Irr(R) = {2, 3} (up to associates). ...
... The two preceding examples provide negative answers to (5) and (6). Likewise, since complete implies connected, we have negative answers to (3) and (4). ...
... In literature such as [5] and [6], frequent references to complete bipartite and star graphs were made to the zerodivisor graph, Γ(R) (as well as complete bipartite reducible and star-shape reducible graphs). In these papers it is shown that the zero-divisor graph Γ(R) may be realized in these forms. ...
Article
In previous literature Coykendall & Maney, as well as Axtell & Stickles, have discussed the concept of irreducible divisor graphs of elements in domains and ring with zero-divisors respectively, with two different definitions. In this paper we seek to look at the irreducible divisor graphs of ring elements under a hybrid definition of the two previous ones—in hopes that this graph will reveal structure concerning irreducible divisors in rings with zero-divisors. We also compare the three graphs and examine in what respects they are related. Other graph-theoretic properties of this graph will also be discussed.
... Zero divisor graphs for different rings has been defined and their properties have been discussed. Warfel [5] studied the zero divisor graph of direct product of two commutative rings. Akbari and Mohammadian [8] have made significant contribution in the study of zero divisor graph of non-commutative ring. ...
... Although some results on the zero-divisor graph do not depend on an identity (see Theorem 2.2) and a few results have been given without assuming an identity (cf. [13]), there has been no systematic study for commutative rings without an identity. The concept of zero-divisor graph of a commutative ring was extended to commutative semigroups by DeMeyer, McKenzie, and Schneider in [20]. ...
Article
Let R be a commutative ring. The zero-divisor graph of R is the (simple) graph Γ(R) with vertices the nonzero zero-divisors of R, and two distinct vertices x and y are adjacent if and only if xy = 0. In this article, we investigate Γ(R) when R does not have an identity, and we determine all such zero-divisor graphs with 14 or fewer vertices.
... Proposition 5.1 [11]: If 1 and 2 are commutative rings with identity and nonzero zero divisors, then ( 1 × 2 )= 3. Using this result and the above theorems we get the following theorem: Proof: It was shown earlier that if = 2 , then ℤ n i is a complete graph. ...
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Let R(+)N be the idealization of the ring R by the R-module N. In this paper, we investigate when Γ(R(+)N) is a Planar graph where R is an integral domain and we investigate when Γ(Z n (+)Z m) is a Planar graph.
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This article gives a comprehensive survey on zero-divisor graphs of finite commutative rings. We investigate the results on structural properties of these graphs.
Conference Paper
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The present paper is focusing on the connection between the Isomorphic groups and the Isomorphic graphs. We established a few results on the necessary and sufficient condition on the dimension of the graphs, which are isomorphic with respect to the isomorphic groups. We proved those results depending on the Isomorphic groups. We discussed the methods to find isomorphic graphs by using isomorphic groups. We derived few results on the isomorphism of graphs, by illustrating few examples.
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Let R be a commutative ring with identity. We consider the idealization of R in R, i.e., the ring R(+) R, and explain the relationship between diamΓ(R) and diamΓ(R(+) R). In this paper, it is proved that R is an integral domain if and only if Γ(R(+) R) is a complete graph and also, we show that diamΓ(R) ≤ diamΓ(R(+) R) by starting with either diamΓ(R) or diamΓ(R(+) R).
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This paper introduces the notions of a zero-divisor labeling and the zero-divisor index of a graph using the zero-divisors of a commutative ring. Viewed in this way, the usual zero-divisor graph is a maximal graph with respect to a zero-divisor labeling. We also study optimal zero-divisor labelings of a finite graph.
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A commutative ring R can be considered as a simple graph whose vertices are the elements of R and two different elements x and y of R are adjacent if and only if xy = 0. Beck conjectured that χ(R) = cl(R). We give a counterexample where R is a finite local ring with cl(R) = 5 but χ(R) = 6. We show that if A is a regular Noetherian ring with maximal ideals N1, ..., Ns, such that each A/Ni is finite, then for R = A/Nn11 ··· Nnss, χ(R) = cl(R). Finally, we give a complete listing of all finite rings R with χ(R) ≤ 4.
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There is a natural graph associated to the zero-divisors of a commutative ring In this article we essentially classify the cycle-structure of this graph and establish some group-theoretic properties of the group of graph-automorphisms We also determine the kernel of the canonical homomorphism from to
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We recall several results of zero divisor graphs of com- mutative rings. We then examine the preservation of diameter and girth of the zero divisor graph under extension to polynomial and power series rings. The concept of the graph of the zero-divisors of a ring was first introduced by Beck in (3) 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. In the first section we provide the reader with a few known results concerning the girth and diameter of the graph of zero-divisors of a commutative ring. In the second section we examine the preservation and lack thereof of the diameter of the zero-divisor graphs under extensions to polynomial and power series rings, while in the third section we consider the eects of the same extensions upon the girth of the graphs. Throughout, R is a commutative ring and, unless otherwise stated, is assumed to have an identity. 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. 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 2 Z (R), there is an edge connecting r and s if and only if rs = 0. For two distinct vertices a and b in a graph , the distance between a and 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 is diam() = sup {d(a,b) | a and b are distinct vertices of }. We will use the notation diam(R) and diam(( R)) interchangeably 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) and g(( R)) interchangeably 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
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For each commutative ring R we associate a (simple) graph Γ(R). We investigate the interplay between the ring-theoretic properties of R and the graph-theoretic properties of Γ(R).
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It is well known that a polynomialf(X) over a commutativering R with identity is nilpotent if and only if each coefficientof f(X) is nilpotent; and that/(X) is a zero divisor in R[X] ifand only if f(X) is annihilated by a nonzero element of R. Thispaper considers the problem of determining when a power seriesg(X) over R is either nilpotent or a zero divisor in R[[X]]. If R isNoetherian, then g(X) is nilpotent if and only if each coefficient ofg(X) is nilpotent; and g(X) is a zero divisor in R[[X]] if and only ifg(X) is annihilated by a nonzero element of R. If R has positive characteristic, then g (X) is nilpotent if and only if each coefficient ofg(X) is nilpotent and there is an upper bound on the orders of nilpotencyof the coefficients of g(X). Examples illustrate, however, that in general g(X) need not be nilpotent if there is an upperbound on the orders of nilpotency of the coefficients of g(X), and that g(X) may be a zerdivisor in R[[X]] while g(X) has a unitcoefficient.
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We consider zero-divisor graphs of idealizations of commutative rings. Specifically, we look at the preservation, or lack thereof, of the diameter and girth of the zero-divisor graph of a ring when extending to idealizations of the ring.