arXiv:1611.03730v1 [] 10 Nov 2016

Some Properties of the Nil-Graphs of
Ideals of Commutative Rings ∗†
R. Nikandisha‡, F. Shaveisib
a

Department of Basic Sciences, Jundi-Shapur University of Technology, Dezful, Iran
P.O. Box 64615-334,

b

Department of Mathematics, Faculty of Sciences, Razi University, Kermanshah, Iran

r.nikandish@ipm.ir

f.shaveisi@ipm.ir

Abstract
Let R be a commutative ring with identity and Nil(R) be the set of nilpotent
elements of R. The nil-graph of ideals of R is defined as the graph AGN (R)
whose vertex set is {I : (0) 6= I ⊳ R and there exists a non-trivial ideal J such
that IJ ⊆ Nil(R)} and two distinct vertices I and J are adjacent if and only
if IJ ⊆ Nil(R). Here, we study conditions under which AGN (R) is complete or
bipartite. Also, the independence number of AGN (R) is determined, where R is
a reduced ring. Finally, we classify Artinian rings whose nil-graphs of ideals have
genus at most one.

1. Introduction
When one assigns a graph with an algebraic structure, numerous interesting algebraic problems arise from the translation of some graph-theoretic parameters such as
clique number, chromatic number, diameter, radius and so on. There are many papers in this topic, see for example [5], [8] and [12]. Throughout this paper, all rings
are assumed to be non-domain commutative rings with identity. By I(R) (I(R)∗ ) and
Nil(R), we denote the set of all proper (non-trivial) ideals of R and the nil-radical of
∗

Key Words: Nil-graph; Complete graph; Bipartite graph; Genus; Independence number.
2010 Mathematics Subject Classification: 05C15; 05C69; 13E05; 13E10.
‡
Corresponding author
†

1

R, respectively. The set of all maximal and minimal prime ideals of R are denoted
by Max(R) and Min(R), respectively. The ring R is said to be reduced, if it has no
non-zero nilpotent element.
Let G be a graph. The degree of a vertex x of G is denoted by d(x). The graph
G is said to be r-regular, if the degree of each vertex is r. The complete graph with
n vertices, denoted by Kn , is a graph in which any two distinct vertices are adjacent.
A bipartite graph is a graph whose vertices can be divided into two disjoint parts U
and V such that every edge joins a vertex in U to one in V . It is well-known that a
bipartite graph is a graph that does not contain any odd cycle. A complete bipartite
graph is a bipartite graph in which every vertex of one part is joined to every vertex
of the other part. If the size of one of the parts is 1, then it is said to be a star graph.
A tree is a connected graph without cycles. Let Sk denote the sphere with k handles,
where k is a non-negative integer, that is, Sk is an oriented surface of genus k. The
genus of a graph G, denoted by γ(G), is the minimal integer n such that the graph can
be embedded in Sn . A genus 0 graph is called a planar graph. It is well-known that
γ(Kn ) = ⌈

(n − 3)(n − 4)
⌉ for all n ≥ 3,
12

(m − 2)(n − 2)
⌉, for all n ≥ 2 and m ≥ 2.
4
For a graph G, the independence number of G is denoted by α(G). For more details
about the used terminology of graphs, see [13].
γ(Km,n ) = ⌈

We denote the annihilator of an ideal I by Ann(I). Also, the ideal I of R is called
an annihilating-ideal if Ann(I) 6= (0). The notation A(R) is used for the set of all
annihilating-ideals of R. By the annihilating-ideal graph of R, AG(R), we mean the
graph with vertex set A(R)∗ = A(R)\{0} and two distinct vertices I and J are adjacent
if and only if IJ = 0. Some properties of this graph have been studied in [1, 2, 3, 5, 6].
In [12], the authors have introduced another kind of graph, called the nil-graph of
ideals. The nil-graph of ideals of R is defined as the graph AGN (R) whose vertex set
is {I : (0) 6= I ⊳ R and there exists a non-trivial ideal J such that IJ ⊆ Nil(R)}
and two distinct vertices I and J are adjacent if and only if IJ ⊆ Nil(R). Obviously,
our definition is slightly different from the one defined by Behboodi and Rakeei in [5]
and it is easy to see that the usual annihilating-ideal graph AG(R) is a subgraph of
AGN (R). In [12], some basic properties of nil-graph of ideals have been studied. In this
article, we continue the study of the nil-graph of ideals. In Section 2, the necessary and
sufficient conditions, under which the nil-graph of a ring is complete or bipartite, are
found. Section 3 is devoted to the studying of independent sets in nil-graph ideals. In
2

Section 4, we classify all Artinian rings whose nil-graphs of ideals have genus at most
one.

2. When Is the Nil-Graph of Ideals Complete or Bipartite?
In this section, we study conditions under which the nil-graph of ideals of a commutative ring is complete or complete bipartite. For instance, we show that if R is a
Noetherian ring, then AGN (R) is a complete graph if and only if either R is Artinian
local or R ∼
= F1 × F2 , where F1 and F2 are fields. Also, it is proved that if AGN (R)
is bipartite, then AGN (R) is complete bipartite. Moreover, if R is non-reduced, then
AGN (R) is star and Nil(R) is the unique minimal prime ideal of R.
We start with the following theorem which can be viewed as a consequence of [12,
Theorem 5] (Here we prove it independently). Note that it is clear that if R is a reduced
ring, then AGN (R) ∼
= AG(R).
Theorem 1. Let R be a Noetherian ring. Then AGN (R) is a complete graph if and
only if either R is Artinian local or R ∼
= F1 × F2 , where F1 and F2 are fields.
Proof. First suppose that AGN (R) is complete. If R is reduced, then by [5, Theorem
2.7], R ∼
= F1 × F2 . Thus we can suppose that Nil(R) 6= (0). We continue the proof in
the following two cases:
Case 1. R is a local ring with the unique maximal ideal m. Since R is non-reduced,
by Nakayama’s Lemma (see [4, Proposition 2.6]), m and m2 are two distinct vertices of
AGN (R). Thus m3 ⊆ Nil(R) and so R is an Artinian local ring.
Case 2. R has at least two maximal ideals. First we show that R has exactly two
maximal ideals. Suppose to the contrary, m, n and p are three distinct maximal ideals
of R. Since AGN (R) is complete, we deduce that mn ⊆ Nil(R) ⊆ p, a contradiction.
Thus R has exactly two maximal ideals, say m and p. Now, we claim that both m
and p are minimal prime ideals. Since m and p are adjacent, we conclude one of the
maximal ideals, say p, is a minimal prime ideal of R. Now, suppose to the contrary, m
properly contains a minimal prime ideal q of R. Since mp ⊆ q, we get a contradiction.
So the claim is proved. Thus R is Artinian. Hence by [4, Theorem 8.7 ], we have
R∼
= R1 × R2 , where R1 and R2 are Artinian local rings. By contrary and with no loss
of generality, suppose that R1 contains a non-trivial ideal, say I. Then the vertices
I × R2 and (0) × R2 are not adjacent, a contradiction. Thus R ∼
= F1 × F2 , where F1
and F2 are fields.
3

Conversely, if R ∼
= F1 × F2 , where F1 and F2 are fields, then it is clear that AGN (R) ∼
=
K2 . Now, suppose that (R, m) is an Artinian local ring. Since m is nilpotent, it follows
that AGN (R) is complete.

The following example shows that Theorem 1 does not hold for non-Noetherian
rings.
Example 2. Let R =
is a complete graph.

k[xi : i≥1]
,
(x2i : i≥1)

where k is a field. Then R is not Artinian and AGN (R)

Remark 3. Let R be a ring. Every non-trivial ideal contained in Nil(R) is adjacent
to every other vertex of AGN (R). In particular, if R is an Artinian local ring, then
AGN (R) is a complete graph.
The next result shows that nil-graphs, whose every vertices have finite degrees, are
finite graphs.
Theorem 4. If every vertex of AGN (R) has a finite degree, then R has finitely many
ideals.
Proof. First suppose that R is non-reduced. Since d(Nil(R)) < ∞, the assertion
follows from Remark 3. Thus we can assume that R is reduced. Choose 0 6= x ∈ Z(R).
Since d(Rx) < ∞ and Rx is adjacent to every ideal contained in Ann(x), we deduce
that Ann(x) is an Artinian R-module. Similarly, one can show that Rx is an Artinian
R
implies that R is an Artinian ring.
R-module. Now, the R-isomorphism Rx ∼
= Ann(x)
Now, since R is reduced, [4, Theorem 8.7] implies that R is a direct product of finitely
many fields and hence we are done.

The next result gives another condition under which AGN (R) is complete.
Theorem 5. If AGN (R) is an r-regular graph, then AGN (R) is a complete graph.
Proof. If Nil(R) 6= (0), then by Remark 3, there is nothing to prove. So, suppose
that R is reduced. Since AGN (R) is an r-regular graph, Theorem 4 and [4, Theorem
8.7] imply that R ∼
= F1 × · · · × Fn , where n ≥ 2 and each Fi is a field. It is not hard
to check that every ideal I = I1 × · · · × In of R has degree 2nI − 1, where nI = |{i :
1 ≤ i ≤ n and Ii = (0)}|. Let I = F1 × (0) × · · · × (0) and J = F1 × · · · × Fn−1 × (0).
Then we have d(I) = 2n−1 − 1 and d(J) = 1. The r-regularity of AGN (R) implies that
2n−1 − 1 = 1 and so n = 2. Therefore R ∼

= F1 × F2 , as desired.
4

In the rest of this section, we study bipartite nil-graphs of ideals of rings.
Theorem 6. Let R be a ring such that AGN (R) is bipartite. Then AGN (R) is complete
bipartite. Moreover, if R is non-reduced, then AGN (R) is star and Nil(R) is the unique
minimal prime ideal of R.
Proof. If R is reduced, then by [6, Corollary 2.5], AGN (R) is a complete bipartite
graph. Now, suppose that R is non-reduced. Then by Remark 3, AGN (R) is a star
graph. So, by Remark 3, either Nil(R) is a minimal ideal or R has exactly two ideals.
In the latter case, R is an Artinian local ring and so Nil(R) is the unique minimal
prime ideal of R. Thus we can assume that Nil(R) = (x) is a minimal ideal of R, for
some x ∈ R. To complete the proof, we show that R has exactly one minimal prime
ideal. Suppose to the contrary, p1 and p2 are two distinct minimal prime ideals of R.
Choose z ∈ p1 \ p2 and set S1 = R \ p1 and S2 = {1, z, z 2 , . . .}. If 0 ∈
/ S1 S2 , then
by [11, Theorem 3.44], there exists a prime ideal p in R such that p ∩ S1 S2 = ∅ and
hence p = p1 , a contradiction. So, 0 ∈ S1 S2 . Therefore, there exist positive integer k
and y ∈ R \ p1 such that yz k = 0. Consider the ideals (x), (y) and (z k ). This is clear
that (x), (y) and (z k ) are three distinct vertices which form a triangle in AGN (R), a
contradiction.

The following corollary is an immediate consequence of Theorem 6 and Remark 3.
Corollary 7. If AGN (R) is a tree, then AGN (R) is a star graph.
We finish this section with the next corollary.
Corollary 8. Let R be an Artinian ring. Then AGN (R) is bipartite if and only if
AGN (R) ∼
= Kn , where n ∈ {1, 2}.
Proof. Let R be an Artinian ring and AGN (R) be bipartite. Then by Theorem 6,
AGN (R) is complete bipartite. If R is local, then Remark 3 implies that AGN (R) is
complete. Since AGN (R) is complete bipartite, we deduce that AGN (R) ∼
= Kn , where
n ∈ {1, 2}. Now, suppose that R is not local. Then by [4, Theorem 8.7], there exists a
positive integer n such that R ∼
= R1 × · · · × Rn , where every Ri is an Artinian local ring.
Since AGN (R) contains no odd cycle, it follows that n = 2. To complete the proof,
we show that both R1 and R2 are fields. By contrary and with no loss of generality,
suppose that R1 contains a non-trivial ideal, say I. Then it is not hard to check that
R1 ×(0), I ×(0) and (0)×R2 forms a triangle in AGN (R), a contradiction. The converse
is trivial.

5

3. The Independence Number of Nil-Graphs of Ideals
In this section, we use the maximal intersecting families to obtain a low bound for
the independence number of nil-graphs of ideals. Let R ∼
= R1 × R2 × · · · × Rn ,
T (R) = {(0) 6= I = I1 × I2 × · · · × In ⊳ R| ∀ 1 ≤ k ≤ n : Ik ∈ {(0), Rk }};
and denote the induced subgraph of AGN (R) on T (R) by GT (R).
Proposition 9. If R ∼
= R1 × R2 × · · · × Rn is a ring, then α(GT (R)) = 2n−1 .
Proof. For every ideal I = I1 × I2 × · · · × In , let
∆I = {k| 1 ≤ k ≤ n and Ik = Rk };
Then two distinct vertices I and J in GT (R) are not adjacent if and only if ∆I ∩∆J 6= ∅.
So, there is a one to one correspondence between the independent sets of GT (R) and
the set of families of pairwise intersecting subsets of the set [n] = {1, 2, . . . , n}. So, [10,
Lemma 2.1] completes the proof.

Using [4, Theorem 8.7], we have the following immediate corollary.
Corollary 10. Let R be an Artinian with n maximal ideals. Then α(AGN (R)) ≥ 2n−1 ;
moreover, the equality holds if and only if R is reduced.
Lemma 11. [9, Proposition 1.5] Let R be a ring and {p1 , . . . , pn } be a finite set of
S
distinct minimal prime ideals of R. Let S = R \ ni=1 pi . Then RS ∼
= Rp1 × · · · × Rpn .

Proposition 12. If |Min(R)| ≥ n, then α(AGN (R)) ≥ 2n−1 .

S
Proof. Let {p1 , . . . , pn } be a subset of Min(R) and S = R \ ni=1 pi . By Lemma 11,
there exists a ring isomorphism RS ∼
= Rp1 × · · · × Rpn . On the other hand, if IS , JS
are two non-adjacent vertices of AGN (RS ), then it is not hard to check that I, J are
two non-adjacent vertices of AGN (R). Thus α(AGN (R)) ≥ α(AGN (RS )) and so by
Proposition 9, we deduce that α(AGN (R)) ≥ 2n−1 .

From the previous proposition, we have the following immediate corollary which
shows that the finiteness of α(AGN (R)) implies the finiteness of number of the minimal
prime ideals of R.
6

Corollary 13. If R contains infinitely many minimal prime ideals, then the independence number of AGN (R) is infinite.
Theorem 14. For every Noetherian reduced ring R, α(AGN (R)) = 2|Min(R)|−1 .
S
Proof. Let Min(R) = {p1 , p2 , . . . , pn } and S = R \ nk=1 pk . Then Lemma 11 implies
that RS ∼
= Rp1 × · · · × Rpn . On the other hand, by using [9, Proposition 1.1], we deduce
that every Rpi is a field. Thus α(AGN (R)) ≥ α(AGN (RS )) = 2n−1 , by Corollary 10.
To complete the proof, it is enough to show that α(AGN (R)) ≤ α(AGN (RS )). To
see this, let I(x1 , x2 , . . . , xr ) and J = (y1 , y2 , . . . , ys ) be two non-adjacent vertices of
AGN (R). By [7, Corollary 2.4], S contains no zero-divisor and so IS , JS are non-trivial
ideals of RS . We show that IS , JS are non-adjacent vertices of AGN (RS ). Suppose
to the contrary, IS JS ⊆ Nil(R)S = (0). Then for every 1 ≤ i ≤ r and 1 ≤ j ≤ s,
Q
there exists sij ∈ S such that sij xi yj = 0. Setting t = i,j sij , we have tIJ = (0).
Since t is not a zero-divisor, we deduce that IJ = (0), a contradiction. Therefore,
α(AGN (R)) ≤ α(AGN (RS )), as desired.

Finally as an application of the nil-graph of ideals in the ring theory we have the
following corollary which shows that number of minimal prime ideals of a Noetherian
reduced ring coincides number of maximal ideals of the total ring of R.
Corollary 15. Let R be a Noetherian reduced ring. Then
|Min(R)| = |Max(T (R))| = log2 (α(AGN (R))).
S
Proof. Setting Min(R) = {p1 , p2 , . . . , pn } and S = R \ p∈Min(R) p, we have T (R) ∼
=
Rp1 × · · · × Rpn , by Lemma 11. Since every Rpi is a field, Corollary 10 and Theorem
14 imply that 2|Min(R)|−1 = 2|Max(T (R))|−1 = α(AGN (R)). So, the assertion follows. 

4. The Genus of Nil-Graphs of Ideals
In [3, Corollary 2.11], it is proved that for integers q > 0 and g ≥ 0, there are
finitely many Artinian rings R satisfying the following conditions:
(1) γ(AG(R)) = g,
(2) | R
m | ≤ q for any maximal ideal m of R.
We begin this section with a similar result for the nil-graph of ideals.

7

Theorem 16. Let g and q > 0 be non-negative integers. Then there are finitely many
Artinian rings R such that γ(AGN (R)) = g and | R
m | ≤ q, for every maximal ideal m of
R.
Proof. Let R be an Artinian ring. Then [4, Theorem 8.7] implies that R ∼
= R1 × · · · ×
Rn , where n is a positive integer and each Ri is an Artinian local ring. We claim that
for every i, |Ri | ≤ q I(Ri ) . Since γ(AGN (R)) < ∞, we deduce that γ(AGN (Ri )) < ∞,
for every i. So by Remark 3 and formula for the genus of complete graphs, every
Ri has finitely many ideals. Therefore, by hypothesis and [3, Lemma 2.9], we have
i I(Ri )
|Ri | ≤ | R
≤ q I(Ri ) and so the claim is proved. To complete the proof, it is
mi |
sufficient to show that |R| is bounded by a constant, depending only on g and q. With
no loss of generality, suppose that |R1 | ≥ |Ri |, for every i ≥ 2. By the formula for the
1 )−5
≤ γ(AGN (R1 )) ≤ g. Hence |I(R1 )| ≤ 12g + 5 and so
genus of complete graphs, I(R12
|R| ≤ |R1 |n ≤ (q I(R1 ) )n ≤ q n(12g+5) .
So, we are done.



Let {Ri }i∈N be an infinite family of Artinian rings such that every Ri is a direct
product of 4 fields. Then it is clear that γ(AGN (Ri )) = 1, for every i. So, the condition
|R
m | ≤ q, for every maximal ideal m of R, in the previous theorem is necessary.
Let R be a Noetherian ring. Then one may ask does γ(AGN (R)) < ∞ imply that R
is Artinian? The answer of this question is negative. To see this, let R ∼
= S × D, where
S is a ring with at most one non-trivial ideal and D is a Noetherian integral domain
which is not a field. Then it is easy to check that AGN (R) is a planar graph and R is
a Noetherian ring which is not Artinian.
Before proving the next lemma, we need the following notation. Let G be a graph
e for the subgraph
and V ′ be the set of vertices of G whose degrees equal one. We use G
′
G \ V and call it the reduction of G.
e where G
e is the reduction of G.
Lemma 17. γ(G) = γ(G),

Remark 18. It is well-known that if G is a connected graph of genus g, with n vertices,
m edges and f faces, then n − m + f = 2 − 2g.
In the following, all Artinian rings, whose nil-graphs of ideals have genus at most
one, are classified.
8

Theorem 19. Let R be an Artinian ring. If γ(AGN (R)) < 2, then |Max(R)| ≤ 4 and
moreover, the following statements hold.
(i) If |Max(R)| = 4, then γ(AGN (R)) < 2 if and only if R is isomorphic to a direct
product of four fields.
(ii) If |Max(R)| = 3, then γ(AGN (R)) < 2 if and only if R ∼
= F1 × F2 × R3 , where
F1 , F2 are fields and R3 is an Artinian local ring with at most two non-trivial
ideals.
(iii) If |Max(R)| = 2, then γ(AGN (R)) < 2 if and only if either R ∼
= F1 × R2 , where
F1 is a field and R2 is an Artinian local ring with at most three non-trivial ideals
or R ∼
= R1 × R2 , where every Ri (i = 1, 2) is an Artinian local ring with at most
one non-trivial ideal.
(iv) If R is local, then γ(AGN (R)) < 2 if and only if R has at most 7 non-trivial
ideals.
Proof. Let γ(AGN (R)) < 2. First we show that |Max(R)| ≤ 4. Suppose to the
contrary, |Max(R)| ≥ 5. By [4, Theorem 8.7], R ∼
= R1 × · · · × R5 , where every Ri is an
Artinian ring. Let
I1 = R1 × (0) × (0) × (0) × (0);

I2 = (0) × R2 × (0) × (0) × (0);

I3 = R1 × R2 × (0) × (0) × (0);

J1 = (0) × (0) × R3 × (0) × (0);

J2 = (0) × (0) × (0) × R4 × (0);

J3 = (0) × (0) × (0) × (0) × R5 ;

J4 = (0) × (0) × R3 × R4 × (0);

J5 = (0) × (0) × (0) × R4 × R5 ;

J6 = (0) × (0) × R3 × (0) × R5 ;

J7 = (0) × (0) × R3 × R4 × R5 .

Then for every 1 ≤ i ≤ 3 and every 1 ≤ j ≤ 7, Ii and Jj are adjacent and so K3,7 is
a subgraph of AGN (R). Thus by the formula for the genus of the complete bipartite
graph, we have γ(AGN (R)) ≥ γ(K3,7 ) ≥ 2, a contradiction.
(i) Let |Max(R)| = 4 and γ(AGN (R)) < 2. By [4, Theorem 8.7], R ∼
= R1 × R2 ×
R3 × R4 , where every Ri is an Artinian local ring. We show that every Ri is a field.
Suppose not and with no loss of generality, R4 contains a non-trivial ideal, say a. Set
I1 = R1 × (0) × (0) × (0); I2 = (0) × R2 × (0) × (0); I3 = R1 × R2 × (0) × (0);
I4 = R1 × R2 × (0) × a; J1 = (0) × (0) × R3 × (0); J2 = (0) × (0) × (0) × R4 ;
9

J3 = (0) × (0) × (0) × a; J4 = (0) × (0) × R3 × R4 ; J5 = (0) × (0) × R3 × a.
It is clear that every Ii , 1 ≤ i ≤ 4, is adjacent to Jj , 1 ≤ j ≤ 5, and so K4,5 is a
subgraph of AGN (R). Thus by the formula for the genus of the complete bipartite
graph, we have γ(AGN (R)) ≥ γ(K4,5 ) ≥ 2, a contradiction. Conversely, assume that
R∼
= F1 × F2 × F3 × F4 , where every Fi is a field. We show that γ(AGN (R)) = 1. By
^
^
Lemma 17, it is enough to prove that γ(AG
N (R)) = 1. We know that AGN (R) has 4
^
vertices of degree 6 and 6 vertices of degree 3. So, AG
N (R) has n = 10 vertices and
^
m = 21 edges. Also, it is not hard to check that AG
N (R) has f = 11 faces. Now,
^
Remark 18 implies that γ(AGN (R)) = 1.
(ii) Let γ(AGN (R)) < 2 and R ∼
= R1 × R2 × R3 , where every Ri is an Artinian local
ring. We show that at least two of the three rings R1 , R2 and R3 are fields. Suppose not
and with no loss of generality, b and c are non-trivial ideals of R2 and R3 , respectively.
Set
I1 = R1 × (0) × (0); I2 = (0) × R2 × (0); I3 = R1 × R2 × (0); I4 = R1 × b × (0);
J1 = (0) × b × (0); J2 = (0) × (0) × c; J3 = (0) × b × c;
J4 = (0) × (0) × R3 ; J5 = (0) × b × R3 .
It is clear that every Ii , 1 ≤ i ≤ 4, is adjacent to Jj , 1 ≤ j ≤ 5, and so K4,5 is a subgraph
of AGN (R). Thus by the formula for the genus of the complete bipartite graph, we
have γ(AGN (R)) ≥ γ(K4,5 ) ≥ 2, a contradiction. Thus with no loss of generality, we
can suppose that R ∼
= F1 × F2 × R3 , where F1 and F2 are fields and R3 is an Artinian
local ring. Now, we prove that R3 has at most two non-trivial ideals. Suppose to the
contrary, a, b and c are three distinct non-trivial ideals of R3 . Let
I1 = (0) × (0) × R3 ; I2 = (0) × (0) × a; I3 = (0) × (0) × b; I4 = (0) × (0) × c;
J1 = F1 × (0) × (0); J2 = (0) × F2 × (0); J3 = F1 × F2 × (0);
J4 = F1 × F2 × a; J5 = F1 × F2 × b; J6 = F1 × F2 × c.
Clearly, every Ii , 1 ≤ i ≤ 4, is adjacent to Jj , 1 ≤ j ≤ 6, and so K4,6 is a subgraph of
AGN (R). Thus by the formula for the genus of the complete bipartite graph, we have
γ(AGN (R)) ≥ γ(K4,6 ) ≥ 2, a contradiction. Conversely, let R ∼
= F1 × F2 × R3 , where
F1 and F2 are fields and R3 be a ring with two non-trivial ideals c and c′ . Set
I1 = (0) × (0) × R3 ; I2 = (0) × (0) × c; I3 = (0) × (0) × c′ ;
J1 = F1 × (0) × (0); J2 = (0) × F2 × (0); J3 = F1 × F2 × (0).
10

Then for every 1 ≤ i, j ≤ 3, we have Ii Jj = (0). Hence γ(AGN (R)) ≥ γ(K3,3 ) ≥ 1.
However, in this case, AGN (R) is a subgraph of AGN (F1 × F2 × F3 × F4 ) (in which
every Fi is a field). Therefore, by (i), γ(AGN (R)) = 1. If R3 contains at most one
non-trivial ideal, then it is not hard to check that AGN (R) is a planar graph. This
completes the proof of (ii).
(iii) Assume that γ(AGN (R)) < 2 and R ∼
= R1 × R2 , where R1 and R2 are Artinian
local rings. We prove the assertion in the following two cases:
Case 1. R ∼
= F1 × R2 , where F1 is a field and R2 is an Artinian local ring. In this
case, we show that R2 has at most three non-trivial ideals. Suppose to the contrary,
R2 has at least four non-trivial ideals. Then for every two non-zero ideals I2 6= R2 and
J2 of R2 , the vertices F1 × I2 and (0) × J2 are adjacent and so K4,5 is a subgraph of
AGN (R). Thus by the formula for the genus of the complete bipartite graph, we have
γ(AGN (R)) ≥ γ(K4,5 ) ≥ 2, a contradiction.
Case 2. Neither R1 nor R2 is a field. We prove that every Ri has at most one nontrivial ideal. Suppose not and with no loss of generality, R2 has two distinct non-trivial
ideals. Then every ideal of the form R1 × J is adjacent to every ideal of the form I × K,
where I and J are proper ideals of R1 and R2 , respectively, and K is an arbitrary ideal
of R2 . So γ(AGN (R)) ≥ γ(K3,7 ) ≥ 2, a contradiction.
Conversely, if R ∼
= F1 × R2 , where F1 is a field and R2 is an Artinian local ring with
n ≤ 3 non-trivial ideals, then one can easily show that
(
1; n = 2, 3
γ(AGN (R)) =
0; n = 1.
∼ R1 × R2 , where R1 and R2 are Artinian local rings with one
Now, suppose that R =
non-trivial ideals. Then it is not hard to show that γ(AGN (R)) = 1. This comletes the
proof of (iii).
(iv) This follows from the formula of genus for the complete graphs and Remark 3.

From the proof of the previous theorem, we have the following immediate corollary.
Corollary 20. Let R be an Artinian ring. Then AGN (R) is a planar graph if and only
if |Max(R)| ≤ 3 and R satisfies one of the following conditions:
(i) R is isomorphic to the direct product of three fields.

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(ii) R ∼
= F1 × R2 , where F1 is a field and R2 is an Artinian local ring with at most
one non-trivial ideal.
(iii) R is a local ring with at most four non-trivial ideals.
We close this paper with the following example.
Z [x]
Example 21. (i) Suppose that R ∼
= (x6m ) , where m ≥ 2. Let I1 = (3), I2 = (3x),
I3 = (3x + 3), J1 = (2), J2 = (4), J3 = (2x), J4 = (4x), J5 = (2x + 2), J6 = (4x + 2)
and J7 = (2x+4). Then one can check that these ideals are distinct vertices of AGN (R).
Also, every Ii (1 ≤ i ≤ 3) is adjacent to every Jk (1 ≤ k ≤ 7). Thus K3,7 is a subgraph
of AGN (R) and so the formula of genus for the complete bipartite graphs implies that
γ(AGN (R)) ≥ 2.
Z [x]
(ii) Let R ∼
= (x4 3 ) . Set I1 = (2x), I2 = (2x2 ), I3 = (2x + 2x2 ), J1 = (2), J2 = (2 + x2 ),
J3 = (2 + 2x2 ), J4 = (2 − x2 ), J5 = (2 + 2x), J6 = (2 + 2x + x2 ) and J7 = (2 + 2x + 2x2 ).
Similar to (i), one can show that every Ii is adjacent to every Jk and so γ(AGN (R)) ≥ 2.

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