7_753_behboodi.dvi


EUROPEAN JOURNAL OF PURE AND APPLIED MATHEMATICS

Vol. 3, No. 4, 2010, 686-694

ISSN 1307-5543 – www.ejpam.com

On the Structure of Commutative Rings with p1
k1 · · ·pn

kn (1≤ ki ≤ 7)

Zero-Divisors II

M. Behboodi1,2,∗and R. Beyranvand 3

1 Department of Mathematical Science, Isfahan University of Technology, Isfahan, Iran
2 School of Mathematics, Institute for Research in Fundamental Sciences (IPM), Tehran, Iran
3 Faculty of Science, Department of Mathematics, Lorestan University, Khorramabad, Iran

Abstract. In this paper, we determine the structure of nonlocal commutative rings with p6 zero-

divisors and characterize the structure of nonlocal commutative rings with p7 zero-divisors. Also, the

structure and classification up to isomorphism all commutative rings with p1
k1 . . . pn

kn zero-divisors,

where n is a positive integer, pi
,s are distinct prime number and 1≤ ki ≤ 4, are determined.

2000 Mathematics Subject Classifications: 16B99; 13A99; 68R10

Key Words and Phrases: Finite ring, Zero-divisor, Local ring

1. Introduction

The present paper is a sequel to [2] and so the notations introduced in Introduction of [2]

will remain in force. In particular, all rings are associative rings with identity elements, J(R)

denotes the jacobson radical of R, Z(R) denotes the set of all zero-divisors of R and for any

finite subset Y of R, we denote |Y | for the cardinality of Y . Also, Fq is the finite field of order

q, Fq
∗ is the group of nonzero elements of Fq and for a prime number p, Σm is a set of coset

representation of (Fp
∗)m in Fp

∗, Σ0
m = Σm∪{0} and GR(pnr , pr) is the Galois ring of order pnr

and characteristic pr .

In [2] the structure and classification up to isomorphism all rings with p1
k1 . . . ps

ks zero-

divisors, where s is a positive integer, pi
,s are distinct prime number and 1 ≤ ki ≤ 3 were

determined. Also we determined the structure of nonlocal rings with pk zero-divisors where

k = 4 or 5. In the paper we develop these results. In fact the structure and classification up

to isomorphism all rings with p1
k1 p2

k2 . . . ps
ks zero-divisors, where s is a positive integer, pi

,s

are distinct prime number and 1≤ ki ≤ 4 are determined. Also we determine the structure of

nonlocal rings with p6 zero-divisors and characterize the structure of nonlocal rings with p7

zero-divisors.

∗Corresponding author.

Email addresses: mbehbood�

.iut.a
.ir (M. Behboodi), beyranvand.r�lu.a
.ir (R. Beyranvand)

http://www.ejpam.com 686 c© 2010 EJPAM All rights reserved.



M. Behboodi, R. Beyranvand / Eur. J. Pure Appl. Math, 3 (2010), 686-694 687

2. On Rings with pk Zero-divisors

We recall the following facts that we will use them in the paper:

(i) An Artinian commutative ring R is called completely primary if R/J(R) is a field. One

can easily see that an Artinian commutative ring R is completely primary if and only if

Z(R) is an ideal of R, if and only if R is a local ring.

(ii) Let Ri (1 ≤ i ≤ t) be a nonzero finite commutative ring with mi elements and ni zero-

divisors. Then by [6, Theorem 2], the ring R1×. . .×R t has m1m2 . . . mt−(m1−n1)(m2−
n2) . . . (mt − nt) zero-divisors.

(iii) Every finite commutative ring is uniquely expressible as a direct sum of completely

primary (local) rings (see for example [7, p.95]).

We need the following two lemmas which are crucial in our investigation.

Lemma 1. [8, Theorem 2] Let R be a finite completely primary ring. Then

1. Z(R) = J(R);

2. |Z(R)|= p(n−1)r and |R|= pnr for some prime number p, and some positive integers n, r;

3. Z(R)n = 0;

4. char(R) = pk for some integer k with 1≤ k ≤ n;

5. R/J(R)∼= Fq, where q = pr .

Lemma 2. [2, Theorem 2] Let R be a commutative ring such that |Z(R)| = pk for some prime

number p and a positive number k. Then either

(i) R is local,

(ii) R is reduced or

(iii) k ≥ 3 and R∼= R1× . . .×Rs× Fq1
× . . .× Fqt

where s and t are positive integers, each Fqi
is

a field, and where each Ri is a commutative finite local ring with |Z(Ri)| = pti , |Ri| = pki

for some positive integers ki and t i with 1≤
∑s

i=1 t i ≤
∑s

i=1 ki − s ≤ k− s− 1 such that

pk−Σs
i=1 ti = q1 . . . qt p

Σs
i=1(ki−ti) − (q1− 1) . . . (qt − 1)Πs

i=1(p
ki−ti − 1). (1)

Consequently, in the latter case, qi ≡ 1 (p) and for each i = 1, . . . , s, t i ≤ k − 2. Moreover, if

t j = k− 2 for some j ∈ {1, . . . , s}, then s = t = 1, i.e., R∼= R1× Fq where |Z(R1)|= pk−2 and so

p2 = p+ q− 1.



M. Behboodi, R. Beyranvand / Eur. J. Pure Appl. Math, 3 (2010), 686-694 688

Also we need the following construction [3, p.5071].

Construction A. Let R0 be the Galois ring GR(p2r , p2) or GR(p3r , p3). Let s, d , t, λ be integers

with either 1≤ t ≤ s2, 1≤ 1+ t ≤ s2 or 1≤ d+ t ≤ s2 if char(R0) = p2 or 1≤ 1+d+ t ≤ 1+s2

if char(R0) = p3, and λ≥ 0. Let V,W be R0/pR0-spaces which when considered as R0-modules

have generating sets {v1, . . . , vλ} and {w1, . . . , wt} respectively. Let U be an R0-module with

an R0-modules generating set {u1, . . . ,us}; and suppose that d ≥ 0 of the ui are such that

pui 6= 0. Since R0 is commutative, we can think of them as left and right R0-module.

Let (al
i j
), for l = 0,1, . . . , t, t + 1 or d + t, be s × s matrices with entries in R0/pR0 if

char(R0) = p2 or l = 0,1, . . . , d + t be (1+ s) × (1 + s) matrices with entries in R0/pR0 if

char(R0) = p3.

Consider the additive group direct sum R= R0 ⊕ U ⊕ V ⊕W and define a multiplication on R

by

(α0,
∑s

i=1αiui ,
∑λ

j=1 β j v j,
∑t

k=1 γkwk).(α
′
0,
∑s

i=1α
′
iui ,
∑λ

j=1 β
′
j v j,
∑t

k=1γ
′
k
wk) =

(α0α
′
0+p f
∑s

i, j=1 a0
i j
[αiα

′
j
+pR0],
∑s

i=1[α0α
′
i
+αiα

′
0+p
∑s

i, j=1 ai
i j
[αiα

′
j
+pR0]]ui ,
∑λ

j=1[(α0+

pR0)β
′
j
+β j(α

′
0+ pR0)]v j,
∑t

k=1[(α0+ pR0)γ
′
k
+γk(α

′
0+ pR0)+
∑s

i, j=1 ad+k
i j
[αiα

′
j
+ pR0]]wk)

where f = 1 or 2, depending on whether char(R) = p2 or p3. Then by [3, Theorem 6.1], this

multiplication turns R into a ring and any local ring with Z(R)3 = 0, Z(R)2 6= 0 of characteris-

tic p2 or p3, is isomorphic to one given by construction A.

Proposition 1. Let R be a commutative ring with |Z(R)| = p4 and |R| = p6 where p is a prime

number. Then R is isomorphic to one of the rings GR(p6, p3), Fp2 ⊕ Fp2 ⊕ Fp2 with multiplication

(r0, r1, r2)(s0, s1, s2) = (r0s0, r0s1 + r1s0, r0s2 + r2s0),S ⊕ F with multiplication (r0, r1)(s0, s1) =

(r0s0, r0s1 + r1s0), where S = GR(p4, p2) and F = S/pS, Fp2 ⊕ Fp2 ⊕ Fp2 with multiplication

(α0,α,γ)(α′0,α′,γ′) = (α0α
′
0,α0α

′+αα′0,α0γ
′+γα′0+αα

′) or R0⊕R0/pR0 with multiplication

(α0,α+ pR0)(α
′
0,α′ + pR0) = (α0α

′
0 +αα

′p,α0α
′ +αα′0 + pR0) where R0 = GR(p4, p2).

Proof. Since R is a ring with |Z(R)| = p4 and |R| = p6, by Lemma 1, Z(R)3 = 0. Thus we

consider the following cases.

Case 1: Z(R)2 = 0 i.e., R is a ring in which the multiplication of any two zero-divisors is zero.

Then by [1, Theorem 1], R is isomorphic to one of the rings S⊕F k, where S is either the

field of pr elements or the Galois ring GR(p2r , p2) and F = S/pS with the multiplication

(r0, r1, . . . , rk)(s0, s1, . . . , sk) = (r0s0, r0s1 + r1s0, . . . , r0sk + rks0)

for some positive integers r and k.

Now since |Z(R)| = p4 and |R| = p6, we can conclude that S = Fp2 and k = 2 or

S = GR(p4, p2) and k = 1. Thus R is isomorphic to one of the rings Fp2 ⊕ Fp2 ⊕ Fp2

with (r0, r1, r2)(s0, s1, s2) = (r0s0, r0s1 + r1s0, r0s2 + r2s0) or S ⊕ F with (r0, r1)(s0, s1) =

(r0s0, r0s1 + r1s0), where S = GR(p4, p2) and F = S/pS.

Case 2: Z(R)2 6= 0. If char(R) = p, then by [ 3, Theorem 4.1], any commutative local ring of

characteristic p in which the multiplication of any two zero-divisors is zero, is isomor-



M. Behboodi, R. Beyranvand / Eur. J. Pure Appl. Math, 3 (2010), 686-694 689

phic to one of the rings F ⊕ U ⊕ V ⊕W with multiplication

(α0,

s∑

i=1

αiui ,

λ∑

j=1

β j v j,

t∑

k=1

γkwk)(α
′
0,

s∑

i=1

α′iui,

λ∑

j=1

β ′j v j ,

t∑

k=1

γ′kwk)

= (α0α
′
0,

s∑

i=1

[α0α
′
i +αiα

′
0]ui,

λ∑

j=1

[α0β
′
j + β jα

′
0]v j ,

t∑

k=1

[α0γ
′
k + γkα

′
0 +

s∑

i, j=1

ak
i, jαiα

′
j]wk)

where F is the field of order r and U , V,W are s,λ, t -dimensional F -spaces respectively,

for some integers s,λ, t with λ ≥ 0 and 1 ≤ t ≤ s2, where {ui}, {vi} and {wi} are bases

for U , V and W respectively, and (ak
i, j
) 1≤ k ≤ t are t matrices of size s× s with entries

in F . Since |R| = p6, we can conclude that t = s = 1 and λ = 0. On the other hand by

[3, Corollary 5.2], we can put a1
11 = 1. Thus R ∼= Fp2 ⊕ Fp2 ⊕ Fp2 with multiplication

(α0,α,γ)(α′0,α′,γ′) = (α0α
′
0,α0α

′ +αα′0,α0γ
′ + γα′0 +αα

′).

Now suppose that char(R) = p2 or p3. Since |R|= p6 and |R/J(R)|= p2, by construction

A we conclude that s = 1 and t = λ = 0 if char(R) = p2 and s = t = λ = 0 if

char(R) = p3. Also by [3, Lemma 7.1], we can put a0
11 = 1 and so the following rings is

obtained.

If char(R) = p2, then R∼= R0⊕R0/pR0 with multiplication (α0,α+pR0).(α
′
0,α′+pR0) =

(α0α
′
0 + αα

′p,α0α
′ + αα′0 + pR0), where R0 = GR(p4, p2) and if char(R) = p3, then

R∼= GR(p6, p3).

Proposition 2. Let R be a commutative ring with |Z(R)| = p4 and |R| = p5 where p is a

prime number. Then R is isomorphic to one of the rings Zp5 , Fp[x]/(x
5), Fp[x , y]/(x4, x y, y2),

Fp[x , y]/(x4, x y, y2 − x3), Zp[x , y, z, t]/(x , y, z, t)2 , Zp2[x]/(px , x4 − ap) where a ∈ Σ0
4,

Zp2[x]/(px2, x3 − bp) where b ∈ Σ3 and p 6= 3, Z9[x]/(3x2, x3 − 3 − 3bx) where b ∈
{−1,0,1}, Zp2[x]/(px2, x3− apx) where a ∈ Σ0

2, Zp2[x , y, z]/(p, x , y, z)2 , Zp3[x]/(p2 x , x2−
ap) where a ∈ Σ2 and p 6= 2, Z8[x]/(4x , x2−2a−2bx) where (a, b) ∈ {(1,0), (1,1), (−1,1)},
Zp3[x]/(px , x3 − ap2) where a ∈ Σ0

3, Zp3[x]/(p2 x , x2 − ap2) where a ∈ Σ0
2 and p 6= 2,

Z8[x]/(4x , x2− 4a− 2bx) where (a, b) ∈ {(0,0), (0,1), (1,1)}, Zp4[x]/(px , x2− ap3) where

a ∈ Σ0
2,

〈1, x1, x2, y1, y2; p1 = 0, x1
2 = y1, x2

2 = 0, x1 x2 = y2, x i yi = yi y j = 0〉,
〈1, x1, x2, y1, y2; p1 = 0, x1

2 = x2
2 = y1, x1 x2 = y2, x i yi = yi y j = 0〉 where p 6= 2,

〈1, x1, x2, y1, y2; p1= 0, x1
2 = y1, x2

2 = ξy1, x1 x2 = y2, x i yi = yi y j = 0〉 where p 6= 2 and

ξ is a non-square in Fp,

〈1, x1, x2, y1, y2; 2.1= 0, x1
2 = y1, x2

2 = y2, x1 x2 = y2, x i yi = yi y j = 0〉,
〈1, x1, x2, y1, y2; 2.1= 0, x1

2 = y1, x2
2 = y1 + y2, x1 x2 = y2, x i yi = yi y j = 0〉,

〈1, x1, x2, x3, y; p1= 0, x1
2 = y, x2

2 = x3
2 = x i x j = x i y = y2 = 0, for i 6= j〉,

〈1, x1, x2, x3, y; p1= 0, x1
2 = x2

2 = y, x3
2 = x i x j = x i y = y2 = 0, for i 6= j〉,

〈1, x1, x2, x3, y; p1= 0, x i
2 = y, x i x j = x i y = y2 = 0, for i 6= j〉,

〈1, x1, x2, x3, y; p1 = 0, x1
2 = y, x2

2 = εy, x3
2 = x i x j = x i y = y2 = 0, for i 6= j〉 where

p 6= 2 and ε is a non-square in Fp,



M. Behboodi, R. Beyranvand / Eur. J. Pure Appl. Math, 3 (2010), 686-694 690

〈1, x1, x2, x3, y; 2.1= 0, x i
2 = x1 x2 = x1 x3 = x i y = y2 = 0, x2 x3 = y〉,

〈1, x , y, z, p; p21 = 0, x2 = αz, xz = p, y2 = δp, z2 = x y = yz = 0〉 where α ∈ Σ3 and

δ ∈ Σ0
2,

〈1, x , y; p21 = p2 x = p y = 0, x2 = αp, y2 = δpx , x y = 0〉 where p 6= 2, α ∈ Σ2 and δ = 0

or α ∈ Σ4 and δ = 1,

〈1, x , y; 4.1= 4x = 2y = 0, x2 = 2, y2 = x y = 0〉,
〈1, x , y; 4.1= 4x = 2y = 0, x2 = 2+ 2x , y2 = x y = 0〉,
〈1, x , y; 4.1= 4x = 2y = 0, x2 = 2, y2 = 2x , x y = 0〉,
〈1, x1, x2, x3, p; p21 = px i = 0, x1

2 = νp, x2
2 = x3

2 = 0, x i x j = 0 for i 6= j〉 where p 6= 2, ε

is a non-square in Fp and ν ∈ {1,ε},
〈1, x1, x2, x3, p; p21 = px i = 0, x1

2 = 1, x2
2 = νp, x3

2 = 0, x i x j = 0 for i 6= j〉 where p 6= 2,

ε is a non-square in Fp and ν ∈ {1,ε} ,

〈1, x1, x2, x3, p; p21 = px i = 0, x1
2 = x2

2 = 1, x3
2 = νp, x i x j = 0 for i 6= j〉 where p 6= 2, ε

is a non-square in Fp and ν ∈ {1,ε},
〈1, x1, x2, x3; 4.1= 2x i = 0, x1

2 = 2, x2
2 = x3

2 = 0, x i x j = 0 for i 6= j〉,
〈1, x1, x2, x3; 4.1= 2x i = 0, x1

2 = 1, x2
2 = 2, x3

2 = 0, x i x j = 0 for i 6= j〉,
〈1, x1, x2, x3; 4.1= 2x i = 0, x1

2 = x2
2 = 1, x3

2 = 2, x i x j = 0 for i 6= j〉,
〈1, x1, x2, y; p21= px i = p y = 0, x1

2 = 1, x2
2 = 0, x1 x2 = x1 y = x2 y = 0〉,

〈1, x1, x2, y; p21= px i = p y = 0, x1
2 = 1, x2

2 = y, x1 x2 = x1 y = x2 y = 0〉,
〈1, x1, x2, y; p21 = px i = p y = 0, x1

2 = 1, x2
2 = ξy, x1 x2 = x1 y = x2 y = 0〉 where p 6= 2,

ξ is a non-square in Fp,

〈1, x1, x2, y; 4.1= 2x i = 2y = 0, x1
2 = x2

2 = x1 y = x2 y = 0, x1 x2 = y〉,
〈1, x , y, p; p21= p2 x = p y = 0, x2 = 0, y2 = δpx , x y = 0〉 where δ ∈ {0,1} and p 6= 2,

〈1, x , y; 4.1= 4x = 2y = 0, x2 = α2x , y2 = δ2x , x y = 0〉where (α,δ) ∈ {(0,0), (1,0), (1,1)},
〈1, x1, x2; p31= px i = 0, x1

2 = x2
2 = 0, x1 x2 = 0〉,

〈1, x1, x2; p31= px i = 0, x1
2 = p2, x2

2 = x1 x2 = 0〉,
〈1, x1, x2; p31= px i = 0, x1

2 = εp2, x2
2 = x1 x2 = 0〉 where p 6= 2,

〈1, x1, x2; p31= px i = 0, x1
2 = x2

2 = p2, x1 x2 = 0〉,
〈1, x1, x2; p31 = px i = 0, x1

2 = p2, x2
2 = εp2, x1 x2 = 0〉 where p 6= 2 and ε is a non-square

in Fp,

〈1, x1, x2; 8.1= 2x i = 0, x1
2 = x2

2 = 0, x1 x2 = 4〉 or

one of the rings given in full in [9]. The number of these rings is 10 or 6 according to whether

p 6= 2 or p = 2.

Proof. By using [5] and [9] one can check that R is isomorphic to one of the above rings.

Theorem 1. Let R be a ring with |Z(R)| = p4, where p is a prime number. Then R is isomor-

phic to one of the rings described in Proposition 1, Proposition 2, the Galois ring GR(p8, p2),

Fp4[x]/(x2), Zp2 × Fq1
× . . .× Fqt

, Zp[x]/(x
2)× Fq1

× . . .× Fqt
where p3 = p2q1 . . . qt − (p

2 −
p)(q1 − 1) . . . (qt − 1), Fq1

× . . . × Fqt
where p4 = q1q2 . . . qt − (q1 − 1)(q2 − 1) . . . (qt − 1) or

R1 × Fq with p2 = p + q− 1 where R1 is isomorphic to one of the rings Zp3 , Fp[x , y]/(x , y)2,

Fp[x]/(x
3) or Zp2[x]/(px , x2− ǫp) where ǫ ∈ Σ0

2.



M. Behboodi, R. Beyranvand / Eur. J. Pure Appl. Math, 3 (2010), 686-694 691

Proof. Suppose that R is a local ring. Then by Lemma 1, |R|= p5, p6 or p8. If |R|= p5 or p6,

then R is isomorphic to one of the rings described in Proposition 1 or Proposition 2. If |R|= p8,

then by [8, Theorem 12], R is isomorphic to the Galois ring GR(p8, p2) or Fp4[x]/(x2). Now

suppose R is a nonlocal ring. If R is reduced, then we are down. Otherwise by Lemma 2,

1≤
∑s

i=1 t i ≤ 2 and hence 1≤ s ≤ 2. Thus we proceed by cases.

Case 1: s = 1. Then t1 = 1 or 2. If t1 = 1, then R∼= R1 × Fq1
× . . .× Fqt

, where R1 is a local ring

of order p2 with p zero-divisors. By [4, p.687], R1 is isomorphic to Zp2 or Zp[x]/(x
2).

If t1 = 2, then by Lemma 2, R ∼= R1 × Fq, where R1 is a local ring of order p3 with

p2 zero-divisors and p2 = p + q− 1. Moreover by [4, p.687], R1 is isomorphic to Zp3 ,

Fp[x , y]/(x , y)2, Fp[x]/(x
3) or Zp2[x]/(px , x2− ǫp) where ǫ ∈ Σ0

2.

Case 2: s = 2, i.e., t1 = t2 = 1. Then R∼= R1× R2× Fq1
× . . .× Fqt

, where each Ri is a local ring

with |Z(Ri)| = p. Now by Lemma 1, |R1| = |R2| = p2. If t > 1, then clearly |Z(R)|> p4,

a contradiction. Therefore t = 1 and hence by relation (1) in Lemma 2, p2 is a divisor

of q1− 1. Thus q1 > p2 and so |Z(R)|> |Z(R1)||R2||Fq1
| ≥ p5, a contradiction.

Corollary 1. Let R be a ring with |Z(R)| = p
k1

1 p
k2

2 . . . p
kn
n , where n ≥ 1, 1 ≤ ki ≤ 4 and pi

,s are

distinct prime numbers. Then there exist 0≤ s ≤ Σn
i=1ki and t ≥ 0 such that

R∼= R1 × . . .× Rs × Fq1
× . . .× Fqt

where Fqi

,s are finite fields and each Ri is local ring with |Z(Ri)| = p
t j

j
for some p j (1 ≤ j ≤ n)

and 1 ≤ t j ≤ k j . Consequently, each Ri is isomorphic to one of the local rings described in [2,

Theorem 5] or Theorem 1.

Proof. We put

R∼= R1× . . .× Rs × Fq1
× . . .× Fqt

,

where Fq1
, . . . , Fqt

are finite fields and each Ri is a commutative finite local ring with identity

that is not a field. By Lemma 2, for each i, |Z(Ri)| = pk for some prime number p and

k ≥ 1 such that pk is a divisor of |Z(R)| and also 0 ≤ s ≤ Σn
i=1

ki. Thus |Z(Ri)| = p
t j

j
where

1≤ t j ≤ k j, 1 ≤ j ≤ n and 1≤ i ≤ s. Hence for each 1≤ i ≤ s, t i = 1,2,3 or 4 and so each Ri

is isomorphic to one of the local rings described in [2, Theorem 5] or Theorem 1.

Theorem 2. Let R be a commutative nonlocal ring with |Z(R)|= p6 where p is a prime number.

Then R is isomorphic to one of the rings Fq1
× . . . × Fqt

with p6 = q1q2 . . . qt − (q1 − 1)(q2 −
1) . . . (qt − 1), Z4 × Z4 × F5,Z2[x]/(x

2) × Z4 × F5, Z2[x]/(x
2) × Z2[x]/(x

2) × F5, R1 ×
Fq1
× . . . × Fqt

, where R1 is isomorphic to Zp2 or Zp[x]/(x
2) and p5 = pq1q2 . . . qt − (p −

1)(q1− 1)(q2− 1) . . . (qt − 1), R1 × Fq1
× . . .× Fqt

where R1 is isomorphic to one the rings Zp3 ,

Fp[x , y]/(x , y)2, Fp[x]/(x
3) or Zp2[x]/(px , x2 − ǫp) where ǫ ∈ Σ0

2 and p4 = pq1q2 . . . qt −
(p− 1)(q1− 1)(q2− 1) . . . (qt − 1), R1× Fq1

× . . .× Fqt
, where R1 is isomorphic to Fp2[x]/(x2)

or GR(p4, p2) and p4 = p2q1q2 . . . qt − (p
2−1)(q1−1)(q2−1) . . . (qt −1), R1× Fq1

× . . .× Fqt
,



M. Behboodi, R. Beyranvand / Eur. J. Pure Appl. Math, 3 (2010), 686-694 692

where R1 is isomorphic to one of the local rings of order p4 described in [2, corollary 3] and

p3 = pq1q2 . . . qt − (p− 1)(q1− 1)(q2− 1) . . . (qt − 1) or R1× Fq where R1 is isomorphic to one

of the rings described in Proposition 2, and p2 = p+ q− 1.

Proof. If R is reduced, then we are down. Now suppose that R is not reduced. Then by

[2, Theorem 4], either R is isomorphic to Z4 × Z4 × F5, Z2[x]/(x
2)×Z4 × F5, Z2[x]/(x

2)×
Z2[x]/(x

2) × F5 or R ∼= R1 × Fq1
× . . . × Fqt

, where R1 is a local ring with |Z(R1)| = pk

(1≤ k ≤ 4) and t ≥ 1. Thus we proceed by cases.

Case 1: |Z(R1)| = p. Then by [4, p.687], R1 is isomorphic to Zp2 or Zp[x]/(x
2) and p5 =

pq1q2 . . . qt − (p− 1)(q1− 1)(q2− 1) . . . (qt − 1).

Case 2: |Z(R1)| = p2. Then by Lemma 1, |R1| = p3 or p4. If |R1| = p3, then by [4, p.687], R1 is

isomorphic to one the rings Zp3 , Fp[x , y]/(x , y)2, Fp[x]/(x
3) or Zp2[x]/(px , x2− ǫp)

where ǫ ∈ Σ0
2 and p4 = pq1q2 . . . qt − (p − 1)(q1 − 1)(q2 − 1) . . . (qt − 1). If |R1| = p4,

then by [8, Theorem 12], R1 is isomorphic to Fp2[x]/(x2) or GR(p4, p2) and p4 =

p2q1q2 . . . qt − (p
2 − 1)(q1− 1)(q2− 1) . . . (qt − 1).

Case 3: |Z(R1)| = p3. Then by Lemma 1, |R1| = p4 or p6. If |R1| = p6, then |Z(R)| > |R1| which

is impossible. Thus |R1| = p4 and so R1 is isomorphic to one of the local rings of order

p4 described in [2, corollary 3] and p3 = pq1q2 . . . qt−(p−1)(q1−1)(q2−1) . . . (qt−1).

Case 4: |Z(R1)|= p4. Then by Lemma 1, |R1| = p5, p6 or p8. If |R1| = p6 or p8, then |Z(R)|> |R1|
which is impossible. Thus |R1| = p5 and so R1 is isomorphic to one of the rings described

in Proposition 2. Also by Lemma 2, R∼= R1× Fq and p2 = p+ q− 1.

Theorem 3. Let R be a commutative nonlocal ring with |Z(R)|= p7 where p is a prime number.

Then R is isomorphic to one of the rings Fq1
× . . . × Fqt

with p7 = q1q2 . . . qt − (q1 − 1)(q2 −
1) . . . (qt−1), Z4×Z4×F3×F3, Z2[x]/(x

2)×Z4×F3×F3, Z2[x]/(x
2)×Z2[x]/(x

2)×F3×F3,

R1 × R2 × F5 where R1 is isomorphic to Z4 or Z2[x]/(x
2) and R2 is isomorphic to one of the

rings Z8, Z2[x , y]/(x , y)2, Z2[x]/(x
3) or Z4[x]/(2x , x2− 2ǫ) where ǫ ∈ Σ0

2, R1 × R2 × Fq1
×

. . .× Fqt
, where p is an odd prime number, each Ri is isomorphic to Zp2 or Zp[x]/(x

2) and p5 =

p2q1q2 . . . qt−(p−1)2(q1−1)(q2−1) . . . (qt−1),R1×Fq1
×. . .×Fqt

where R1 is isomorphic to Zp2

or Zp[x]/(x
2)with p6 = pq1q2 . . . qt−(p−1)(q1−1)(q2−1) . . . (qt−1), R1×Fq1

×. . .×Fqt
where

R1 is isomorphic to one the rings Zp3 , Fp[x , y]/(x , y)2, Fp[x]/(x
3) or Zp2[x]/(px , x2 − ǫp)

where ǫ ∈ Σ0
2 with p5 = pq1q2 . . . qt−(p−1)(q1−1)(q2−1) . . . (qt−1), R1×Fq1

×. . .×Fqt
where

R1
∼= Fp2[x]/(x2) or GR(p4, p2) with p5 = p2q1q2 . . . qt − (p

2 − 1)(q1− 1)(q2− 1) . . . (qt − 1),

R1 × Fq1
× . . .× Fqt

where R1 is isomorphic to one of the local rings of order p4 described in [2,

Corollary 3] and p4 = pq1q2 . . . qt − (p − 1)(q1 − 1)(q2 − 1) . . . (qt − 1), R1 × Fq where R1
∼=

Fp3[x]/(x2) or GR(p6, p2) with p4 = p3+p−1, R1×Fq1
×. . .×Fqt

where R1 is isomorphic to one

of the rings described in Proposition 2, with p3 = pq1q2 . . . qt−(p−1)(q1−1)(q2−1) . . . (qt−1),

R1×Fq where R1 is isomorphic to one of the rings described in Proposition 1, with p3 = p2+q−1

or R1× Fq , where R1 is a local ring of order p6 with p5 zero-divisors and p2 = p+ q− 1.



M. Behboodi, R. Beyranvand / Eur. J. Pure Appl. Math, 3 (2010), 686-694 693

Proof. If R is reduced or R∼= R1×R2× F5 where R1 is isomorphic to Z4 or Z2[x]/(x
2) and

R2 is isomorphic to one of the rings Z8, Z2[x , y]/(x , y)2, Z2[x]/(x
3) or Z4[x]/(2x , x2− 2ǫ)

where ǫ ∈ Σ0
2, then we are done. Otherwise by [2, Theorem 4], we have the following two

cases.

Case 1: R∼= R1× Fq1
× . . .× Fqt

, where each Fqi
(1≤ i ≤ t) is a finite field and R1 is a local ring

with |Z(R1)| = pm, |R1| = pn such that 0< m < n≤ 6 and

p7 = pnq1q2 . . . qt − (p
n − pm)(q1− 1)(q2− 1) . . . (qt − 1).

Case 2: R ∼= R1 × R2 × Fq1
× . . . × Fqt

, where each Fqi
(1 ≤ i ≤ t) is a finite field, each Ri is

isomorphic to Zp2 or Zp[x]/(x
2) and

p5 = p2q1q2 . . . qt − (p− 1)2(q1− 1)(q2− 1) . . . (qt − 1). (2)

In case 1, as in the proof of Theorem 2, R is isomorphic to one of the rings R1 × Fq1
×

. . . × Fqt
where R1 is isomorphic to Zp2 or Zp[x]/(x

2) with p6 = pq1q2 . . . qt − (p −
1)(q1 − 1)(q2 − 1) . . . (qt − 1), R1 × Fq1

× . . . × Fqt
, where R1 is isomorphic to one the

rings Zp3 , Fp[x , y]/(x , y)2, Fp[x]/(x
3) or Zp2[x]/(px , x2−ǫp) where ǫ ∈ Σ0

2 with p5 =

pq1q2 . . . qt−(p−1)(q1−1)(q2−1) . . . (qt−1), R1×Fq1
×. . .×Fqt

, where R1 is isomorphic

to Fp2[x]/(x2) or GR(p4, p2) with p5 = p2q1q2 . . . qt−(p
2−1)(q1−1)(q2−1) . . . (qt−1),

R1×Fq1
× . . .×Fqt

, where R1 is isomorphic to one of the local rings of order p4 described

in [2, Corollary 3] with p4 = pq1q2 . . . qt − (p− 1)(q1 − 1)(q2 − 1) . . . (qt − 1), R1 × Fq

where R1 is isomorphic to Fp3[x]/(x2) or GR(p6, p2) with p4 = p3 + p− 1, R1 × Fq1
×

. . . × Fqt
where R1 is isomorphic to one of the rings described in Proposition 2, with

p3 = pq1q2 . . . qt − (p−1)(q1−1)(q2−1) . . . (qt −1), R1× Fq where R1 is isomorphic to

one of the rings described in Proposition 1, with p3 = p2+ q− 1 or R1× Fq, where R1 is

a local ring of order p6 with p5 zero-divisors.

In case 2, If p = 2, then Since |Z(R)| > |R1||R2|q1 . . . qt−1, t ≤ 3. We claim that t = 2.

If t = 1, then the relation (2) implies that q1 = 31/3, a contradiction. If t = 3, then

the relation (2) implies that 4 is a divisor of (q1 − 1)(q2 − 1)(q3 − 1). Without loss of

generality we can assume that either 4 is a divisor of (q1 − 1) or 2 is a divisor of both

(q1 − 1) and (q2 − 1). Therefore either q1 ≥ 5 or q1 ≥ 3 and q2 ≥ 3. This implies that

either 27 = |Z(R)| > 5|R1||R2|q2 = 80q2 or 27 = |Z(R)| > 9|R1||R2| = 144. But it is

impossible in any case. Thus t = 2 and by the relation (2) we have

25 = 22q1q2− (q1 − 1)(q2− 1). (3)

If 4 is a divisor of (qi − 1) for some i, then qi ≥ 5 and hence 25 = 3q1q2+ q1+ q2− 1≥
30+7−1= 36 , a contradiction. Thus 2 is a divisor of both (q1−1) and (q2−1). Then

q1−1= 2k1, q2−1= 2k2 for some positive integers k1, k2 and put them into (3). Then

we obtain 8= 3k1k2+ 2k1+ 2k2+ 1. It yields k1 = k2 = 1 and hence q1 = q2 = 3. Thus

R∼= R1 × R2× F3 × F3 where R1 and R2 are isomorphic to Z4 or Z2[x]/(x
2).



REFERENCES 694

ACKNOWLEDGEMENTS This work was partially supported by IUT (CEAMA). The research

of the first author was in part supported by a grant from IPM (No. 87160026).

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