(m, n)-Prime Ideals of Commutative Rings

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(m, n)-Prime Ideals of Commutative Rings

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03 January 2024

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05 January 2024

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Abstract

Let R be a commutative ring with identity and m, n be positive integers. In this paper, we introduce the class of (m,n)-prime ideals which lies properly between the classes of prime and (m,n)-closed ideals. A proper ideal I of R is called (m,n)-prime if for a,b∈R, a^{m}b∈I implies either aⁿ∈I or b∈I. Several characterizations of this new class with many examples are given. Analougus to primary decomposition, we define the (m,n)-decomposition of ideals and show that every ideal in an n-Noetherian ring has an (m,n)-decomposition. Furthermore, the (m,n)-prime avoidance theorem is proved.

Keywords:

Subject: Computer Science and Mathematics - Algebra and Number Theory

1. Introduction

Throughout, all rings are assumed to be a commutative with identity. For such a ring R, by

I d (R)

N (R),

U (R)

and

dim (R)

, we denote the set of idempotent, the set of nilpotent, the set of unit elements and the Krull dimension of R. Moreover, for a proper ideal I of R,

\sqrt{I}

denotes the prime radical of

I .

As a more general concept than prime ideals, in 2011, Anderson and Badawi defined n-absorbing ideals, [1]. A proper ideal I of a ring R is called n-absorbing if whenever

a_{1} \dots a_{n + 1} \in I

for

a_{1}, \dots, a_{n + 1} \in R

, then there are n of the

a_{i}

’s whose product is in I. We also recall that a proper ideal I of a ring R is said to be semiprime if whenever

x^{2} \in I

for some

x \in R

, then

x \in I .

Generalizing this concept, for a positive integer n, a proper ideal I of a ring R is called semi-n-absorbing if for

x \in R

x^{n + 1} \in I

implies

x^{n}

\in I

[2]. It is clear that every n-absorbing ideal is semi n-absorbing. Let m and n be positive integers. Afterwards, in [2], the structure of

(m, n)

-closed ideals is first introduced. A proper ideal I of a ring R is called an

(m, n)

-closed ideal of R if whenever

a^{m} \in I

for some

a \in I

, then

a^{n} \in I

. On the other hand, in 2020, Badawi and Yetkin Celikel introduced the concept of 1-absorbing primary ideals. According to [8], a proper ideal I of a ring R is said to be a 1-absorbing primary if for non-unit elements

a, b, c \in R

such that

a b c \in I

, then either

a b \in I

c \in \sqrt{I}

. Following this paper, a subclass of 1-absorbing primary ideals is given in [15]. A proper ideal I of R is called 1-absorbing prime if for non-unit elements

a, b, c \in R

with

a b c \in I

, then either

a b \in I

c \in I

. For the related papers, the reader may consult [4,5,6,7,8] and [13].

Motivated and inspired from the ideal notions above, in this paper, we introduce

(m, n)

-prime ideals which is a structure lies between a prime and primary ideals, i.e. Prime ideal ⇒

(m, n)

-prime ideal ⇒ primary ideal. We call a proper ideal of a ring R a

(m, n)

-prime ideal where

m, n

are positive integers if for

a, b \in R

a^{m} b \in I

implies either

a^{n} \in I

b \in I

. In Section 2, we discuss all relationships among the ideal types listed above and the new one by supporting many examples (Remark 2 and Example 3). Furthermore, several characterizations of

(m, n)

-prime ideals of rings are given. We determine all

(m, n)

-prime ideals of some special rings such as integral domains and zero dimensional rings. Among many other results in this paper, a characterization for rings in which every ideal is

(m, n)

-prime is given (Theorem 11). Let I be an ideal of a ring R and n a positive integer. We define I to be of maximum length n if any ascending chain

I = I_{0} \subseteq I_{1} \subseteq I_{2} \subseteq \dots

of ideals of a ring R terminates and n is the largest integer such that

I_{n} = I_{n + 1} = \dots

. Moreover, a commutative ring R is called n-Noetherian if every ideal of R has a maximum length at most n. Analogous to primary ideal case, we introduce the

(m, n)

-decomposition of I which is an expression for I as a finite intersection of

(m, n)

-prime ideals. It is proved that every ideal in an n-Noetherian ring has

(m, n)

-decomposition (Theorem 23). In Section 3, we justify the behavior of

(m, n)

-prime ideals in localizations, quotient rings, finite direct product of rings, idealization rings and amalgamation rings. For an ideal I of R, we introduce the set

ℑ (I) = \{(m, n) \in N \times N : I is (m, n) - prime\}

and study some of its properties (Theorem 37). Analogous to prime avoidance theorem, the last section is devoted to state and prove the

(m, n)

-prime avoidance theorem (Theorem 40).

2. $(m, n)$ -Prime Ideals

In this section, we give some basic properties of

(m, n)

-prime ideals and investigate

(m, n)

-prime ideals in several classes of rings. Especially, we determine the

(m, n)

-prime ideals of rings in which every power of a prime ideal is primary. Among many other results, we characterize rings in which every ideal is

(m, n)

-prime.

Definition 1.

Let I be a proper ideal of a ring R and

m, n

be positive integers. Then I is called a

(m, n)

-prime in R if for

a, b \in R

a^{m} b \in I

implies either

a^{n} \in I

b \in I .

It is clear that any

(m, n)

-prime ideal I in a ring R is both primary and

(m, n)

-closed. Hence,

P = \sqrt{I}

is the smallest prime ideal of R containing I. In this case, we call I a P-

(m, n)

-prime ideal of R. Moreover, in the following remark, we justify the relationship between

(m, n)

-prime ideals and some other kinds of ideals.

Remark 2.

Let I be a proper ideal of R and

m, n

be positive integers.

I is a prime ideal of R if and only if I is a $(1, 1)$ -prime ideal.
If I is a 1-absorbing prime (resp. if I is a prime) ideal of R, then I is a $(m, n)$ -prime ideal for $n \geq 2$ (resp. for all $n)$ . Indeed, let $a, b \in R$ with $a^{m} b \in I$ and $b \notin I$ . Then a is nonunit. If b is unit, then $a^{m} = a \cdot a^{m - 2} \cdot a \in I$ and since I is 1-absorbing prime, we have $a^{m - 1} = a \cdot a^{m - 2} \in I$ or $a \in I$ . Continue this process to get $a^{2} \in I$ and so $a^{n} \in I$ for all $n \geq 2$ , (if I is prime, then $a \in I$ ) as required. The converse is also true if I is a semi-prime (radical) ideal.
In general, we may find an n-absorbing ideal that is not $(m, n)$ -prime for all integers m and n. For example, the ideal $18 Z$ is 3-absorbing in $Z$ which is not $(m, n)$ -prime for all integers m and n (as it is not primary).
If I is $(m, n)$ -prime in R, then I is a semi n-absorbing ideal of R. Indeed, let $a \in R$ such that $a^{n + 1} \in I$ . Suppose $m ≩ n$ so that $a^{m} \in I$ . Then $a^{n} \in I$ as I is $(m, n)$ -closed in R. On the other hand, suppose $m \leq n$ and note that $a^{m} a^{n + 1 - m} \in I$ . Then by assumption, either $a^{n} \in I$ or $a^{n + 1 - m} \in I$ and the result follows as $n + 1 - m \leq n$ .
If I is $(m, n)$ -prime in $R,$ then it is $(m^{'}, n^{'})$ -prime where $n \leq n^{'}$ and $m^{'} \leq m$ .
If I is a $(m, n)$ -prime in R, then $(I : x)$ is $(m, n)$ -prime ideal in R for all $x \in R ∖ I$ .

We illustrate the place of the class of

(m, n)

-prime ideals for all positive integers m and n by the following diagram:

\begin{matrix} semiprime & ⟶ & semi n - absorbing & ⟵ & n - absorbing \\ ↑ & ↘ & ↑ & ↙ \\ prime & ⟶ & (m, n) - closed & ⟶ & Quasi (m, n) - closed \\ ↓ & ↘ & ↑ \\ 1 - absorbing prime & ⟶ & (m, n) - prime \end{matrix}

However, the arrows in the above diagram are irreversible as we can see in the following example.

Example 3.

The ideal $I = 8 Z$ is a $(5, 3)$ -prime that is not prime in $Z$ . Indeed, let $a, b \in Z$ such that $a^{5} b \in I$ . Then $a b \in 2 Z$ and so $a \in 2 Z$ or $b \in 2 Z$ . if $a \in 2 Z$ , then $a^{3} \in I$ . If $a \notin 2 Z$ , then clearly we have $b \in 8 Z = I$ .
The ideal $I = 16 Z$ is primary and clearly $(3, 2)$ -closed in $Z$ . However, I is not $(3, 2)$ -prime since for example, $2^{3} \cdot 2 \in I$ but $2^{2},$ $2 \notin I$ .
The ideal $M = 〈X, Y〉$ is a maximal ideal of $F [x, y]$ where F is a field and so $M^{2} = 〈X^{2}, X Y, Y^{2}〉$ is M-primary. On the other hand, $M^{2}$ is not $(2, 1)$ -prime in $F [x, y]$ since for example, ${(X - Y)}^{2} \in M^{2}$ but $(X - Y) \notin M^{2}$ .
In general, if $(I : x) \neq I$ and $(I : x)$ is a $(m, n)$ -prime ideal in R for all $x \in R ∖ I$ , then I need not be $(m, n)$ -prime. Consider the ideal $I = \bar{8} Z_{16}$ in the ring $Z_{16}$ . Then for all $x \in Z_{16}$ such that $(I : x) \neq I$ , we have $(I : x) = \bar{4} Z_{16}$ or $\bar{2} Z_{16}$ are clearly $(3, 2)$ -prime ideals of $Z_{16}$ . But, I is not $(3, 2)$ -prime as ${\bar{2}}^{3} . \bar{2} \in I$ where ${\bar{2}}^{2}, \bar{2} \notin I$ .
Unlike the case of $(m, n)$ -closed ideals, if $n \geq m$ , then a proper ideal need not be $(m, n)$ -prime. For example, the ideal $I = 32 Z$ is not $(3, 4)$ -prime in $Z$ as $2^{3} . 2^{2} \in I$ but $2^{4}, 2^{2} \notin I$ .

In general, if I is an ideal of a ring R and n is a positive integer, then

P = \{a \in R : a^{n} \in I\}

need not be an ideal of R. For example, consider the ideal

I = 〈X^{2}, Y^{2}〉

in the ring

R [X, Y]

. Then

X, Y \in \{f \in R [X, Y] : f^{2} \in I\}

but

X - Y \notin \{f \in R [X, Y] : f^{2} \in I\}

{(X - Y)}^{2} \notin I

. However, for certain types of ideals I such as

(m, n)

-prime (in particular, radical) ideals, the set P is an ideal of R.

Lemma 4.

Let m and n be positive integers and I be a P-

(m, n)

-prime ideal of a ring R. Then

P = \{a \in R : a^{n} \in I\}

Proof.

Let

a \in P = \sqrt{I}

and let k be the smallest positive integer such that

a^{k} \in I

. Now,

a \cdot a^{k - 1} \in I

implies

a^{m} \cdot a^{k - 1} \in I

. Since I is

(m, n)

-prime and

a^{k - 1} \notin I

, then

a^{n} \in I

and so

\sqrt{I} \subseteq \{a \in R : a^{n} \in I\}

. The other containment is clear. □

Proposition 5.

Let m and n be positive integers and I be an ideal of a ring R. If

M = \{a \in R : a^{n} \in I\}

is a maximal ideal of R, then I is an M-

(m, n)

-prime in R.

Proof.

It is clear that I is proper in R. Let

a^{m} b \in I

for

a, b \in R

such that

a^{n} \notin I

. Then

a \notin M

and so

a^{m} \notin M

. Since M is maximal in R, then

M + R a^{m} = R

and so

1 = t + r a^{m}

for some

t \in M

and

r \in R

. Thus,

1 = 1^{n} = {(t + r a^{m})}^{n} = t^{n} + s a^{m}

for some

s \in R

. Hence,

b = b \cdot 1 = b t^{n} + b s a^{m} \in I

and I is

(m, n)

-prime in R. Moreover,

\sqrt{I} = M

by Lemma 4. □

Corollary 6.

Let

m, n, k

be positive integers. If

I = M^{k}

for a maximal ideal M of R and

k \leq n

, then I is M-

(m, n)

-prime in R.

Proof.

Clearly, for

k \leq n

we have

\{a \in R : a^{n} \in I = M^{k}\} = M

. Thus, I is M-

(m, n)

-prime in R by Proposition 5. □

However, if

\{a \in R : a^{n} \in I\}

is a non-maximal prime ideal of R, then I need not be

(m, n)

-prime. Indeed, for a field F and the ideal

I = P^{3}

R = F [x, y] / 〈x^{2} y〉

where

P = 〈\bar{x}〉

, we have

\{a \in R : a^{3} \in I\} = P

is a (non-maximal) prime ideal of R. But, I is not

(m, 3)

-prime in

R = F [x, y] / 〈x^{2} y〉

, see Example 15. Also, if

k ≩ n

, then Corollary 6 may not be true, see Example 3.

Following [10], a proper ideal Q of a ring R is called uniformly primary, if there exists a positive integer k such that whenever

a, b \in R

such that

a b \in Q

and

b \notin Q

, then

a^{k} \in Q

. Moreover, a uniformly primary ideal Q has order n and write

o (Q) = n

if n is the smallest positive integer for which the aforementioned property holds. While clearly every uniformly primary ideal is primary, the converse is not true. For example, in the ring

K [X_{1}, X_{2}, . . .]

where K is a field, the ideal

({\{X_{i}^{2}\}}_{i = 1}^{\infty}, {\{X_{1} X_{i}\}}_{i = 1}^{\infty}) K [X_{1}, X_{2}, . . .]

is a primary ideal that is not uniformly primary, [10].

For positive integers m and n, if I is

(m, n)

-prime in R, then clearly I is uniformly primary. Moreover, the two concepts coincide if

o (I) \leq n

Proposition 7.

Let

{\{m_{i}, n_{i}\}}_{i = 1}^{k}

be positive integers and let

{\{I_{i}\}}_{i = 1}^{k}

be P-

(m_{i}, n_{i})

-prime ideals of a ring R. Then

⋂_{i = 1}^{k} I_{i}

is a P-

(m, n)

-prime ideal of R for all

m \leq min \{m_{1}, m_{2}, \dots, m_{k}\}

and

n \geq max \{n_{1}, n_{2}, \dots, n_{k}\}

Proof.

Suppose

I_{i}

is P-

(m_{i}, n_{i})

-prime in R for all

i \in \{1, 2, \dots, k\}

. Let

a^{m} b \in

⋂_{i = 1}^{k} I_{i}

and

b \notin

⋂_{i = 1}^{k} I_{i}

for

a, b \in R

. Then

b \notin I_{j}

for some

j \in \{1, 2, \dots, k\}

. Since

a^{m j} b \in I_{j}

, then by assumption

a^{n_{j}} \in I_{j}

and so

a \in P

. By Lemma 4, we have for all

i \in \{1, 2, \dots, k\}

P = \{a \in R : a^{n_{i}} \in I_{i}\}

. Thus,

a^{n} \in ⋂_{i = 1}^{k} I_{i}

n \geq max \{n_{1}, n_{2}, \dots, n_{k}\}

. Since also

\sqrt{⋂_{i = 1}^{k} I_{i}} = ⋂_{i = 1}^{k} {\sqrt{I}}_{i} = P

, then

⋂_{i = 1}^{k} I_{i}

is a P-

(m, n)

-prime ideal of R. □

Remark 8.

In general, if I and J are two $(m, n)$ -prime ideals with $\sqrt{I} \neq \sqrt{J}$ , then $I \cap J$ need not be $(m, n)$ -prime. For example, the ideals $2 Z$ and $3 Z$ are $(m, n)$ -prime ideal for all positive integers n and m (since they are prime), but $2 Z \cap 3 Z = 6 Z$ is not $(m, n)$ -prime (since it is not primary).
If I and J are two P- $(m, n)$ -prime ideals, then $I J$ or $I^{k}$ $(k \leq m)$ need not be P- $(m, n)$ -prime. For instance, consider the ring $R = Z + p X Z [X]$ where p is a prime integer and the ideal $P = p X Z [X]$ of R. Since P is prime, it is P- $(m, n)$ -prime for all $m, n .$ However, $P^{k}$ $(k \leq m)$ is not P- $(m, n)$ -prime as $p^{m} X^{m} \in P^{k}$ but neither $p^{n} \in P^{k}$ nor $X^{m} \in P^{k}$ .

Next, we give more characterizations of

(m, n)

-prime ideals.

Theorem 9.

Let I be a proper ideal of a ring R and let m and n be positive integers. Then the following statements are equivalent.

I is a $(m, n)$ -prime ideal of $R .$
$I = (I : a^{m})$ for all $a \in R$ such that $a^{n} \notin I$ .
If whenever $a \in R$ and K is an ideal of R with $a^{m} K \subseteq I$ , then $a^{n} \in I$ or $K \subseteq I$ .

Proof.

(1)⇒(2) Let

a \in R

such that

a^{n} \notin I

and let

b \in (I : a^{m})

. Then

a^{m} b \in I

implies

b \in I

as I is

(m, n)

-prime in R. Thus,

(I : a^{m}) \subseteq I

and so

I = (I : a^{m})

(2)⇒(3) Let

a \in R

and K be an ideal of R with

a^{m} K \subseteq I

and suppose

a^{n} \notin I

. Then by (2)

K \subseteq (I : a^{m}) = I

as needed.

(3)⇒(1) It is straightforward. □

In view of the above theorem, several equivalent characterizations of

(m, n)

-prime ideals of a principal ideal domain is given in the following.

Corollary 10.

Let R be a principal ideal domain and let m, n be positive integers. Then the following are equivalent.

I is a $(m, n)$ -prime ideal of $R .$
$I = (I : a^{m})$ for all $a \in R$ such that $a^{n} \notin I$ .
If $a \in R$ and K is an ideal of R with $a^{m} K \subseteq I$ , then $a^{n} \in I$ or $K \subseteq I$ .
If J and K are ideals of R with $J^{m} K \subseteq I$ , then $J^{n} \subseteq I$ or $K \subseteq I$ .
$I = (I : J^{m})$ for all ideals J of R such that $J^{n} ⊈ I .$
If J is an ideal of R and $b \in R$ with $J^{m} b \subseteq I$ , then $J^{n} \subseteq I$ or $b \in I$ .

Proof.

(1)⇒(2)⇒(3) Clear by Theorem 9.

(3)⇒(4) Since J is principal,

J = < a >

for some

a \in R .

Hence, the claim is clear.

(4)⇒(5) is straightforward.

(5)⇒(6) Assume that

J^{m} b \subseteq I

and

J^{n} ⊈ I

Then

b \in (I : J^{m}) = I

(5)

, as needed.

(6)⇒(1) Let

a^{m} b \in I

and

a^{n} \notin I .

Put

J = < a > .

Hence

J^{m} b

and

J^{n} ⊈ I

which imply by (6) that

b \in I

. Thus I is a

(m, n)

-prime ideal of

R .

□

In the next theorem, we characterize rings in which every ideal is

(m, n)

-prime.

Theorem .

Let R be a ring and

m,

n \in N

. The following are equivalent.

Every proper ideal of R is $(m, n)$ -prime.
R has no non-trivial idempotents (for example, R is a quasi local ring or an integral domain), $dim (R) = 0$ and $x^{n} = 0$ for all $x \in N (R)$ .

Proof.

(1) \Rightarrow (2)

Suppose that every proper ideal of R is a

(m, n)

-prime. Suppose there is an idempotent element

e \notin \{0, 1\}

in R. Since by assumption,

〈0〉

(m, n)

-prime in R,

e^{m} (1 - e) = 0

and

e \neq 1

, then

e = e^{n} = 0

, a contradiction. Therefore, R has no non-trivial idempotents. If

n < m

, then

dim (R) = 0

and

x^{n} = 0

for all

x \in N (R)

by [2, Theorem 2.14]. Suppose

n \geq m

and

x \in N (R)

. Then

〈x^{n + m}〉

is a

(m, n)

-prime ideal of R and

x^{m} x^{n} \in 〈x^{n + m}〉

. Thus,

x^{n} \in 〈x^{n + m}〉

and so

x^{n} = x^{n + m} y

for some

y \in R

. Hence,

x^{n} (1 - x^{m} y) = 0

and so

x^{n} = 0

1 - x^{m} y \in U (R)

. Moreover, suppose in the case

n \geq m

that

dim (R) ≩ 0

and choose two prime ideals

P_{1}

and

P_{2}

in R such that

P_{1} P_{2}

. If

x \in P_{2} ∖ P_{1}

, then similar to the above argument, we get

x^{n} = x^{n + m} y

and so

x^{n} (1 - x^{m} y) = 0 \subseteq P_{1}

. Thus,

1 - x^{m} y \in P_{1} P_{2}

x \notin P_{1}

. Since

x^{m} y \in P_{2}

, we conclude that

1 \in P_{2}

, a contradiction. Therefore,

dim (R) = 0

as required.

(2) \Rightarrow (1)

Let I be a proper ideal of R and let

a^{m} b \in I

for

a, b \in R

such that

b \notin I

. Since

dim (R) = 0

, then R is

π

-regular and so

a = e u + c

where

e \in I d (R)

u \in U (R)

and

c \in N (R)

by [14, Theorem 13]. Therefore, as

I d (R) = \{0, 1\}

, we have either

a = c \in N (R)

a = u + c \in U (R) .

In the first case, we conclude by assumption that

a^{n} = 0 \in I

. Otherwise, we have

b = {(a^{m})}^{- 1} a^{m} b \in I

. Therefore, every proper ideal of R is

(m, n)

-prime. □

Note that the condition "R has no non-trivial idempotents" in Theorem 11 can not be discarded. For example, the ring

Z_{6}

has non-trivial idempotents. Moreover,

dim (Z_{6}) = 0

and

x^{n} = 0

for all

x \in N (Z_{6})

and

n \in N

. However, the zero ideal of

Z_{6}

is not

(m, n)

-prime for any

m \in N

as it is not primary.

It is well-known that a field is characterized as a ring in which every proper ideal is prime (

(1, 1)

-prime). Recall also that in a von Neumann regular ring every element is of the form

u e

for

u \in U (R)

and

e \in I d (R)

. In the following corollary, we generalize this result.

Corollary 12.

Let R be a ring and

m \in N

. Then every proper ideal of R is

(m, 1)

-prime if and only if R is a field.

Proof.

If every proper ideal of R is

(m, 1)

-prime, then by Theorem 11, R is a reduced zero dimensional ring and so von Neumann regular. Thus, every element of R is of the form

u e

for some

u \in U (R)

and

e \in I d (R)

. Since also R has no non-trivial idempotents, then R is a field. The converse part is obvious. □

In the following theorem, we determine when the powers of a principal prime ideal of rings in which every power of a prime ideal is primary are

(m, n)

-prime.

Theorem 13.

Let R be a ring such that every power of a prime ideal is primary. Let m, n and k be positive integers and

I = 〈p^{k}〉

where p is a prime element of R. Then I is a

(m, n)

-prime ideal of R if and only if

n \geq k

Proof.

Suppose

I = 〈p^{k}〉

is a

(m, n)

-prime ideal of

R .

Suppose on contrary that

n ≨ k

. If

k \leq m

, then

p^{m} \in I

but

p^{n} \notin I

, a contradiction. If

k ≩ m

, then

p^{m} p^{k - m} \in I

but

p^{n} \notin I

and

p^{k - m} \notin I

which is also a contradiction. Therefore,

n \geq k

. Conversely, suppose

n \geq k

and let

a, b \in R

such that

a^{m} b \in I

and

b \notin I

. Since by assumption I is primary, then

a^{m} \in \sqrt{I} = 〈p〉

. It follows that

a \in 〈p〉

and so

a^{n} \in 〈p^{n}〉 \subseteq 〈p^{k}〉 = I

. Thus, I is a

(m, n)

-prime ideal of R. □

Corollary 14.

Let R be either an integral domain or a zero dimensional ring and m, n, k and I as in Theorem 13. Then I is a

(m, n)

-prime ideal of R if and only if

n \geq k

If some power of a prime ideal of R is not primary, then Theorem 24 need not be true in general.

Example 15.

Consider the non integral domain

R = F [x, y] / 〈x^{2} y〉

where F is any field. Then the ideal

P = 〈\bar{x}〉

is prime in R as

〈x〉

is prime in

F [x, y]

containing

〈x^{2} y〉

. Now, we prove that

I = P^{3}

is not primary in R. Indeed,we have

\bar{x^{2}} \bar{y} = \bar{0} \in I

but

\bar{y} \notin \sqrt{I}

y \notin 〈x〉

F [x, y]

. If

\bar{x^{2}} \in I

and

φ : F [x, y] \to R

is the projection mapping, then

x^{2} = φ^{- 1} (\bar{x^{2}}) \in φ^{- 1} (I) = 〈x^{3}, x^{2} y〉

which is impossible. Thus, also

\bar{x^{2}} \notin I

and

I = P^{3}

is not primary in R. Hence, I is not

(m, n)

-prime in R for all positive integers m and n (and so in particular for all

n \geq k = 3

In view of the above theorem and [2, Theorem 3.1], we have the following corollary.

Corollary 16.

Let R be an integral domain, m and n positive integers and

I = 〈p^{k}〉

where p is a prime element of R and k is a positive integer. Then I is an

(m, n)

-closed ideal of R that is not a

(m, n)

-prime ideal of R if and only if the following hold.

$n ≨ k$ .
$k = m a + r$ , where $a, r \in N$ such that $a \geq 0$ and $1 \leq r \leq n$ , $a (m$ $m o d n) + r \leq n$ , and if $a \neq 0$ , then $m = n + c$ for an integer c with $1 \leq c \leq n - 1$ .

Remark 17.

Let R be a ring such that every power of a prime ideal is primary (e.g. an integral domain or a zero dimensional) and m and n are positive integers. If

I = 〈p_{1}^{k_{1}} p_{2}^{k_{2}} \dots p_{t}^{k_{t}}〉

where

p_{1}, p_{2}, \dots, p_{t}

are non-associate prime elements of R and

k_{1}, k_{2}, \dots, k_{t}

are positive integers, then clearly, I is not primary in R. Thus, I is not

(m, n)

-prime in R.

We note that [2, Theorem 3.4] and Remark 17 give plenty examples of

(m, n)

-closed ideals that are not

(m, n)

-prime.

Corollary 18.

Let R be a principal ideal domain, I a proper ideal of R and m and n positive integers. Then I is

(m, n)

-prime in R if and only if I is generated by a power less than or equal n of a prime element in R.

Next, we define a new subclass of Noetherian rings.

Definition 19.

Let I be an ideal of a ring R. Then I is said to be of maximum length n if any ascending chain

I = I_{0} \subseteq I_{1} \subseteq I_{2} \subseteq \dots

of ideals of R terminates and n is the largest integer such that

I_{n} = I_{n + 1} = \dots

. Moreover, R is called n-Noetherian if every ideal of R has a maximum length at most n.

Clearly, any n-Noetherian ring is Noetherian. But the converse need not be true as for example the Noetherian ring

Z

is not n-Noetherian for any positive integer n. Moreover, a 1-Noetherian ring is a field clearly as every ideal is prime.

If we consider the ideal

24 Z

of the the ring

Z

, then

24 Z \subseteq 12 Z \subseteq 6 Z \subseteq 2 Z \subseteq Z

is the chain of maximum length

n = 4

. In general, we have:

Example 20.

Let R be a principal ideal domain and

I = 〈p_{1}^{k_{1}} p_{2}^{k_{2}} \dots p_{t}^{k_{t}}〉

where

p_{1}, p_{2}, \dots, p_{t}

are non-associate prime elements R. Then I is of maximal length

k_{1} + k_{2} + \dots + k_{t}

Proof.

We use mathematical induction on t. If

t = 1

, then

I = 〈p_{1}^{k_{1}}〉 \subseteq 〈p_{1}^{k_{1} - 1}〉 \subseteq \dots \subseteq 〈p_{1}〉 \subseteq R

is the chain of maximum length

n = k_{1}

. Suppose the result is true for

t - 1

. Then

\begin{matrix} I & = 〈p_{1}^{k_{1}} p_{2}^{k_{2}} \dots p_{t}^{k_{t}}〉 \subseteq 〈p_{1}^{k_{1}} p_{2}^{k_{2}} \dots p_{t}^{k_{t} - 1}〉 \subseteq 〈p_{1}^{k_{1}} p_{2}^{k_{2}} \dots p_{t}^{k_{t} - 2}〉 \subseteq \dots \\ \subseteq 〈p_{1}^{k_{1}} p_{2}^{k_{2}} \dots p_{t - 1}^{k_{t - 1}}〉 \subseteq_{1} \dots \subseteq_{k_{1} + k_{2} + \dots + k_{t - 1}} R \end{matrix}

is the chain of maximum length

n = k_{1} + k_{2} + \dots + k_{t}

as needed. □

Thus, if

k = p_{1}^{k_{1}} p_{2}^{k_{2}} \dots p_{t}^{k_{t}}

for distinct prime

p_{1}, p_{2}, \dots, p_{t}

elements, then the ring

Z_{k}

is n-Noetherian where

n = k_{1} + k_{2} + \dots + k_{t}

Recall that an ideal I of a ring R is called irreducible if whenever

I = K \cap L

for ideals K and L of R, then either

I = K

I = L

. Next, we prove that for

m, n \in N

, if I is an irreducible ideal of length n in a ring R, then I is

(m, n)

-prime in R.

Proposition 21.

Let

m, n

be positive integers and I be a proper ideal of R of maximum length n. If I is irreducible in R, then it is

(m, n)

-prime.

Proof.

Let

a, b \in R

such that

a^{m} b \in I

. For each i consider the ideal

I_{i} = \{x \in R : a^{i} x \in I\}

. Then

I = I_{0} \subseteq I_{1} \subseteq I_{2} \subseteq \dots

and so

I_{n} = I_{n + 1} = \dots

as I is of maximum length n. Thus, whenever

k \geq n

and

a^{k} x \in I

, then

a^{n} x \in I

for any

x \in R

. Now, let

Q = I + b R

and

L = I + a^{n} R

. Then clearly

I \subseteq Q \cap L

. Let

y \in Q \cap L

, say,

y = x_{1} + r_{1} b = x_{2} + r_{2} a^{n}

where

x_{1}, x_{2} \in I

. Then

r_{2} a^{n} - r_{1} b \in I

and so

r_{2} a^{n + m} - r_{1} b a^{m} \in I

. Since

a^{m} b \in I

, then

r_{2} a^{n + m} \in I

and so

r_{2} a^{n} \in I

. Therefore,

y = x_{2} + r_{2} a^{n} \in I

and so

Q \cap L \subseteq I

. Thus,

I = Q \cap L

and by assumption, either

I = Q

I = L

. If

I = Q

, then

b \in I

and if

I = L

, then

a^{n} \in Q

and so I is

(m, n)

-prime. □

Definition 22.

Let I be a proper ideal of a ring R and

m, n

positive integers. An

(m, n)

-decomposition of I is an expression for I as a finite intersection of

(m, n)

-prime ideals, say

I = ⋂_{i = 1}^{k} Q_{i}

where

Q_{i}

P_{i}

(m, n)

-prime for all i. Moreover, such an

(m, n)

-decomposition of I is called minimal if

$P_{1}, P_{2}, \dots, P_{k}$ are different prime ideals of R, and
For all $j = 1, 2, \dots, n$ , we have $I \neq ⋂_{\begin{matrix} i = 1 \\ i \neq j \end{matrix}}^{k} Q_{i}$ .

We say that I is

(m, n)

-decomposable in R precisely when it has an

(m, n)

-decomposition. By Proposition 7, the intersection of P-

(m, n)

-prime ideals is P-

(m, n)

-prime. Thus, similar to the case of primary decomposition of ideals, any

(m, n)

-decomposition of an ideal can be reduced to a minimal one.

Since any

(m, n)

-prime ideals is primary, then any

(m, n)

-decomposable ideal is decomposable. However, the converse is not true as for example, the ideal

72 Z = 2^{3} Z \cap 3^{2} Z

is decomposable in

Z

but not

(3, 2)

-decomposable. Indeed,

2^{3} Z

is not

(3, 2)

-prime by Theorem 13 and any

(3, 2)

-prime ideal in

Z

is a power of a prime.

Let

I = ⋂_{i = 1}^{k} Q_{i}

be a minimal primary decomposition of an ideal I of a ring R where

\sqrt{Q_{i}} = P_{i}

for each

i = 1, 2, \dots, k

. Recall that

\{P_{1}, P_{2}, \dots, P_{k}\}

is called the set of associated prime ideals of I (denoted by

a s s (I)

) which is independent of the choice of minimal primary decomposition of I. Moreover, it is well-known that a prime ideal P of R is a minimal prime ideal of I if and only if P is a minimal member of

a s s (I)

,3].

Now, clearly any minimal

(m, n)

-decomposition of I is a minimal primary decomposition. Thus, if

I = ⋂_{i = 1}^{k} Q_{i}

is any minimal

(m, n)

-decomposition of I where

\sqrt{Q_{i}} = P_{i}

for each

i = 1, 2, \dots, k

, then

a s s (I) =

\{P_{1}, P_{2}, \dots, P_{k}\}

Theorem 23.

Let

m, n

be positive integers. If a ring R is n-Noetherian, then any ideal of R is

(m, n)

-decomposable.

Proof.

Suppose R is is n-Noetherian and let I be a proper ideal of R. Then I is of maximal length n. Since R is Noetherian, it is well-known that I is a finite intersection of irreducible ideals. Now, the result follows since every irreducible ideal is

(m, n)

-prime by Proposition 21. □

3. $(m, n)$ -Prime Ideals in Extensions of Rings, Idealization and Amalgamation Rings

This section is devoted to justify the behavior of

(m, n)

-prime ideals in localizations, quotient rings, direct product of rings, idealization rings and amalgamation rings. Moreover, for an ideal I of a ring R, we study some properties of the set

ℑ (I) = \{(m, n) \in N \times N : I is (m, n) - prime\}

Proposition 24.

Let

f : R_{1} \to R_{2}

be a ring homomorphism and

m, n

be positive integers.

If J is a $(m, n)$ -prime ideal of $R_{2}$ , then $f^{- 1} (J)$ is a $(m, n)$ -prime ideal of $R_{1}$
If f is an epimorphism and I is a $(m, n)$ -prime ideal containing $K e r$ $f,$ then $f (I)$ is a $(m, n)$ -prime ideal of $R_{2}$ .

Proof.

(1) Let

a, b \in R_{1}

such that

a^{m} b \in f^{- 1} (J)

and

b \notin f^{- 1} (J)

. Then

f (a^{m} b) = f {(a)}^{m} f (b) \in J

and

f (b) \notin J

imply

f {(a)}^{n} = f (a^{n}) \in J .

Hence

a^{n} \in f^{- 1} (J)

, as required.

(2) Let

a : = f (x)

b : = f (y) \in R_{2}

such that

a^{m} b \in f (I)

and

b \notin f (I)

.Then clearly we have

f (x^{m} y) \in f (I)

and so

x^{m} y \in I

K e r (f) \subseteq I .

Since I is

(m, n)

-prime, we conclude that

x^{n} \in I

y \in I

. Therefore,

a^{n} = f (x^{n}) \in f (I)

b = f (y) \in f (I)

. □

In view of Proposition 24, we have the following.

Corollary 25.

Let R be a ring and

m, n

positive integers. Then the following statements hold.

If I is a $(m, n)$ -prime ideal of an overring $R^{'}$ of $R,$ then $I \cap R$ is a $(m, n)$ -prime ideal of $R .$
If $I \subseteq J$ are be proper ideals of R, then $J / I$ is a $(m, n)$ -prime ideal of $R / I$ if and only if J is a $(m, n)$ -prime ideal of $R .$

Corollary 26.

Let I be a proper ideal of a ring R, X be an indeterminate and

m, n

be positive integers. Then the following statements hold.

$< I, X >$ is a $(m, n)$ -prime ideal of $R [X]$ if and only if I is a $(m, n)$ -prime ideal of R.
If $I [X]$ is a $(m, n)$ -prime ideal of $R [X],$ then I is a $(m, n)$ -prime ideal of R.

Proof.

(1) Keeping in mind the isomorphisms

R [X] / < X > ≅ R

and

< I, X > / < X > ≅ I

, we conclude by Corollary 25(2) that

< I, X >

is a

(m, n)

-prime ideal of

R [X]

if and only if I is a

(m, n)

-prime ideal of R.

(2) Clear by Corollary 25(1). □

In the following,

Z_{I} (R)

denotes the set

{x \in R : x y \in I

for some

y \in R ∖ I} .

Next, we discuss the relationship between

(m, n)

-prime ideals and their localizations.

Proposition 27.

Let I be a proper ideal of a ring

R, S

a multiplicatively closed subset of R such that

I \cap S = \emptyset

and

m,

n be positive integers.

If I is a P- $(m, n)$ -prime ideal of R, then $S^{- 1} I$ is an $S^{- 1} P$ - $(m, n)$ -prime ideal of $S^{- 1} R .$
If $S^{- 1} I$ is a $\bar{P}$ - $(m, n)$ -prime ideal of $S^{- 1} R$ and $S \cap Z_{I} (R) = \emptyset,$ then I is a $(\bar{P} \cap R)$ - $(m, n)$ -prime ideal of R.

Proof.

(1) Let

{(\frac{a}{s_{1}})}^{m} (\frac{b}{s_{2}}) \in S^{- 1} I

for

\frac{a}{s_{1}}, \frac{b}{s_{2}} \in S^{- 1} R

. Then

{(u a)}^{m} b \in I

for some

u \in S

which implies either

{(u a)}^{n} \in I

b \in I .

Hence, either

{(\frac{a}{s_{1}})}^{n} = \frac{u^{n} a^{n}}{u^{n} s_{1}^{n}} \in S^{- 1} I

\frac{b}{s_{2}} \in S^{- 1} I .

Now, since

\sqrt{I} = P

, then

\sqrt{S^{- 1} I} = S^{- 1} \sqrt{I} = S^{- 1} P

(2) Let

a, b \in R

with

a^{m} b \in I

. Then

\frac{a^{m} b}{1} = {(\frac{a}{1})}^{m} (\frac{b}{1}) \in S^{- 1} I

. Since

S^{- 1} I

(m, n)

-prime, we conclude either

{(\frac{a}{1})}^{n} \in S^{- 1} I

(\frac{b}{1}) \in S^{- 1} I .

Thus, there are some elements

u, v \in S

such that

u a^{n} \in I

v b \in I .

Our assumption yields

a^{n} \in I

b \in I

. Moreover, as

\sqrt{I}

is a prime ideal of R, we have

S^{- 1} \sqrt{I} = \sqrt{S^{- 1} I} = \bar{P}

implies

\sqrt{I} = S^{- 1} \sqrt{I} \cap R = \sqrt{S^{- 1} I}

\cap R = \bar{P} \cap R

. □

Corollary 28.

Let I be a proper ideal of a ring R, P a prime ideal of R with

I \subseteq P

and

m, n

positive integers. Then I is a Q-

(m, n)

-prime ideal of R if and only if

I_{P}

is a

Q_{P}

(m, n)

-prime ideal of

R_{P}

Proof.

\Rightarrow)

Follows by Proposition 27(1).

\Leftarrow)

Let

a, b \in R

such that

a^{m} b \in I

. Consider the ideals

J_{1} = \{r \in R : r a^{n} \in I\}

J_{2} = \{r \in R : r b \in I\}

. Now,

{(\frac{a}{1})}^{m} (\frac{b}{1}) \in I_{P}

implies

{(\frac{a}{1})}^{n} \in I_{P}

(\frac{b}{1}) \in I_{P}

as I is

(m, n)

-prime. Hence, there are

u, v \in R ∖ P

such that

u a^{n} \in I

v b \in I

. If

u a^{n} \in I

, then

J_{1} ⊈ P

. Moreover,

J_{1} ⊈ L

for every prime ideal L such that

I ⊈ L

I \subseteq J_{1}

. Thus,

J = R

and

a^{n} \in I

. If

v b \in I

, then similarly,

J_{2} = R

and

b \in I

. Since also clearly

\sqrt{I_{P}} = Q_{P}

, then I is a

Q_{P}

(m, n)

-prime ideal of R. □

Let R be a ring and P a prime ideal of R. For a positive integer n, the kth symbolic power of P is the ideal

P^{(k)} = P^{k} R_{P} \cap R = φ^{- 1} (P^{k} R_{P})

where

φ : R \to R_{P}

is the natural canonical map. Thus,

P^{(k)} = {a \in R : s a \in P^{k}

for some

s \in R / P}

. It is well-known that if P is prime, then

P^{(k)}

is the smallest P-primary ideal containing

P^{k}

Corollary 29.

Let

m, k

be a positive integers and P be a prime ideal of a ring R. Then for all

n \geq k

P^{(k)}

is the smallest P-

(m, n)

-prime ideal containing

P^{k}

Proof.

Since

P R_{P}

is maximal in

R_{P}

and

n \geq k

, then

P^{k} R_{P} = {(P R_{P})}^{k}

is a

(m, n)

-prime ideal of

R_{P}

for any positive integer m by Corollary 6. Thus,

P^{(k)} = P^{k} R_{P} \cap R

is a

(m, n)

-prime ideal of R by Proposition 24(1).

Now, clearly

P^{k} \subseteq P^{(k)}

since

1 \in R ∖ P

. Let J be another P-

(m, n)

-prime ideal with

P^{k} \subseteq J

and let

r \in P^{(k)}

. Then

s r \in P^{k}

for some

s \in R ∖ P

. Since

P^{k} \subseteq J

, then

s r \in J

, and so

s^{m} r \in J

. Hence, either

s \in P = \{x \in R : x^{n} \in J\}

r \in J

as J is P-

(m, n)

-prime. Since we chose

s \in R ∖ P

, then

r \in J

. Therefore,

P^{(k)} \subseteq J

and

P^{(k)}

is the smallest P-

(m, n)

-prime ideal containing

P^{k}

. □

Theorem 30.

Let

R_{1}, R_{2}, \dots, R_{k}

be rings,

R = R_{1} \times R_{2} \times \dots \times R_{k}

and

I_{1}, I_{2}, \dots, I_{k}

be ideals of

R_{1}, R_{2}, \dots, R_{k},

respectively. For any positive integers m and n, we have

I_{1} \times I_{2} \times \dots \times I_{k}

is a

(m, n)

-prime ideal of R if and only if there exists

i \in \{1, 2, \dots, k\}

such that

I_{i}

is a

(m, n)

-prime ideal of

R_{i}

and

I_{j} = R_{j}

for all

j \neq i

Proof.

Suppose

I_{1} \times I_{2} \times \dots \times I_{k}

is a

(m, n)

-prime in R. Assume, say,

I_{1}

and

I_{2}

are proper and choose

a_{1} \in I_{1}

and

a_{2} \in I_{2}

. Then

{(a_{1}, 1, 0, . . ., 0)}^{m} (1, a_{2}, 0, . . ., 0) \in I_{1} \times I_{2} \times \dots \times I_{k}

but neither

{(a_{1}, 1, 0, . . ., 0)}^{n} \in I_{1} \times I_{2} \times \dots \times I_{k}

nor

(1, a_{2}, 0, . . ., 0) \in I_{1} \times I_{2} \times \dots \times I_{k} .

Thus, there is

i \in \{1, 2, \dots, k\}

such that

I_{j} = R_{j}

for all

j \neq i

. Without loss of generality, assume

I_{j} = R_{j}

for all

j \neq 1

. We show that

I_{1}

is a

(m, n)

-prime ideal of

R_{1}

. Let

a,

b \in R_{1}

and

a^{m} b \in I_{1} .

Then

{(a, 0, . . ., 0)}^{m} (b, 0, . . ., 0) \in I_{1} \times R_{2} \times \dots \times R_{k}

which implies that

{(a, 0, . . ., 0)}^{n} \in I_{1} \times R_{2} \times \dots \times R_{k}

(b, 0, . . ., 0) \in I_{1} \times R_{2} \times \dots \times R_{k}

. Thus

a^{n} \in I_{1}

b \in I_{1}

and

I_{1}

is a

(m, n)

-prime ideal of

R_{1} .

Conversely, suppose, say,

I_{1}

is a

(m, n)

-prime ideal of

R_{1}

and

I_{j} = R_{j}

for all

j \neq 1

. Suppose

{(a_{1}, a_{2}, . . ., a_{k})}^{m} (b_{1}, b_{2}, . . ., b_{k}) \in I_{1} \times R_{2} \times \dots \times R_{k}

but

(b_{1}, b_{2}, . . ., b_{k}) \notin I_{1} \times R_{2} \times \dots \times R_{k}

. Then

a_{1}^{m} b_{1} \in I_{1}

and

b_{1} \notin I_{1}

imply that

a_{1}^{n} \in I .

Thus

{(a_{1}, a_{2}, . . ., a_{k})}^{n} \in I_{1} \times R_{2} \times \dots \times R_{k}

, as needed. □

In particular, we have:

Corollary 31.

Let

R_{1}

and

R_{2}

be rings,

R = R_{1} \times R_{2}

and

I, J

be be ideals of

R_{1}

R_{2},

respectively. For any positive integers m and n, we have

I \times J

is a

(m, n)

-prime ideal of R if and only if one of the following statements is satisfied:

I is a $(m, n)$ -prime ideal of $R_{1}$ and $J = R_{2} .$
J is a $(m, n)$ -prime ideal of $R_{2}$ and $I =$ $R_{1} .$

Note that if I and J are

(m, n)

-prime ideals of

R_{1}

and

R_{2},

respectively, then I and J are proper and so

I \times J

is never

(m, n)

-prime ideal in

R_{1} \times R_{2} .

Recall that the idealization of an R-module M denoted by

R (+) M

, is the commutative ring

R \times M

with coordinate-wise addition and multiplication defined as

(r_{1}, m_{1}) (r_{2}, m_{2}) = (r_{1} r_{2}, r_{1} m_{2} + r_{2} m_{1})

. For an ideal I of R and a submodule N of M,

I (+) N

is an ideal of

R (+) M

if and only if

I M \subseteq N

Proposition 32.

Let I be a proper ideal of a ring R, N be a proper submodule of an R-module M and

m, n

be positive integers. Then

I is a $(m, n)$ -prime ideal of R if and only if $I (+) M$ is a $(m, n)$ -prime ideal of $R (+) M .$
If $I (+) N$ is a $(m, n)$ -prime ideal of $R (+) M,$ then I is a $(m, n)$ -prime ideal of $R .$

Proof.

(1) Let I be a

(m, n)

-prime ideal of R and

{(a, x)}^{m} (b, y) \in I (+) M

for some

(a, x), (b, y) \in R (+) M .

Then

a^{m} b \in I

which implies either

a^{n} \in I

b \in I

. Hence, either

{(a, x)}^{n} \in I (+) M

(b, y) \in I (+) M

. Conversely, if

a^{m} b \in I

for some

a, b \in R

, then

{(a, 0)}^{m} (b, 0) \in I (+) M

which implies

{(a, 0)}^{n} \in I (+) M

(b, 0) \in I (+) M

, and so

a^{n} \in I

b \in I

, we are done.

(2) Similar to the converse part of (1). □

We note that the converse of (2) of Proposition 32 is not true in general. For example, while

2 Z

is a

(2, 1)

-prime ideal in

Z

, the ideal

2 Z (+) 2 Z

is not so in

Z (+) Z

. Indeed,

{(2, 1)}^{2} = (4, 4) \in 2 Z (+) 2 Z

but

(2, 1) \notin 2 Z (+) 2 Z

Let R and S be two rings, J be an ideal of S and

f : R \to S

be a ring homomorphism. As a subring of

R \times S,

the amalgamation of R and S along J with respect to f is defined by

R ⋈^{f} J = (a, f (a) + j) : a \in R

j \in J} .

If f is the identity homomorphism on R, then we get the amalgamated duplication of R along an ideal J,

R ⋈ J = \{(a, a + j) : a \in R, j \in J\}

. For more related definitions and several properties of this kind of rings, one can see [11]. If I is an ideal of R and K is an ideal of

f (R) + J

, then

I ⋈^{f} J = \{(i, f (i) + j) : i \in I, j \in J\}

and

{\bar{K}}^{f} = {(a, f (a) + j) : a \in R

j \in J

f (a) + j \in K}

are ideals of

R ⋈^{f} J

,12].

For positive integers m and n, in the next result, we give a characterization about when the ideals

I ⋈^{f} J

and

{\bar{K}}^{f}

are

(m, n)

-prime ideals of

R ⋈^{f} J

Theorem 33.

Let R, S, f, J, I and K be as above. For positive integers m and n, we have:

$I ⋈^{f} J$ is a $(m, n)$ -prime ideal of $R ⋈^{f} J$ if and only if I is a $(m, n)$ -prime ideal of R.
${\bar{K}}^{f}$ is a $(m, n)$ -prime ideal of $R ⋈^{f} J$ if and only if K is a $(m, n)$ -prime ideal of $f (R) + J$ .

Proof.

(1) Suppose

I ⋈^{f} J

(m, n)

-prime in

R ⋈^{f} J

and let

a, b \in R

such that

a^{m} b \in I

. Then

{(a, f (a))}^{m} (b, f (b)) \in I ⋈^{f} J

and so either

{(a, f (a))}^{n} \in I ⋈^{f} J

(b, f (b)) \in I ⋈^{f} J

. Thus, either

a^{n} \in I

b \in I

and I is

(m, n)

-prime in R. Conversely, suppose I is

(m, n)

-prime in R. Let

(a, f (a) + j_{1}), (b, f (b) + j_{2}) \in R ⋈^{f} J

such that

{(a, f (a) + j_{1})}^{m} (b, f (b) + j_{2}) \in I ⋈^{f} J

. Then

a^{m} b \in I

and so either

a^{n} \in I

b \in I

. It follows that

{(a, f (a) + j_{1})}^{n} \in I ⋈^{f} J

(b, f (b) + j_{2}) \in I ⋈^{f} J

as needed.

(2) Suppose

{\bar{K}}^{f}

(m, n)

-prime ideal in

R ⋈^{f} J

. Let

f (a) + j_{1},

f (b) + j_{2} \in f (R) + J

such that

{(f (a) + j_{1})}^{m} (f (b) + j_{2}) \in K

. Then

{(a, f (a) + j_{1})}^{m} (b, f (b) + j_{2}) \in {\bar{K}}^{f}

and hence by assumption,

{(a, f (a) + j_{1})}^{n} \in {\bar{K}}^{f}

(b, f (b) + j_{2}) \in {\bar{K}}^{f}

. It follows that

{(f (a) + j_{1})}^{n} \in K

(f (b) + j_{2}) \in K

. Conversely, suppose K is

(m, n)

-prime in

f (R) + J

. Suppose

{(a, f (a) + j_{1})}^{m} (b, f (b) + j_{2}) \in {\bar{K}}^{f}

for

(a, f (a) + j_{1}),

(b, f (b) + j_{2}) \in R ⋈^{f} J

. Then

{(f (a) + j_{1})}^{m} (f (b) + j_{2}) \in K

and so

{(f (a) + j_{1})}^{n} \in K

(f (b) + j_{2}) \in K

. Therefore,

{(a, f (a) + j_{1})}^{n} \in {\bar{K}}^{f}

(b, f (b) + j_{2}) \in {\bar{K}}^{f}

and the result follows. □

In particular, we have:

Corollary 34.

Let I and J be an ideal of a ring R. Then

I ⋈ J

is a

(m, n)

-prime ideal of

R ⋈ J

if and only if I is a

(m, n)

-prime ideal of R.

Lemma 35.

([9,12]). Let

f : R \to S

be a ring homomorphism and J be an ideal of S. Then

$N (R ⋈^{f} J) = \{(a, f (a) + j) : a \in N (R)), j \in N (S) \cap J\} .$
$dim (R ⋈^{f} J) = max \{dim (R), dim (f (R) + J)\}$
$I d (R ⋈^{f} J) = \{(a, f (a) + j) : a \in I d (R), f (a) + j \in I d (R) + J\}$ .

Next, we use Lemma 35 and Theorem 11, to determine when every proper ideal of the amalgamation

R ⋈^{f} J

(m, n)

-prime.

Theorem 36.

Let

m,

n be positive integers and R, S, f and J be as above where J is proper in S. Then every proper ideal of

R ⋈^{f} J

(m, n)

-prime if and only if the following statements hold

Every proper ideal of R is $(m, n)$ -prime.
Every proper ideal of $f (R) + J$ is $(m, n)$ -prime.

Proof.

Suppose every proper ideal of

R ⋈^{f} J

(m, n)

-prime. If there is a proper ideal I of R which is not

(m, n)

-prime, then

I ⋈^{f} J

is proper in

R ⋈^{f} J

which is not a

(m, n)

-prime ideal of

R ⋈^{f} J

by Theorem 33(1), a contradiction. Similarly, if K is a proper non

(m, n)

-prime ideal of

f (R) + J

, then

{\bar{K}}^{f}

is a proper non

(m, n)

-prime ideal of

R ⋈^{f} J

by Theorem 33(2), which is also a contradiction.

Conversely, suppose (1) and (2) hold. Then

dim (R) = dim (f (R) + J) = 0

by Theorem 11 and so

dim (R ⋈^{f} J) = max \{dim (R), dim (f (R) + J)\} = 0

by Lemma 35(2). Now, let

(a, f (a) + j) \in N (R ⋈^{f} J)

. Then

a \in N (R)

and

f (a) + j \in N (f (R) + J)

by Lemma 35(1). Thus,

a^{n} = {(f (a) + j)}^{n} = 0

by Theorem 11 and so

{(a, f (a) + j)}^{n} = (0, 0)

. Again by Theorem 11, we have

I d (R) = \{0_{R}, 1_{R}\}

and

I d (f (R) + J) = \{0_{S}, 1_{S}\}

. Since J is proper in S and by the definition of

R ⋈^{f} J

, we have

(0_{R}, 1_{S}), (1_{R}, 0_{S}) \notin R ⋈^{f} J

. By Lemma 35(3),

I d (R ⋈^{f} J) = \{(0_{R}, 0_{S}), (1_{R}, 1_{S})\}

. It follows that every proper ideal of

R ⋈^{f} J

(m, n)

-prime by Theorem 11. □

Following [2], for an ideal I of a ring R,

ℜ (I) = \{(m, n) \in N \times N : I is (m, n) - closed\}

Similarly, we let

ℑ (I) = \{(m, n) \in N \times N : I is (m, n) - prime\}

and assume

ℑ (R) = N \times N

. It is clear that

ℑ (I) \subseteq ℜ (I)

and this containment in general is proper as we have seen in Example 3. Moreover, we have

(1, 1) \in ℑ (I)

if and only if I is prime.

For an ideal I of a ring R, the following are some properties concerning

ℑ (I)

. Theses properties are analogous to those of

(m, n)

-closed ideals, [2].

Theorem 37.

Let I and J be ideals of a ring R, and m, n, k and t be positive integers.

If $(m, n) \in ℑ (I)$ , then $(m^{'}, n^{'}) \in ℑ (I)$ for all positive integers $m^{'}$ and $n^{'}$ with $m^{'} \leq m$ and $n^{'} \geq n$ .
If $(m, n) \in ℑ (I)$ , then $(k m, t n) \in ℑ (I)$ for all $t \geq k$ .
If $(m, n) \in ℑ (I)$ and $(n, k) \in ℜ (I)$ , then $(m, k) \in ℑ (I)$ .
$(m, n) \in ℑ (I)$ if and only if $(m + 1, n) \in ℑ (I)$ . Hence, $(m, n) \in ℑ (I)$ if and only if $(t, n) \in ℑ (I)$ for all $t \geq m$ .
If I and J are proper, then $ℑ (I \times J) = ϕ$ . If only one of I and J is proper, then $ℑ (I \times J) = ℑ (I) \cap ℑ (J)$ .

Proof.

(1), (2) and (3): Clear.

(4) Suppose

(m, n) \in ℑ (I)

and let

a, b \in R

such that

a^{m + 1} b \in I

and

b \notin I

. Then

{(a^{2})}^{m} b \in I

2 m \geq m + 1

. Since I is

(m, n)

-prime, then

a^{2 n} \in I

. Thus,

a \in \sqrt{I}

and so

a^{n} \in I

by Lemma 4. The converse is clear by (1).

(5) If I and J are proper, then

ℑ (I \times J) = \emptyset

by Theorem 31. Suppose, say,

I \neq R

and

J = R

. Then

ℑ (I \times J) = ℑ (I) \cap ℑ (J)

since

ℑ (R) = N \times N

and by using Corollary 31. □

The converse of (2) of Theorem 37 is not true in general. For example, the ideal

I = 〈p^{k}〉

where p is a prime element of any integral domain R, is

(k, k)

-prime by Theorem 13. But, I is not

(1, 1)

-prime as it is not prime in R.

4. $(m, n)$ -Prime Avoidance Theorem

In this section, we prove the

(m, n)

-prime avoidance theorem analogous to prime avoidance theorem. Recall that a covering

I \subseteq I_{1} \cup I_{2} \cup \dots \cup I_{n}

is said to be efficient if no

I_{k}

is superfluous. Also,

I = I_{1} \cup I_{2} \cup \dots \cup I_{n}

is an efficient union if none of the

I_{k}

may be excluded. Here, it is easy to see that a covering

I \subseteq I_{1} \cup I_{2} \cup \dots \cup I_{n}

automatically implies a union

I = (I \cap I_{1}) \cup (I \cap I_{2}) \cup

···

\cup (I \cap I_{n})

. First, we need to state a very useful lemma.

Lemma 38.

(McCoy) Let

I = I_{1} \cup I_{2} \cup \dots \cup I_{n}

be an efficient union of ideals where

n ≩ 1

. Then

⋂_{i \neq k} I_{i} = ⋂_{i = 1}^{n} I_{i}

for all k.

Theorem 39.

Let

I \subseteq I_{1} \cup I_{2} \cup \dots \cup I_{n}

be an efficient covering of ideals

I_{1}, I_{2}, . . ., I_{n}

of R where

n ≩ 2 .

Suppose that

\sqrt{I_{i}} ⊈ \sqrt{(I_{j} : x)}

for all

x \in R ∖ \sqrt{I_{j}}

whenever

i \neq j

. Then no

I_{i}

(

1 \leq i \leq n

) is a

(m, n)

-prime ideal of R for all

n \leq m .

Proof.

Suppose on the contrary that

I_{k}

is a

(m, n)

-prime ideal of R for some

1 \leq k \leq n

. First, note that as

I \subseteq ⋃_{i = 1}^{n} I_{i}

is an efficient covering, then

I \subseteq ⋃_{i = 1}^{n} (I_{i} \cap I)

is also an efficient covering. It follows that

(*) (⋂_{i \neq k} I_{i}) \cap I = (⋂_{i = 1}^{n} I_{i}) \cap I \subseteq I_{k} \cap I

by Lemma 38. For all

x \in R ∖ \sqrt{I_{k}}

and

i \neq k

, we have

\sqrt{I_{i}} ⊈ \sqrt{(I_{k} : x)}

and so we can choose

a_{i} \in \sqrt{I_{i}} ∖ \sqrt{(I_{k} : x)}

. Then, there exists the least positive integer

m_{i}

such that

a_{i}^{m_{i}} \in I_{i}

for each

i \neq k

. Write

a = a_{1} a_{2} \dots a_{k - 1}

b = a_{k + 1} a_{k + 2} \dots a_{n}

and

m = max {m_{1}, m_{2}, . . ., m_{k - 1}, m_{k + 1}, . . ., m_{n}}

. Then

a^{m} b^{m} x \in (⋂_{i \neq k} I_{i}) \cap I .

In the rest of the proof, we show that

a^{m} b^{m} x \in ((⋂_{i \neq k} I_{i}) \cap I) ∖ (I_{k} \cap I)

. For this purpose, assume on the opposite that

a^{m} b^{m} x \in I_{k} \cap I .

Then

a^{m} b^{m} \in (I_{k} : x) \subseteq \sqrt{(I_{k} : x)} .

Since

\sqrt{(I_{k} : x)}

is a prime ideal by Proposition 4 (1) and (2), we get either

a = a_{1} a_{2} \dots a_{k - 1} \in \sqrt{(I_{k} : x)}

b = a_{k + 1} a_{k + 2} \dots a_{n} \in \sqrt{(I_{k} : x)} .

Again, since

\sqrt{(I_{k} : x)}

is prime,

a_{i} \in \sqrt{(I_{k} : x)}

for some

i \neq k

, a contradiction. Consequently,

a^{m} b^{m} x \notin (I_{k} \cap I),

and so

a^{m} b^{m} x \in ((⋂_{i \neq k} I_{i}) \cap I) ∖ (I_{k} \cap I)

which contradicts

(*)

. Therefore, no

I_{i}

is a

(m, n)

-prime ideal for

1 \leq i \leq n

and we are done. □

Theorem 40.

(

(m, n)

-prime Avoidance Theorem) Let

I,

I_{1},

I_{2}, . . ., I_{n}

(n \geq 2)

be ideals of R such that at most two of

I_{1}, I_{2}, . . ., I_{n}

are not

(m, n)

-prime and

\sqrt{I_{i}} ⊈ \sqrt{(I_{j} : x)}

for all

x \in R ∖ \sqrt{I_{j}}

whenever

i \neq j

. If

I \subseteq I_{1} \cup I_{2} \cup \dots \cup I_{n}

, then

I \subseteq I_{k}

for some

1 \leq k \leq n .

Proof.

Assume that

I ⊈ I_{k}

for all

1 \leq k \leq n .

Without loss of generality, we may assume that

I \subseteq I_{1} \cup I_{2} \cup \dots \cup I_{n}

is an efficient covering of ideals of R as any covering can be reduced to an efficient one by omitting any unnecessary terms. It is well-known that a covering of an ideal by two ideals is never efficient. If

n \geq 3

, then no

I_{k}

is a

(m, n)

-prime ideal of R by Theorem 39. But our assumption implies that at most two of

I_{1}, I_{2}, . . ., I_{n}

are not

(m, n)

-prime. Thus,

I \subseteq I_{k}

for some

1 \leq k \leq n

. □

Corollary 41.

Let I be a proper ideal of a ring R. If

(m, n)

-prime avoidance theorem holds for R, then the

(m, n)

-prime avoidance theorem holds for

R / I

Proof.

Let

J / I,

I_{1} / I,

I_{2} / I, . . ., I_{n} / I

(n \geq 2)

be ideals of

R / I

such that at most two of

I_{1} / I,

I_{2} / I, . . ., I_{n} / I

are not

(m, n)

-prime and

J / I \subseteq (I_{1} / I) \cup (I_{2} / I) \cup \dots \cup (I_{n} / I) .

Then, Corollary 25 implies that

J \subseteq I_{1} \cup I_{2} \cup \dots \cup I_{n}

and at most two of

I_{1}, I_{2}, . . ., I_{n}

are not

(m, n)

-prime. Suppose that

\sqrt{I_{i} / I} ⊈ \sqrt{(I_{j} / I : x + I)}

for all

x + I \in (R / I) ∖ \sqrt{(I_{j} / I)}

whenever

i \neq j

. It is easy to verify that if

\sqrt{I_{i}} \subseteq \sqrt{(I_{j} : x)}

for some

x \in R,

then

\sqrt{(I_{i} / I)} \subseteq \sqrt{(I_{j} / I : x + I)}

for some

x + I \in R / I .

Also observe that if

x + I \in (R / I) ∖ \sqrt{(I_{j} / I)} = (R / I) ∖ (\sqrt{I_{j}} / I)

, then

x \in R ∖ \sqrt{I_{j}} .

Thus, by our assumption

\sqrt{(I_{i} / I)} ⊈ \sqrt{(I_{j} / I : x + I)}

for all

x + I \in (R / I)) ∖ \sqrt{(I_{j} / I)}

whenever

i \neq j

. Hence, we conclude that

\sqrt{I_{i}} ⊈ \sqrt{(I_{j} : x)}

for all

x \in R ∖ \sqrt{I_{j}}

whenever

i \neq j .

Therefore, Theorem 40 implies

J \subseteq I_{k}

for some

1 \leq k \leq n .

. Consequently,

J / I \subseteq I_{k} / I

for some

1 \leq k \leq n

; so we are done. □

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(m, n)-Prime Ideals of Commutative Rings

Abstract

1. Introduction

2. ( m , n ) -Prime Ideals

3. ( m , n ) -Prime Ideals in Extensions of Rings, Idealization and Amalgamation Rings

4. ( m , n ) -Prime Avoidance Theorem

References

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2. $(m, n)$ -Prime Ideals

3. $(m, n)$ -Prime Ideals in Extensions of Rings, Idealization and Amalgamation Rings

4. $(m, n)$ -Prime Avoidance Theorem