On the enumeration of permutations avoiding chains of patterns

Robin D.P. Zhou∗,††∗Corresponding author. ††E-mail address<EMAIL_ADDRESS>Yongchun Zang

College of Mathematics Physics and Information

Shaoxing University

Shaoxing 312000, P.R. China

Abstract. In 2019, Bóna and Smith introduced the notion of strong pattern
avoidance, saying that a permutation $\pi$ strongly avoids a pattern $\sigma$
if $\pi$ and $\pi^{2}$ both avoid $\sigma$. Recently, Archer and Geary
generalized the idea of strong pattern avoidance to chain avoidance, in which
a permutation $\pi$ avoids a chain of patterns
$(\tau^{(1)}:\tau^{(2)}:\cdots:\tau^{(k)})$ if the $i$-th power of the
permutation avoids the pattern $\tau^{(i)}$ for $1\leq i\leq k$. In this
paper, we give explicit formulae for the number of sets of permutations
avoiding certain chains of patterns. Our results give affirmative answers to
two conjectures proposed by Archer and Geary.

Keywords: pattern avoidance, strong pattern avoidance, chain avoidance.

AMS Subject Classifications: 05A05, 05C30

##  1 Introduction

Let $\mathcal{S}_{n}$ denote the symmetry group consisting of all permutations
of $[n]=\\{1,2,\ldots,n\\}$, which we can view as bijections from $[n]$ to
$[n]$. A permutation $\pi$ can also be viewed as a word by associating $\pi$
with the word $\pi(1)\pi(2)\cdots\pi(n)$. Throughout this paper, we always
write a permutation $\pi\in\mathcal{S}_{n}$ as a word
$\pi=\pi_{1}\pi_{2}\cdots\pi_{n}$, where $\pi_{i}=\pi(i)$ for $1\leq i\leq n$.

Given a permutation $\pi\in\mathcal{S}_{n}$ and a permutation
$\sigma\in\mathcal{S}_{k}$, an occurrence of $\sigma$ in $\pi$ is a
subsequence $\pi_{i_{1}}\pi_{i_{2}}\cdots\pi_{i_{k}}$ of $\pi$ that is order
isomorphic to $\sigma$. We say $\pi$ contains the pattern $\sigma$ if $\pi$
contains an occurrence of $\sigma$. Otherwise, we say $\pi$ avoids the pattern
$\sigma$ or $\pi$ is $\sigma$-avoiding. For example, the permutation $1534627$
avoids the pattern $3142$ while it contains the pattern $12345$ corresponding
to the subsequence $13467$.

The study of pattern avoidance can be traced back to the work of MacMahon [8]
and has become a research focus in enumerative combinatorics over the past
half century. A recent survey on permutation patterns can be found in the
books [2] and [6]. The notion of pattern avoidance has arisen many further
variations. Bóna and Smith [3] initiated the study of strong pattern
avoidance, in which a permutation $\pi$ strongly avoids a pattern $\sigma$ if
$\pi$ and $\pi^{2}$ both avoid $\sigma$. The study of strong pattern avoidance
has attracted the interest of combinatorial scholars such as [4, 9].

Recently, Archer and Geary [1] generalized the idea of strong pattern
avoidance to chain avoidance, in which a permutation $\pi$ avoids a chain (of
patterns) $(\tau^{(1)}:\tau^{(2)}:\cdots:\tau^{(k)})$ if the $i$-th power of
the permutation (i.e., $\pi^{i}$) avoids the pattern $\tau^{(i)}$ for $1\leq
i\leq k$. Let $\mathcal{S}_{n}(\tau_{1}:\tau_{2}:\cdots:\tau_{k})$ denote the
set of $(\tau_{1}:\tau_{2}:\cdots:\tau_{k})$-avoiding permutations in
$\mathcal{S}_{n}$. This definition can be extended to include sets of
patterns, which we will separate the patterns in a set with commas. For
example, if $\pi$ avoids the chain $(\sigma,\rho:\tau)$, then $\pi$ avoids
both $\sigma$ and $\rho$ and $\pi^{2}$ avoids $\tau$. For example, let
$\pi=1325467$. Then we have $\pi\in\mathcal{S}_{7}(231,1432:231)$ because
$\pi$ avoids both $231$ and $1432$, $\pi^{2}=1234567$ avoids $231$.

It would also be natural to consider consecutive patterns as part of the
chain. We say $\pi$ contains a consecutive pattern
$\bar{\sigma}=\overline{\sigma_{1}\sigma_{2}\cdots\sigma_{k}}$ if there is an
occurrence of $\sigma$ corresponding to a consecutive subsequence
$\pi_{i}\pi_{i+1}\cdots\pi_{i+k-1}$ ($1\leq i\leq n-k+1)$ of $\pi$. Otherwise,
we say $\pi$ avoids the (consecutive) pattern $\bar{\sigma}$ or $\pi$ is
$\bar{\sigma}$-avoiding. For example, the permutation $\pi=1534627$ avoids the
pattern $\overline{321}$ while $\pi$ contains the pattern $\overline{213}$
corresponding to the consecutive subsequence $627$.

Archer and Geary [1] obtained the explicit formulae for the number of sets
$\mathcal{S}_{n}(213,312:\tau)$ for $\tau\in\mathcal{S}_{3}$ and further posed
the following conjectures.

###### Conjecture 1.1

For $n\geq 1$, we have
$|\mathcal{S}_{n}(231,1432:231)|=L_{n+1}-\lceil\frac{n}{2}\rceil-1$, where
$L_{n+1}$ is the $(n+1)$-th Lucas number (A000032, [7]).

###### Conjecture 1.2

For $n\geq 2$, we have
$|\mathcal{S}_{n}(213,312:\overline{213})|=2^{n-2}+n-1$.

The objective of this paper is to prove Conjecture 1.1 and Conjecture 1.2. Our
proofs are based on the recursive decompositions of the permutations in
$\mathcal{S}_{n}(231,1432:231)$ and the permutations in
$\mathcal{S}_{n}(213,312:\overline{213})$, respectively.

##  2 Avoiding the chain $(231,1432:231)$

This section is devoted to the proof of Conjecture 1.1. This will be
accomplished by showing that the both sides of the equation in Conjecture 1.1
satisfy the same recurrence and initial conditions. The following lemma is
crucial for our proof.

###### Lemma 2.1

Let $n\geq 3$ and
$\pi=\pi_{1}\pi_{2}\cdots\pi_{n}\in\mathcal{S}_{n}(231,1432:231)$. Then we
have either $\pi_{1}=n$, $\pi_{n}=n$ or $\pi_{n-1}\pi_{n}=n(n-1)$.

Proof. The result is trivial for $n=3$. Now we let $n\geq 4$. Assume that
$\pi_{1}\neq n$, $\pi_{n}\neq n$ and $\pi_{n-1}\pi_{n}\neq n(n-1)$. Let
$\pi_{k}=n$. Then $2\leq k\leq n-1$. Combining the fact $\pi$ is
$231$-avoiding, we have that $\pi$ can be written in the form $\pi=\sigma
n\tau$, where $\sigma$ and $\tau$ are not empty and all the elements in
$\sigma$ are less than the elements in $\tau$. Notice that
$\pi_{n-1}\pi_{n}\neq n(n-1)$. It is easily seen that $\tau$ contains at least
two elements, namely, $k\leq n-2$. We claim that the elements in $\tau$ are
increasing. If not, there exists some $i>k$ such that $\pi_{i}>\pi_{i+1}$. It
follows that the subsequence $\pi_{1}n\pi_{i}\pi_{i+1}$ forms an occurrence of
the pattern $1432$ in $\pi$, contradicting to the fact that $\pi$ is
$1432$-avoiding. This proves the claim. Hence $\pi$ is of the form $\pi=\sigma
nk(k+1)\cdots(n-1)$. Now consider $\pi^{2}$. It is easily checked that
$\pi^{2}(k)=\pi(\pi(k))=\pi(n)=n-1$, $\pi^{2}({k+1})=\pi(\pi(k+1))=\pi(k)=n$
and $\pi^{2}(n)=\pi(\pi(n))=\pi(n-1)=n-2$. Then
$\pi^{2}(k)\pi^{2}(k+1)\pi^{2}(n)=(n-1)n(n-2)$ forms an occurrence of the
pattern $231$ in $\pi^{2}$, contradicting to the fact that $\pi^{2}$ is
$231$-avoiding. This completes the proof.

Given $n\geq 3$, Lemma 2.1 enables us to divide the set
$\mathcal{S}_{n}(231,1432:231)$ into the following three disjoint subsets:

$\displaystyle\mathcal{P}_{n}^{1}$
$\displaystyle:=\\{\pi\mid\pi=\pi_{1}\pi_{2}\cdots\pi_{n}\in\mathcal{S}_{n}(231,1432:231)\text{
and }\pi_{1}=n\\},$ $\displaystyle\mathcal{P}_{n}^{2}$
$\displaystyle:=\\{\pi\mid\pi=\pi_{1}\pi_{2}\cdots\pi_{n}\in\mathcal{S}_{n}(231,1432:231)\text{
and }\pi_{n}=n\\},$ $\displaystyle\mathcal{P}_{n}^{3}$
$\displaystyle:=\\{\pi\mid\pi=\pi_{1}\pi_{2}\cdots\pi_{n}\in\mathcal{S}_{n}(231,1432:231)\text{
and }\pi_{n-1}\pi_{n}=n(n-1)\\}.$

We shall characterize the permutations in $\mathcal{P}_{n}^{1}$,
$\mathcal{P}_{n}^{2}$ and $\mathcal{P}_{n}^{3}$, respectively. The following
theorem is needed for the characterization of $\mathcal{P}_{n}^{1}$.

###### Theorem 2.2

([3], Theorem 3.1) For any permutation $\pi$ ending in $1$, the following two
statements are equivalent.

  * (A)

The permutation $\pi$ is strongly $312$-avoiding.

  * (B)

The permutation $\pi$ has form $\pi=(k+1)(k+2)\cdots nk(k-1)(k-2)\cdots 1$
where $\lceil\frac{n}{2}\rceil\leq k\leq n-1$.

###### Lemma 2.3

Let $n\geq 3$ and $\pi=\pi_{1}\pi_{2}\cdots\pi_{n}$. Then we have

$\mathcal{P}_{n}^{1}=\\{\pi\mid\pi=n(n-1)\cdots(n-k+1)12\cdots(n-k),\,\lceil\frac{n}{2}\rceil\leq
k\leq n-1\\}.$

Proof. Let
$\mathcal{A}_{n}:=\\{\pi\mid\pi=\pi_{1}\pi_{2}\cdots\pi_{n}\in\mathcal{S}_{n}(312:312),\,\pi_{n}=1\\}$
and
$\mathcal{B}_{n}:=\\{\pi\mid\pi=\pi_{1}\pi_{2}\cdots\pi_{n}\in\mathcal{S}_{n}(231:231),\,\pi_{1}=n\\}$.
By the definition of $\mathcal{P}_{n}^{1}$, we have
$\mathcal{P}_{n}^{1}\subseteq\mathcal{B}_{n}$. As the pattern $231$ is the
inverse of $312$, it is straightforward to see that $\pi$ is strongly
$231$-avoiding if and only if $\pi^{-1}$ is strongly $312$-avoiding. Note that
a permutation $\pi$ begins with $n$ if and only if $\pi^{-1}$ ends in $1$.
Therefore, the inverse map is a bijection between $\mathcal{A}_{n}$ and
$\mathcal{B}_{n}$. From Theorem 2.2, we obtain that

$\mathcal{A}_{n}=\\{\pi\mid\pi=(k+1)(k+2)\cdots nk(k-1)(k-2)\cdots
1,\,\lceil\frac{n}{2}\rceil\leq k\leq n-1\\}.$

It can be checked that the permutation $(k+1)(k+2)\cdots nk(k-1)(k-2)\cdots 1$
is the inverse of $n(n-1)\cdots(n-k+1)12\cdots(n-k)$. It yields that

$\mathcal{B}_{n}=\\{\pi\mid\pi=n(n-1)\cdots(n-k+1)12\cdots(n-k),\,\lceil\frac{n}{2}\rceil\leq
k\leq n-1\\}.$

Observe that all the permutations of the form
$n(n-1)\cdots(n-k+1)12\cdots(n-k)$ are $1432$-avoiding. We derive that
$\mathcal{B}_{n}\subseteq\mathcal{P}_{n}^{1}$. Hence we have
$\mathcal{P}_{n}^{1}=\mathcal{B}_{n}$. This completes the proof.

Given $\sigma\in\mathcal{S}_{k}$ and $\tau\in\mathcal{S}_{m}$, let
$\sigma\oplus\tau$ denote the direct sum of $\sigma$ and $\tau$ defined by

$(\sigma\oplus\tau)(i)=\begin{cases}\sigma(i),\text{\quad if $1\leq i\leq
k$},\\\ \tau(i-k)+k,\text{\quad if $k+1\leq i\leq k+m$.}\end{cases}$

For example, 231 $\oplus$ 21 = 23154. The following fact is straightforward
from the definition of direct sum.

$\displaystyle(\sigma\oplus\tau)^{2}=\sigma^{2}\oplus\tau^{2}.$ (2.1)

###### Lemma 2.4

Let $n\geq 3$ and $\pi=\pi_{1}\pi_{2}\cdots\pi_{n}\in\mathcal{S}_{n}$. Then
$\pi\in\mathcal{P}_{n}^{2}$ if and only if $\pi=\sigma\oplus 1$ for some
$\sigma\in\mathcal{S}_{n-1}(231,1432:231)$.

Proof. If $\pi\in\mathcal{P}_{n}^{2}$, then we have $\pi_{n}=n$. By the
definition of the direct sum, we have that $\pi=\sigma\oplus 1$ for some
$\sigma\in\mathcal{S}_{n-1}$. From (2.1) we have $\pi^{2}=\sigma^{2}\oplus 1$.
This means that $\sigma$ (resp. $\sigma^{2}$) is a subsequence of $\pi$ (resp.
$\pi^{2}$). Since $\pi$ avoids the chain $(231,1432:231)$, we deduce that
$\sigma$ also avoids the chain $(231,1432:231)$, that is,
$\sigma\in\mathcal{S}_{n-1}(231,1432:231)$.

Conversely, if $\pi=\sigma\oplus 1$ for some
$\sigma\in\mathcal{S}_{n-1}(231,1432:231)$, we have $\pi_{n}=n$ and
$\pi^{2}=\sigma^{2}\oplus 1$. To prove $\pi\in\mathcal{P}_{n}^{2}$, it
suffices to show that $\pi$ avoids the chain $(231,1432:231)$. Observe that
the element $\pi_{n}=n$ can not appear in any occurrences of the pattern $231$
(resp. $1432$) in $\pi$. Combining the fact that $\sigma$ avoids the chain
$(231,1432:231)$, we deduce that $\pi$ avoids both $231$ and $1432$. By a
similar argument as above, we have $\pi^{2}$ is $231$-avoiding. In conclusion,
we have $\pi$ avoids the chain $(231,1432:231)$, as desired.

###### Lemma 2.5

Let $n\geq 3$ and $\pi=\pi_{1}\pi_{2}\cdots\pi_{n}\in\mathcal{S}_{n}$. Then
$\pi\in\mathcal{P}_{n}^{3}$ if and only if $\pi=\sigma\oplus 21$ for some
$\sigma\in\mathcal{S}_{n-2}(231,1432:231)$.

Proof. If $\pi\in\mathcal{P}_{n}^{3}$, then we have $\pi_{n-1}\pi_{n}=n(n-1)$.
By the definition of the direct sum, we have that $\pi=\sigma\oplus 21$ for
some $\sigma\in\mathcal{S}_{n-2}$. Then we obtain $\pi^{2}=\sigma^{2}\oplus
12$ from (2.1). This means that $\sigma$ (resp. $\sigma^{2}$) is a subsequence
of $\pi$ (resp. $\pi^{2}$). Since $\pi$ avoids the chain $(231,1432:231)$, we
deduce that $\sigma$ also avoids the chain $(231,1432:231)$, that is,
$\sigma\in\mathcal{S}_{n-2}(231,1432:231)$.

Conversely, assume that $\pi=\sigma\oplus 21$ for some
$\sigma\in\mathcal{S}_{n-2}(231,1432:231)$. By the definition of direct sum
and (2.1), we have $\pi_{n-1}\pi_{n}=n(n-1)$ and $\pi^{2}=\sigma^{2}\oplus
12$. To prove $\pi\in\mathcal{P}_{n}^{3}$, it suffices to show that $\pi$
avoids the chain $(231,1432:231)$. As $\pi_{n-1}\pi_{n}=n(n-1)$, we deduce
that neither $n$ nor $n-1$ can appear in any occurrences of the pattern $231$
(resp. $1432$) in $\pi$. Combining the fact that $\sigma$ avoids the chain
$(231,1432:231)$, we deduce that $\pi$ avoids both $231$ and $1432$. By a
similar argument as above, we have $\pi^{2}$ is $231$-avoiding. Therefore
$\pi$ avoids the chain $(231,1432:231)$, as desired.

Proof of Conjecture 1.1. Let $n\geq 1$ and
$f(n)=|\mathcal{S}_{n}(231,1432:231)|$. Then we have

$f(n)=|\mathcal{P}_{n}^{1}|+|\mathcal{P}_{n}^{2}|+|\mathcal{P}_{n}^{3}|$

for $n\geq 3$. Lemmas 2.3, 2.4 and 2.5 tell us that
$|\mathcal{P}_{n}^{1}|=n-\lceil\frac{n}{2}\rceil=\lceil\frac{n-1}{2}\rceil$,
$|\mathcal{P}_{n}^{2}|=f(n-1)$ and $|\mathcal{P}_{n}^{3}|=f(n-2)$,
respectively. Therefore we have

$f(n)=\lceil\frac{n-1}{2}\rceil+f(n-1)+f(n-2)$

for $n\geq 3$ with the initial conditions $f(1)=1$ and $f(2)=2$. Notice that
the Lucas numbers $L_{n}$ satisfy that $L_{n}=L_{n-1}+L_{n-2}$ for $n\geq 3$
with the initial conditions $L_{1}=1$ and $L_{2}=3$. It is routine to check
that $f(n)$ and $L_{n+1}-\lceil\frac{n}{2}\rceil-1$ satisfy the same
recurrence and initial conditions. Thus
$f(n)=L_{n+1}-\lceil\frac{n}{2}\rceil-1$ for $n\geq 1$. This completes the
proof.

##  3 Avoiding the chain $(213,312:\overline{213})$

This section is devoted to the proof of Conjecture 1.2. To this end, we need
to give a characterization of the permutations in
$\mathcal{S}_{n}(213,312:\overline{213})$. Note that the permutations in
$\mathcal{S}_{n}(213,312)$ are exactly the set of unimodal permutations, i.e.,
those permutations $\pi=\pi_{1}\pi_{2}\cdots\pi_{n}$ with
$\pi_{1}<\pi_{2}<\cdots<\pi_{k}>\pi_{k+1}>\cdots>\pi_{n}$ for some $1\leq
k\leq n$ [5]. For a unimodal permutation $\pi=\pi_{1}\pi_{2}\cdots\pi_{n}$, we
have either $\pi_{1}=1$ or $\pi_{n}=1$. Hence, for $n\geq 2$, we can divide
the set $\mathcal{S}_{n}(213,312:\overline{213})$ into the following two
disjoint subsets:

$\displaystyle\mathcal{Q}_{n}^{1}$
$\displaystyle:=\\{\pi\mid\pi=\pi_{1}\pi_{2}\cdots\pi_{n}\in\mathcal{S}_{n}(213,312:\overline{213})\text{
and }\pi_{1}=1\\},$ $\displaystyle\mathcal{Q}_{n}^{2}$
$\displaystyle:=\\{\pi\mid\pi=\pi_{1}\pi_{2}\cdots\pi_{n}\in\mathcal{S}_{n}(213,312:\overline{213})\text{
and }\pi_{n}=1\\}.$

We will give characterizations of the unimodal permutations in
$\mathcal{Q}_{n}^{1}$ and $\mathcal{Q}_{n}^{2}$, respectively.

###### Lemma 3.1

Let $n\geq 3$ and $\pi=\pi_{1}\pi_{2}\cdots\pi_{n}\in\mathcal{S}_{n}$. Then we
have $\pi\in\mathcal{Q}_{n}^{1}$ if and only if $\pi=1\oplus\sigma$ for some
$\sigma\in\mathcal{S}_{n-1}(213,312:\overline{213})$.

Proof. If $\pi\in\mathcal{Q}_{n}^{1}$, then we have $\pi_{1}=1$. By the
definition of direct sum, the permutation $\pi$ can be written as
$\pi=1\oplus\sigma$ for some $\sigma\in\mathcal{S}_{n-1}$. From (2.1), we have
$\pi^{2}=1\oplus\sigma^{2}$. Hence $\sigma$ (resp. $\sigma^{2}$) is order
isomorphic to $\pi_{2}\pi_{3}\cdots\pi_{n}$ (resp.
$\pi^{2}(2)\pi^{2}(3)\cdots\pi^{2}(n)$). Then
$\sigma\in\mathcal{S}_{n-1}(213,312:\overline{213})$ follows directly from the
fact that $\pi$ avoids the chain $(213,312:\overline{213})$.

Conversely, if $\pi=1\oplus\sigma$ for some
$\sigma\in\mathcal{S}_{n-1}(213,312:\overline{213})$. Then we have $\pi_{1}=1$
and $\pi^{2}=1\oplus\sigma^{2}$. To prove $\pi\in\mathcal{Q}_{n}^{1}$, it
suffices to show that $\pi$ avoids the chain $(213,312:\overline{213})$.
Observe that the element $\pi_{1}=1$ can not appear in any occurrences of the
pattern $213$ (resp. $312$) in $\pi$. Combining the fact that $\sigma$ avoids
the chain $(213,312:\overline{213})$, we deduce that $\pi$ avoids both $213$
and $312$. By a similar argument as above, we have $\pi^{2}$ is
$\overline{213}$-avoiding. Therefore the permutation $\pi$ avoids the chain
$(213,312:\overline{213})$, as desired.

The following lemma is needed for the characterization of permutations in
$\mathcal{Q}_{n}^{2}$.

###### Lemma 3.2

Let $n\geq 3$ and
$\pi=\pi_{1}\pi_{2}\cdots\pi_{n}\in\mathcal{S}_{n}(213,312)$. If
$\pi^{2}(i-1)\pi^{2}(i)\pi^{2}(i+1)$ is an occurrence of the consecutive
pattern $\overline{213}$ in $\pi^{2}$, then we have $\pi_{i}=n$.

Proof. We have known that the permutations in $\mathcal{S}_{n}(213,312)$ are
unimodal permutations. If $\pi_{i}\neq n$, then we have either
$\pi_{i-1}<\pi_{i}<\pi_{i+1}$ or $\pi_{i-1}>\pi_{i}>\pi_{i+1}$. In the former
case, we have
$\pi(\pi_{i-1})\pi(\pi_{i})\pi(\pi_{i+1})=\pi^{2}(i-1)\pi^{2}(i)\pi^{2}(i+1)$
is a subsequence of $\pi$. Since $\pi^{2}(i-1)\pi^{2}(i)\pi^{2}(i+1)$ is an
occurrence of the consecutive pattern $\overline{213}$ in $\pi^{2}$, we obtain
that $\pi$ contains the pattern $213$, contradicting to the fact that $\pi$ is
$213$-avoiding. In the later case, we have
$\pi(\pi_{i+1})\pi(\pi_{i})\pi(\pi_{i-1})=\pi^{2}(i+1)\pi^{2}(i)\pi^{2}(i-1)$
is a subsequence of $\pi$. Since $\pi^{2}(i-1)\pi^{2}(i)\pi^{2}(i+1)$ is an
occurrence of the consecutive pattern $\overline{213}$ in $\pi^{2}$, we have
$\pi^{2}(i+1)\pi^{2}(i)\pi^{2}(i-1)$ is an occurrence of the pattern $312$ in
$\pi$, contradicting to the fact that $\pi$ is $312$-avoiding. This completes
the proof.

###### Lemma 3.3

For $n\geq 3$, we have

$\mathcal{Q}_{n}^{2}=\\{\pi=\pi_{1}\pi_{2}\cdots\pi_{n}\mid\pi=23\cdots
n1\text{ or }\pi=\sigma n(n-1)\tau 1\\},$

where $\sigma$ (possibly empty) is increasing and $\tau$ (possibly empty) is
decreasing.

Proof. Let $\pi=\pi_{1}\pi_{2}\cdots\pi_{n}$ be a unimodal permutation with
$\pi_{k}=n$ and $\pi_{n}=1$. Then we have either $\pi_{k-1}=n-1$ or
$\pi_{k+1}=n-1$.

Case (i). $\pi_{k-1}=n-1$.
First we consider the permutation $\pi=23\cdots n1$. We have $\pi^{2}=34\cdots
n12$. By the definition of $\mathcal{Q}_{n}^{2}$, one can easily check that
$\pi\in\mathcal{Q}_{n}^{2}$. Now assume that $\pi\neq 23\cdots n1$. It yields
that there exists at least one element between $n$ and $1$ in $\pi$. We claim
that $\pi\nin\mathcal{Q}_{n}^{2}$. We will prove the claim by showing that
$\pi^{2}(k-1)\pi^{2}(k)\pi^{2}(k+1)$ is an occurrence of the consecutive
pattern $\overline{213}$ in $\pi^{2}$. Since $\pi$ is unimodal with
$\pi_{k}=n$ and $\pi_{n}=1$, we have $\pi_{i}>i$ for $i\in[1,k]$ and
$\pi_{k+1}\geq\pi_{j}\geq\pi_{n-1}$ for $j\in[k+1,n-1]$. It is routine to
check that $\pi^{2}(k)=\pi(\pi_{k})=\pi_{n}=1$ and
$\pi^{2}(k-1)=\pi(\pi_{k-1})=\pi_{n-1}$. If $\pi_{k+1}\leq k$, we have
$\pi^{2}(k+1)=\pi(\pi_{k+1})>\pi_{k+1}$. As
$\pi_{k+1}\geq\pi_{n-1}=\pi^{2}(k-1)$, we have $\pi^{2}(k+1)>\pi^{2}(k-1)$. It
follows that $\pi^{2}(k-1)\pi^{2}(k)\pi^{2}(k+1)$ is an occurrence of the
pattern $\overline{213}$ in $\pi^{2}$, as desired. Now suppose that
$k<\pi_{k+1}<n-1$. We have
$\pi^{2}(k-1)=\pi(n-1)<\pi(\pi_{k+1})=\pi^{2}(k+1)$. Again
$\pi_{k-1}^{2}\pi_{k}^{2}\pi_{k+1}^{2}$ is an occurrence of the pattern
$\overline{213}$ in $\pi^{2}$, as desired. Hence, the claim is verified. In
conclusion, there exists only one permutation $\pi=23\cdots
n1\in\mathcal{Q}_{n}^{2}$ if $\pi_{k-1}=n-1$.

Case (ii). $\pi_{k+1}=n-1$.
Recall that $\pi$ is unimodal. In this case, the permutation $\pi$ can be
written in the form $\pi=\sigma n(n-1)\tau 1$ where $\sigma$ (possibly empty)
is increasing and $\tau$ (possibly empty) is decreasing. We need to show that
all such permutations are contained in the set $\mathcal{Q}_{n}^{2}$. If
$\sigma$ is empty, it is straightforward to check that $\pi=n(n-1)\cdots
1\in\mathcal{Q}_{n}^{2}$, as desired. Now assume that $\sigma$ is nonempty. To
prove $\pi\in\mathcal{Q}^{2}_{n}$, we need to show that $\pi^{2}$ avoids the
consecutive pattern $\overline{213}$. From Lemma 3.2, it is sufficient to show
that $\pi^{2}(k-1)\pi^{2}(k)\pi^{2}(k+1)$ is not an occurrence of
$\overline{213}$ in $\pi^{2}$. Observe that $k\leq\pi_{k-1}<n-1$. Since $\pi$
is unimodal, we have $\pi^{2}(k-1)=\pi(\pi_{k-1})>\pi(n-1)=\pi^{2}(k+1)$.
Thus, $\pi^{2}(k-1)\pi^{2}(k)\pi^{2}(k+1)$ is not an occurrence of
$\overline{213}$ in $\pi^{2}$, as desired.

Proof of Conjecture 1.2. Let $n\geq 2$ and
$g(n)=|\mathcal{S}_{n}(213,312:\overline{213})|$. Then we have

$g(n)=|\mathcal{Q}_{n}^{1}|+|\mathcal{Q}_{n}^{2}|.$

By Lemma 3.1, we have $|\mathcal{Q}_{n}^{1}|=g(n-1)$ for $n\geq 3$. Lemma 3.3
tells us that

$\mathcal{Q}_{n}^{2}=\\{\pi=\pi_{1}\pi_{2}\cdots\pi_{n}\mid\pi=23\cdots
n1\text{ or }\pi=\sigma n(n-1)\tau 1\\},$

where $\sigma$ (possibly empty) is increasing and $\tau$ (possibly empty) is
decreasing. A permutation $\pi\in\mathcal{Q}_{n}^{2}$ with the form
$\pi=\sigma n(n-1)\tau 1$ is uniquely determined by the set consisting of the
elements in $\sigma$. We can choose the elements of $\sigma$ in
$\binom{n-3}{r}$ way if $\sigma$ contains $r$ elements. Then we deduce that

$|\mathcal{Q}_{n}^{2}|=1+\sum_{\mathrm{r}=0}^{n-3}\binom{n-3}{r}=2^{n-3}+1$

for $n\geq 3$. Therefore we have

$g(n)=g(n-1)+2^{n-3}+1$

for $n\geq 3$. Taking into account the initial conditions $g(2)=2$, the unique
solution is $g(n)=2^{n-2}+n-1$. This completes the proof.

## Acknowledgments

The work was supported by the National Natural Science Foundation of China
(11801378).

## References

  * [1] K. Archer, A. Geary, Powers of permutations that avoid chains of patterns, arXiv:2312.14351v1 [math.CO].
  * [2] M. Bóna, Combinatorics of Permutations, second ed., CRC Press, 2012.
  * [3] M. Bóna, R. Smith, Pattern avoidance in permutations and their squares, Discrete Math., 342(11) (2019), 3194–3200.
  * [4] A. Burcroff and C. Defant, Pattern-avoiding permutation powers, Discrete Math., 343(11) (2020), 112017.
  * [5] T. Gannon, The cyclic structure of unimodal permutations, Discrete Math., 237 (2001), 149–161.
  * [6] S. Kitaev, Patterns in permutations and words. Springer Verlag (EATCS monographs in Theoretical Computer Science book series), 2011.
  * [7] N.J.A. Sloane, The on-line encyclopedia of integer sequences, http://oeis.org.
  * [8] P.A. MacMahon, Combinatory Analysis, Cambridge University Press, London, 1915/16. 1.
  * [9] J. Pan, On a conjecture about strong pattern avoidance, Graphs and Combin., 39(2) (2023), https://doi.org/10.1007/s00373-022-02602-y.


Journal of Algebra and Related Topics
Vol. XX, No XX, (201X), pp XX-XX

# I-prime ideals

Ismael Akray

###### Abstract.

In this paper, we introduce a new generalization of weakly prime ideals called
$I$-prime. Suppose $R$ is a commutative ring with identity and $I$ a fixed
ideal of $R$. A proper ideal $P$ of $R$ is $I$-prime if for $a,b\in R$ with
$ab\in P-IP$ implies either $a\in P$ or $b\in P$. We give some
characterizations of $I$-prime ideals and study some of its properties.
Moreover, we give conditions under which $I$-prime ideals becomes prime or
weakly prime and we construct the view of $I$-prime ideal in decomposite
rings.

MSC(2010): Primary: 13A15;
Keywords: Prime ideal, weakly prime ideal, almost prime ideal, radical of the
ideal.

##  1\. Introduction

Throughout this article, $R$ will be a commutative ring with identity. Prime
ideals play an essential role in ring theory. A prime ideal $P$ of $R$ is a
proper ideal $P$ with the property that for $a,b\in R$, $ab\in P$ implies
$a\in P$ or $b\in P$. There are several ways to generalize the notion of a
prime ideal. We could either restrict or enlarge where $a$ and/or $b$ lie or
restrict or enlarge where $ab$ lies. In this article we will interested in a
generalization obtained by restricting where $ab$ lies. A proper ideal $P$ of
$R$ is weakly prime if for $a,b\in R$ with $ab\in P-0$, either $a\in P$ or
$b\in P$. Weakly prime ideals were studied in some detail by Anderson and
Smith (2003) in [1]. Thus any prime ideal is weakly prime. Bhatwadekar and
Sharma (2005) in [2] recently deﬁned a proper ideal $I$ of an integral domain
$R$ to be almost prime if for $a,b\in R$ with $ab\in I-I^{2}$, then either
$a\in I$ or $b\in I$. This deﬁnition can obviously be made for any commutative
ring $R$. Thus a weakly prime ideal is almost prime and any proper idempotent
ideal is almost prime. Moreover, an ideal $I$ of $R$ is almost prime if and
only if $I/I^{2}$ is a weakly prime ideal of $R/I^{2}$. Also almost prime
ideals were generalized to $n$-almost prime as follows; a proper ideal $I$ is
called $n$-almost prime ideal if for any $a,b\in R$ with $ab\in I-I^{n}$, then
either $a\in I$ or $b\in I$. With weakly prime ideals and almost prime ideals
in mind, we make the following deﬁnition. Let $R$ be a commutative ring and
$I$ be a fixed ideal of $R$. Then a proper ideal $P$ of $R$ is called
$I$-prime ideal if for $a,b\in R$, $ab\in P-IP$, implies $a\in P$ or $b\in P$.
So every weakly prime and $n$-almost prime ideal is $I$-prime where $I$ taken
to be zero or $P^{n-1}$ respectively. If $I=R$, then every ideal is $I$-prime,
so we can assume $I$ to be proper ideal of $R$. For more details see [3].

######  Example 1.1.

Consider the ring $\textbf{Z}_{12}$ and take $P=I=<4>$. Then $P$ is $I$-prime
which is not prime nor weakly prime.

##  2\. Main Results

We begin with the following lemma.

######  Lemma 2.1.

Let $P$ be a proper ideal of a ring $R$. Then $P$ is an $I$-prime ideal if and
only if $P/IP$ is weakly prime in $R/IP$.

###### Proof.

$(\Rightarrow)$ Let $P$ be an $I$-prime in $R$. Let $a,b\in R$ with
$0\neq(a+IP)(b+IP)=ab+IP\in P/IP$. Then $ab\in P-IP$ implies $a\in P$ or $b\in
P$, hence $a+IP\in P/IP$ or $b+IP\in P/IP$. So $P/IP$ is weakly prime ideal in
$R/IP$.

$(\Leftarrow)$ Suppose that $P/IP$ is weakly prime in $R/IP$ and take $r,s\in
R$ such that $rs\in P-IP$. Then $0\neq rs+IP=(r+IP)(s+IP)\in P/IP$ so $r+IP\in
P/IP$ or $s+IP\in P/IP$. Therefore $r\in P$ or $s\in P$. Thus $P$ is an
$I$-prime ideal in $R$. ∎

######  Theorem 2.2.

(1) Let $I\subseteq J$. If $P$ is $I$-prime ideal of a ring $R$, then it is
$J$-prime.
(2) Let $R$ be commutative ring and $P$ an $I$-prime ideal that is not prime,
then $P^{2}\subseteq IP$. Thus, an $I$-prime ideal $P$ with $P^{2}\nsubseteq
IP$ is prime.

###### Proof.

(1) The proof come from the fact that if $I\subseteq J$, then $P-JP\subseteq
P-IP$. (2) Suppose that $P^{2}\nsubseteq IP$, we show that $P$ is prime. Let
$ab\in P$ for $a,b\in R$. If $ab\notin IP$, then $P$ $I$-prime gives $a\in P$
or $b\in P$. So assume that $ab\in IP$. First, suppose that $aP\nsubseteq IP$;
say $ax\notin IP$ for some $x\in P$. Then $a(x+b)\in P-IP$. So $a\in P$ or
$x+b\in P$ and hence $a\in P$ or $b\in P$. So we can assume that $aP\subseteq
IP$ in similar way we can assume that $bP\subseteq IP$. Since $P^{2}\nsubseteq
IP$, there exist $y,z\in P$ with $yz\notin IP$. Then $(a+y)(b+z)\in P-IP$. So
$P$ $I$-prime gives $a+y\in P$ or $b+z\in P$; Hence $a\in P$ or $b\in P$.
Therefore $P$ is prime. ∎

In the following we give a counter example on the converse of part (1) of
Theorem 2.2.

######  Example 2.3.

In the ring $\textbf{Z}_{12}[x]$, put $I=0$, $J=<4>$ and $P=<4x>$. Then
$P-IP=<4x>-0$ and $P-JP=<4x>-<4><4x>=\phi$. Hence $P$ is $J$-prime but not
$I$-prime.

######  Corollary 2.4.

Let $P$ be an $I$-prime ideal of a ring $R$ with $IP\subseteq P^{3}$. Then $P$
is $\cap_{i=1}^{\infty}P^{i}$-prime.

###### Proof.

If $P$ is prime, then $P$ is $\cap_{i=1}^{\infty}P^{i}$-prime. Assume that $P$
is not prime. By Theorem 2.2 $P^{2}\subseteq IP\subseteq P^{3}$. Thus
$IP=P^{n}$ for each $n\geq 2$. So $\cap_{i=1}^{\infty}P^{i}=P\cap P^{2}=P^{2}$
and $(\cap_{i=1}^{\infty}P^{i})P=P^{2}P=P^{3}=IP$. Hence $P$ $I$-prime implies
$P$ is $\cap_{i=1}^{\infty}P^{i}$-prime. ∎

######  Remark 2.5.

Let $P$ be $I$-prime ideal. Then $P\subseteq\sqrt{IP}$ or $\sqrt{IP}\subseteq
P$. If $P\subsetneqq\sqrt{IP}$, then $P$ is not prime since otherwise
$IP\subseteq P$ implies $\sqrt{IP}\subseteq\sqrt{P}=P$. While if
$\sqrt{IP}\subsetneqq P$, then $P$ is prime.

######  Corollary 2.6.

Let $P$ be $I$-prime ideal of a ring $R$ which is not prime. Then
$\sqrt{P}=\sqrt{IP}$

###### Proof.

By Theorem 2.2, $P^{2}\subseteq IP$ and hence
$\sqrt{P}=\sqrt{P^{2}}\subseteq\sqrt{IP}$. The other containment always holds.

∎

Now we give a way to construct $I$-prime ideals $P$ when
$\cap_{i=1}^{\infty}P^{i}\subseteq IP\subseteq P^{3}$.

######  Remark 2.7.

Assume that $P$ is $I$-prime, but not prime. Then by Theorem 2.2, if
$IP\subseteq P^{2}$, then $P^{2}=IP$. In particular, if $P$ is weakly prime
($0$-prime) but not prime, then $P^{2}=0$. Suppose that $IP\subseteq P^{3}$.
Then $P^{2}\subseteq IP\subseteq P^{3}$; So $P^{2}=P^{3}$ and thus $P^{2}$ is
idempotent.

######  Theorem 2.8.

(1) Let $R,S$ be commutative rings and $P$ $0$-prime ideal of $R$. Then
$P\times S$ is $I$-prime ideal of $R\times S$ for each ideal $I$ of $R\times
S$ with $\cap_{i=1}^{\infty}(P\times S)^{i}\subseteq I(P\times S)\subseteq
P\times S$.
(2) Let $P$ be finitely generated proper ideal of commutative ring $R$. Assume
$P$ is $I$-prime where $IP\subseteq P^{3}$. Then either $P$ is $0$-prime or
$P^{2}\neq 0$ is idempotent and $R$ decomposes as $T\times S$ where $S=P^{2}$
and $P=J\times S$ where $J$ is $0$-prime. Thus $P$ is $I$-prime for
$\cap_{i=1}^{\infty}P^{i}\subseteq IP\subseteq P$.

###### Proof.

(1) Let $R$ and $S$ be commutative rings and $P$ be weakly prime ($0$-prime)
ideal of $R$. Then $P\times S$ need not be a $0$-prime ideal of $R\times S$;
In fact, $P\times S$ is $0$-prime if and only if $P\times S$ (or equivalently
$P$) is prime [see Anderson 2003]. However, $P\times S$ is $I$-prime for each
$I$ with $\cap_{i=1}^{\infty}(P\times S)^{i}\subseteq I(P\times S)$. If $P$ is
prime , then $P\times S$ is prime ideal and thus is $I$-prime for all $I$.
Assume that $P$ is not prime. Then $P^{2}=0$ and $(P\times S)^{2}=0\times S$.
Hence $\cap_{i=1}^{\infty}(P\times S)^{i}=\cap_{i=1}^{\infty}P^{i}\times
S=0\times S$. Thus $P\times S-\cap_{i=1}^{\infty}(P\times S)^{i}=P\times
S-0\times S=(P-0)\times S$. Since $P$ is weakly prime, $P\times S$ is
$\cap_{i=1}^{\infty}(P\times S)^{i}$-prime and as $\cap_{i=1}^{\infty}(P\times
S)^{i}\subseteq I(P\times S)$, $P\times S$ is $I$-prime.

(2) If $P$ is prime, then $P$ is $0$-prime. So we can assume that $P$ is not
prime. Then $P^{2}\subseteq IP$ and hence $P^{2}\subseteq IP\subseteq P^{3}$.
So $P^{2}=P^{3}$. Hence $P^{2}$ is idempotent. Since $P^{2}$ is finitely
generated, $P^{2}=<e>$ for some idempotent $e\in R$. Suppose $P^{2}=0$. Then
$IP\subseteq P^{3}=0$. So $IP=0$ and hence $P$ is $0$-prime. So assume
$P^{2}\neq 0$. Put $S=P^{2}=Re$ and $T=R(1-e)$, so $R$ decomposesas $T\times
S$ where $S=P^{2}$. Let $J=P(1-e)$, so $P=J\times S$ where
$J^{2}=(P(1-e))^{2}=P^{2}(1-e)^{2}=<e>(1-e)=0$. We show that $J$ is $0$-prime.
Let $ab\in J^{2}-0$, so $(a,1)(b,1)=(ab,1)\in J\times S-(J\times
S)^{2}=J\times S-0\times S\subseteq P-IP$ since $IP\subseteq P^{3}$ implies
$IP\subseteq P^{3}=(J\times S)^{3}=0\times S$. Hence $(a,1)\in P$ or $(b,1)\in
P$ so $a\in J$ or $b\in J$. Therefore $J$ is weakly prime. ∎

######  Corollary 2.9.

Let $R$ be an indecomposable commutative ring and $P$ a finitely generated
$I$-prime ideal of $R$, where $IP\subseteq P^{3}$. Then $P$ is weakly prime.

######  Corollary 2.10.

Let $R$ be a Noetherian integral domain. A proper ideal $P$ of $R$ is prime if
and only if $P$ is $P^{2}$-prime.

######  Theorem 2.11.

Let $a$ be a non-unit element in $R$.
(1) Let $(0:a)\subseteq(a)$. Then $(a)$ is $I$-prime for
$I(a)\subseteq(a)^{2}$ if and only if $(a)$ is prime.
Let $(R,m)$ be quasi-local ring. Then
(2) $(a)$ is $I$-prime for $I(a)\subseteq(a)^{3}$ if and only if $(a)$ is
$0$-prime.
(3) $(a)$ is $m$-prime if and only if $a$ is irreducible.

###### Proof.

(1) Suppose that (a) is $I$-prime and $bc\in(a)$. If $bc\notin I(a)$, then
$b\in(a)$ or $c\in(a)$. So suppose that $bc\in I(a)$. Now $(b+a)c\in(a)$. If
$(b+a)c\notin I(a)$, then $b+a\in(a)$ or $c\in(a)$ and hence $b\in(a)$ or
$c\in(a)$. So assume that $(b+a)c\in I(a)$. Then $ac\in I(a)$ and hence
$ac\in(a)^{2}$. So $ac=za^{2}$ and hence $c-za\in(0:a)$. Thus
$c\in(0:a)+(a)=(a)$. The converse part is trivial since every prime ideal is
$I$-prime.
(2) If $(a)$ is $0$-prime, then $(a)$ is $I$-prime for each $I$ with
$I(a)\subseteq(a)^{3}$. Conversely, let $(a)$ be $I$-prime for
$I(a)\subseteq(a)^{3}$. Since a quasi local ring has no nontrivial
idempotents, $(a)$ is $0$-prime by Theorem 2.8 part(2).
(3) If $a$ is irreducible means that $a=xy$ implies that $x\in(a)$ or
$y\in(a)$ and $(a)$ is $m$-prime means that $xy\in(a)-m(a)$ which implies that
$x\in(a)$ or $y\in(a)$. But $xy\in(a)-m(a)$ if and only if $xy=za$ for some
unit $z\in R$ if and only if $a=z^{-1}xy$ for some unit $z^{-1}\in R$. Thus
$(a)$ is $m$-prime if and only if $a=xy$ implies $x\in(a)$ or $y\in(a)$. ∎

We now give some characterizations of $I$-prime ideals.

######  Theorem 2.12.

Let $P$ be a proper ideal of $R$. Then the following are equivalent:
(1) $P$ is $I$-prime.
(2) For $x\in R-P$, $(P:x)=P\cup(IP:x)$
(3) For $x\in R-P$, $(P:x)=P$ or $(P:x)=(IP:x)$
(4) For ideals $J$ and $K$ of $R$, $JK\subseteq P$ and $JK\nsubseteq IP$ imply
$J\subseteq P$ or $K\subseteq P$.

###### Proof.

(1) $\Rightarrow$ (2) Suppose $r\in R-P$. Let $s\in(P:r)$, so $rs\in P$. If
$rs\in P-IP$, then $s\in P$. If $rs\in IP$, then $s\in(IP:r)$, So
$(P:r)\subseteq P\cup(IP:r)$. The other containment always holds.
(2) $\Rightarrow$ (3) Note that if an ideal is a union of two ideals, then it
is equal to one of them.
(3) $\Rightarrow$ (4) Let $J$ and $K$ be two ideals of $R$ with $JK\subseteq
P$. Assume that $J\nsubseteq P$ and $K\nsubseteq P$. We claim that
$JK\subseteq IP$. Suppose $r\in J$. First, Let $r\notin P$. Then $rK\subseteq
P$ gives $K\subseteq(P:r)$. Now $K\nsubseteq P$, so $(P:r)=(IP:r)$. Thus
$rK\subseteq IP$. Next, let $r\in J\cap P$. Choose $s\in J-P$. Then $r+s\in
J-P$. So by the first case $sK\subseteq IP$ and so $(r+s)K\subseteq IP$. Let
$t\in K$. Then $rt=(r+s)t-st\in IP$. So $rK\subseteq IP$. Hence $JK\subseteq
IP$.
(4) $\Rightarrow$ (1) Let $rs\in P-IP$. Then $(r)(s)\subseteq P$ but
$(r)(s)\nsubseteq IP$. So $(r)\subseteq P$ or $(s)\subseteq P$; i-e. $r\in P$
or $s\in P$. ∎

######  Corollary 2.13.

Suppose $P$ is $I$-prime ideal that is not prime. Then $P\sqrt{IP}\subseteq
IP$.

###### Proof.

Let $r\in\sqrt{IP}$. If $r\in P$, then $rP\subseteq P^{2}\subseteq IP$ by
Theorem 2.2. So assume that $r\notin P$ by Theorem 2.12, $(P:r)=P$ or
$(P:r)=(IP:r)$. As $P\subseteq(P:r)$, the last gives $rP\subseteq IP$. So
assume that $(P:r)=P$. Let $r^{n}\in IP$, but $r^{n-1}\notin IP$. Then
$r^{n}\in P$, so $r^{n-1}\in(P:r)=P$. Thus $r^{n-1}\in P-IP$, so $r\in P$ a
contradiction. ∎

It is known that if $S$ is a multiplicatively closed subset of a commutative
ring $R$ and $P$ as a prime ideal of $R$ with $P\cap S=\phi$, then $S^{-1}P$
is a prime ideal of $S^{-1}R$ and $S^{-1}P\cap R=P$. The first result extended
to weakly prime ideals in [2, Proposition 13] and to almost prime ideals in
[5, Lemma 2.13]. Fix an ideal $I$ of $R$ we prove that if $P$ is $I$-prime
with $P\cap S=\phi$, then $S^{-1}P$ is $S^{-1}I$-prime. Note that for an ideal
$J$ of $R$ with $J\subseteq P$, $I(P/J)=(IP+J)/J$. If $P$ is prime
(respectively, weakly prime, $n$-almost prime), then so is $P/J$. We
generalize this result to $I$-prime ideals in the following proposition.

######  Proposition 2.14.

Let $R$ be a ring and $I$ be an ideal of $R$. Let $P$ be $I$-prime ideal of
$R$. Then the following are true.
(1) If $J$ is an ideal of $R$ with $J\subseteq P$, then $P/J$ is $I$-prime
ideal of $R/J$.
(2) Assume that $S$ is multiplicatively closed subset of $R$ with $P\cap
S=\phi$. Then $S^{-1}P$ is a $S^{-1}I$-prime ideal of $S^{-1}R$. If
$S^{-1}P\neq S^{-1}(IP)$, then $S^{-1}P\cap R=P$.

###### Proof.

(1) Let $x,y\in R$ with $\bar{x}\bar{y}\in P/J-I(P/J)=P/J-(IP+J)/J$. Thus
$xy\in P-(IP+J)$. Hence $xy\in P-IP$, so $x\in P$ or $y\in P$. Therefore
$\bar{x}\in P/J$ or $\bar{y}\in P/J$; so $P/J$ is $I$-prime.
(2) Let $\frac{a}{s}\frac{b}{t}\in S^{-1}P-S^{-1}I.S^{-1}P\subseteq
S^{-1}P-S^{-1}(IP)=S^{-1}(P-IP)$. So $abk\in P-IP$ for some $k\in S$. Thus $P$
$I$-prime gives $a\in P$ or $bk\in P$. So $a/s\in S^{-1}P$ or $b/t\in
S^{-1}P$. Hence $S^{-1}P$ is $S^{-1}I$-prime. Let $a\in S^{-1}P\cap R$, so
there exists $u\in S$ with $au\in P$. If $au\notin IP$, then $au\in P-IP$, so
$a\in P$. If $au\in IP$, then $a\in S^{-1}(IP)\cap R$. So $S^{-1}P\cap
R\subseteq P\cup(S^{-1}(IP)\cap R)$. Hence $S^{-1}P\cap R=P$ or $S^{-1}P\cap
R=S^{-1}(IP)\cap R$. But the second case gives $S^{-1}P=S^{-1}(IP)$. ∎

Let $R_{1}$ and $R_{2}$ be two rings. It is known that the prime ideals of
$R_{1}\times R_{2}$ have the form $P\times R_{2}$ or $R_{1}\times Q$, where
$P$ is a prime ideal of $R_{1}$ and $Q$ is a prime ideal of $R_{2}$. We next,
generalize this result to $I$-primes.

######  Theorem 2.15.

For $i=1,2$ let $R_{i}$ be ring and $I_{i}$ ideal of $R_{i}$. Let
$I=I_{1}\times I_{2}$. Then the $I$-prime ideals of $R_{1}\times R_{2}$ have
exactly one of the following three types:
(1) $P_{1}\times P_{2}$, where $P_{i}$ is a proper ideal of $R_{i}$ with
$I_{i}P_{i}=P_{i}$.
(2) $P_{1}\times R_{2}$ where $P_{1}$ is an $I_{1}$-prime of $R_{1}$ and
$I_{2}R_{2}=R_{2}$.
(3) $R_{1}\times P_{2}$, where $P_{2}$ is an $I_{2}$-prime of $R_{2}$ and
$I_{1}R_{1}=R_{1}$.

###### Proof.

We first prove that an ideal of $R_{1}\times R_{2}$ having one of these three
types is $I$-prime. The first type is clear since $P_{1}\times
P_{2}-I(P_{1}\times P_{2})=P_{1}\times P_{2}-(I_{1}P_{1}\times
I_{2}P_{2})=\phi$. Suppose that $P_{1}$ is $I_{1}$-prime and
$I_{2}R_{2}=R_{2}$. Let $(a,b)(x,y)\in P_{1}\times R_{2}-I_{1}P_{1}\times
I_{2}R_{2}=P_{1}\times R_{2}-I_{1}P_{1}\times R_{2}=(P_{1}-I_{1}P_{1})\times
R_{2}$. Then $ax\in P_{1}-I_{1}P_{1}$ implies that $a\in P_{1}$ or $x\in
P_{1}$, so $(a,b)\in P_{1}\times R_{2}$ or $(x,y)\in P_{1}\times R_{2}$. Hence
$P_{1}\times R_{2}$ is $I$-prime. Similarly we can prove the last case.

Next, let $P_{1}\times P_{2}$ be $I$-prime and $ab\in P_{1}-I_{1}P_{1}$. Then
$(a,0)(b,0)=(ab,0)\in P_{1}\times P_{2}-I(P_{1}\times P_{2})$, so $(a,0)\in
P_{1}\times P_{2}$ or $(b,0)\in P_{1}\times P_{2}$, i-e, $a\in P_{1}$ or $b\in
P_{1}$. Hence $P_{1}$ is $I_{1}$-prime. Likewise, $P_{2}$ is $I_{2}$-prime.

Assume that $P_{1}\times P_{2}\neq I_{1}P_{1}\times I_{2}P_{2}$. Say
$P_{1}\neq I_{1}P_{1}$. Let $x\in P_{1}-I_{1}P_{1}$ and $y\in P_{2}$. Then
$(x,1)(1,y)=(x,y)\in P_{1}\times P_{2}$. So $(x,1)\in P_{1}\times P_{2}$ or
$(1,y)\in P_{1}\times P_{2}$. Thus $P_{2}=R_{2}$ or $P_{1}=R_{1}$. Assume that
$P_{2}=R_{2}$. So $P_{1}\times R_{2}$ is $I$-prime, where $P_{1}$ is
$I_{1}$-prime. ∎

Acknowledgments

The author is deeply grateful to referee for his careful reading of the
manuscript. Whose comments made the paper more readable.

## References

  * [1] D. D. Anderson and E. Smith, Weakly prime ideals, Houston J. Math 29 (2003), 831–840.
  * [2] S. M. Bhatwadekar and P. K. Sharma, Unique factorization and birth of almost primes, Comm. Algebra 33 (2005), 43–49.
  * [3] I. Kaplansky, Commutative Rings (Revised Edition), Chicago, University of Chicago press, 1974.

Ismael Akray
Department of Mathematics, University of Soran Kurdistan region-Erbil, Iraq.
Email<EMAIL_ADDRESS><EMAIL_ADDRESS>