On 2-nil Primary Ideals of Commutative Rings

1. Some Properties of 2-nil Primary Ideals

In this section, a new class of ideals, the 2-nil primary ideal, will be introduced. We then delve into the properties and characterizations of 2-nil primary ideals. We aim to thoroughly understand their defining characteristics by presenting various theorems and proofs. Examples are included to illustrate the practical implications of these theoretical results.

Definition 1. 

Let R be a ring and I a proper ideal of R. We call I a 2-nil primary ideal if whenever $abc\in I$ then either $ab\in \sqrt{0}$ or $ac\in \sqrt{I}$ or $bc\in \sqrt{I}.$

From the definition 1 one can immediately infer that if I is 2-nil ideal, then I is 2-nil primary ideal. The converse of this implication is not true in general, as demonstrated by the following example. However, if I is a radical ideal, then I is a 2-nil primary if and only if I is a 2-nil ideal.

Example 1. 

Consider $R=\mathbb{Z}$ and $I=4\mathbb{Z}.$ Then I is not 2-nil ideal as $2·2·1\in I$ but $2.2\notin \sqrt{0}$ and $2.1\notin I.$ However, I is 2-nil primary ideal. To verify this claim, consider $abc\in 4\mathbb{Z}$. Then $abc=4k,\mathrm{where}k\in \mathbb{Z}.$ It is clear that at least one of $a,b,$ and c should be even. If $ab\notin \sqrt{0}$, then $ac\in \sqrt{I}=2\mathbb{Z}$ or $bc\in \sqrt{I}=2\mathbb{Z}.$

The following diagram illustrates the relationship of 2-nil primary ideal with some well-known ideal classes in the literature.

However, as the reader can see in the following, there is a strict relationship between 2-nil primary ideal and 2-nil ideals. By the following proposition, we can define 2-nil primary ideal as an ideal whose radical is 2-nil, that is; a proper ideal I of a ring R is called a 2-nil primary ideal if $\sqrt{I}$ is a 2-nil ideal of R.

Proposition 1. 

Let I be a proper ideal of a ring R. Then I is a 2-nil primary ideal if and only if $\sqrt{I}$ is a 2-nil ideal.

Proof. 

Let $a,b,c\in R$ such that $abc\in \sqrt{I}$ and $ab\notin \sqrt{0}$. Then ${\left(abc\right)}^n=a^n{b}^n{c}^n\in I$ for some $n\in \mathbb{N}.$ Since I is 2-nil primary, we have either $a^n{c}^n\in \sqrt{I}$ or $b^n{c}^n\in \sqrt{I}.$ This means that either $ac\in \sqrt{I}$ or $bc\in \sqrt{I}.$ Hence, $\sqrt{I}$ is a 2-nil ideal of R. Conversely, suppose that $\sqrt{I}$ is a 2-nil ideal and $a,b,c\in R$ such that $abc\in I$. Since $I\subseteq \sqrt{I}$, we have either $ab\in \sqrt{0}$ or $bc\in \sqrt{I}$ or $ac\in \sqrt{I}$. Thus I is a 2-nil primary ideal of R. □

Next, we discuss the connection between the concepts of 2-nil primary ideal and quasi primary ideal. The next example indicates that a 2-nil primary ideal need not be a quasi primary ideal in general.

Example 2. 

Let $R={\mathbb{Z}}_{12}$ and consider the ideal $I=6{\mathbb{Z}}_{12}.$I is not a quasi primary ideal since $2.3\in I$ but neither $2\in \sqrt{I}$ nor $3\in \sqrt{I}=6{\mathbb{Z}}_{12}.$ However, I is a 2-nil primary ideal of R. To show that let $abc\in I$ for some $a,b,c\in R$. Then $2|abc$ and $3|abc$ which implies that at least one of the elements $ab$, $ac$ or $bc$ is a multiple of 6. Hence, either $ab\in \sqrt{0}=6{\mathbb{Z}}_{12}$ or $ac\in \sqrt{I}=6{\mathbb{Z}}_{12}$ or $bc\in \sqrt{I}=6{\mathbb{Z}}_{12}.$

We conclude some conditions for a 2-nil primary ideal to be quasi primary.

Proposition 2. 

In any ring R with $dim\left(R\right)=1$ (in particular, in Dedekind domains that are not fields; for example, in discrete valuation rings) every 2-nil primary ideals are quasi primary.

Proof. 

Let R be a ring with Krull dimension 1, and I be a 2-nil primary ideal of R. Assume on the contrary that there are some $a,b\in R-\sqrt{I}$ such that $ab\in I$. Then $a.b.1\in I$. Since $a,b\notin \sqrt{0}$ and $\sqrt{0}$ is prime, $ab\notin \sqrt{0}$. Hence, $a=a.1\in \sqrt{I}$ or $b=b.1\in \sqrt{I}$, a contradiction. Thus, I is a quasi primary ideal of R. □

In the following, we determine 2-nil primary ideals of the ring of integer $\mathbb{Z}.$

Theorem 1. 

In $\mathbb{Z},$ the following statements are equivalent:

(1)

I is a 2-nil primary ideal of $\mathbb{Z}.$

(2)

$I=0$ or $I=\left(p^n\right)$ for some prime integer p and $n\ge 1.$

(3)

I is a primary ideal of $\mathbb{Z}.$

Proof. 

$\left(1\right)\Rightarrow \left(2\right)$ Let I be a nonzero 2-nil primary ideal of $\mathbb{Z}.$ Then $I=(p_1^{k_1}{p}_2^{k_2}...p_m^{k_m})$ for some prime integers $p_1,p_2,...,p_m$. Assume that $m\ge 2$. Then $p_1^{k_1}(p_2^{k_2}...p_m^{k_m})1\in I$ but $p_1^{k_1}(p_2^{k_2}...p_m^{k_m})\notin \sqrt{0}=0$, $p_1^{k_1}.1\notin \sqrt{I}$, $(p_2^{k_2}...p_m^{k_m})1\notin \sqrt{I}$, a contradiction. Thus $m=1$ and we are done.

$\left(2\right)\Rightarrow \left(3\right)$ is well-known.

$\left(3\right)\Rightarrow \left(1\right)$ It is straightforward. □

By the proposition below, we give a characterization for fields in terms of 2-nil primary ideals.

Proposition 3. 

For any commutative ring R, the following are equivalent:

(1)

R is a field.

(2)

The zero ideal is the only 2-nil primary ideal of R.

Proof. 

$\left(1\right)\Rightarrow \left(2\right)$ is straightforward. $\left(2\right)\Rightarrow \left(1\right)$ Let R be a commutative ring. Suppose that I is a proper ideal of R. Then there exists a prime ideal P containing I. Hence P is 2-nil primary, so $I\subseteq P=0$ by our assumption. Thus $I=0$ and R is a field. □

In the following, we present a characterization of 2-nil primary ideal in terms of ideal of $R.$

Theorem 2. 

Let $R={\mathbb{Z}}_n$ where $n=p_1^{k_1}{p}_2^{k_2}...p_m^{k_m}$ for some prime integers $p_1,...,p_m$. Every proper ideal of R is 2-nil primary if and only if $m\le 2$.

Proof. 

Let $n=p_1^{k_1}{p}_2^{k_2}...p_m^{k_m}$. Then an ideal of R is of the form $I=p_1^{t_1}{p}_2^{t_2}...p_m^{t_m}$ where $t_i\le k_i$ for all $i=1,...,m$. Assume that $m\ge 3$. Then $p_1^{t_1}{p}_2^{t_2}(p_3^{t_3}...p_m^{t_m})\in I$ but neither $p_1^{t_1}{p}_2^{t_2}\in \sqrt{0}$ nor $p_2^{t_2}(p_3^{t_3}...p_m^{t_m})\in \sqrt{I}=(p_1{p}_2...p_m)$ nor $p_1^{t_1}(p_3^{t_3}...p_m^{t_m})\in \sqrt{I}$, a contradiction. Thus $m\le 2$. Conversely, let $m=1$. Then $R={\mathbb{Z}}_{p^k}$ and every ideal of R is of the form $I=\left(p^t\right)$ for some $1\le t\le k$. Since any primary ideal is 2-nil primary, we are done. Now, let $m=2$. Then $R={\mathbb{Z}}_{{p_i}^{k_i}{p_j}^{k_j}}$$I=\left({p_i}^{t_i}{p_j}^{t_j}\right)$ for some prime integers $p_i,p_j$ and $t_i\le k_i,$$t_j\le k_j$. The claim follows from the fact that $I\subseteq \sqrt{0}$, Remark 1 and Corollary 2.12 in [5]. □

We recall from [6] that a proper ideal I of a ring R is said to be 2-absorbing quasi-primary ideal if its radical is a 2-absorbing ideal of R. Here, we have the following observations.

Remark 1. 

Any 2-absorbing quasi primary ideal contained in $\sqrt{0}$ is 2-nil primary. To see that, let $I\subseteq \sqrt{0}$ be an ideal of a ring R and $abc\in I$ such that $ab\notin \sqrt{0}$. Then $ab\notin I$ and since I is 2-absorbing primary, we have either $ac\in \sqrt{I}$ or $bc\in \sqrt{I}.$ Hence, I is a 2-nil primary ideal of R.

Note that if $\sqrt{0}=Z\left(R\right)$, then $\sqrt{0}$ is a 2-nil primary ideal. Moreover, the following statements are equivalent in general:

(1)

$\sqrt{0}$ is a 2-absorbing ideal.

(2)

$\sqrt{0}$ is a 2-absorbing primary ideal.

(3)

$\sqrt{0}$ is a 2-absorbing quasi primary ideal.

(4)

$\sqrt{0}$ is a 2-nil ideal.

(5)

$\sqrt{0}$ is a 2-nil primary ideal.

(6)

$\sqrt{0}$ is a $(2,n)$-ideal.

We enclosed the above relations between different classes of ideals, in which all elements are nilpotent, in the following diagram. Let R be a commutative ring with unity, and let I be an ideal of R such that $I\subset \sqrt{0}$.

The following example shows that the condition $I\subset \sqrt{0}$ is an essential for a 2-absorbing primary ideal to be 2-nil primary ideal.

Example 3. 

Consider the ideal $I=〈pq〉$ in $\mathbb{Z}$ where p and q are prime numbers, then $p.q.1\in I.$ However, neither $pq\in \sqrt{0}$ nor $p·1\in \sqrt{〈pq〉}$ nor $q·1\in \sqrt{〈pq〉}.$ Hence, it is not a 2-nil primary ideal. But I is 2-absorbing primary ideal. To show this, let $abc\in I.$ Then $p|abc$ and $q|abc,$ thus at least one of $ab,$$ac$ or $bc$ is a multiple of $pq,$ which implies that either $ab\in I$ or $ac\in \sqrt{I}$ or $bc\in \sqrt{I}.$

From the above example, it is evident that the intersection of a family of 2-nil primary ideals is not necessarily a 2-nil primary ideal. However, through the next definition, we can conclude a condition for this case.

Definition 2. 

Let R be a ring and I be a 2-nil primary ideal of R. By Proposition 1, $Q=\sqrt{I}$ is a 2-nil ideal. In this case, we call I a Q-2-nil ideal of R.

Following this definition we have:

(1)

Let R be a Noetherian ring. If I is a Q-2-nil primary ideal of R, then $Q^k\subseteq I\subseteq Q$ for some $k\ge 1$.

(2)

Let I be an ideal of a ring R providing $Q^k\subseteq I\subseteq Q$. Then I is a 2-nil primary ideal if and only if Q is a 2-nil primary ideal.

(3)

Let J and K be Q-2-nil primary ideals of a ring R with the order $J\subseteq I\subseteq K$. Then I is a Q-2-nil primary ideal.

(4)

If J and K are Q-2-nil primary and $Q^{\prime }$-2-nil primary ideals of a ring R with $Q\subseteq Q^{\prime }$, then $I=JK$ is a Q-2-nil primary ideal of R.

(5)

If $I_1,I_2,...,I_n$ are P-2-nil primary ideal of a ring R where P is a prime ideal, then ${\bigcap }_{j=1}^n{I}_j$ is a P-2-nil primary ideal of R.

In terms of ideals of a ring, a characterization of the class of 2-nil primary ideals is presented below.

Theorem 3. 

Let I be a proper ideal of $R.$ Then the following statements are equivalent:

(1)

I is a 2-nil primary ideal of R.

(2)

For any $a,b\in R$ and $ab\notin \sqrt{0},$ then $(I:ab)\subseteq (\sqrt{I}:a)\cup (\sqrt{I}:b).$

(3)

For any $a,b\in R$ and $ab\notin \sqrt{0},$ then $(I:ab)\subseteq (\sqrt{I}:a)$ or $(I:ab)\subseteq (\sqrt{I}:b).$

(4)

If $abJ\subseteq I$ for some $a,b\in R$ and J is an ideal of R then either $ab\in \sqrt{0}$ or $aJ\subseteq \sqrt{I}$ or $bJ\subseteq \sqrt{I}.$

(5)

If $JKL\subseteq I$ for some ideals $J,K,L$ of R then either $JK\subseteq \sqrt{0}$ or $JL\subseteq \sqrt{I}$ or $KL\subseteq \sqrt{I}.$

Proof. 

$\left(1\right)\Rightarrow \left(2\right)$ Let $x\in (I:ab)$ and $ab\notin \sqrt{0}$. Then we have either $ax\in \sqrt{I}$ or $bx\in \sqrt{I}.$ This implies that either $x\in (\sqrt{I}:a)$ or $x\in (\sqrt{I}:b).$ Hence, $x\in (\sqrt{I}:a)\cup (\sqrt{I}:b).$

$\left(2\right)\Rightarrow \left(3\right)$ Since $(I:ab)$ is an ideal which is a subset of the union of two ideals $(\sqrt{I}:a),(\sqrt{I}:b).$ Then either $(I:ab)\subseteq (\sqrt{I}:a)$ or $(I:ab)\subseteq (\sqrt{I}:b).$

$\left(3\right)\Rightarrow \left(4\right)$ Assume that $abJ\subseteq I$ but neither $ab\in \sqrt{0}$ nor $aJ\subseteq \sqrt{I}$ nor $bJ\subseteq \sqrt{I}.$ Then there exist $c_1,c_2\in J$ such that $a{c}_1\notin \sqrt{I}$ and $b{c}_2\notin \sqrt{I}.$ But $ab{c}_1,ab{c}_2\in I$ and $a{c}_1\notin \sqrt{I}$ and $b{c}_2\notin \sqrt{I}$ this means that $(I:ab)⊈(\sqrt{I}:a)$ and $(I:ab)⊈(\sqrt{I}:b)$ which is contradicts to $\left(3\right).$

$\left(4\right)\Rightarrow \left(5\right)$ Suppose that $JKL\subseteq I$ but neither $JK\subseteq \sqrt{0}$ nor $JL\subseteq \sqrt{I}$ nor $KL\subseteq \sqrt{I}$. Then there exist $c_1,c_2\in J$ and $d_1,d_2\in K$ such that $c_1{d}_1\notin \sqrt{0}$ and $d_{2}L⊈\sqrt{I}$ and $c_{2}L⊈\sqrt{I}.$ Since $c_2{d}_{2}L\subseteq I$ and $d_{2}L⊈\sqrt{I}$ and $c_{2}L⊈\sqrt{I}$, we have $c_2{d}_2\in \sqrt{0}.$ From $c_1{d}_{1}L\subseteq I$ and $c_1{d}_1\notin \sqrt{0},$ we have either $c_{1}L\subseteq \sqrt{I}$ or $d_{1}L\subseteq \sqrt{I}.$ Hence, we have to consider three cases as follows:

Case 1.

Suppose that $c_{1}L\subseteq \sqrt{I}$ but $d_{1}L⊈\sqrt{I}$. Since $c_2{d}_{1}L\subseteq I$ and $c_{2}L⊈\sqrt{I}$ and $d_{1}L⊈\sqrt{I}$, we have $c_2{d}_1\in \sqrt{0}.$ But $c_{1}L\subseteq \sqrt{I}$ and $c_{2}L⊈\sqrt{I}$ imply $c_{1}L+c_{2}L⊈\sqrt{I}$. Now, from $(c_1+c_2)d_{1}L\subseteq I$ and $d_{1}L⊈\sqrt{I}$, we conclude $(c_1+c_2)d_1\in \sqrt{0}.$ Thus, $c_1{d}_1\in \sqrt{0}$ which is a contradiction.

Case 2.

If $d_{1}L\subseteq \sqrt{I}$ and $c_{1}L⊈\sqrt{I}$, then as in Case 1 we have a contradiction.

Case 3.

Suppose that $c_{1}L\subseteq \sqrt{I}$ and $d_{1}L\subseteq \sqrt{I}$. Since $d_{1}L\subseteq \sqrt{I}$ and $d_{2}L⊈\sqrt{I}$ we have $(d_1+d_2)L⊈\sqrt{I}$. Now, since $c_2(d_1+d_2)L\subseteq I$ and $c_{2}L⊈\sqrt{I}$, then we have $c_2(d_1+d_2)\in \sqrt{0}.$ Since $c_2{d}_2\in \sqrt{0},$ we have $c_2{d}_1\in \sqrt{0}.$ On the other hand, as $c_{1}L\subseteq \sqrt{I}$ and $c_{2}L⊈\sqrt{I}$, we have $(c_1+c_2)L⊈\sqrt{I}.$ Since $(c_1+c_2)d_{2}L\subseteq I,$ we conclude $(c_1+c_2)d_2\in \sqrt{0}.$ Since $c_2{d}_2\in \sqrt{0},$ we get $c_1{d}_2\in \sqrt{0}.$ Finally, since $(c_1+c_2)(d_1+d_2)L\subseteq I$, $(c_1+c_2)L⊈\sqrt{I}$ and $(d_1+d_2)L⊈\sqrt{I}$, we conclude $(c_1+c_2)(d_1+d_2)=c_1{d}_1+c_1{d}_2+c_2{d}_1+c_2{d}_2\in \sqrt{0}$. As $c_2{d}_2\in \sqrt{0},c_1{d}_2\in \sqrt{0}$ and $c_2{d}_1\in \sqrt{0}$, we obtain $c_1{d}_1\in \sqrt{0},$ a contradiction.

$\left(5\right)\Rightarrow \left(1\right)$ Take $J=\left(a\right),K=\left(b\right)$ and $L=\left(c\right)$ in $\left(5\right).$

□

2. Extensions of 2-nil Primary Ideals

In this section, we are going to study quotients, subrings, localization, Cartesian products and idealizations of 2-nil primary ideals.

Theorem 4. 

Let $\phi :R_1\to R_2$ be a homomorphism of a commutative rings. Then the following statements hold:

(1))

If φ is a monomorphism, then the inverse image ${\phi }^{-1}\left(I_2\right)$ of a 2-nil primary ideal $I_2\subset R_2$ is 2-nil primary ideal of $R_1.$

(2)

If φ is a epimorphism, then the image $\phi \left(I_1\right)$ of 2-nil primary ideal $I_1\subset R_1$ is 2-nil primary in $R_2$ if $Ker\left(\phi \right)\subseteq I_1\cap \sqrt{0_{R_1}}$

Proof. 

(1) Let $a,b,c\in R_1$ and assume that $abc\in {\phi }^{-1}\left(I_2\right)$. Then $\phi \left(abc\right)\in I_2$ which implies either $\phi \left(a\right)\phi \left(b\right)\in \sqrt{0_{R_2}}$ or $\phi \left(a\right)\phi \left(c\right)\in \sqrt{I_2}$ or $\phi \left(b\right)\phi \left(c\right)\in \sqrt{I_2}.$ Since $Ker\left(\phi \right)=0$, we get $ab\in {\phi }^{-1}\left(\sqrt{0_{R_2}}\right)\subseteq \sqrt{0_{R_1}}$ or $ac\in {\phi }^{-1}\left(\sqrt{I_2}\right)\subseteq \sqrt{{\phi }^{-1}\left(I_2\right)}$ or $bc\in {\phi }^{-1}\left(\sqrt{I_2}\right)\subseteq \sqrt{{\phi }^{-1}\left(I_2\right)},$ as needed.

(2) Let $a=\phi \left(x\right),b=\phi \left(y\right),c=\phi \left(z\right)\in R_2$ such that $abc=\phi \left(xyz\right)\in \phi \left(I_1\right).$ It follows from $Ker\left(\phi \right)\subseteq I_1$ that $xyz\in I_1,$ which implies either $xy\in \sqrt{0_{R_1}}$ or $xz\in \sqrt{I_1}$ or $yz\in \sqrt{I_1}.$ It is easy to verify that since $Ker\left(\phi \right)\subseteq I_1\cap \sqrt{0_{R_1}}$, we have $\phi \left(\sqrt{0_{R_1}}\right)\subseteq \sqrt{0_{R_2}}$ and $\phi \left(\sqrt{I_1}\right)\subseteq \sqrt{\phi \left(I_1\right)}$. Hence, $\phi \left(x\right)\phi \left(y\right)\in \phi \left(\sqrt{0_{R_1}}\right)\subseteq \sqrt{0_{R_2}}$ or $\phi \left(x\right)\phi \left(z\right)\in \phi \left(\sqrt{I_1}\right)\subseteq \sqrt{\phi \left(I_1\right)}$ or $\phi \left(y\right)\phi \left(z\right)\in \phi \left(\sqrt{I_1}\right)\subseteq \sqrt{\phi \left(I_1\right)}.$ □

The following corollary is an immediate consequence of Theorem (4).

Corollary 1. 

Let $J\subseteq I\subset R$ be ideals. Then

(1)

If I is a 2-nil primary ideal of R, then $I/J$ is a 2-nil primary of $R/J$.

(2)

If A is a subring of R and I is a 2-nil primary ideal of R, then $I\cap A$ is a 2-nil primary ideal of $A.$

(3)

If $I/J$ is a 2-nil primary of $R/J$ and $J\subseteq \sqrt{0},$ then I is a 2-nil primary of $R.$

Let us define the set $Z_I\left(R\right)=\{r\in R:rs\in I,s\in R-I\}$

Theorem 5. 

Let S be a multiplicatively closed subset of a ring $R.$ Then the following assertions hold

1.

If I is a 2-nil primary ideal of R and $I\cap S=\varnothing ,$ then $S^{-1}I$ is a 2-nil primary ideal of $S^{-1}R.$

2.

If $S^{-1}I$ is a 2-nil primary ideal of $S^{-1}R$ and $S\cap Z\left(R\right)=S\cap Z_I\left(R\right)=\varnothing ,$ then I is a 2-nil primary ideal of $R.$

Proof. 

(1) Suppose that ${\displaystyle \frac{a}{s_1}}{\displaystyle \frac{b}{s_2}}{\displaystyle \frac{c}{s_3}}\in S^{-1}I.$ Then $uabc\in I$ for some $u\in S.$ Hence, $ab\in \sqrt{0}$ or $uac\in \sqrt{I}$ or $ubc\in \sqrt{I}.$ This implies that ${\displaystyle \frac{a}{s_1}}{\displaystyle \frac{b}{s_2}}\in S^{-1}\sqrt{0}\subseteq \sqrt{0_{S^{-1}R}}$ or ${\displaystyle \frac{a}{s_1}}{\displaystyle \frac{c}{s_3}}={\displaystyle \frac{uac}{u{s}_1{s}_3}}\in S^{-1}\sqrt{I}\subseteq \sqrt{S^{-1}I}$ or ${\displaystyle \frac{b}{s_2}}{\displaystyle \frac{c}{s_3}}={\displaystyle \frac{ubc}{u{s}_2{s}_3}}\in S^{-1}\sqrt{I}\subseteq \sqrt{S^{-1}I}.$ Therefore, $S^{-1}I$ is a 2-nil primary ideal of $S^{-1}R.$

(2) Let $abc\in I$ for some $a,b,c\in R.$ Then ${\displaystyle \frac{a}{1}}{\displaystyle \frac{b}{1}}{\displaystyle \frac{c}{1}}\in S^{-1}I$ which yields ${\displaystyle \frac{a}{1}}{\displaystyle \frac{b}{1}}\in \sqrt{0_{S^{-1}R}}$ or ${\displaystyle \frac{a}{1}}{\displaystyle \frac{c}{1}}\in \sqrt{S^{-1}I}$ or ${\displaystyle \frac{b}{1}}{\displaystyle \frac{c}{1}}\in \sqrt{S^{-1}I}.$ If ${\displaystyle \frac{a}{1}}{\displaystyle \frac{b}{1}}\in \sqrt{0_{S^{-1}R}},$ then there exist $u\in S$ and $n\in \mathbb{N}$ such that $u{a}^n{b}^n=0.$ But $S\cap Z\left(R\right)=\varnothing$ then $a^n{b}^n=0$ and $ab\in \sqrt{0}.$ If ${\displaystyle \frac{a}{1}}{\displaystyle \frac{c}{1}}\in \sqrt{S^{-1}I}$ then there exist $v\in S$ such that $v{a}^n{c}^n\in I$ but $Z_I\left(R\right)\cap S=\varnothing ,$ which implies that ${\left(ac\right)}^n\in I.$ Hence, $ac\in \sqrt{I}.$ Similarly, one can show that if ${\displaystyle \frac{b}{1}}{\displaystyle \frac{c}{1}}\in \sqrt{S^{-1}I},$ then $bc\in \sqrt{I}.$ □

Let A be a ring and E an A-module. Then $A⋉E$, the trivial (ring) extension of A by E, is the ring whose additive structure is that of the external direct sum $A\oplus E$ and whose multiplication is defined by

$$(a,e)(b,f):=(ab,af+be)\mathrm{for}\mathrm{all}a,b\in A\mathrm{and}\mathrm{all}e,f\in E.$$

The basic properties of trivial ring extensions are summarized in the celebrated books [7] and [8]. Recall from [7], that

$$\sqrt{0_{A⋉E}}=\sqrt{0_A}⋉E$$

We note that

$$\sqrt{I_{A⋉E}}=\sqrt{I}⋉E.$$

Now, we investigate the possible transfer of 2-nil primary ideals to the trivial ring extension.

Theorem 6. 

Let A be a ring, E be an A-module and $R:=A⋉E.$ Then I is a 2-nil primary ideal of $A,$ if and only if $I⋉E$ is a 2-nil primary ideal of $R.$

Proof. 

Let I be a 2-nil primary ideal of A and assume that $(a,e)(b,f),(c,g)\in I⋉E$ for some elements of $R.$ Thus, $abc\in I$ with $a,b$ and c are elements of $A.$ Furthermore, the fact that I is 2-nil primary ideal of A implies that either $ab\in \sqrt{0_A}$ or $ac\in \sqrt{I}$ or $bc\in \sqrt{I}.$ Hence, one has that either $(ab,af+be)=(a,e)(b,f)\in \sqrt{0_{A⋉E}}$ or $(ac,ag+ce)=(a,e)(c,g)\in \sqrt{I}⋉E$ or $(bc,bg+cf)=(b,f)(c,g)\in \sqrt{I}⋉E.$ Therefore, $I⋉E$ is a 2-nil primary ideal of $A⋉E.$

Conversely, assume that I is an ideal of A such that $I⋉E$ is a 2-nil primary ideal of $A⋉E.$ We are going to prove that I is a 2-nil primary ideal of $A.$ Let $abc\in I$ for some elements $a,b,c$ in A. Then $(a,0)(b,0)(c,0)\in I⋉E$ which yields $(a,0)(b,0)\in \sqrt{0_{A⋉E}}=\sqrt{0_A}⋉E$ or $(a,0)(c,0)\in \sqrt{I_{A⋉E}}=\sqrt{I}⋉E$ or $(b,0)(c,0)\in \sqrt{I_{A⋉E}}=\sqrt{I}⋉E.$ Then one has that $ab\in \sqrt{0_A}$ or $ac\in \sqrt{I}$ or $bc\in \sqrt{I}.$ Hence, I is a 2-nil primary ideal of $A.$ □

It is natural to ask the following question: Suppose that I is a 2-nil primary ideal of $R_1$. What can we say about the ideal $I\times R_2$ of $R_1\times R_2$? Is this ideal also a 2-nil primary ideal? The following example will address this issue.

Example 4. 

Let $R_1$ and $R_2$ be commutative rings and let I be 2-nil primary ideal of $R_1.$ Then $I\times R_2$ is not necessarily 2-nil primary of $R_1\times R_2.$ To show this, let $R_1=R_2={\mathbb{Z}}_{12}$ and $I=6{\mathbb{Z}}_{12}$. Then I is a 2-nil primary ideal by Theorem 2. However, $(2,1)(3,1)(1,1)\in 6{\mathbb{Z}}_{12}\times {\mathbb{Z}}_{12}$ but neither $(2,1)(3,1)\in \sqrt{0_{{\mathbb{Z}}_{12}\times {\mathbb{Z}}_{12}}}$ nor $(2,1)(1,1)\in \sqrt{6{\mathbb{Z}}_{12}\times {\mathbb{Z}}_{12}}$ nor $(3,1)(1,1)\in \sqrt{6{\mathbb{Z}}_{12}\times {\mathbb{Z}}_{12}}.$

Proposition 4. 

Let $R_1$ and $R_2$ be commutative rings and I a proper ideal of $R_1$. Then the following assertions are equivalent:

(1)

$I\times R_2$ is a 2-nil primary ideal of $R_1\times R_2.$

(2)

I is a quasi primary ideal of $R_1.$

(3)

$I\times R_2$ is a quasi primary ideal of $R_1\times R_2.$

Proof. 

$\left(1\right)\Rightarrow \left(2\right)$ Let $a,b\in R$ such that $ab\in I.$ Then $(a,1)(b,1)(1,1)\in I\times R_2.$ Since $(a,1)(b,1)\notin \sqrt{0_{R_1\times R_2}}$ then either $(a,1)(1,1)\in \sqrt{I\times R_2}\subseteq \sqrt{I}\times R_2$ or $(b,1)(1,1)\in \sqrt{I\times R_2}\subseteq \sqrt{I}\times R_2.$ This implies that either $a\in \sqrt{I}$ or $b\in \sqrt{I}$ which proves that I is a quasi primary ideal of $R_1.$

$\left(2\right)\Rightarrow \left(3\right)$ From Lemma 2.2 in [6]

$\left(3\right)\Rightarrow \left(1\right)$ It is clear. □

Theorem 7. 

Let $I_1$ and $I_2$ be proper ideals of $R_1$ and $R_2,$ respectively. Then the following statements are equivalent:

(1)

$I_1\times I_2$ is a 2-nil primary ideal of $R_1\times R_2.$

(2)

$I_1\subseteq \sqrt{0_{R_1}}$ and $I_2\subseteq \sqrt{0_{R_2}}$ are quasi primary ideals of $R_1$ and $R_2$, respectively.

(3)

$I_1\times I_2$ is a 2-absorbing quasi primary ideal of $R_1\times R_2$ and $I_1\times I_2\subseteq \sqrt{0_{R_1\times R_2}}.$

Proof. 

$\left(1\right)\Rightarrow \left(2\right)$ Let $I_1⊈\sqrt{0_{R_1}},$ then there exists $a\in I_1-\sqrt{0_{R_1}}.$ Then $(a,1)(1,0)(1,1)\in I_1\times I_2$ and $(a,1)(1,0)\notin \sqrt{0_{R_1}}\times \sqrt{0_{R_2}}$. Hence, $(a,1)(1,1)\in \sqrt{I_1\times I_2}$ or $(1,0)(1,1)\in \sqrt{I_1\times I_2}$ which in both cases is a contradiction. Thus, $I_1\subseteq \sqrt{0_{R_1}}$ and similarly, $I_2\subseteq \sqrt{0_{R_2}}.$ Now, if $I_1$ is not quasi primary, then there exist $a,b\in R_1-I_1$ such that $ab\in I_1$, but neither $a\in \sqrt{I_1}$ nor $b\in \sqrt{I_1}$. Hence, $(a,1)(b,1)(1,0)\in I_1\times I_2$ and since $(a,1)(b,1)\notin \sqrt{0_{R_1\times R_2}}$ and $(a,1)(1,0)\notin \sqrt{I_1\times I_2}$ and $(b,1)(1,0)\notin \sqrt{I_1\times I_2}$, a contradiction. Thus, $I_1$ is quasi primary in $R_1.$ The same argument shows that $I_2$ is a quasi primary ideal of $R_2.$

$\left(2\right)\Rightarrow \left(3\right)$ Suppose that $I_1\subseteq \sqrt{0_{R_1}}\subset R_1$ and $I_2\subseteq \sqrt{0_{R_2}}\subset R_2$ are quasi primary ideals. Hence, by Proposition 4, one has that $I_1\times R_2$ and $R_1\times I_2$ are quasi primary ideals of $R_1\times R_2.$ By Theorem 2.17(ii) in [6], the intersection of two quasi primary ideals is 2-absorbing quasi primary ideal. So, we conclude that $I_1\times I_2=(I_1\times R_2)\cap (R_1\times I_2)$ is a 2-absorbing quasi primary ideal of $R_1\times R_2$.

$\left(3\right)\Rightarrow \left(1\right)$ It follows from Remark 1. □

Theorem 8. 

Let $R_1,R_2,...,R_n(n\ge 3)$ be commutative rings and $R=R_1\times R_2\times \cdots \times R_n$. The following assertions are equivalent:

(1)

$I=I_1\times I_2\times \cdots \times I_n$ is a 2-nil primary ideal of R.

(2)

$I_k$ is a quasi primary ideal of $R_k$ for some $k\in \{1,2,...,n\}$ and $I_j=R_j$ for all $j\in \{1,2,...,n\}-\left\{k\right\}$.

(3)

$I=I_1\times I_2\times \cdots \times I_n$ is a quasi primary ideal of R.

Proof. 

$\left(1\right)\Rightarrow \left(2\right)$ Suppose that $I=I_1\times I_2\times \cdots \times I_n$$(n\ge 3)$ is a 2-nil primary ideal of R. Without loss of generality, assume on the contrary that $I_1$ and $I_2$ are proper ideals of $R_1$ and $R_2$, respectively. Since $(0,1,1,...,1)(1,0,1,1,...,1)(1,1,...,1)\in I$ and $(0,1,1,...,1)(1,0,1,1,...,1)\notin \sqrt{0}$, we have either $(0,1,1,...,1)(1,1,...,1)\in \sqrt{I}$ or $(1,0,1,1,...,1)(1,1,...,1)\in \sqrt{I}$. Thus, $I_1=R_1$ or $I_2=R_2$, a contradiction. Now, we may suppose that $I_1$ is proper and $I=I_1\times R_2\times \cdots \times R_n$. Let $ab\in I_1$. Since $(a,1,1)(b,1,1)(1,1,1)\in I$ but $(a,1,1)(b,1,1)\notin \sqrt{0}$, we have either $(a,1,1)(1,1,1)\in \sqrt{I}$ or $(b,1,1)(1,1,1)\in \sqrt{I}$ which means that $I_1$ is a 2-nil quasi primary ideal of $R_1$.

$\left(2\right)\Rightarrow \left(3\right)$ It is clear by Theorem 2.3 in [6].

$\left(2\right)\Rightarrow \left(3\right)$ Straightforward. □