On $\mathcal{H}_Y$-Ideals

TL;DR

This paper extends the theory of $\\mathcal{H}_Y$-ideals by introducing fixed and free variants, establishing their properties, and characterizing the compactness of Y and regularity of R through these ideals.

Contribution

It introduces fixed and free $\\mathcal{H}_Y$-ideals and their relative versions, unifies previous results, and characterizes topological and algebraic properties via these ideals.

Findings

Y is compact iff every proper $\\mathcal{H}_Y$-ideal is fixed.

Every proper ideal is a relative $\\mathcal{H}_Y$-ideal iff R is regular.

Characterization of fixed and strong $\\mathcal{H}_Y$-ideals in terms of properties of Y and R.

Abstract

In this article, we continue the studying of $\mathcal{H}_Y$-ideals. We introducing two notions fixed and free $\mathcal{H}_Y$-ideals as an extension of fixed and free z-ideals in C(X) and relative $\mathcal{H}_Y$-ideals as an extension of relative z-ideals. It has been shown that a large amount of the results of the mentioned papers and generally the papers in the literature about these topics, are special cases of the results of this paper. We prove that Y is compact if and only if every proper $\mathcal{H}_Y$-ideal is a fixed $\mathcal{H}_Y$-ideal; if and only if every proper strong H_Y-ideal is a fixed ideal. Also, we show that every proper ideal is a relative $\mathcal{H}_Y$-ideal, if and only if every proper ideal is a strong $\mathcal{H}_Y$-ideal; if and only if R is regular.

Equations

$$P\in h_Y(I)\iff I\subseteq P\iff \forall S\in F_I,S\subseteq P$$

$$\iff P\in \bigcap _{S\in F_I}{h}_Y(S)=\bigcap {\mathcal{H}}_Y(I)\iff P\in \bigcap {\mathcal{H}}_Y(I).$$

$$\big{(}\bigcap\mathcal{H}_{Y}(I)\big{)}\cap S=h_{Y}(I)\cap S=h_{S}(I)=\bigcap\mathcal{H}_{S}(I)$$

$${\mathcal{F}}^{\prime }=\{h_{Y^{\prime }}(F):h_Y(F)\in \mathcal{F}\text{ and }F\text{ is a finite subset of }{R}^{\prime }\},$$

$$kk_{Y}(a)\cap kh_{Y}(b)\cap J=k\big{(}h_{Y}(a)\cup h_{Y}(b)\big{)}\cap J=kh_{Y}(ab)\cap J=kh_{Y}(ab)^{n}\cap J\subseteq I\subseteq P$$

$$kh_{Y}(b)\cap kh_{Y}(F)\cap J=k\big{(}h_{Y}(b)\cup h_{Y}(F)\big{)}\cap J=k(h_{Y}(E))\cap J\subseteq I\subseteq P$$

$$k{h}_Y(a)=k{h}_Y(a)\cap K=k{h}_Y(a^n)\cap K\subseteq I$$

$$I_{\mathcal{H}}\cap J$$

$$\displaystyle\subseteq\big{(}\bigcap_{P\in\mathrm{Min}(I)}P_{\mathcal{H}}\big{)}\cap J=\big{(}\bigcap_{P\in\mathrm{Min}(I)}P\big{)}\cap J=\sqrt{I}\cap J$$

$$(\sqrt{I}{)}_{\mathcal{H}}\cap \sqrt{J}=I_{\mathcal{H}}\cap \sqrt{J}=\sqrt{I_{\mathcal{H}}}\cap \sqrt{J}=\sqrt{I_{\mathcal{H}}\cap J}=\sqrt{I\cap J}=\sqrt{I}\cap \sqrt{J}$$

$$I_{\mathcal{H}}\cap J=I_{\mathcal{H}}\cap J_{\mathcal{H}}\cap J=(I\cap J{)}_{\mathcal{H}}\cap J\subseteq K_{\mathcal{H}}\cap J\subseteq K$$

$$(I\cap K{)}_{\mathcal{H}}\cap J=(IK{)}_{\mathcal{H}}\cap J\subseteq IK\subseteq I\cap K$$

$$I_{\mathcal{H}}\cap P=\big{(}(I_{\mathcal{H}}\setminus I)\cup I\big{)}\cap P\big{)}=I\cap P=I$$

$$\big{<}c\big{>}\cap kh_{Y}(F)\subseteq J\cap kh_{Y}(F)=(K:I)\cap kh_{Y}(F)\subseteq K\subseteq I$$

$$I\cap (J+K)=(I\cap J)+(I\cap K)$$

$$\{P:P\text{ is a prime (resp., strong) }{\mathcal{H}}_Y\text{-ideal and }J⊈P\}.$$

$$K=\bigcap \{P\in \mathrm{Min}(I):I\text{ is not a (resp., strong) }{\mathcal{H}}_Y\text{-ideal }\},$$

$$kh_{Y}(a)\cap K=kh_{Y}(a)\cap\big{(}\bigcup_{x\in K}\big{<}x\big{>}\big{)}=\bigcup_{x\in K}kh_{Y}(a)\cap\big{<}x\big{>}\subseteq I$$

$$I_{\mathcal{H}}\cap K$$

$$\displaystyle=\big{(}\bigcap_{P\in\mathrm{Min}(I)}P_{\mathcal{H}}\big{)}\cap K=\big{(}\bigcap_{P\in\cal{A}}P\big{)}\cap\big{(}\bigcap_{P\in\cal{B}}P_{H}\big{)}\cap\big{(}\bigcap_{P\in\cal{B}}P\big{)}$$

$$\displaystyle=\big{(}\bigcap_{P\in\cal{A}}P\big{)}\cap\big{(}\bigcap_{P\in\cal{B}}P\big{)}=\sqrt{I}$$

$$I_{\mathcal{H}}\cap (J+K)‌=I_{\mathcal{H}}\cap J+‌I_{\mathcal{H}}\cap J\subseteq I$$

$$J=\bigcap \{P\in \mathrm{Min}(I):I\text{ is not a (resp., strong) }{\mathcal{H}}_Y\text{-ideal }\}.$$

Full text

1. Introduction

The concept of $z$-ideal, was first studied in the rings of continuous functions as an ideal $I$ of $C(X)$ that $Z(f)\subseteq Z(g)$ and $f\in I$ implies that $g\in I$, see [9]. Then this concept was studied more generally for the commutative rings, in [12], as an ideal $I$ of $R$ that whenever two elements of $R$ are contained in the same family of maximal ideals and $I$ contains one of them, then it follows that $I$ contains the other one. If we use $(Z(f))^{\circ}\subseteq(Z(g))^{\circ}$ instead of the above inclusion relation and the minimal prime ideals instead of the maximal ideals in the above definitions, then we obtain the concept of $z^{\circ}$-ideal ($d$-ideal) in $C(X)$ and the commutative rings, which are introduced and carefully studied in [7, 8, 10]. The concepts of $z$-ideal and $z^{\circ}$-ideal can be generalized to the concepts of $sz$-ideal and $sz^{\circ}$-ideal ($\xi$-ideal), respectively, based on the finite subsets of the ideals instead of the single points in the ideal, and are studied in [4, 5, 12].

The concepts of $\mathcal{H}_{Y}$-ideals and the strong $\mathcal{H}_{Y}$-ideals are generalizations of $z$-ideals, strong $z$-ideals, $z^{\circ}$-ideals and strong $z^{\circ}$-ideals. These concepts have been introduced and carefully studied in [3]. It has been shown that a large amount of the results of the above mentioned papers and generally the papers in the literature about these topics, are special cases of the results of [3]. In this paper, we continue the studying of these concepts.

In the next section we recall some pertinent definitions. In Section 3, we introduce and study the fixed and free $\mathcal{H}_{Y}$-ideals and free and fixed $\mathcal{H}_{Y}$-ideals respect to a subset of $Y$. In this section we show that $Y$ is compact if and only if every proper $\mathcal{H}_{Y}$-ideal is a fixed $\mathcal{H}_{Y}$-ideal; if and only if every proper strong $\mathcal{H}_{Y}$-ideal is a fixed ideal. Also, we give a compactification of $Y$. The Section 4, devoted to introducing and studying of relative $\mathcal{H}_{Y}$-ideal as an extension of relative $z$-ideals. In this section, we show that every proper ideal is a relative $\mathcal{H}_{Y}$-ideal, if and only if every proper ideal is a strong $\mathcal{H}_{Y}$-ideal; if and only if $R$ is regular.

2. Prelementary

In this article, any ring $R$ is commutative with unity. A semi-prime ideal is an ideal which is an intersection of prime ideals. For each ideal $I$ of $R$ and each element $a$ of $R$, we denote the ideal $\{x\in R:ax\in I\}$ by $(I:a)$. When $I=\{0\}$ we write $\mathrm{Ann}(a)$ instead of $(\{0\}:a)$ and call this the annihilator of $a$. If $\mathrm{Ann}(a)$ is maximal in the set of all annihilator of nonzero elements of $R$, then $\mathrm{Ann}(a)$ is a prime ideal, and it is called an affiliated prime ideal. A prime ideal $P$ containing an ideal $I$ is said to be a minimal prime over $I$, if there are not prime ideals strictly contained in $P$ that contains $I$. $\mathrm{Spec}(R)$, $\mathrm{Min}(R)$, and $\mathrm{Rad}(R)$ denote the set of all prime ideals, all minimal prime ideals, all maximal ideals of $R$ and their intersections, respectively. By $\mathrm{Min}(I)$ we mean the set of minimal prime ideals of $I$. In fact $\mathrm{Min}((0))=\mathrm{Min}(R)$. A ring $R$ is said to be reduced if $\mathrm{Rad}(R)=(0)$.

A prime ideal $P$ is called a Bourbaki associated prime divisor of an ideal $I$ if $(I‌:x)=P$, for some $x\in R$. We denote the set of all Bourbaki associated prime divisors of an ideal $I$ by $\cal{B}(I)$. We use $\cal{B}(R)$ instead of $\cal{B}(\{0\})$. A representation $I=\bigcap_{P‌\in\cal{P}}P$ of $I$ as an intersection of prime ideals is called irredundant if no $P\in\cal{P}$ may be omitted. Let $I$ be a semi-prime ideal, $P_{\circ}\in\mathrm{Min}(I)$ is called irredundant with respect to $I$, if $I\neq\bigcap_{P_{\circ}\neq P\in\mathrm{Min}(I)}P$. If $I$ is equal to the intersection of all irredundant ideals with respect to $I$, then we call $I$ a fixed-place ideal, exactly, by [2, Theorem 2.1], we have $I=\bigcap\cal{B}(I)$.

In this paper, all $Y\subseteq\mathrm{Spec}(R)$ is considered by Zariski topology; i.e., by assuming as a base for the closed sets of $Y$, the sets $h_{Y}(a)$ where $h_{Y}(a)=\{P\in Y:a\in P\}$. Hence, closed sets of $Y$ are of the form $h_{Y}(I)=\bigcap_{a\in I}h_{Y}(a)=\{P\in Y:I\subseteq P\}$, for some ideal $I$ in $R$. Also, we set $h_{Y}^{c}(I)=Y\backslash h_{Y}(I)$. For any subset $S$ of $Y$, we show the kernel of $S$ by $k(S)=\bigcap_{P\in S}P$ and we have $\overline{S}=cl_{Y}S=h_{Y}k(S)$. When $Y=\mathrm{Spec}(R)$, we omit the index $Y$ and when $Y=\mathrm{Max}(R)$ ($Y=\mathrm{Min}(R)$) we write $M$ (m) instead of $Y$ in the index. By these notations, for every $S\subseteq R$, we can use the notations $kh_{m}(S)$ and $kh_{M}(S)$ instead of $P_{S}$ and $M_{S}$ (which is usually used in the context of $C(X)$), respectively.

The reader is referred to [6], [9], [11] and [13] for undefined terms and notations.

Let $R$ be a ring, $Y\subseteq\mathrm{Spec}(R)$ and $I$ be an ideal of $R$. Then, by [3, Proposition3.2] the following are equivalent:

• (a)

For every $a\in I$ and $S\subseteq R$, it follows from $h_{Y}(a)\subseteq h_{Y}(S)$ that $S\subseteq I$.

• (b)

For every $a\in I$ and $S\subseteq R$, it follows from $h_{Y}(a)=h_{Y}(S)$ that $S\subseteq I$.

• (c)

For every $a\in I$ and $b\in R$, it follows from $h_{Y}(a)=h_{Y}(b)$ that $b\in I$.

• (d)

For every $a\in I$ and $b\in R$, it follows from $h_{Y}(a)\subseteq h_{Y}(b)$ that $b\in I$.

• (e)

If $a\in I$, then $kh_{Y}(a)\subseteq I$.

• (f)

For every $a\in I$ and $S\subseteq R$, it follows from $kh_{Y}(S)\subseteq kh_{Y}(a)$ that $S\subseteq I$.

• (g)

For every $a\in I$ and $S\subseteq R$, it follows from $kh_{Y}(S)=kh_{Y}(a)$ that $S\subseteq I$.

• (h)

For every $a\in I$ and $b\in R$, it follows from $kh_{Y}(b)=kh_{Y}(a)$ that $b\in I$.

• (k)

For every $a\in I$ and $b\in R$, it follows from $kh_{Y}(b)\subseteq kh_{Y}(a)$ that $b\in I$.

An ideal $I$ of $R$ is said to be an $\mathcal{H}_{Y}$-ideal if it satisfies in the above equivalent conditions.

Let $R$ be a ring, $Y\subseteq\mathrm{Spec}(R)$ and $I$ be an ideal of $R$. Then, by [3, Proposition 3.4], the following are equivalent:

• (a)

For every finite subset $F$‌of$I$andevery$S⊆R$,itfollowsfrom$h_Y(F)=h_Y(S)$that$S⊆I$.\par\itemize@item@[(b)]Foreveryfinitesubset$F$‌ of $I$ and every finite subset $G$ of $R$, it follows from $h_{Y}(F)=h_{Y}(G)$ that $G\subseteq I$.

• (c)

For every finite subset $F$‌of$I$andeveryfinitesubset$G$of$R$,itfollowsfrom$h_Y(F)⊆h_Y(G)$that$G⊆I$.\par\itemize@item@[(d)]Itfollowsfrom$h_Y(a)∈H_Y(I)$that$a∈I$.\par\itemize@item@[(e)]Foreveryfinitesubset$F$‌of$R$,itfollowsfrom$h_Y (F)∈H_Y (I)$that$F⊆I$.\par\itemize@item@[(f)]Foreveryfinitesubset$F$of$I$and$a∈R$‌, it follows from $h_{Y}(F)=h_{Y}(a)$ that $a\in I$.

• (g)

For every finite subset $F$ of $I$ and $a\in R$, it follows from $h_{Y}(F)\subseteq h_{Y}(a)$ that $a\in I$.

• (k)

For every finite subset $F\subseteq I$, we have $kh_{Y}(F)\subseteq I$.

• (l)

For every finite subset $F$ of $I$ and $a\in R$‌,itfollowsfrom$kh_Y(a)=kh_Y(F)$that$a∈I$.\par\itemize@item@[(m)]Foreveryfinitesubset$F$of$I$and$a∈R$,itfollowsfrom$kh_Y(a)⊆kh_Y(F)$that$a∈I$.\par\itemize@item@[(n)]Foreveryfinitesubset$F$of$I$andany$S⊆R$‌, it follows from $kh_{Y}(S)=kh_{Y}(F)$ that $S\subseteq I$.

• (o)

For every finite subset $F$ of $I$ and any $S\subseteq R$, it follows from $kh_{Y}(S)\subseteq kh_{Y}(F)$ that $S\subseteq I$.

An ideal $I$ of $R$ is said to be a strong $\mathcal{H}_{Y}$-ideal if it satisfies in the above equivalent conditions. An ideal $I$ of $R$ is called a $Y$-Hilbert ideal, if $I$ is an intersection of elements of some subfamily of $Y$; i.e., $I=kh_{Y}(I)$. By $\mathrm{maxl}(E)$, we mean the set of all maximal elements of $E$. Also, we denote by $S\mathcal{H}_{Y}({\cal{A}})$ ($PS\mathcal{H}_{Y}({\cal{A}})$) the set of all strong $H_{Y}$-ideals (proper strong $H_{Y}$-ideals) of $A$. See [3], for more information about the $\mathcal{H}_{Y}$-ideals and strong $\mathcal{H}_{Y}$-ideals.

3. $\mathcal{H}_{Y}$-fixed ideals and $\mathcal{H}_{Y}$-free ideals

In this section, we introduce and study the fixed and free $\mathcal{H}_{Y}$-ideals and free and fixed $\mathcal{H}_{Y}$-ideals respect to a subset of $Y$ as extensions of fixed and free ideals in $C(X)$. Also, we give some equivalent properties to compaction of $Y$

Definition 3.1**.**

Let $Y\subseteq\mathrm{Spec}(R)$ and $I$ be an ideal of $R$. Then $I$ is called an $\mathcal{H}_{Y}$-fixed ideal, if $\mathcal{H}_{Y}(I)$ is a fixed $\mathcal{H}_{Y}$-filter; i.e. $\bigcap\mathcal{H}_{Y}(I)\neq\emptyset$. If $I$ is not an $\mathcal{H}_{Y}$-fixed ideal, then $I$ is called $\mathcal{H}_{Y}$-free ideal.

Proposition 3.2**.**

Let $Y\subseteq\mathrm{Spec}(R)$ and $I$ be an ideal of $R$. Then the following statements hold.

• (a)

$\bigcap\mathcal{H}_{Y}(I)=h_{Y}(I)$.

• (b)

$I$* is an $\mathcal{H}_{Y}$-fixed ideal if and only if $h_{Y}(I)\neq\emptyset$.*

Proof.

(a). Let $F_{I}$ be the set of all finite subsets of $I$, then we can write

[TABLE]

[TABLE]

(b). By (a), it is clear. ∎

Now we can conclude the following corollary from the above proposition.

Corollary 3.3**.**

Let $Y\subseteq\mathrm{Spec}(R)$ and $I$ be an ideal of $R$. The following hold.

• (a)

If $I$ is a $Y$-Hilbert ideal, then $I$ is a fixed $\mathcal{H}_{Y}$-ideal.

• (b)

$I$* is $\mathcal{H}_{Y}$-fixed (resp., $\mathcal{H}_{Y}$-free) if and only if $\mathcal{H}_{Y}^{-1}\mathcal{H}_{Y}(I)$ is an $\mathcal{H}_{Y}$-fixed ideal ($\mathcal{H}_{Y}$-free ideal).*

• (c)

$\mathcal{H}_{Y}^{-1}\mathcal{H}_{Y}(I)=R$* if and only if $I\not\subseteq\bigcup Y$.*

It is clear that every $\mathcal{H}_{Y}$-fixed ideal is proper. Moreover the following corollary is immediate.

Corollary 3.4**.**

Let $Y\subseteq\mathrm{Spec}(R)$ and $I$ be an ideal of $R$. Then $I$ is a maximal $\mathcal{H}_{Y}$-fixed ideal if and only if $I\in\mathrm{maxl}(Y)$.

Note that a maximal fixed $\mathcal{H}_{Y}$-filter is a also an $\mathcal{H}_{Y}$-ultrafilter. Therefore, $I$ is a maximal $\mathcal{H}_{Y}$-fixed ideal if and only if $\mathcal{H}_{Y}(I)$ is a fixed $\mathcal{H}_{Y}$-ultrafilter. Hence, clearly, an $\mathcal{H}_{Y}$-filter $\mathscr{U}$ is a fixed $\mathcal{H}_{Y}$-ultrafilter if and only if $\mathcal{H}_{Y}^{-1}(\mathscr{U})$ is a maximal element of $Y$.

Theorem 3.5**.**

Let $Y\subseteq\mathrm{Spec}(R)$. The following statements are equivalent.

• (a)

Every proper $\mathcal{H}_{Y}$-ideal is a fixed $\mathcal{H}_{Y}$-ideal.

• (b)

Every proper strong $\mathcal{H}_{Y}$-ideal is a fixed $\mathcal{H}_{Y}$-ideal.

• (c)

Every ideal contained in $\bigcup Y$ is an $\mathcal{H}_{Y}$-fixed ideal.

• (d)

$\mathrm{maxl}(PS\mathcal{H}_{Y})\subseteq Y$.

• (e)

$Y$* is compact.*

Proof.

(a)$\Rightarrow$(b). It is clear.

(b)$\Rightarrow$(c). It follows from Corollary 3.3.

(c)$\Rightarrow$(d). Suppose that $M\in\mathrm{maxl}(PS\mathcal{H}_{Y})$, then $M$ is an $\mathcal{H}_{Y}$-fixed ideal, so $h_{Y}(M)\neq\emptyset$, by Proposition 3.2, so $M\in Y$, consequently $\mathrm{Max}(R)\subseteq Y$.

(d)$\Rightarrow$(e). Suppose that $\{h_{Y}(a_{\alpha})\}_{\alpha\in A}$ is a family of closed basic with the finite intersection. Then there is a proper $\mathcal{H}_{Y}$-filter $\mathscr{F}$ containing $\{h_{Y}(a_{\alpha})\}_{\alpha\in A}$. Then an ultrafilter $\mathscr{U}$ on $Y$ exists such that $\mathscr{F}\subseteq\mathscr{U}$. Since $\mathcal{H}_{Y}^{-1}(\mathscr{U})\in\mathrm{maxl}(PS\mathcal{H}_{Y})$, $\mathcal{H}_{Y}^{-1}(\mathscr{U})$ is $\mathcal{H}_{Y}$-fixed and therefore $\mathscr{U}$ is fixed, thus $\mathscr{F}$ is fixed, hence $\emptyset\neq\bigcap\mathscr{F}\subseteq\bigcap_{\alpha\in A}h_{Y}(a_{\alpha})$. Consequently $Y$ is compact.

(e)$\Rightarrow$(a). Since $\mathcal{H}_{Y}(I)=\{h_{Y}(F):F\text{ is a finite subset }I\}$ is a family closed sets with the finite intersection property, It is clear. ∎

Corollary 3.6**.**

Suppose that $Y\subseteq\mathrm{Min}(R)$ and $\bigcap Y=\{0\}$. $Y$ is compact if and only if $\mathcal{B}(R)\subseteq Y$.

Proof.

It concludes from [2] and Theorem 3.5. ∎

Theorem 3.7**.**

Suppose that $Y\subseteq\mathrm{Spec}(R)$. If $Z=\mathrm{maxl}(PS\mathcal{H}_{Y})\cup Y$, then $Z$ is a compactification of $Y$.

Proof.

$Y$ is dense in $Z$ and $\mathrm{maxl}(PS\mathcal{H}_{Z})\subseteq\mathrm{maxl}(PS\mathcal{H}_{Y})\subseteq Z$, so $Z$ is a compactification of $Y$. ∎

Suppose that $X$ is a topological space and $Y=\mathrm{Max}(C(X))$. For every $p\in\beta X\setminus X$, the maximal ideal $M^{p}$ is a free $z$-ideal which is an $\mathcal{H}_{Y}$-fixed ideal. This example shows that the notion of fixed ideals and notion of $\mathcal{H}_{Y}$-fixed ideals need not coincide in the rings of continuous functions literature. In the following definition we introduce new notion which coincides with the fixed ideals notion in the rings of continuous functions literature.

Definition 3.8**.**

Suppose $Y\subseteq\mathrm{Spec}(R)$ and $S\subseteq Y$. An ideal $I$ is called an $\mathcal{H}_{Y}$-fixed ideal with respect to $S$ if \big{(}\bigcap\mathcal{H}_{Y}(I)\big{)}\cap S\neq\emptyset and $I$ is called an $\mathcal{H}_{Y}$-free ideal with respect to $S$ if $I$ is not an $\mathcal{H}_{Y}$-fixed ideal with respect to $S$; i.e. \big{(}\bigcap\mathcal{H}_{Y}(I)\big{)}\cap S=\emptyset

Suppose that $S\subseteq Y\subseteq\mathrm{Spec}(R)$. Clearly if an ideal $I$ is $\mathcal{H}_{Y}$-fixed with respect to $S$, then it is also an $\mathcal{H}_{Y}$-fixed; but the converse is not true in general. If $Y=\mathrm{Max}(C(X))$ and $S=\{M_{p}:p\in X\}$, then the family of all $\mathcal{H}_{Y}$-fixed ideals (in fact, fixed $z$-ideals) and the family of all $\mathcal{H}_{Y}$-fixed ideal with respect to $S$ coincide.

Theorem 3.9**.**

Let $S\subseteq Y\subseteq\mathrm{Spec}(R)$ and $I$ be an ideal of $R$. $I$ is an $\mathcal{H}_{Y}$-fixed ideal with respect to $S$ if and only if $I$ is a $\mathcal{H}_{S}$-fixed ideal.

Proof.

We can conclude from Proposition 3.2, that

[TABLE]

This completes the proof. ∎

Now Corollary 3.4, and Theorem 3.9, conclude the following corollary.

Corollary 3.10**.**

If $S\subseteq Y\subseteq\mathrm{Spec}(R)$, then $\mathrm{maxl}(Y)\cap S$ and the family of maximal $\mathcal{H}_{Y}$-fixed ideal with respect to $S$ coincide.

Proposition 3.11**.**

Suppose $R^{\prime}$ is a subring of a ring $R$ and $Y\subseteq\mathrm{Spec}(R)$ , then $Y^{\prime}=\{P\cap R^{\prime}:P\in Y\}\subseteq\mathrm{Spec}(R^{\prime})$. Set

[TABLE]

for every $\mathcal{H}_{Y}$-filter $\mathscr{F}$. If $I$ is an $\mathcal{H}_{Y}$-fixed ideal, then $I\cap R^{\prime}$ is an $\mathcal{H}_{Y^{\prime}}$-fixed ideal.

Proof.

The proof is straightforward. ∎

4. Relative $\mathcal{H}_{Y}$-ideals

In this section we introduce and study relative $\mathcal{H}_{Y}$-ideals as an extension of relative $z$-ideals and we show that the most important results about the relative $z$-ideals can be extended to relative $\mathcal{H}_{Y}$-ideals.

Definition 4.1**.**

Let $Y\subseteq\mathrm{Spec}(R)$ and $I$ and $J$ be ideals of $R$. $I$ is called (resp., strong) $\mathcal{H}_{YJ}$-ideal if for every $a\in I$ (resp., finite subset $F$ of $I$) we have $kh_{Y}(a)\cap J\subseteq I$ (resp., $kh_{Y}(F)\cap J\subseteq I$). If $I$ is (resp., strong) $H_{YJ}$ ideal, for some ideal $J\not\subseteq I$, then $I$ is called relative (resp., strong) $\mathcal{H}_{Y}$-ideal and $J$ is called a (resp., strong) $\mathcal{H}_{Y}$-factor of $I$. A (resp., strong) $\mathcal{H}_{YJ}$-ideal $I$ is called a trivial (resp., strong) $\mathcal{H}_{YJ}$-ideal, if $J\subseteq I$, otherwise $I$ is called a nontrivial $\mathcal{H}_{YJ}$-ideal.

Proposition 4.2**.**

If $Y\subseteq\mathrm{Spec}(R)$, then the following hold.

• (a)

If $J$ is an ideal of $R$, then every strong $\mathcal{H}_{YJ}$-ideal is an $\mathcal{H}_{YJ}$-ideal.

• (b)

Every strong relative $\mathcal{H}_{Y}$-ideal is a relative $\mathcal{H}_{Y}$-ideal.

• (c)

Every ideal $I$ is a strong $\mathcal{H}_{YI}$-ideal.

• (d)

Every (resp., strong) $\mathcal{H}_{Y}$-ideal is a (resp., strong) $\mathcal{H}_{YJ}$-ideal, for every ideal $J$ of $R$.

• (e)

Every (resp., strong) $\mathcal{H}_{Y}$-ideal is a relative (resp., strong) $\mathcal{H}_{Y}$-ideal.

• (f)

Suppose that $J$ is a (resp., strong) $\mathcal{H}_{Y}$-ideal. Every (resp., strong) $\mathcal{H}_{YJ}$-subideal of $J$ is a (resp., strong) $\mathcal{H}_{Y}$-ideal.

Proof.

It is straightforward. ∎

Theorem 4.3**.**

Let $Y\subseteq\mathrm{Spec}(R)$ and $I$ and $J$ be ideals of $R$. Then the following hold

• (a)

If $I$ is a (resp., strong) $\mathcal{H}_{YJ}$-ideal and $P\in\mathrm{Min}(R)$, then $P$ is a (resp., strong) $\mathcal{H}_{YJ}$-ideal.

• (b)

A prime ideal $P$ is a (resp., strong) $\mathcal{H}_{YJ}$-ideal if and only if $P$ is either a (resp., strong) $\mathcal{H}_{Y}$-ideal or $J\subseteq P$.

• (c)

If $I$ is a relative (resp., strong) $H_{Y}$-ideal, then $\mathrm{Min}(I)$ has a (resp., strong) $\mathcal{H}_{Y}$-ideal.

Proof.

(a). Suppose that $I$ is an $\mathcal{H}_{YJ}$-ideal. If $a\in P$, then $b\notin P$ exists such that $(ab)^{n}\in I$, for some $n\in\mathbb{N}$. Since $I$ is a $\mathcal{H}_{YJ}$-ideal, it follows that

[TABLE]

Since $kh_{Y}(b)\not\subseteq P$ and $P$ is prime, it follows that $kh_{Y}(a)\cap J\subseteq P$. Consequently, $P$ is an $\mathcal{H}_{YJ}$-ideal.

Now Suppose that $I$ is a strong $\mathcal{H}_{Y}$-ideal. If $F=\{a_{1},a_{2},\ldots,a_{n}\}\subseteq P$, then $b_{i}\notin P$ and $m_{i}\in\mathbb{N}$ exist such that $(a_{i}b_{i})^{m_{i}}\in I$, for every $1\leqslant i\leqslant n$. Set $b=b_{1}b_{2}\ldots b_{n}\notin P$ and $m=\max\{m_{1},m_{2},\ldots,m_{n}\}$, then $(a_{i}b)^{m}\in I$, for each $1\leqslant i\leqslant n$. Supposing $E=\{a_{i}^{m}b^{m}:1\leqslant i\leqslant n\}$, $kh_{Y}(E)\cap J\subseteq I$, because $I$ is a strong $\mathcal{H}_{YJ}$-ideal. Thus

[TABLE]

Since $kh_{Y}(b)\not\subseteq P$ and $P$ is prime, it follows that $kh_{Y}(F)\cap J\subseteq P$. Consequently, $P$ is a strong $\mathcal{H}_{YJ}$-ideal.

(b). It is clear if $J\subseteq P$, then $P$ is (resp., strong) $\mathcal{H}_{YJ}$-ideal. Now suppose that $J\not\subseteq P$ and $a\in P$ (resp., $F$ is a finite subset of $P$), then $kh_{Y}(a)\cap J\subseteq P$ (resp., $kh_{Y}(F)\cap J\subseteq P$) implies that $kh_{Y}(a)\subseteq P$ ($kh_{Y}(F)\subseteq P$). Thus $P$ is (resp., strong) $\mathcal{H}_{YJ}$-ideal.

(c). Since $I$ is a relative (resp., strong) $\mathcal{H}_{Y}$-ideal, there is some ideal $J\supseteq I$ such that $I$ is a (resp., strong) $\mathcal{H}_{YJ}$-ideal. Thus $a\in J\setminus I$ exists, Set K=\big{<}a\big{>}, clearly, $I$ is a strong $\mathcal{H}_{YK}$-ideal. If $K\subseteq\sqrt{I}$, then $a^{n}\in I$, for some $n\in\mathbb{N}$. Thus

[TABLE]

This shows that $a\in I$, which is a contradiction. Hence $K\not\subseteq\sqrt{I}$, so $K\not\subseteq P$, for some $P\in\mathrm{Min}(R)$. Now (a) and (b) follow that $P$ is a (resp., strong) $\mathcal{H}_{Y}$-ideal. ∎

Theorem 4.4**.**

Let $Y\subseteq\mathrm{Spec}(R)$ and $I$ and $J$ be ideals of $R$. The following statements are equivalent.

• (a)

$I$* is a (resp., strong) $\mathcal{H}_{YJ}$-ideal.*

• (b)

$I_{\mathcal{H}}\cap J\subseteq I$* ($I_{S\mathcal{H}}\cap J\subseteq I$).*

• (c)

$I_{\mathcal{H}}\cap J=I\cap J$* ($I_{S\mathcal{H}}\cap J=I\cap J$).*

• (d)

A (resp., strong) $\mathcal{H}_{Y}$-ideal $K$ containing $I$ exists such that $K\cap J=I\cap J$.

• (e)

A (resp., strong) $\mathcal{H}_{Y}$-ideal $K$ containing $I$ exists such that $K\cap J\subseteq I$.

• (f)

For each $a\in I$ and $b\in J$ (resp., finite subset $F$ of $I$ and finite subset $E$ of $J$), $h_{Y}(a)\subseteq h_{Y}(b)$ (resp., $h_{Y}(F)\subseteq h_{Y}(E)$) implies that $b\in I$ (resp., $E\subseteq I$).

Proof.

We prove the theorem for $\mathcal{H}_{Y}$-ideals, similarly one can prove it for strong $\mathcal{H}_{Y}$-ideals.

(a) $\Rightarrow$ (b). By Theorem 4.3, either $P_{\mathcal{H}}=P$ or $J\subseteq P$, for each $P\in\mathrm{Min}(I)$, so [3, Proposition 7.11], follows that

[TABLE]

If $a\in I_{\mathcal{H}}\cap J$, then $a\in\sqrt{I}\cap J$, so $\in J$ and $a^{n}\in J$, for some $n\in\mathbb{N}$. Hence $a\in kh_{Y}(a)\cap J=kh_{Y}(a^{n})\cap J\subseteq I$ and therefore $I_{\mathcal{H}}\cap J\subseteq I$.

(b) $\Rightarrow$ (c). It is clear.

(c) $\Rightarrow$ (d). Set $K=I_{\mathcal{H}}$.

(e) $\Rightarrow$ (f). It is clear.

(e) $\Rightarrow$ (f). $h_{Y}(a)\subseteq h_{Y}(b)$ concludes that $kh_{Y}(b)\subseteq kh_{Y}(a)$, since $a\in I\subseteq K$, it follows that $b\in kh_{Y}(b)\cap J\subseteq kh_{Y}(a)\cap J\subseteq K\cap J\subseteq I$.

(f) $\Rightarrow$ (a). Suppose $a\in I\cap J$. If $b\in kh_{Y}(a)$, then $h_{Y}(a)\subseteq h_{Y}(b)$, so $b\in I$, by the assumption. Hence $kh_{Y}(a)\subseteq I$. It follows that $I$ is an $\mathcal{H}_{YJ}$-ideal. ∎

Proposition 4.5**.**

Let $Y\subseteq\mathrm{Spec}(R)$. The following statements hold.

• (a)

Suppose that $\{I_{\alpha}\}_{\alpha\in A}$ and $\{J_{\alpha}\}_{\alpha\in A}$ are two families of ideals of $R$. If $I=\bigcap_{\alpha\in A}I_{\alpha}$, $J=\bigcap_{\alpha\in A}J_{\alpha}$ and $I_{\alpha}$ is a (resp., strong) $\mathcal{H}_{YJ}$-ideal, then $I$ is a (resp., strong) $\mathcal{H}_{YJ}$-ideal.

• (b)

If $J\subseteq K$ are two ideals of $R$ and $I$ is a (resp., strong) $\mathcal{H}_{YK}$-ideal, then $I$ is a (resp., strong) $\mathcal{H}_{YJ}$-ideal.

• (c)

If $\{I_{\alpha}\}_{\alpha\in A}$ is a family of (resp., strong) $\mathcal{H}_{YJ}$-ideals, then $\bigcap_{\alpha\in A}I_{\alpha}$ is a (resp., strong) $\mathcal{H}_{YJ}$-ideal.

• (d)

If $I$ is a (resp., strong) $\mathcal{H}_{YJ}$-ideal and $J$ is a (resp., strong) $\mathcal{H}_{YK}$-ideal, then $I$ is a (resp., strong) $\mathcal{H}_{YK}$-ideal.

• (e)

If $I$ is a (resp., strong) $\mathcal{H}_{YJ}$-ideal, then $\sqrt{I}$ is a (resp., strong) $\mathcal{H}_{Y\sqrt{J}}$-ideal.

• (f)

$I$* is a (resp., strong) $\mathcal{H}_{YJ}$-ideal if and only if $I\cap J$ is a $\mathcal{H}_{YJ}$-ideal.*

• (g)

$I$* is a (resp., strong) $\mathcal{H}_{YJ}$-ideal if and only if $I$ is a (resp., strong) $\mathcal{H}_{Y(I+J)}$-ideal.*

• (h)

Suppose that $J$ is a (resp., strong) $\mathcal{H}_{Y}$-ideal containing an ideal $I$. $I$ is a (resp., strong) $\mathcal{H}_{YJ}$-ideal if and only if $I$ is a (resp., strong) $\mathcal{H}_{Y}$-ideal.

• (i)

Suppose that $J$ is a (resp., strong) $\mathcal{H}_{Y}$-ideal. $I$ is a (resp., strong) $\mathcal{H}_{YJ}$-ideal if and only if $I\cap J$ is a (resp., strong) $\mathcal{H}_{Y}$-ideal.

• (j)

Suppose that $I$ and $J$ are ideals of $R$. $I\cap J$ is both (resp., strong) $\mathcal{H}_{YI}$-ideal and $\mathcal{H}_{YJ}$-ideal if and only if $I$ is a (resp., strong) $\mathcal{H}_{YJ}$-ideal and $J$ is a (resp., strong) $\mathcal{H}_{YI}$-ideal.

• (k)

Suppose that $I$ and $J$ are ideals of $R$. $I_{\mathcal{H}}\cap J$ (resp., $I_{S\mathcal{H}}\cap J$) is the smallest (resp., strong) $\mathcal{H}_{YJ}$-ideal containing $I\cap J$.

• (l)

Suppose that $I\subseteq K$ and $J$ are ideals of $R$ and $I_{\mathcal{H}}=K_{\mathcal{H}}$ (resp., $I_{S\mathcal{H}}=K_{S\mathcal{H}}$). If $I$ is a (resp., strong) $\mathcal{H}_{YJ}$-ideal, then $K$ is a (resp., strong) $\mathcal{H}_{YJ}$-ideal.

• (m)

Suppose that $I$, $J$ and $K$ are ideals of $R$. If $I\subseteq K\subseteq I_{\mathcal{H}}$ (resp., $I\subseteq K\subseteq K\subseteq I_{S\mathcal{H}}$) and $I$ is a (resp., strong) $\mathcal{H}_{YJ}$-ideal, then $K$ is a (resp., strong) $\mathcal{H}_{YJ}$-ideal.

• (n)

If $I$ is a (resp., strong) $\mathcal{H}_{YJ}$-ideal, then $\sqrt{I}$ is a (resp., strong) $\mathcal{H}_{YJ}$-ideal.

• (o)

Suppose that $I$, $J$ and $K$ are ideals of $R$. If $IK$ is a (resp., strong) $\mathcal{H}_{YJ}$-ideal, then $I\cap K$ is a (resp., strong) $\mathcal{H}_{YJ}$-ideal.

• (p)

$I$* is (resp., strong) $\mathcal{H}_{YP}$-ideal, for some prime ideal $P$ if and only if (resp., $I_{S\mathcal{H}}\setminus I$) $I_{\mathcal{H}}\setminus I$ is a multiplicatively closed set.*

• (q)

If $J$ is an ideals of $R$ which not a (resp., strong) $\mathcal{H}_{Y}$-ideal and $J\not\subseteq\bigcap Y$, then there is some strong $\mathcal{H}_{YJ}$-ideal $I\subset J$ which is not a (resp., strong) $\mathcal{H}_{Y}$-ideal.

• (r)

Suppose that $I$ and $J$ are ideals of $R$ and $P$ is a prime ideal of $R$. If $I\cap P$ is a (resp., strong) $\mathcal{H}_{YJ}$-ideal, then either $I$ or $P$ is a (resp., strong) $\mathcal{H}_{YJ}$-ideal.

• (s)

Suppose that $P$ and $Q$ are prime ideals of $R$ and $J$ is an ideal of $R$. If $P\not\subseteq Q$, $Q\not\subseteq P$ and $P\cap Q$ is a (resp., strong) $\mathcal{H}_{YJ}$-ideal, then both $P$ and $Q$ are (resp., strong) $\mathcal{H}_{YJ}$-ideal.

Proof.

(a) and (c). It is clear, according to the fact that the intersection of (resp., strong) $\mathcal{H}_{Y}$-ideals is a (resp., strong) $\mathcal{H}_{Y}$-ideal and Theorem 4.4.

(b), (d), (f), (h), (l). They follow easily from Theorem 4.4.

(e). [3, Lemma 3.2], [3, Proposition 7.11] and Theorem 4.4, conclude

[TABLE]

Now Theorem 4.4, follows that $\sqrt{I}$ is an $\mathcal{H}_{Y\sqrt{J}}$-ideal.

(g). [1, Lemma 2.4], and Theorem 4.4, conclude that $I_{\mathcal{H}}\cap(I+J)=I_{\mathcal{H}}\cap I+I_{\mathcal{H}}\cap J\subseteq I$. Hence $I$ is an $H_{Y(I+J)}$-ideal, by Theorem 4.4. The similar proof states for strong $H_{YJ}$-ideals.

(i $\Rightarrow$). Since $I$ is a (resp., strong) $\mathcal{H}_{YJ}$-ideal and $J$ is a (resp., strong) $\mathcal{H}_{Y}$-ideal, $I\cap J$ is a (resp., strong) $\mathcal{H}_{YJ}$-ideal. Now part (g) follows that $I\cap J$ is a (resp., strong) $\mathcal{H}_{Y}$-ideal.

(i $\Leftarrow$). Since $I\cap J$ is a (resp., strong) $\mathcal{H}_{YJ}$-ideal, $I$ is a (resp., strong) $\mathcal{H}_{YJ}$-ideal, by part (f).

(j). It follows immediately from (f).

(k). It is clear that $I_{\mathcal{H}}\cap J$ is an $\mathcal{H}_{YJ}$-ideal containing $I\cap J$. Now suppose that $K$ is an $\mathcal{H}_{YJ}$-ideal containing $I\cap J$, [3, Proposition 5.1] and Theorem 4.4, follow that

[TABLE]

Similarly, we can prove it for strong $\mathcal{H}_{YJ}$-ideals.

(m). Since $I_{\mathcal{H}}\subseteq K_{\mathcal{H}}\subseteq$, $I_{\mathcal{H}}=K_{\mathcal{H}}$. Now part (k) implies that $K$ is an $\mathcal{H}_{YJ}$-ideal. Similarly, we can prove it for strong $\mathcal{H}_{YJ}$-ideals.

(n). By [3, Proposition 7.11], $I\subseteq\sqrt{I}\subseteq I_{\mathcal{H}}$, now part (m) completes the proof.

(o). Proposition [3, Proposition 5.1], and Theorem 4.4, deduce that

[TABLE]

Thus $I\cap K$ is an $\mathcal{H}_{YJ}$-ideal. Similarly, we can prove it for strong $\mathcal{H}_{YJ}$-ideals.

(p $\Rightarrow$). Theorem 4.4, follows that $I_{\mathcal{H}}\cap P\subseteq I$. If $a,b\in I_{\mathcal{H}}\setminus I$, then $a,b\notin P$, so $ab\notin P$, thus $ab\notin P$, since $ab\in I_{\mathcal{H}}$, it follows that $ab\in I_{\mathcal{H}}\setminus I$. Consequently $I_{\mathcal{H}}\setminus I$ is a multiplicatively closed set.

(p $\Leftarrow$). Since $(I_{\mathcal{H}}\setminus I)\cap I=\emptyset$, there is some prime ideal $P\in\mathrm{Min}(I)$ such that $(I_{\mathcal{H}}\setminus I)\cap P=\emptyset$, thus

[TABLE]

Now Theorem 4.4, concludes that $I$ is an $\mathcal{H}_{YP}$-ideal.

(q). Since $J\not\subseteq\bigcap Y$, there is some $P\in Y$ such that $J\not\subseteq P$. Set $I=P\cap J$, it is clear that $I\subset J$ and $I$ is a strong $\mathcal{H}_{YJ}$-ideal. If $I$ is a (resp., strong) $\mathcal{H}_{YJ}$-ideal, then Proposition 4.5, follows that $J$ is a (resp., strong), which contradicts our assumption.

(r). It clear that if $I\subseteq P$, then $I$ is a $\mathcal{H}_{YJ}$-ideal. Now suppose $I\not\subseteq P$, so $b\in I\setminus P$ exists. For each $a\in P$, $ab\in I\cap J$, so $kh_{Y}(a)\cap kh_{Y}(b)\cap J=kh_{Y}(ab)\cap J\subseteq I\cap P$, since $kh_{Y}(b)\not\subseteq P$, it follows that $kh_{Y}(a)\cap J\subseteq P$. Consequently, $P$ is $\mathcal{H}_{YJ}$-ideal. The similar proof states for strong $\mathcal{H}_{YJ}$-ideals.

(s). By the assumption, $P,Q\in\mathrm{Min}(I\cap P)$, so $P$ and $Q$ are $\mathcal{H}_{YJ}$-ideals, by Theorem 4.3(a). ∎

We say that a ring $R$ has the root property if for each $x\in R$ there are $y\in R$ and $2\leqslant n\in\mathbb{N}$ such that $y^{n}=x$.

Proposition 4.6**.**

Let $Y\subseteq\mathrm{Spec}(R)$ and $K=\bigcap Y$. The following hold.

• (a)

An ideal $I$ of $R$ is a relative (resp., strong) $\mathcal{H}_{Y}$-ideal if and only if there is some ideal $J\supset I$ such that $I$ is a (resp., strong) $\mathcal{H}_{YJ}$-ideal.

• (b)

Suppose that $I\subseteq K$ are ideals of $R$ and $I_{\mathcal{H}}=K_{\mathcal{H}}$ (resp., $I_{S\mathcal{H}}=K_{S\mathcal{H}}$). If $I$ is a relative (resp., strong) $\mathcal{H}_{Y}$-ideal, then $K$ is a relative (resp., strong) $\mathcal{H}_{Y}$-ideal.

• (c)

Suppose that $I$ and $K$ are ideals of $R$. If $I\subseteq K\subseteq I_{\mathcal{H}}$ (resp., $I\subseteq K\subseteq K\subseteq I_{S\mathcal{H}}$) and $I$ is a relative (resp., strong) $\mathcal{H}_{Y}$-ideal, then $K$ is a relative (resp., strong) $\mathcal{H}_{Y}$-ideal.

• (d)

If $I$ is an ideal of $R$ and $P$ is a prime ideal of $R$. If $I\cap P$ is a relative (resp., strong) $\mathcal{H}_{Y}$-ideal, then either $I$ or $P$ is a relative (resp., strong) $\mathcal{H}_{Y}$-ideal.

• (e)

An ideal $I$ of a ring $R$ is a relative (resp., strong) $\mathcal{H}_{Y}$-ideal if and only if there is some (resp., finite subset $F\subseteq R$ which $F\cap I=\emptyset$ ) $c\in R\setminus I$ such that (\big{<}c\big{>}\cap kh_{Y}(F)\subseteq I) \big{<}c\big{>}\cap kh_{Y}(a)\subseteq I.

• (f)

Suppose that $I$ is an ideal of $R$. If $(K:I)\not\subseteq I$ and $K\subseteq I$, then $I$ is a relative strong $\mathcal{H}_{Y}$-ideal.

• (g)

Suppose that $a\in R$, K\subseteq\big{<}a\big{>} and $R$ has the root property. The principal ideal \big{<}a\big{>} is a relative strong $\mathcal{H}_{Y}$-ideal if and only if (K:a)\not\subseteq\big{<}a\big{>}.

Proof.

(a). It is clear, by Proposition 4.5(g).

(b). By the assumption, there is some ideal $J‌\not\subseteq I$ such that $I$ is a (resp., strong) $\mathcal{H}_{YJ}$-ideal, then Theorem 4.5(l), concludes that $K$ is a (resp., strong) $\mathcal{H}_{YJ}$-ideal. If $J\subseteq K‌\subseteq K_{\mathcal{H}}$, then $J‌\subseteq K_{\mathcal{H}}=I_{\mathcal{H}}\cap J\subseteq I$, which is a contradiction, so $J‌\not\subseteq K$ and therefore $K$ is a relative $\mathcal{H}_{Y}$-ideal.

(c). It follows immediately from part (b).

(d). It concludes immediately from Proposition 4.5(r).

We prove part (e) for $\mathcal{H}_{Y}$-ideals, it is clear the similar proof states for strong $\mathcal{H}_{Y}$-ideals.

(e $\Rightarrow$). Since $I$ is a relative $\mathcal{H}_{Y}$-ideal, there is an ideal $J\supseteq I$ such that $I$ is an $\mathcal{H}_{YJ}$-ideal. Thus $c\in J\setminus I$ exists and $I_{H}\cap J\subseteq I$, by Theorem 4.4. For each $a\in I\subseteq I_{\mathcal{H}}$ we have, kh_{Y}(a)\cap\big{<}c\big{>}\subseteq I_{\mathcal{H}}\cap J\subseteq I.

(e $\Leftarrow$). Set J=\big{<}c\big{>}. Theorem 4.4 follows that $I$ is a $\mathcal{H}_{YJ}$-ideal, since $J‌\not\subseteq I‌$, it follows that $I$ is a relative $\mathcal{H}_{Y}$-ideal.

(f). First we show that $kh_{Y}(F)\cap(K:I)\subseteq K$, for each finite subset $F$ of $R$. Suppose that $F$ is a finite subset of $I$ and $a\in kh_{Y}(F)\cap(K:I)$, then $kh_{Y}(a)\subseteq kh_{Y}(F)$ and $aI\subseteq K$, thus $kh_{Y}(a)=kh_{Y}(aF)$ and $aF$ is a finite subset of $K$, since $K$ is a strong $\mathcal{H}_{Y}$-ideal, it follows that $a\in kh_{Y}(aF)\subseteq K$. Consequently, $kh_{Y}(F)\cap(K:I)\subseteq K$. Now take $J=(K:I)$, then $c\in J\setminus I$ exists. For each finite subset $F$ of $I$, we have

[TABLE]

Now part (c) follows that $I$ is a relative strong $\mathcal{H}_{Y}$-ideal.

(g $\Rightarrow$). Since \big{<}a\big{>} is a relative strong $\mathcal{H}_{Y}$-ideal, there is some ideal J\not\subseteq\big{<}a\big{>} such that \big{<}a\big{>} is a strong $\mathcal{H}_{YJ}$-ideal, so kh_{Y}(a)\cap J\subseteq\big{<}a\big{>} and c\in J\setminus\big{<}a\big{>} exists, thus $c\in J\setminus kh_{Y}(a)$. Since $R$ has the root property, there is some $b\in R$ and $2\leqslant n\in\mathbb{N}$ such that $a=b^{n}$. Then bc\in J\cap kh_{Y}(a)\subseteq\big{<}a\big{>}, so $bc=ar=b^{n}r$, thus $b(c-n^{n-1}r)=0$. We claim c-b^{n-1}\notin\big{<}a\big{>}, on contrary, c\in\big{<}a\big{>}\subseteq kh_{Y}(a), which is a contradiction. Hence $c-b^{n-1}\in(K:b)=(K:a)$ and therefore c\in(K:a)\setminus\big{<}a\big{>}.

(g $\Leftarrow$). It follows from part (f). ∎

A ring $R$ is called arithmetical, if for every ideal $I$, $J$ and $K$ of $R$, we have

[TABLE]

Proposition 4.7**.**

Suppose that $Y\subseteq\mathrm{Spec}(R)$ and $J$ is an ideal of $R$.

• (a)

$I$* is a maximal element of $\{I:J\text{ is a (resp., strong) }\mathcal{H}_{J}\text{-factor of }I\}$ if and only if $I$ is a maximal element of*

[TABLE]

• (b)

Every maximal element of $\{I\subset J:I\text{ is a (resp., strong) }\mathcal{H}_{YJ}\text{-ideal }\}$ is of the form $P\cap I$, in which $P$ is a maximal element of

[TABLE]

• (c)

Suppose that $I$ is an ideal of $R$. If $I$ has a (resp., strong) $\mathcal{H}_{Y}$-factor, then the family of all (resp., strong) $H_{Y}$-factors of $I$ has a maximal element, which contains $I$.

• (d)

If $J$ is a minimal (resp., strong) $\mathcal{H}_{Y}$-factor of an ideal $I$, then $I+J$ is a minimal (resp., strong) $\mathcal{H}_{Y}$-factor of $I$ containing $I$.

• (e)

If the largest (resp., strong) $\mathcal{H}_{Y}$-factor of an ideal $I$ exists, then it is of the form (resp., \{x\in R:kh_{Y}(x)\cap\big{<}F\big{>}\subseteq I\text{ for all finite subset }F\text{ of }I\}) \{x\in R:kh_{Y}(x)\cap\big{<}a\big{>}\subseteq I\text{ for all }a\in I\}.

• (f)

Suppose $I$ is an ideal of $R$. If

[TABLE]

then (resp., $I_{S\mathcal{H}}\cap K\subseteq\sqrt{I}$) $I_{\mathcal{H}}\cap K\subseteq\sqrt{I}$. Also if $I$ is a relative semi-prime ideal, then $K$ is the greatest (resp., strong) $\mathcal{H}_{Y}$-factor of $I$.

• (g)

If $I$ is a relative (resp., strong) $\mathcal{H}_{Y}$-ideal, then the greatest (resp., strong)‌ $\mathcal{H}_{Y}$-factor of $I$ exists.

• (h)

If $R$ is arithmetical, then every relative (resp., strong)‌ $\mathcal{H}_{Y}$-ideal has the greatest (resp., strong) $\mathcal{H}_{Y}$-factor.

• (i)

Suppose $K=\bigcap Y$. ideal $J$ is a minimal (resp., strong) $\mathcal{H}_{Y}$-factor of an ideal $I$ if and only if J=\big{<}e\big{>}, for some $e\notin\sqrt{I}$, such that I\cap\big{<}K,e\big{>} is a (resp., strong) $\mathcal{H}_{Y}$-ideal and \big{<}e\big{>} is a minimal principal ideal which is not contained in $I$.

Proof.

(a). We show this part for $\mathcal{H}_{Y}$-ideals, the similar proof states for strong $\mathcal{H}_{YJ}$-ideals. Suppose that $I\in\mathrm{maxl}\{I:J\text{ is a }\mathcal{H}_{Y}\text{-factor of }I\}$. Since $J$ is an $\mathcal{H}_{Y}$-factor of every prime $\mathcal{H}_{Y}$-ideal which not contains $J$, it is sufficient to $I$ is a prime $\mathcal{H}_{Y}$-ideal. We claim $J\not\subseteq I_{\mathcal{H}}$, otherwise, Theorem 4.4, concludes $J=I_{\mathcal{H}}\cap J\subseteq I$, which is a contradiction. Since $I_{\mathcal{H}}$ is an $\mathcal{H}_{YJ}$-ideal, it follows that $I_{\mathcal{H}}=I$, by the maximality of $I$, hence $I$ is an $\mathcal{H}_{Y}$-ideal, consequently, $I$ is semi-prime, by [3, Lemma 3.2]. Since $I\not\subseteq J$, there is some $P\in\mathrm{Min}(I)$ such that $J\not\subseteq P$. $P$ is an $\mathcal{H}_{YJ}$-ideal, by Theorem 4.3. Hence $P=I$, by the maximality of $I$.

(b). Suppose that $I$ a maximal element of $\{I\subset J:I\text{ is a }\mathcal{H}_{YJ}\text{-ideal }\}$. If $J\subseteq I_{\mathcal{H}}$, then Theorem 4.4, concludes $J=J\cap I_{\mathcal{H}}\subseteq I$, which is a contradiction. Hence $J\not\subseteq I_{\mathcal{H}}$, so there is some $Q\in\mathrm{Min}(I_{\mathcal{H}})$, such that $J\not\subseteq P$. By [3, Theorem 3.13], $Q$ is a prime $\mathcal{H}_{Y}$-ideal. So there is some maximal element $P$ of $\{P:P\text{ is a prime }\mathcal{H}_{Y}\text{-ideal and }J\not\subseteq P\}$ such that contains $Q$. Thus $I=I\cap J\subseteq Q\cap J\subseteq P\cap J\subseteq J$, since $P\cap J$ is an $\mathcal{H}_{YJ}$-ideal, it follows that $I=P\cap J$.

(c). If $\mathscr{C}$ is a chain of $H_{Y}$-factor of $I$, then \big{(}\bigcup_{J\in\mathscr{C}}J\big{)}\cap I_{\mathcal{H}}=\bigcup_{J\in\mathscr{C}}(J\cap I_{\mathcal{H}})\subseteq I and clearly, $\bigcup_{J\in\mathscr{C}}J\not\subseteq I$, so $\bigcup_{J\in\mathscr{C}}J$ is an $\mathcal{H}_{Y}$-factor of $I$, thus the family of all $\mathcal{H}_{Y}$-factor of $I$ has a maximal element, by Zorn’s lemma. Now Proposition 4.6(a), follows that $I\subset J$.

(d). Since $J$ is a (resp., strong) $\mathcal{H}_{Y}$-factor of $I$, $I+J$ is a $\mathcal{H}_{Y}$-factor of $I$, by Proposition 4.5(g). If $K\subseteq I+J$ is an $\mathcal{H}_{Y}$-factor of $I$ containing $I$, then $k\in K\setminus I$ exists, hence there are $i\in I$ and $j\in J$ such that $k=i+j$, so $j=k-i\in K\setminus I$, and therefore $j\in(K\cap J)\setminus I$, so $K\cap J$ is an $\mathcal{H}_{Y}$-factor of $I$, by Proposition 4.5(a). Now the minimality of $J$ concludes that $K\cap J=J$, hence $J\subseteq K$, and therefore $I+J\subseteq K$.

(e). Suppose $J$ is the largest $\mathcal{H}_{Y}$-factor of $I$ and K=\{x\in R:kh_{Y}(x)\cap\big{<}a\big{>}\subseteq I\quad\forall a\in I\}. $x,y\in K$ implies that kh_{Y}(a)\cap\big{<}x\big{>}\subseteq I and kh_{Y}(a)\cap\big{<}y\big{>}\subseteq I, for all $a\in I$, so \big{<}x\big{>} and \big{<}y\big{>} are $\mathcal{H}_{Y}$-factors of $I$, since $J$ is the greatest $\mathcal{H}_{Y}$-factor of $I$, it follows that $x,y\in J$, thus kh_{Y}(a)\cap\big{<}x+y\big{>}\subseteq kh_{Y}(a)\cap J\subseteq I, for all $a\in I$, hence $x+y\in K$. Now suppose $x\in K$ and $r\in R$, then kh_{Y}(a)\cap\big{<}rx\big{>}\subseteq kh_{Y}(a)\cap\big{<}x\big{>}\subseteq I and therefore $rx\in K$. Consequently, $K$ is an ideal of $I$. Also for every $a\in I$,

[TABLE]

Hence $I$ is an $\mathcal{H}_{YK}$-ideal. Since for each $x\in J$ we have kh_{Y}(a)\cap\big{<}x\big{>}\subseteq kh_{Y}(a)\cap J\subseteq I, for all $a\in I$, it follows that $J\subseteq K$, and thus $J=K$. Similarly, we can show this part for strong $\mathcal{H}_{Y}$-factors.

(f). Set $\cal{A}=\{P\in\mathrm{Min}(I):P\text{ is an }\mathcal{H}_{Y}\text{-ideal}\}$ and $\cal{B}=\mathrm{Min}(I)\setminus\cal{A}$. By Proposition [3, Proposition 7.11],

[TABLE]

Now suppose that $I$ is a semi-prime ideal, then $I_{H}\cap K=\sqrt{I}=I$, and therefore $I$ is an $\mathcal{H}_{YK}$-ideal. If $J$ is an $\mathcal{H}_{Y}$-factor of $I$, then for each $P\in\cal{B}$, then every $P\in\cal{B}$ is an $\mathcal{H}_{YJ}$-ideal, by Theorem 4.3(a), since every $P\in\cal{B}$ is a not $H_{Y}$-ideal, it follows that $J\subseteq P$, by Theorem 4.3(b), hence $J\subseteq K$. Since $I$ is a relative $\mathcal{H}_{Y}$-ideal, it has a $\mathcal{H}_{Y}$-factor $J\subseteq K$ and therefore $K\not\subseteq I$. Consequently, $K$ is the greatest $\mathcal{H}_{Y}$-factor of $I$.

(g). Since $I$ is a relative (resp., strong) $\mathcal{H}_{Y}$ -ideal, $\sqrt{I}$ is $\mathcal{H}_{Y}$-ideal, by Proposition 4.6(b). Now part (f) deduces that $\sqrt{I}$ has the greatest (resp., strong) $\mathcal{H}_{Y}$-factor.

(h). Suppose that $I$ is a relative (resp., strong)‌ $\mathcal{H}_{Y}$-ideal. Part (c) deduces that $I$ has a maximal (resp., strong) $\mathcal{H}_{Y}$-factor $J$. If $K$ is a $\mathcal{H}_{Y}$-factor of $I$, then

[TABLE]

Thus $J+‌K$ is a $\mathcal{H}_{Y}$-factor of $I$, and therefore $K\subseteq J$, by the maximality of $J$. Hence $J$ is the greatest $\mathcal{H}_{Y}$-factor of $I$.

We prove part (i) for $\mathcal{H}_{Y}$-ideals, similarly one can show it for strong $\mathcal{H}_{Y}$-ideals

(i$\Rightarrow$). Set $e\in J\setminus I$. Proposition 4.5(b), concludes that $I$ is an \mathcal{H}_{Y\big{<}e\big{>}}-ideal, hence J=\big{<}e\big{>}, by the minimality of $J$. If for some $n\in\mathbb{N}$, $e^{n}\in I$, then $e\in kh_{Y}(e)\cap J=kh_{Y}(e^{n})\cap J\subseteq I$, which is a contradiction. Hence $e\notin\sqrt{I}$. Since \big{<}e^{2}\big{>}\subseteq\big{<}e\big{>}, $I$ is an \mathcal{H}_{Y\big{<}e^{2}\big{>}}-ideal, by Proposition 4.5(b). The minimality of $J$ follows that \big{<}e\big{>}=\big{<}e^{2}\big{>}, thus for some $r\in R$, $e=e^{2}r$. Now we show that \big{<}e,K\big{>}=\bigcap_{1-er\notin P\in Y}P. If $1-er\notin P$, then $e(1-er)=e-e^{2}r=0\in P$, so $e\in R$, hence \big{<}e,K\big{>}\subseteq P and therefore \big{<}e,K\big{>}\subseteq\bigcap_{1-er\notin P\in Y}P. If $y\in\bigcap_{1-er\notin P\in Y}P$, since $\bigcap_{1-er\notin P\in Y}P=(K:1-er)$, it follows $y-yer=y(1-er)\in K$ and thus y\in\big{<}K,e\big{>}. Consequently, \big{<}K,e\big{>}=\bigcap_{1-er\notin P\in Y}P. Since $\bigcap_{1-er\notin P\in Y}P$ is a $Y$-Hilbert, \big{<}e\big{>} is an $\mathcal{H}_{Y}$-ideal. Proposition 4.5(i), deduces that I\cap\big{<}K,e\big{>} is an $\mathcal{H}_{Y}$-ideal. If \big{<}a\big{>}\subseteq\big{<}e\big{>}, then $I$ is an \mathcal{H}_{Y\big{<}a\big{>}}-ideal, by Proposition 4.5(b). Now the minimality of $J$ implies that \big{<}a\big{>}\subseteq I, hence \big{<}e\big{>} is a minimal principal ideal which not contained in $I$.

(i$\Leftarrow$). Since J=\big{<}e\big{>} and \big{<}K,e\big{>}\cap I are $\mathcal{H}_{Y}$-ideals, $I$ is an \mathcal{H}_{Y\big{<}K,e\big{>}}-ideal, by Proposition 4.5(i). Proposition 4.5(b), implies that $I$ is a $H_{YJ}$-ideal, since $e\notin I$, it follows that $J$ is an $\mathcal{H}_{Y}$-factor of $I$. If K\subset\big{<}e\big{>} is an $\mathcal{H}_{Y}$-factor of $I$, then $k\in K\setminus I$ exists, then \big{<}k\big{>}\subseteq K\subset\big{<}e\big{>} and \big{<}k\big{>}\not\subseteq I, which contradicts the minimality of \big{<}e\big{>}. Hence $J$ is a minimal $\mathcal{H}_{Y}$-factor of $I$. ∎

Corollary 4.8**.**

Suppose $Y\subseteq\mathrm{Spec}(R)$ and $I$ is a semi-prime ideal of $R$. $I$ is a relative (resp., strong) $\mathcal{H}_{Y}$-ideal if and only if in each representation of $I$ as the intersection of prime ideals, there is a (resp., strong) $\mathcal{H}_{Y}$-ideal.

Proof.

($\Rightarrow$). By Proposition 4.7(f), $I$ has the greatest $\mathcal{H}_{Y}$-factor $J$. Proposition 4.5(e), concludes $J$ is a semi-prime. If $I=\bigcap_{P\in\cal{A}}$ is a representation of prime ideals, then Theorem 4.3(a), deduces that $P_{\alpha}$ is an $\mathcal{H}_{YJ}$-ideal, for each $\alpha\in A$. If for every $\alpha\in\cal{A}$, $P_{\alpha}$ is not an $\mathcal{H}_{Y}$-ideal, then Theorem 4.3(b), concludes $P_{\alpha}\supseteq J$ and therefore $I=\bigcap_{P\in\cal{A}}P\supseteq J$, which is a contradiction.

($\Leftarrow$). Set

[TABLE]

Then $J\not\subseteq I$ and therefore Theorem 4.7(f), concludes $I$ is a relative $\mathcal{H}_{Y}$-ideal. ∎

Theorem 4.9**.**

Suppose that $X,Y\subseteq\mathrm{Spec}(R)$ and $J$ is an ideal of $R$. Every (resp., strong) $\mathcal{H}_{XJ}$-ideal is a (resp., strong) $\mathcal{H}_{YJ}$-ideal if and only if every prime (resp., strong) $\mathcal{H}_{X}$-ideal which is not containing $J$ is a (resp., strong) $\mathcal{H}_{Y}$-ideal.

Proof.

$\Rightarrow$). Suppose that $P$ is a prime (resp., strong) $\mathcal{H}_{X}$-ideal which is not containing $J$, then $P$ is a (resp., strong) $\mathcal{H}_{XJ}$-ideal and therefore $P$ is a (resp., strong) $\mathcal{H}_{YJ}$-ideal, by the hypothesis. Since $J\not\subseteq P$, $P$ is $\mathcal{H}_{Y}$-ideal, by Theorem 4.3.

$\Leftarrow$). Set $Z$ the family of all prime (resp., strong) $\mathcal{H}_{X}$-ideals which are not containing $J$. Suppose that $I$ is a (resp., strong) $\mathcal{H}_{XJ}$-ideal. We claim $I$ is a (resp., strong) $\mathcal{H}_{ZJ}$-ideal, on contrary $a\in I$ and $b\in J$ exist such that $h_{Z}(a)\subseteq h_{Z}(b)$ and $b\notin I$, then $P\in\mathrm{Min}(I)$ exists such that $b\notin P$, so $J\not\subseteq P$, hence $P$ is a (resp., strong) $\mathcal{H}_{X}$-ideal, by Theorem 4.3. Thus $P\in Z$, since $h_{Z}(a)\subseteq h_{Z}(b)$ and $a\in I\subseteq P$, it follows that $b\in kh_{Z}(b)\subseteq kh_{Z}(a)\subseteq P$, which is a contradiction. Hence $I$ is a (resp., strong) $\mathcal{H}_{ZJ}$-ideal, so there is some (resp., strong) $\mathcal{H}_{Z}$-ideal $K$ such that $K\cap J\subseteq I$, by Theorem 4.4. Since every element of $Z$ is a (resp., strong) $\mathcal{H}_{Y}$-ideal, $K$ is a (resp., strong) $\mathcal{H}_{Y}$-ideal, by [3, Corollary 6.4]. Now Theorem 4.4, concludes $I$ is a (resp., strong) $\mathcal{H}_{YJ}$-ideal. ∎

Theorem 4.10**.**

Suppose $Y\subseteq\mathrm{Spec}(R)$ and $\bigcap Y=\{0\}$. The following statements are equivalent.

• (a)

Every proper ideal of $R$ is a relative strong $\mathcal{H}_{Y}$-ideal.

• (b)

Every proper ideal of $R$ is a relative $\mathcal{H}_{Y}$-ideal.

• (c)

$R$* is a regular ring.*

Proof.

(a $\Rightarrow$ b). It is clear.

(b $\Rightarrow$ c). Suppose that $I$ is a proper ideal of $R$, then $I$ is a relative $\mathcal{H}_{Y}$-ideal, so it has a $\mathcal{H}_{Y}$-factor and therefore it has a maximal $\mathcal{H}_{Y}$-factor $J$, by Proposition 4.7. If $J\neq R$, then $J$ is a $H_{Y}$-ideal, thus has a $\mathcal{H}_{Y}$-factor $K$ containing $J$, by Proposition 4.6. Now Proposition 4.5, concludes that $K$ is a $\mathcal{H}_{Y}$-factor of $I$, which contradicts the maximality of $J$. Hence $J=R$, and therefore $I$ is a $\mathcal{H}_{Y}$-ideal. Thus $R$ is a regular ring, by [3, Proposition 5.8].

(c $\Rightarrow$ a). Suppose that $I$ is a proper ideal of $R$, then [3, Proposition 5.8], concludes that $I$ is a strong $\mathcal{H}_{Y}$-ideal and therefore $I$ is a strong $\mathcal{H}_{Y}$-ideal, by Proposition 4.2. ∎