# On ideals in quantales, I

Amartya Goswami  [1] Department of Mathematics and Applied Mathematics,
University of Johannesburg, South Africa; [2] National Institute for
Theoretical and Computational Sciences (NITheCS), South Africa.
<EMAIL_ADDRESS>
###### Abstract.

Taking a ring-theoretic perspective as our motivation, the main aim of this
series is to establish a comprehensive theory of ideals in commutative
quantales with an identity element. This particular article focuses on an
examination of several key properties related to ideals in quantale, including
prime, semiprime, radical, primary, irreducible, and strongly irreducible
ideals. Furthermore, we investigate the primary decomposition problem for
quantale ideals. In conclusion, we present a set of future directions for
further exploration, serving as a natural continuation of this article.

###### Key words and phrases:

Quantale; Multiplicative lattice; Semiprime ideal; Primary decomposition;
Strongly irreducible ideal.

###### 1991 Mathematics Subject Classification:

06F07; 13A15; 06B23.

###### Contents

  1. 1 Introduction
  2. 2 Ideals
    1. 2.1 Operations on ideals
    2. 2.2 Prime and semiprime ideals
    3. 2.3 Primary ideals and primary decompositions
    4. 2.4 Irreducible and strongly irreducible ideals

##  1\. Introduction

As pointed out in [36], a quantale can be regarded as a special case within
various algebraic and categorical perspectives. Here are a few examples:

  1. $\bullet$

complete multiplicative lattices,

  2. $\bullet$

complete residuated lattices,

  3. $\bullet$

semirings with infinitary sums,

  4. $\bullet$

thin closed monoidal categories,

  5. $\bullet$

monoids in complete semilattices.

For the purpose of developing an ideal theory of quantales, we will adopt the
first approach mentioned above. Ideal theory in an algebraic structure
provides a powerful framework for understanding the structural properties and
behaviour of ideals within the algebraic system. It focuses on the
investigation of properties and characteristics of ideals in algebraic
structures, such as their generation, factorization, and intersection. It
explores the interplay between ideals and other fundamental algebraic
concepts, such as modules, prime ideals, and quotient structures. One of the
primary goals of ideal theory is to establish connections between ideals and
the overall structure of the algebraic system in which they reside.

Although abstract ideal theory has been extensively studied (see, e.g., [1, 3,
4, 11, 12, 44, 45]) for various classes of lattices by introducing different
types of elements (replacing various types of ideals), there is a notable
absence of detailed study on semiprime, primary, strongly irreducible, etc.,
types of elements in the context of multiplicative lattices111 however,
definitions of semiprime and primary ideals have been mentioned in [46,
Definition 2.6] (see also [37, Definition 2.2]) in order to in order to
introduce rough semiprime and rough primary ideals in a quantale. Also, a
definition of prime ideal and completely prime ideal has been introduced in
[10, Definition 3 and Definition 4] to show when they coincide.. It is worth
noting that in a lattice without a multiplication operation, the notions of
prime and strongly irreducible elements coincide. However, in the context of
rings, these two types of ideals are distinct from each other.

Surprisingly, despite the wealth of research on ideal theory, we have not come
across a well-defined notion of an ideal in the context of quantales.
Motivated by this observation, the primary objective of this series of papers
is to explore the extent to which the ideal theory of rings can be extended to
quantales. In many instances, the reader find that the proofs of various
properties for different classes of ideals in rings can be adapted almost
identically for ideals in quantales. However, we shall present complete proofs
for all results unless they are trivial. This approach serves two main
purposes:

Firstly, providing comprehensive proofs contribute to the understanding and
advancement of this field of study. By presenting detailed arguments, we aim
to enhance the readers’ comprehension and facilitate further developments in
this area.

Secondly, we intend to showcase the reliance on multiplicative ideal theory
principles by explicitly demonstrating the application of these principles in
quantale analysis. Through the presentation of proofs, we aim to emphasize the
significance of ideal-theoretic aspects in the study of quantales.

Here is an explanation of the terminology and notation used in this paper. The
symbol $\mathds{N}$ represents the set of natural numbers, i.e.,
${1,2,3,\ldots}$. When we have a set $X$ and $S$ is a subset of $X$, we denote
the complement of $S$ with respect to $X$ as $X\neg S=\\{x\in X\mid x\notin
S\\}$. The empty set is denoted by $\emptyset$, while the top and bottom
elements of a lattice $(\mathcal{L},\preccurlyeq,\vee,\wedge)$ are represented
as $\top$ and $\bot$, respectively. For any element $x\in\mathcal{L}$ with
$x\neq\bot$, the notation $nx$ signifies the repeated join (supremum) of $x$
with itself $n$ times, i.e., $x\vee\cdots\vee x$. Throughout the paper, the
default ordering relation for the poset of subsets of a set be the set
inclusion relation, denoted by $\preccurlyeq$.

The paper is organized as follows. In §2.1, we provide an introduction to the
concept of ideals in quantales and explore the construction of new ideals
using operations between them. We examine the general properties of ideals and
investigate their behavior under quantale homomorphisms. Additionally, we
briefly discuss some key results pertaining to maximal ideals in quantales.

In §2.2, we study prime and semiprime ideals in quantales. We examine the
notion of radicals of ideals and establish an equivalence between semiprime
ideals and radical ideals in a quantale. Throughout this analysis, the
significance of multiplicatively closed subsets becomes apparent as they play
a crucial role in the development of the theory.

Moving forward, §2.3 focuses on primary ideals in quantales and primary
decompositions. We explore the properties of primary ideals and investigate
their decompositions. This section sheds light on the fundamental aspects of
primary ideals within the context of quantales.

Lastly, in §2.4, we delve into the discussion of irreducible and strongly
irreducible ideals in quantales. We present a representation theorem, stating
that any ideal in a Noetherian quantale can be expressed as an intersection of
a finite number of irreducible ideals. Furthermore, we partially characterize
arithmetic quantales by examining their strongly irreducible ideals.

##  2\. Ideals

###  2.1. Operations on ideals

Quantales can be seen as a generalization of rings, where the underlying
additive abelian groups are replaced by sup-lattices and multiplication part
as a semigroup. Since in our investigation, we aim to extend the ideal theory
associated with commutative rings with identity, our focus shall be on a
specific type of quantales that align with this goal. For a more comprehensive
exploration of the algebraic aspects of quantales in general, we recommend
[41].

###### Definition 2.1.1. —

An _unital commutative quantale_ is a complete lattice
$(\mathcal{Q},\preccurlyeq,\bot,\top)$ endowed with an operation $\&$,
satisfying the following axioms:

  1. (1)

$x\mbox{{\small{$\&$}}}(y\mbox{{\small{$\&$}}}z)=(x\mbox{{\small{$\&$}}}y)\mbox{{\small{$\&$}}}z$,

  2. (2)

$x\mbox{{\small{$\&$}}}y=y\mbox{{\small{$\&$}}}x$,

  3. (3)

$x\mbox{{\small{$\&$}}}(\bigvee_{\lambda\in\Lambda}y_{\lambda})=\bigvee_{\lambda\in\Lambda}(x\mbox{{\small{$\&$}}}y_{\lambda})$,

  4. (4)

$x\mbox{{\small{$\&$}}}\top=x$,

for all $x,$ $y,$ $y_{\lambda}$, $z\in\mathcal{Q}$, and for all
$\lambda\in\Lambda$, where $\Lambda$ is an index set.

Throughout this paper, unless otherwise stated, by a “quantale,” we shall
always refer to a commutative quantale with the top element $\top$ as the
identity with respect to $\&$. We shall also use the notation $x^{n}$ to
denote $x\mbox{{\small{$\&$}}}\cdots\mbox{{\small{$\&$}}}x$ (repeated $n$
times).

###### Remark 2.1.1. —

The aforementioned Definition 2.1.1 was first introduced in [45] (also see
[11]), where the authors referred to it as a _multiplicative lattice_.

Unital commutative quantales offer numerous examples that illustrate their
versatility and applicability. Below, we provide a few of them.

  1. (1)

The quantale of non-negative real numbers with the operation of maximum
(supremum) and the identity element 0. This quantale is generally denoted as
$([0,\infty],\max,0)$.

  2. (2)

The quantale $(\mathcal{P}(X),\cup,X)$ of non-empty subsets of a set $X$,
ordered by inclusion and equipped with the operation of set union as the
multiplication and the full set $X$ as the identity element.

  3. (3)

The quantale $(\mathcal{C}(X),\vee,0)$ of continuous real-valued functions on
a topological space $X$, ordered pointwise and equipped with the operation of
pointwise maximum (supremum) as the multiplication and the constant function
$0$ as the identity element.

  4. (4)

The quantale $(\mathcal{I}(R),\cap,{0})$ of ideals in a commutative unital
ring $R$, ordered by inclusion and equipped with the operation of ideal
intersection as the multiplication and the zero ideal ${0}$ as the identity
element.

In the following lemma, we compile a number of elementary results on quantales
that be utilized in sequel.

###### Lemma 2.1.2. —

Let $\mathcal{Q}$ be a quantale. Then the following hold.

  1. (1)

$x\mbox{{\small{$\&$}}}y\preccurlyeq x\wedge y$, for all $x,y\in\mathcal{Q}$.

  2. (2)

If $x\preccurlyeq y$, then $x\mbox{{\small{$\&$}}}z\preccurlyeq
y\mbox{{\small{$\&$}}}z,$ for all $z\in\mathcal{Q}$.

  3. (3)

If $x\preccurlyeq y$ and $u\preccurlyeq v$, then
$x\mbox{{\small{$\&$}}}u\preccurlyeq y\mbox{{\small{$\&$}}}v$, for all $x,$
$y,$ $u,$ $v\in\mathcal{Q}$.

  4. (4)

If $n\in\mathds{N}$ and if $x,$ $y\in\mathcal{Q}$, then

(2.1) $(x\vee
y)^{n}=\bigvee_{k=0}^{n}\binom{n}{k}a^{n-k}\mbox{{\small{$\&$}}}b^{k},\qquad\text{where,}\;\;\;\binom{n}{k}=\dfrac{n!}{k!(n-k)!}.$

###### Proof.

(1) Notice that
$x=x\mbox{{\small{$\&$}}}\top=x\mbox{{\small{$\&$}}}(y\vee\top)=(x\mbox{{\small{$\&$}}}y)\vee
x,$ and this implies $x\mbox{{\small{$\&$}}}y\preccurlyeq x$. Similarly, we
have $x\mbox{{\small{$\&$}}}y\preccurlyeq y,$ and hence
$x\mbox{{\small{$\&$}}}y\preccurlyeq x\wedge y.$

(2) Since $x\preccurlyeq y$, we have $y\mbox{{\small{$\&$}}}z=(x\vee
y)\mbox{{\small{$\&$}}}z=(x\mbox{{\small{$\&$}}}z)\vee(y\mbox{{\small{$\&$}}}z)$,
which implies that $x\mbox{{\small{$\&$}}}z\preccurlyeq
y\mbox{{\small{$\&$}}}z.$

(3) Since $x\preccurlyeq y$ and $u\preccurlyeq v$, by (2), we have
$x\mbox{{\small{$\&$}}}u\preccurlyeq y\mbox{{\small{$\&$}}}u$ and
$y\mbox{{\small{$\&$}}}u\preccurlyeq y\mbox{{\small{$\&$}}}v$. From these, we
obtain the claim.

(4) For $n=1$, the formula (2.1) is trivially true. Suppose the formula is
true for $n=k$. Using the conditions of Definition 2.1.1, it is a routine
exercise to check that the formula is also true for $n=k+1$. Hence by
induction we have the desired claim. ∎

We shall now present the definition of the central concept in this paper,
which pertains to the notion of ideals in a quantale.

###### Definition 2.1.2. —

A nonempty subset $I$ in $\mathcal{Q}$ is called an _ideal_ , if for all $x,$
$y,$ $l\in\mathcal{Q}$,

  1. (1)

$x,$ $y\in I$ implies that $x\vee y\in I$, and

  2. (2)

$x\in I$ and $l\preccurlyeq x$ implies that $l\in I$.

We shall denote the set of all ideals in $\mathcal{Q}$ by
$\mathcal{I}_{\mathcal{Q}}$, and by $0$, the zero ideal in $\mathcal{Q}$.

###### Remark 2.1.3. —

One may immediately notice that this definition does not involve an any axiom
related to the operation $\oplus$. If $l\preccurlyeq x$, where
$l\in\mathcal{Q}$ and $x\in I$, then according to Lemma 2.1.2(1), we can
deduce that $l\mbox{{\small{$\&$}}}x\in I$. This implies that in the context
of rings, the property of an ideal mentioned above is automatically satisfied
in a quantale, whereas it needs to be explicitly defined as an axiom. In
simpler terms, the definitions of an ideal in a complete lattice and in a
quantale are identical.

Note that a definition of an ideal in a (commutative) quantale without
identity requires the following additional axiom:

(2.2) $\text{If}\;\;l\preccurlyeq
x,\;\text{where}\;l\in\mathcal{Q}\;\text{and}\;x\in
I,\;\text{and}\;l\mbox{{\small{$\&$}}}x\in I,$

(see [46, Definition 2.2] and [10, Definition 2(2)]). This is because we have
obtained (2.2) using the fact that $x\mbox{{\small{$\&$}}}y\preccurlyeq
x\wedge y$ holds for all $x,y\in\mathcal{Q}$, which we proved assuming the
existence of an identity element in $\mathcal{Q}$ (see Lemma 2.1.2(1)).

Based on Definition 2.1.2, we shall now introduce a set of operations on
ideals in a quantale. These operations play a fundamental role in the
development of our theory.

###### Definition 2.1.3. —

Let $\mathcal{Q}$ be a quantale.

  1. (1)

The _meet_ $\bigwedge\limits_{\lambda\in\Lambda}I_{\lambda}$ of ideals
$\\{I\\}_{\lambda\in\Lambda}$ in $\mathcal{Q}$ is given by their intersection.

  2. (2)

The _join_ of ideals $\\{I\\}_{\lambda\in\Lambda}$ of $\mathcal{Q}$ is defined
by

$\bigvee_{\lambda\in\Lambda}I_{\lambda}=\left\\{l\in\mathcal{Q}\mid
l\preccurlyeq\bigvee_{\text{finite}}x_{\lambda}\;\text{for}\;x_{\lambda}\in
I_{\lambda}\right\\}.$

  3. (3)

The _product_ of two ideals $I$ and $J$ in $\mathcal{Q}$ is defined as

$I\mbox{{\small{$\&$}}}J=\left\\{l\in\mathcal{Q}\mid
l\preccurlyeq\bigvee_{\text{finite}}\left\\{x\mbox{{\small{$\&$}}}y\mid x\in
I,y\in J\right\\}\right\\}.$

  4. (4)

The _residual_ of an ideal $I$ by an ideal $J$ is defined by

$(I:J)=\\{x\in\mathcal{Q}\mid x\mbox{{\small{$\&$}}}J\preccurlyeq
I\\}=\\{x\in\mathcal{Q}\mid x\mbox{{\small{$\&$}}}j\in I,\;\text{for
all}\;j\in J\\}.$

Through the operations defined above, we can establish that the resulting
entities are indeed ideals. This be demonstrated in the following lemma.

###### Lemma 2.1.4. —

Let $\mathcal{Q}$ be a quantale. If $I$ and $J$ are ideals in $\mathcal{Q}$,
then so are $(\mathrm{a})\;I\wedge J,$ $(\mathrm{b})\;I\vee J,$
$(\mathrm{c})\;I\mbox{{\small{$\&$}}}J$, and $(\mathrm{d})\;(I:J)$.

###### Proof.

(a) Since $I$ and $J$ are ideals in $\mathcal{Q}$, if $x,$ $y\in I\wedge J,$
then $x\vee y\in I\wedge J.$ Moreover, if $z\in I\wedge J$ then for any
$x\leqslant z$, $x\in I\wedge J.$ Thus $I\wedge J$ is an ideal.

(b) To show $I\vee J$ is an ideal, let $l,l^{\prime}\in I\vee J.$ This implies
that $l\preccurlyeq x\vee y$ and $l^{\prime}\preccurlyeq x^{\prime}\vee
y^{\prime}$ for some $x,$ $x^{\prime}\in I$ and $y,$ $y^{\prime}\in J$.
Therefore,

$l\vee l^{\prime}\preccurlyeq(x\vee y)\vee(x^{\prime}\vee y^{\prime})=(x\vee
x^{\prime})\vee(y\vee y^{\prime})\in I\vee J.$

Also, if $l\in I\vee J$ and $l^{\prime}\preccurlyeq l$, then
$l^{\prime}\preccurlyeq l\preccurlyeq x\vee y$ for some $x\in I$ and $y\in J$.
This implies that $l^{\prime}\in I\vee J$.

(c) The proof of $I\mbox{{\small{$\&$}}}J$ is an ideal is similar to (b).

(d) Finally, to show $(I:J)$ is an ideal, let $l,$ $l^{\prime}\in(I:J).$ Then
$l\mbox{{\small{$\&$}}}J\preccurlyeq I$ and
$l^{\prime}\mbox{{\small{$\&$}}}J\preccurlyeq I$. This by Definition 2.1.1(3)
implies that

$(l\vee
l^{\prime})\mbox{{\small{$\&$}}}J=(l\mbox{{\small{$\&$}}}J)\vee(l^{\prime}\mbox{{\small{$\&$}}}J)\preccurlyeq
I.$

Suppose $l\in(I:J)$ and $l^{\prime}\preccurlyeq l.$ Then by Lemma 2.1.2(2), we
have $l^{\prime}\mbox{{\small{$\&$}}}j\preccurlyeq l\mbox{{\small{$\&$}}}j$
for all $j\in J$. Hence $l^{\prime}\mbox{{\small{$\&$}}}J\preccurlyeq
l\mbox{{\small{$\&$}}}J\preccurlyeq I$. In other words, $l^{\prime}\in(I:J)$
as required. ∎

###### Definition 2.1.4. —

If $S$ is a nonempty subset in a $\mathcal{Q}$, then the _ideal generated by_
$S$ is defined by

$\langle S\rangle=\left\\{x\in\mathcal{Q}\mid
x\preccurlyeq\bigvee_{i=1}^{n}a_{i},\;\text{for
some}\;n\in\mathds{N}\;\text{and}\;a_{i}\in\mathcal{Q}\mbox{{\small{$\&$}}}S\right\\}.$

In particular, if $S=\\{s\\}$, then the ideal $\langle\\{s\\}\rangle$ is
called _principal_. We shall write $\langle s\rangle$ for
$\langle\\{s\\}\rangle$. The following lemma is going to be useful in the
sequel.

###### Lemma 2.1.5. —

Let $\mathcal{Q}$ be a quantale. If $S$ and $T$ are nonempty subsets of a
quantale $\mathcal{Q}$, then $\langle S\wedge T\rangle=\langle
S\rangle\wedge\langle T\rangle.$

###### Proof.

Since $S\wedge T\preccurlyeq S$ and $S\wedge T\preccurlyeq T,$ we immediately
have $\langle S\wedge T\rangle\preccurlyeq\langle S\rangle\wedge\langle
T\rangle.$ For teh converse, let $x\in\langle S\rangle\wedge\langle T\rangle.$
Then $x\preccurlyeq\bigvee\limits_{i=1}^{n}a_{i}$ and
$x\preccurlyeq\bigvee\limits_{j=1}^{m}b_{i}$, for some
$a_{i}\in\mathcal{Q}\mbox{{\small{$\&$}}}S$ and
$b_{j}\in\mathcal{Q}\mbox{{\small{$\&$}}}T$, where $1\leqslant i\leqslant n$
and $1\leqslant j\leqslant m$. This implies that
$x\preccurlyeq\left(\bigvee\limits_{i=1}^{n}a_{i}\right)\wedge\left(\bigvee\limits_{j=1}^{m}b_{i}\right),$
i.e., $x\in\langle S\wedge T\rangle.$ ∎

The algebraic manipulation of ideals in a quantale closely resembles that of a
commutative ring. We can draw analogies between the following ideal-theoretic
relations and their counterparts in the elementwise setting in a
multiplicative lattice (see [45, 11]).

###### Proposition 2.1.6. —

Let $I$, $\\{I_{\lambda}\\}_{\lambda\in\Lambda},$ $J$,
$\\{J_{\lambda}\\}_{\lambda\in\Lambda},$ and $K$ be ideals in a quantale
$\mathcal{Q}$. Then the following hold.

  1. (1)

$I\mbox{{\small{$\&$}}}(J\mbox{{\small{$\&$}}}K)=(I\mbox{{\small{$\&$}}}J)\mbox{{\small{$\&$}}}K$.

  2. (2)

$I\mbox{{\small{$\&$}}}J=J\mbox{{\small{$\&$}}}I$.

  3. (3)

$\mathcal{Q}\mbox{{\small{$\&$}}}I=I.$

  4. (4)

$0\mbox{{\small{$\&$}}}I=0.$

  5. (5)

$I\mbox{{\small{$\&$}}}\left(\bigvee\limits_{\lambda\in\Lambda}J_{\lambda}\right)=\bigvee\limits_{\lambda\in\Lambda}(I\mbox{{\small{$\&$}}}J_{\lambda}).$

  6. (6)

$I\mbox{{\small{$\&$}}}J\preccurlyeq I\wedge J$.

  7. (7)

$I\mbox{{\small{$\&$}}}(J\wedge
K)\preccurlyeq(I\mbox{{\small{$\&$}}}J)\wedge(I\mbox{{\small{$\&$}}}K).$

  8. (8)

$(I\vee J)\mbox{{\small{$\&$}}}(J\vee
K)\preccurlyeq(I\mbox{{\small{$\&$}}}J)\vee K.$

  9. (9)

If $I\vee K=J\vee K=\mathcal{Q}$, then $((I\mbox{{\small{$\&$}}}J)\vee
K)=\mathcal{Q}.$

  10. (10)

If $I\vee K=\mathcal{Q}$, then $(I\wedge J)\vee K=J\vee K.$

  11. (11)

$(I:J)\mbox{{\small{$\&$}}}J\preccurlyeq I.$

  12. (12)

$I\preccurlyeq(I:J).$

  13. (13)

$J\preccurlyeq I$ if and only if $(I:J)=\mathcal{Q}$.

  14. (14)

$(I:\mathcal{Q})=I.$

  15. (15)

$I\preccurlyeq((I\mbox{{\small{$\&$}}}J):J).$

  16. (16)

$\left(\bigwedge\limits_{\lambda\in\Lambda}I_{\lambda}:J\right)=\bigwedge\limits_{\lambda\in\Lambda}(I_{\lambda}:J).$

  17. (17)

$\bigwedge\limits_{\lambda\in\Lambda}(I:J_{\lambda})\leqslant\left(I:\bigvee\limits_{\lambda\in\Lambda}J_{\lambda}\right).$

  18. (18)

$((I:J):K)=(I:(J\mbox{{\small{$\&$}}}K)).$

  19. (19)

$(I:J)=(I:(I\vee J)).$

  20. (20)

$(I:J)=((I\wedge J):J).$

###### Proof.

(1) Follows from Definition 2.1.1(1).

(2) Follows from Definition 2.1.1(2).

(3) From Remark 2.1.3, it follows that
$\mathcal{Q}\mbox{{\small{$\&$}}}I\preccurlyeq I$. Indeed, $x\in\mathcal{Q},$
$i\in I$ implies $x\mbox{{\small{$\&$}}}i\preccurlyeq x\wedge i\preccurlyeq
i\in I$. By Definition 2.1.1(4),
$i=\top\mbox{{\small{$\&$}}}i\in\mathcal{Q}\mbox{{\small{$\&$}}}I$ for all
$i\in I$, proving that $I\preccurlyeq\mathcal{Q}\mbox{{\small{$\&$}}}I$.

(4) Let $i\in I$. Then $\bot\mbox{{\small{$\&$}}}i\preccurlyeq\bot\wedge
i=\bot\preccurlyeq i$ implies that $\bot\mbox{{\small{$\&$}}}i\in I$, and
hence, $0\mbox{{\small{$\&$}}}I\preccurlyeq I$. Since $\bot\preccurlyeq x$ for
all $x\in\mathcal{Q}$, in particular,
$\bot\preccurlyeq\bot\mbox{{\small{$\&$}}}i$ for all $i\in I$, showing that
$0\preccurlyeq 0\mbox{{\small{$\&$}}}I$.

(5) Follows from Definition 2.1.1(3).

(6) Follows from Remark 2.1.3.

(7) Since $J\wedge K\preccurlyeq J$ and $J\wedge K\preccurlyeq K,$ by Lemma
2.1.2(2), we obtain $I\mbox{{\small{$\&$}}}(J\wedge
K)\preccurlyeq(I\mbox{{\small{$\&$}}}J)\wedge(I\mbox{{\small{$\&$}}}K).$

(8) For all $i\in I,$ $j\in J$, and $k\in K$, applying Definition 2.1.1(3), we
have

$(i\vee k)\mbox{{\small{$\&$}}}(j\vee k)=(i\mbox{{\small{$\&$}}}j)\vee(i\vee
j\vee k)\mbox{{\small{$\&$}}}k\in(I\mbox{{\small{$\&$}}}J)\vee K.$

(9) From (8), it follows that
$\mathcal{Q}\preccurlyeq((I\mbox{{\small{$\&$}}}J)\vee K),$ whereas the other
inclusion is trivial.

(10) Since $I\wedge J\preccurlyeq J$, we have $(I\wedge J)\vee K\preccurlyeq
J\vee K$. For the other half of the inclusion, we have

$\displaystyle J\vee K$ $\displaystyle=(I\vee K)\mbox{{\small{$\&$}}}(J\vee
K)$ $\displaystyle=I\mbox{{\small{$\&$}}}(J\vee K)\vee
K\mbox{{\small{$\&$}}}(J\vee K)$ $\displaystyle\preccurlyeq
I\mbox{{\small{$\&$}}}(J\vee K)\vee K$
$\displaystyle=(I\mbox{{\small{$\&$}}}J)\vee((I\mbox{{\small{$\&$}}}K)\vee K)$
$\displaystyle=(I\mbox{{\small{$\&$}}}J)\vee K$
$\displaystyle\preccurlyeq(I\wedge J)\vee K.$

(11) If $x\in(I:J)\mbox{{\small{$\&$}}}J,$ then for some finite set
$\\{1,\ldots,n\\}$, we have
$x\preccurlyeq\bigvee\limits_{i=1}^{n}l_{i}\mbox{{\small{$\&$}}}j_{i}\in J$,
where $l_{i}\in\mathcal{Q}$ and $j_{i}\in J$.

(12) If $i\in I$, then by Lemma 2.1.2(2), $i\mbox{{\small{$\&$}}}J\preccurlyeq
I\mbox{{\small{$\&$}}}J$, and since $I$ is an ideal, by Remark 2.1.3,
$i\mbox{{\small{$\&$}}}J\preccurlyeq I$.

(13) If $J\preccurlyeq I$ then by applying (3),we have
$\mathcal{Q}\mbox{{\small{$\&$}}}J=J\preccurlyeq I$. Hence
$(I:J)=\mathcal{Q}$. Conversely, if $(I:J)=\mathcal{Q}$, then by Definition
2.1.1(4), $\top\mbox{{\small{$\&$}}}j=j\in I$ for all $j\in J$.

(14) Follows from (3).

(15) Follows by Lemma 2.1.2(2).

(16) The desired identity follows from the following chain of equivalent
statements:

$l\in\left(\bigwedge_{\lambda\in\Lambda}I_{\lambda}:J\right)\Leftrightarrow
l\mbox{{\small{$\&$}}}J\preccurlyeq\bigwedge_{\lambda\in\Lambda}I_{\lambda}\Leftrightarrow
l\mbox{{\small{$\&$}}}J\preccurlyeq
I_{\lambda},\forall\lambda\in\Lambda\Leftrightarrow
l\in\bigwedge_{\lambda\in\Lambda}(I_{\lambda}:J).$

(17) Note that

$\displaystyle l\in\bigwedge_{\lambda\in\Lambda}(I:J_{\lambda})\Rightarrow
l\mbox{{\small{$\&$}}}J_{\lambda}\subseteq
I,\forall\lambda\in\Lambda\Rightarrow\bigvee_{\lambda\in\Lambda}l\mbox{{\small{$\&$}}}J_{\lambda}\subseteq
I\Rightarrow
l\mbox{{\small{$\&$}}}\left(\bigvee_{\lambda\in\Lambda}J_{\lambda}\right)\subseteq
I\Rightarrow l\in(I:\bigvee_{\lambda\in\Lambda}J_{\lambda}).$

(18) Observe that $l\in((I:J):K)\Leftrightarrow
l\mbox{{\small{$\&$}}}K\preccurlyeq(I:J)\Leftrightarrow(l\mbox{{\small{$\&$}}}K)\mbox{{\small{$\&$}}}J\preccurlyeq
I\Leftrightarrow l\mbox{{\small{$\&$}}}((J\mbox{{\small{$\&$}}}K))\subseteq
I\Leftrightarrow l\in(I:J\mbox{{\small{$\&$}}}K).$

(19) Note that

$\displaystyle x\in(I:J)\Rightarrow x\mbox{{\small{$\&$}}}J\preccurlyeq
I\Rightarrow x\mbox{{\small{$\&$}}}I\vee x\mbox{{\small{$\&$}}}J\preccurlyeq
x\mbox{{\small{$\&$}}}I\vee I\Rightarrow x\mbox{{\small{$\&$}}}(I\vee
J)\preccurlyeq I\Rightarrow x\in(I:(I\vee J)),$

and for the other half of the inclusion,

$\displaystyle x\in(I:(I\vee J))\Rightarrow x\mbox{{\small{$\&$}}}(I\vee
J)\preccurlyeq I\Rightarrow x\mbox{{\small{$\&$}}}I\vee
x\mbox{{\small{$\&$}}}J\preccurlyeq I\Rightarrow
x\mbox{{\small{$\&$}}}J\preccurlyeq I.$

(20) Finally, we have $x\in(I:J)\Leftrightarrow
x\mbox{{\small{$\&$}}}J\preccurlyeq I\Leftrightarrow
x\mbox{{\small{$\&$}}}J\preccurlyeq I\wedge J\Leftrightarrow x\in((I\wedge
J):J).$ ∎

The well-known fact is that the set of ideals in a commutative ring with
identity forms a quantale, following the definition provided in Definition
2.1.1. Now, we shall proceed to demonstrate in the following theorem that the
set of ideals in a quantale possesses an analogous structure.

###### Theorem 2.1.7. —

If $\mathcal{Q}$ is a quantale, then
$(\mathcal{I}_{\mathcal{Q}},\preccurlyeq,\vee,\wedge,\bot,\top,\mbox{{\small{$\&$}}})$
is a quantale.

###### Proof.

By (1) and (2) of Definition 2.1.3, it follows that arbitrary joins and
arbitrary meets exist in the poset $(\mathcal{I}_{\mathcal{Q}},\preccurlyeq),$
where $\preccurlyeq$ is the usual subset inclusion relation. It is easy to see
that the bottom element $\bot$ and the top element $\top$ of the poset
$(\mathcal{I}_{\mathcal{Q}},\preccurlyeq)$ are respectively $0$ (the zero
ideal) and $\mathcal{Q}$. Hence $(\mathcal{I}_{\mathcal{Q}},\preccurlyeq)$ is
a complete lattice. Taking the multiplication operation between ideals in
$\mathcal{Q}$ as in Definition 2.1.3(3), it follows respectively from (1),
(2), (5), and (3) of Proposition 2.1.6 that the operation $\&$ satisfies
Axioms (1), (2), (3), and (4) of Definition 2.1.1. ∎

Similar to rings, the concept of an annihilator of a subset in a quantale is
also established as a special case of the residual operation.

###### Definition 2.1.5. —

If $S$ is a nonempty subset in a quantale $\mathcal{Q}$, then the _annihilator
of $S$_ is defined by

$\mathcal{A}_{\mathcal{Q}}(S)=\\{x\in\mathcal{Q}\mid
x\mbox{{\small{$\&$}}}s=\bot,\;\text{for all}\;s\in S\\}.$

In the following lemma, we discuss a few elementary properties of
annihilators.

###### Lemma 2.1.8. —

Let $S$ and $T$ be subsets in a quantale $\mathcal{Q}$. Then the following
hold.

  1. (1)

$S\preccurlyeq T$ implies that
$\mathcal{A}_{\mathcal{Q}}(T)\preccurlyeq\mathcal{A}_{\mathcal{Q}}(S).$

  2. (2)

$S\preccurlyeq\mathcal{A}_{\mathcal{Q}}(\mathcal{A}_{\mathcal{Q}}(S)).$

  3. (3)

$\mathcal{A}_{\mathcal{Q}}(S)=\mathcal{A}_{\mathcal{Q}}(\mathcal{A}_{\mathcal{Q}}(\mathcal{A}_{\mathcal{Q}}(S))).$

###### Proof.

(1) If $x\in\mathcal{A}_{\mathcal{Q}}(T),$ then $x\mbox{{\small{$\&$}}}t=\bot$
for all $t\in T$. Since $S\preccurlyeq T,$ this implies that
$x\mbox{{\small{$\&$}}}s=\bot$ for all $s\in S$. Hence
$\mathcal{A}_{\mathcal{Q}}(T)\preccurlyeq\mathcal{A}_{\mathcal{Q}}(S).$

(2) If $s\in S$, then $x\mbox{{\small{$\&$}}}s=s\mbox{{\small{$\&$}}}x=\bot$
for all $x\in\mathcal{A}_{\mathcal{Q}}(S),$ and hence
$s\in\mathcal{A}_{\mathcal{Q}}(\mathcal{A}_{\mathcal{Q}}(S)).$

(3) Since by (2),
$S\preccurlyeq\mathcal{A}_{\mathcal{Q}}(\mathcal{A}_{\mathcal{Q}}(S))$, by
(1), it follows that
$\mathcal{A}_{\mathcal{Q}}(\mathcal{A}_{\mathcal{Q}}(\mathcal{A}_{\mathcal{Q}}(S)))\preccurlyeq\mathcal{A}_{\mathcal{Q}}(S).$
By (2), it follows that
$\mathcal{A}_{\mathcal{Q}}(S)\preccurlyeq\mathcal{A}_{\mathcal{Q}}(\mathcal{A}_{\mathcal{Q}}(\mathcal{A}_{\mathcal{Q}}(S))).$
∎

###### Definition 2.1.6. —

Let $\mathcal{Q}$ be a quantale.

  1. (1)

An ideal $I$ of $\mathcal{Q}$ is called _proper_ if $I\neq\mathcal{Q}$, and we
shall denote the set of all proper ideals in $\mathcal{Q}$ by
$\mathcal{I}^{+}_{\mathcal{Q}}$.

  2. (2)

A proper ideal $M$ of $\mathcal{Q}$ is called _maximal_ if there is no other
proper ideal $I$ of $\mathcal{Q}$ properly containing $M$. A quantale
$\mathcal{Q}$ with exactly one maximal ideal is called _local_.

  3. (3)

An ideal $I$ of $\mathcal{Q}$ is called _minimal_ if $I\neq 0$ and $I$
properly contains no other nonzero ideals in $\mathcal{Q}$.

The subsequent result ensures that the set of maximal ideals in a quantale is
not empty.

###### Theorem 2.1.9. —

Every quantale $\mathcal{Q}$ with $0\neq 1$ has a maximal ideal.

###### Proof.

Since $0\in\mathcal{I}^{+}_{\mathcal{Q}}$,
$\mathrm{Prp}(\mathcal{Q})\neq\emptyset.$ Consider a chain
$\\{I_{\lambda}\\}_{\lambda\in\Lambda}$ of ideals in
$\mathcal{I}^{+}_{\mathcal{Q}}$. By Definition 2.1.3(1), it follows that
$\bigvee\limits_{\lambda\in\Lambda}I_{\lambda}\in\mathcal{I}^{+}_{\mathcal{Q}}$
and $I_{\lambda}\preccurlyeq\bigvee\limits_{\lambda\in\Lambda}I_{\lambda}$ for
all $\lambda\in\Lambda$. Hence, by Zorn’s lemma, $\mathrm{Prp}(\mathcal{Q})$
has a maximal element, which is our desired maximal ideal in $\mathcal{Q}$. ∎

###### Corollary 2.1.10. —

Every proper ideal in a quantale $\mathcal{Q}$ is contained in a maximal ideal
in $\mathcal{Q}$.

The following proposition presents a condition that is sufficient for a
quantale to be local.

###### Proposition 2.1.11. —

Let $\mathcal{Q}$ be a quantale and $M\neq\mathcal{Q}$ an ideal in
$\mathcal{Q}$ such that every $x\in\mathcal{Q}\neg M$ is a unit in
$\mathcal{Q}$. Then $\mathcal{Q}$ is local and $M$ is its maximal ideal.

###### Proof.

Let $I$ be an ideal in $\mathcal{Q}$ such that $M\subsetneq
I\preccurlyeq\mathcal{Q}.$ By hypothesis, every element $x\in I\neg M$ is a
unit and since $M\neq\mathcal{Q}$, we must have
$\top=x\mbox{{\small{$\&$}}}y\in I$ for some $y\in\mathcal{Q}$. Hence
$I=\mathcal{Q}$, showing that $M$ is maximal. Since every element of
$\mathcal{Q}\neg M$ is a unit, if $M^{\prime}$ be another maximal ideal in
$\mathcal{Q}$, then we must have

$(\mathcal{Q}\neg M)\wedge M^{\prime}=\emptyset,$

which implies that $M^{\prime}\preccurlyeq M$. But $M^{\prime}$ is a maximal
ideal. Therefore, $M^{\prime}=M.$ ∎

Our next objective is to examine the properties of ideals and their products,
meet, and residual operations under quantale homomorphisms. These properties
serve as the lattice-theoretic counterparts to their respective ring-theoretic
versions (see [6, Proposition 1.17 and Exercise 1.18]). To this end, recall
that a map $\phi\colon\mathcal{Q}\to\mathcal{Q}^{\prime}$ from $\mathcal{Q}$
to $\mathcal{Q}^{\prime}$ is called a _quantale homomorphism_ if

  1. (1)

$x\preccurlyeq x^{\prime}$ implies that $\phi(x)\preccurlyeq\phi(x^{\prime});$

  2. (2)

$\phi(x\vee x^{\prime})=\phi(l)\vee\phi(l^{\prime});$

  3. (3)

$\phi(x\wedge x^{\prime})=\phi(x)\wedge\phi(x^{\prime});$

  4. (4)

$\phi(x\mbox{{\small{$\&$}}}x^{\prime})=\phi(x)\mbox{{\small{$\&$}}}\phi(x^{\prime}),$

for all $x,$ $x^{\prime}\in\mathcal{Q}$. Suppose that
$\phi\colon\mathcal{Q}\to\mathcal{Q}^{\prime}$ is a quantale homomorphism. If
$J$ is an ideal in $\mathcal{Q}^{\prime}$, then the _contraction of_ $J$,
denoted by $J^{c}$, is defined by
$\phi^{\raisebox{0.60275pt}{$\scriptscriptstyle-1$}}(J).$ If $I$ is an ideal
in $\mathcal{Q}$, then the _extension of_ $I$, denoted by $I^{e}$, is defined
by $\langle\phi(I)\rangle.$

###### Theorem 2.1.12. —

Let $\phi\colon\mathcal{Q}\to\mathcal{Q}^{\prime}$ be a quantale homomorphism.
For $I,$ $I_{1}$, $I_{2}\in\mathcal{I}_{\mathcal{Q}}$, and $J,$ $J_{1}$,
$J_{2}\in\mathcal{I}(\mathcal{Q}^{\prime})$, the following hold.

  1. (1)

$J^{c}$ is an ideal in $\mathcal{Q}$.

  2. (2)

$I^{e}$ is an ideal in $\mathcal{Q}^{\prime}$.

  3. (3)

$\mathrm{(a)}\;I\preccurlyeq I^{ec}$. $\mathrm{(b)}\;J^{ce}\preccurlyeq J$.
$\mathrm{(c)}\;J^{c}=J^{cec}$. $\mathrm{(d)}\;I^{e}=I^{ece}.$

  4. (4)

There is a bijection between the sets $\\{I\mid I^{ec}=I\\}$ and $\\{J\mid
J^{ce}=J\\}$.

  5. (5)

$\mathrm{(a)}\;(I_{1}\wedge I_{2})^{e}\preccurlyeq I_{1}^{e}\wedge I_{2}^{e}$.
$\mathrm{(b)}\;(I_{1}\mbox{{\small{$\&$}}}I_{2})^{e}=I_{1}^{e}\mbox{{\small{$\&$}}}I_{2}^{e}$.
$\mathrm{(c)}\;(I_{1}:I_{2})^{e}\preccurlyeq(I_{1}^{e}:I_{2}^{e})$.

  6. (6)

$\mathrm{(a)}\;(J_{1}\wedge J_{2})^{c}=J_{1}^{c}\wedge J_{2}^{c}$.
$\mathrm{(b)}\;J_{1}^{c}\mbox{{\small{$\&$}}}J_{2}^{c}\preccurlyeq(J_{1}\mbox{{\small{$\&$}}}J_{2})^{c}$.
$\mathrm{(c)}\;(J_{1}:J_{2})^{c}\preccurlyeq(J_{1}^{c}:J_{2}^{c})$.

###### Proof.

(1) Let $x,$ $x^{\prime}\in J^{c}$. Then $\phi(x),$ $\phi(x^{\prime})\in J$.
Since $J$ is an ideal, we have $\phi(x\vee
x^{\prime})=\phi(x)\vee\phi(x^{\prime})\in J,$ implying that $x\vee
x^{\prime}\in J^{c}$. If $x^{\prime}\preccurlyeq x$ and $x\in J^{c}$, then
$\phi(x^{\prime})\preccurlyeq\phi(x)\in J$, proving that $x^{\prime}\in
J^{c}$, as required.

(2) Follows from Definition 2.1.4.

(3)–(4) Follows by usual set-theoretic arguments.

5(a) Observe that

$(I_{1}\wedge I_{2})^{e}=\langle\phi(I_{1}\wedge
I_{2})\rangle\preccurlyeq\langle\phi(I_{1})\wedge\phi(I_{2})\rangle=\langle\phi(I_{1})\rangle\wedge\langle\phi(I_{2})\rangle=I_{1}^{e}\wedge
I_{2}^{e},$

where the second equality follows from Lemma 2.1.5.

5(b) First, we show that
$\langle\phi(I_{1}\mbox{{\small{$\&$}}}I_{2})\rangle=\langle\phi(I_{1})\rangle\mbox{{\small{$\&$}}}\langle\phi(I_{2})\rangle.$
Since $I_{1}\mbox{{\small{$\&$}}}I_{2}\preccurlyeq I_{1}$ and
$I_{1}\mbox{{\small{$\&$}}}I_{2}\preccurlyeq I_{2},$ we immediately obtain
that
$\langle\phi(I_{1}\mbox{{\small{$\&$}}}I_{2})\rangle\preccurlyeq\langle\phi(I_{1})\rangle\mbox{{\small{$\&$}}}\langle\phi(I_{2})\rangle.$
For the other inclusion, let
$x\in\langle\phi(I_{1})\rangle\mbox{{\small{$\&$}}}\langle\phi(I_{2})\rangle.$
This implies $x=x_{1}\mbox{{\small{$\&$}}}x_{2}$ for some $x_{1},$
$x_{2}\in\mathcal{Q}$ such that
$x_{1}\preccurlyeq\bigvee\limits_{i=1}^{n}a_{i}$ and
$x_{2}\preccurlyeq\bigvee\limits_{j=1}^{m}b_{j},$ where
$a_{i}\in\mathcal{Q}\mbox{{\small{$\&$}}}\phi(I_{1})$ and
$b_{j}\in\mathcal{Q}\mbox{{\small{$\&$}}}\phi(I_{2})$ for all $1\leqslant
i\leqslant n$ and $1\leqslant j\leqslant m$. So,

$x=x_{1}\mbox{{\small{$\&$}}}x_{2}\preccurlyeq\left(\bigvee_{i=1}^{n}a_{i}\right)\mbox{{\small{$\&$}}}\left(\bigvee_{j=1}^{m}b_{j}\right)=\bigvee_{i=1}^{n}\bigvee_{j=1}^{m}a_{i}\mbox{{\small{$\&$}}}b_{j},$

where

$a_{i}\mbox{{\small{$\&$}}}b_{j}\in\mathcal{Q}\mbox{{\small{$\&$}}}\phi(I_{1})\mbox{{\small{$\&$}}}\mathcal{Q}\mbox{{\small{$\&$}}}\phi(I_{2})=\mathcal{Q}\mathcal{Q}\mbox{{\small{$\&$}}}\phi(I_{1})\mbox{{\small{$\&$}}}\phi(I_{2})=\mathcal{Q}\mbox{{\small{$\&$}}}\phi(I_{1})\mbox{{\small{$\&$}}}\phi(I_{2}).$

Therefore, $x\in\langle\phi(I_{1}\mbox{{\small{$\&$}}}I_{2})\rangle.$ Now,
$(I_{1}\mbox{{\small{$\&$}}}I_{2})^{e}=\langle\phi(I_{1}\mbox{{\small{$\&$}}}I_{2})\rangle=\langle\phi(I_{1})\rangle\mbox{{\small{$\&$}}}\langle\phi(I_{2})\rangle=I_{1}^{e}\mbox{{\small{$\&$}}}I_{2}^{e}.$

5(c) By Proposition 2.1.6(11) and 5(b), we obtain

$(I_{1}:I_{2})^{e}\mbox{{\small{$\&$}}}I_{2}^{e}=((I_{1}:I_{2})\mbox{{\small{$\&$}}}I_{2})^{e}\preccurlyeq
I_{1}^{e}.$

6\. The proofs are similar to (5). ∎

###  2.2. Prime and semiprime ideals

In this section, our objective is to present the concept of prime and
semiprime ideals in the context of a quantale. We shall demonstrate that their
definitions based on elements are equivalent to the definitions rooted in
ideals. Additionally, we shall explore the topic of radical ideals and delve
into their some basic properties.

###### Definition 2.2.1. —

Let $\mathcal{Q}$ be a quantale.

  1. (1)

A proper ideal $P$ of $\mathcal{Q}$ is called _prime_ if
$x\mbox{{\small{$\&$}}}y\in P$ implies $x\in P$ or $y\in P$ for all $x,$
$y\in\mathcal{Q}$. By $\mathrm{Spec}_{\mathcal{Q}}$, we denote the set of all
prime ideals in $\mathcal{Q}$. An ideal $P$ of $\mathcal{Q}$ is called
_minimal prime_ if $P$ is both a minimal ideal and a prime ideal.

  2. (2)

A proper ideal $P$ of $\mathcal{Q}$ is called _semiprime_ if $x^{2}\in P$
implies that $x\in P$ for all $x\in\mathcal{Q}$.

  3. (3)

Two ideals $I$ and $J$ of $\mathcal{Q}$ is said to be _coprime_ if $I\vee
J=\mathcal{Q}$.

  4. (4)

$\mathcal{Q}$ is called _Noetherian_ if every ascending chain of ideals in
$\mathcal{Q}$ is eventually stationary.

###### Proposition 2.2.1. —

Let $\mathcal{Q}$ be quantale.

  1. (1)

An ideal $P\in\mathrm{Spec}_{\mathcal{Q}}$ if and only if
$I\mbox{{\small{$\&$}}}J\preccurlyeq P$ implies $I\preccurlyeq P$ or
$J\preccurlyeq P,$ for all $I,J\in\mathcal{I}_{\mathcal{Q}}$.

  2. (2)

Any prime ideal $P$ of $\mathcal{Q}$ contains a prime ideal $P^{\prime}$ such
that $S\preccurlyeq P^{\prime}$.

  3. (3)

If $I$ is a proper ideal in a Noetherian quantale $\mathcal{Q}$, then
$\mathcal{Q}$ has only a finite number of minimal prime ideals over $I$.

  4. (4)

An ideal $I$ of $\mathcal{Q}$ is semiprime if and only if
$J\mbox{{\small{$\&$}}}J\preccurlyeq I$ implies $J\preccurlyeq I,$ for all
$J\in\mathcal{I}_{\mathcal{Q}}.$

  5. (5)

Let $\mathcal{Q}$ be a quantale. If $I$ and $J$ are coprime ideals in
$\mathcal{Q}$, then $I\wedge J=IJ.$

###### Proof.

(1) Suppose $P$ is a prime ideal, and $I\mbox{{\small{$\&$}}}J\preccurlyeq P$
with $J\neq P$. This implies the existence of an element $j\in J$ such that
$j\notin P$. For any $i\in I$, we then have $i\mbox{{\small{$\&$}}}j\in P$ and
based on the assumption, this implies $i\in P$. Since $i$ was chosen
arbitrarily from $I$, we conclude that $I\preccurlyeq P$, as required.
Conversely, suppose $I\mbox{{\small{$\&$}}}J\preccurlyeq P$ implies either
$I\preccurlyeq P$ or $J\preccurlyeq P$. Now consider
$x\mbox{{\small{$\&$}}}y\in P$ for some $x,$ $y\in\mathcal{Q}$. Let
$a\mbox{{\small{$\&$}}}b\in\langle x\rangle\mbox{{\small{$\&$}}}\langle
y\rangle,$ where $a\in\langle x\rangle$ and $b\in\langle y\rangle$. This
implies that

$a\mbox{{\small{$\&$}}}b\preccurlyeq\left(\bigvee_{i=1}^{n}x_{i}\right)\mbox{{\small{$\&$}}}\left(\bigvee_{j=1}^{m}y_{j}\right)=\bigvee_{i=1}^{n}\left(\bigvee_{j=1}^{m}(x_{i}\mbox{{\small{$\&$}}}y_{j})\right),$

where
$x_{i}\mbox{{\small{$\&$}}}y_{j}\in(x\mbox{{\small{$\&$}}}y)\mathcal{Q},$ for
all $i=1,\ldots,n$ and $j=1,\ldots,m.$ Thus,
$x_{i}\mbox{{\small{$\&$}}}y_{j}\in P$, and hence $a\mbox{{\small{$\&$}}}b\in
P$, implying that $\langle x\rangle\mbox{{\small{$\&$}}}\langle
y\rangle\preccurlyeq P.$ By the assumption, this implies either $\langle
x\rangle\preccurlyeq P$ or $\langle y\rangle\preccurlyeq P;$ in other words,
either $x\in P$ or $y\in P$.

(2) Suppose $\Omega=\\{P^{\prime}\in\mathrm{Spec}_{\mathcal{Q}}\mid
S\preccurlyeq P^{\prime}\preccurlyeq P\\}.$ Since $P\in\Omega$, the set
$\Omega$ is nonempty. Consider a subset
$\\{P^{\prime}_{\lambda}\\}_{\lambda\in\Lambda}$ of decreasing chain of prime
ideals of $\Omega$. Then by Zorn’s lemma, there exists a minimal element
$\bigwedge\limits_{\lambda\in\Lambda}P^{\prime}_{\lambda}$ of that chain, and
this minimal element is our desired prime ideal.

(3) First we show that every ideal containing $I$ contains a finite product of
prime ideals each containing $I$. Suppose that this is not the case, and let
$S$ be the set of ideals containing $I$ which do not contain a finite product
of prime ideals each containing $I$. By hypothesis, $S$ is not empty. Let $C$
be a chain in $S$. As $\mathcal{Q}$ is Noetherian, $C$ has a maximum element.
From Zorn’s lemma, $S$ has a maximal element $M$. As $\mathcal{Q}$ has a prime
ideal containing $I$, $M\neq\mathcal{Q}$. Also, $M$ is not prime. There exist
$a$, $b\in\mathcal{Q}$ such that $a\mbox{{\small{$\&$}}}b\in M$ and $a\notin
M$, $b\notin M$. Setting

$A=\langle M,a\rangle,\quad B=\langle M,b\rangle,$

we obtain $A\mbox{{\small{$\&$}}}B\preccurlyeq M$, with $A$ and $B$ strictly
included in $M$. Since $M$ is maximal, $A$ and $B$ both contain a finite
product of prime ideals each containing $I$, hence so does $M$, which
contradicts the fact that $M\in S$. It follows that every ideal containing I
contains a finite product of prime ideals each containing $I$. We now apply
this result to the ideal $I$: there exist prime ideals $P_{1},\ldots,P_{n}$,
each containing $I$, whose product is contained in $I$. We claim that any
minimal prime $P$ over $I$ is among the $P_{i}$. Indeed,

$P_{1}\mbox{{\small{$\&$}}}\cdots\mbox{{\small{$\&$}}}P_{n}\preccurlyeq
I\preccurlyeq P.$

We deduce that $P_{i}\preccurlyeq P$, for some $i$. However, $P$ is minimal,
so $P_{i}=P$ and it follows that there is only a finite number of minimal
prime ideals over $I$.

(4) Suppose $I$ is a semiprime ideal in $\mathcal{Q}$. Let $j\in J$. Then
$j^{2}\in J\mbox{{\small{$\&$}}}J\preccurlyeq I$. This implies that $j^{2}\in
I$ and since $I$ is semirpime, $j\in I$, i.e., $J\preccurlyeq I$. For the
converse, we first show that $\langle x\rangle\mbox{{\small{$\&$}}}\langle
x\rangle\preccurlyeq\langle x^{2}\rangle$, for any $x\in\mathcal{Q}$. Suppose
$l\in\langle x\rangle\mbox{{\small{$\&$}}}\langle x\rangle.$ In particular,
this implies that $l\preccurlyeq x^{2},$ and hence $l\in\langle x^{2}\rangle.$
Now, let $x^{2}\in I$. Then

$\langle x\rangle\mbox{{\small{$\&$}}}\langle x\rangle\preccurlyeq\langle
x^{2}\rangle\preccurlyeq I.$

By assumption, this means $x\in\langle x\rangle\preccurlyeq I.$ Hence, $I$ is
a semiprime ideal.

(5) Notice that

$(I\vee J)\mbox{{\small{$\&$}}}(I\wedge J)=I\mbox{{\small{$\&$}}}(I\wedge
J)\vee J(I\wedge
J)\preccurlyeq(I\wedge(I\mbox{{\small{$\&$}}}J))\vee((I\mbox{{\small{$\&$}}}J)\wedge
J)\preccurlyeq I\mbox{{\small{$\&$}}}J,$

where the equality is obtained by Definition 2.1.1(3), and the first inclusion
follows from Proposition 2.1.6(7). If $I\vee J=\mathcal{Q},$ then by
Proposition 2.1.6(3), $\mathcal{Q}(I\wedge J)=I\wedge J\preccurlyeq
I\mbox{{\small{$\&$}}}J$. ∎

The following proposition is adapted from the realm of rings (see [5, Lemma
3.19]).

###### Proposition 2.2.2 (Prime avoidance lemma). —

Let $I$ be a subset in a quantale $\mathcal{Q}$ that is stable under join and
the operation $\&$. Let $P_{1},\ldots,P_{n}$ be ideals in $\mathcal{Q}$ such
that $P_{3},\ldots,P_{n}\in\mathrm{Spec}_{\mathcal{Q}}$. If $I\neq P_{j}$ for
all $j,$ then there is an $x\in I$ such that $x\notin P_{j}$ for all $j$.

###### Proof.

We prove the claim by induction. If $n=1$, the claim is trivially true.
Suppose $n\geqslant 2$ and suppose for every $i$, there exists an $x_{i}\in I$
such that $x_{i}\notin P_{j}$ for all $j\neq i$. Assume that $x_{i}\in P_{i}$
for all $i$. If $n=2$, then $x_{1}\vee x_{2}\notin P_{j}$ for $j=1,$ $2$.
Indeed, $x_{2}\preccurlyeq x_{1}\vee x_{2}\in P_{1}$ implies $x_{2}\in P_{1},$
a contradiction, and similarly, $x_{1}\preccurlyeq x_{1}\vee x_{2}\in P_{2}$
implies $x_{1}\in P_{2}$, a contradiction. For $n\geqslant 3$,

$(x_{1}\mbox{{\small{$\&$}}}\cdots\mbox{{\small{$\&$}}}x_{n-1})\vee
x_{n}\notin P_{j}.$

For the case, $n=j,$ we have $x_{n}\in P_{n}$ and since $P_{n}$ is a prime
ideal, $x_{k}\in P$ for some $k\in\\{1,\ldots,n-1\\},$ a contradiction. For,
$j<n$, $x_{n}\in P_{j}$, again leads to a contradiction. ∎

Our following aim is to explore the radicals of ideals and their associations
with prime and semiprime ideals.

###### Definition 2.2.2. —

The _radical_ of an ideal $I$ in a quantale $\mathcal{Q}$ is defined as
follows:

$\mathcal{R}(I)=\\{x\in\mathcal{Q}\mid x^{n}\in I,\;\text{for
some}\;n\in\mathds{N}\\}.$

An ideal $I$ is said to be _radical ideal_ if $\mathcal{R}(I)=I.$

The following proposition gives an equivalent definition of radical of an
ideal and it extends Proposition 1.14 of [6].

###### Proposition 2.2.3. —

If $I$ is an ideal in a quantale $\mathcal{Q}$, then
$\mathcal{R}(I)=\bigwedge\limits_{P}\\{P\in\mathrm{Spec}_{\mathcal{Q}}\mid
I\preccurlyeq P\\}.$

###### Proof.

Let $l\in\mathcal{R}(I).$ Then there exists $n\in\mathds{N}$ such that
$l^{n}\in I$, and also $l^{n}\in P,$ for all $P\in\mathrm{Spec}_{\mathcal{Q}}$
with $I\preccurlyeq P$. This implies that $l\in P$ for all such
$P\in\mathrm{Spec}_{\mathcal{Q}}.$ Hence
$l\in\bigwedge\limits_{P}\left\\{P\in\mathrm{Spec}_{\mathcal{Q}}\mid
I\preccurlyeq P\right\\}.$ Conversely, assume that $l\notin\mathcal{R}(I),$
for some $l\in\mathcal{Q}$. Consider the set

$\Omega=\\{J\in\mathcal{I}_{\mathcal{Q}}\mid I\preccurlyeq
J\;\text{and}\;l^{n}\notin J,\forall n\in\mathds{N}\\}.$

It is easy to see that $\Omega$ is nonempty and by Zorn’s lemma
$(\Omega,\leqslant)$ has a maximal element, say $P$ such that $I\preccurlyeq
P$. It suffices to show that $P$ is a prime ideal. Suppose $x,$ $y\notin P$
for some $x,$ $y\in\mathcal{Q}$. This implies $\langle x,P\rangle\notin\Omega$
$\langle x,P\rangle\notin\Omega,$ which followingly implies $\langle
x\mbox{{\small{$\&$}}}y,P\rangle\notin\Omega$. Hence
$x\mbox{{\small{$\&$}}}y\notin P$. ∎

In the next lemma, we compile some elementary properties of the radical of an
ideal in a quantale.

###### Lemma 2.2.4. —

For any ideals $I$, $J$, $\\{I_{\lambda}\\}_{\lambda\in\Lambda}$ in a quantale
$\mathcal{Q}$, the following hold.

  1. (1)

$\mathcal{R}(I)$ is an ideal containing $I$.

  2. (2)

If $I\preccurlyeq J$, then $\mathcal{R}(I)\preccurlyeq\mathcal{R}(J).$

  3. (3)

$\mathcal{R}(\mathcal{R}(I))=\mathcal{R}(I).$

  4. (4)

$\mathcal{R}(I)=\mathcal{R}(I\mbox{{\small{$\&$}}}\cdots\mbox{{\small{$\&$}}}I)$
$($repeated $n$-times$)$.

  5. (5)

$\mathcal{R}(I\wedge
J)=\mathcal{R}(I)\wedge\mathcal{R}(J)=\mathcal{R}(I\mbox{{\small{$\&$}}}J).$

  6. (6)

$\bigvee\limits_{\lambda\in\Lambda}\mathcal{R}(I_{\lambda})\preccurlyeq\mathcal{R}\left(\bigvee\limits_{\lambda\in\Lambda}I_{\lambda}\right).$

  7. (7)

$\mathcal{R}(I)=\mathcal{Q}$ if and only if $I=\mathcal{Q}.$

  8. (8)

$\mathcal{R}(I\vee J)=\mathcal{R}(\mathcal{R}(I)\vee\mathcal{R}(J)).$

###### Proof.

(1) By taking $n=1$, it follows that $I\preccurlyeq\mathcal{R}(I).$ To show
$\mathcal{R}(I)$ is an ideal, let $x\in\mathcal{R}(I)$ and let $y\preccurlyeq
x$, for some $y\in\mathcal{Q}$. Then $x^{n}\in I$ for some $n\in\mathds{N}.$
By Lemma 2.1.2(2), this implies $y^{n}\preccurlyeq x^{n}$, and since $I$ is an
ideal, we obtain that $y^{n}\in I$. Hence $y\in\mathcal{R}(I),$ showing that
the condition (2) of Definition 2.1.2 holds. To check condition (1) of
Definition 2.1.2, let $x,$ $y\in\mathcal{R}(I)$. Then $x^{n},$
$y^{m}\in\mathcal{R}(I)$ for some $n$, $m\in\mathds{N}$. It suffices to show
that $(x\vee y)^{m+n}\in I.$ Applying formula (2.1), we obtain

$(x\vee
y)^{m+n}=\bigvee_{k=0}^{m+n}\binom{m+n}{k}\,x^{m+n-k}\mbox{{\small{$\&$}}}y^{k}.$

Since $0\leqslant k\leqslant m+n$, we have two possibilities: either
$n\leqslant i$, in which case, $x^{i}\in I$ (since $x^{n}\in I$); or
$m\leqslant n+m-i,$ in which case $y^{n+m-i}\in I$ (since $y^{m}\in I$). This
proves that $(x\vee y)^{m+n}\in I.$

(2) Follows from Proposition 2.2.3.

(3) Applying 2. on $I\preccurlyeq\mathcal{R}(I)$ (which follows from (1))
gives $\mathcal{R}(I)\preccurlyeq\mathcal{R}(\mathcal{R}(I)).$ Conversely,
suppose that $x\in\mathcal{R}(\mathcal{R}(I)).$ This implies
$x^{n}\in\mathcal{R}(I)$ for some $n\in\mathds{N}$, which further implies that
$(x^{n})^{m}\in I,$ for some $m\in\mathds{N}$. However, $(x^{n})^{m}=x^{nm}\in
I.$ Hence $x\in\mathcal{R}(I).$

(4) Since $I\mbox{{\small{$\&$}}}\cdots\mbox{{\small{$\&$}}}I\preccurlyeq I$,
by (2), we have
$\mathcal{R}(I\mbox{{\small{$\&$}}}\cdots\mbox{{\small{$\&$}}}I)\preccurlyeq\mathcal{R}(I).$
If $x\in\mathcal{R}(I)$, then $x^{m}\in I$, for some $m\in\mathds{N}$. But
then $x^{mn}=(x^{m})^{n}\in
I\mbox{{\small{$\&$}}}\cdots\mbox{{\small{$\&$}}}I,$ implying that
$x\in\mathcal{R}(I\mbox{{\small{$\&$}}}\cdots\mbox{{\small{$\&$}}}I).$

(5) For the first equality, $\mathcal{R}(I\wedge
J)\preccurlyeq\mathcal{R}(I)\wedge\mathcal{R}(J)$ follows from (2). If
$x\in\mathcal{R}(I)\wedge\mathcal{R}(J),$ then $x^{n}\in\mathcal{R}(I)$ and
$x^{m}\in\mathcal{R}(J)$ for some $n,$ $m\in\mathds{N}$. Let
$k=\max\\{n,m\\}.$ Then $x^{k}\in I\wedge J,$ and hence
$x\in\mathcal{R}(I\wedge J).$ For the second equality, note that

$(I\wedge J)\mbox{{\small{$\&$}}}(I\wedge J)\preccurlyeq
I\mbox{{\small{$\&$}}}J\preccurlyeq I\wedge J.$

Therefore, by (2) and (4), we obtain

$\mathcal{R}((I\wedge J)\mbox{{\small{$\&$}}}(I\wedge J))=\mathcal{R}(I\wedge
J)\preccurlyeq\mathcal{R}(I\mbox{{\small{$\&$}}}J)\preccurlyeq\mathcal{R}(I\wedge
J).$

(6) Follows from (2).

(7) Follows from Proposition 2.2.3.

(8) By (2) and (3), it follows that

$\mathcal{R}(\mathcal{R}(I)\vee\mathcal{R}(J))\preccurlyeq\mathcal{R}(\mathcal{R}(I\vee
J))=\mathcal{R}(I\vee J).$

Since by (1), $I\preccurlyeq\mathcal{R}(I)$ and $J\preccurlyeq\mathcal{R}(J),$
we obtain $I\vee J\preccurlyeq\mathcal{R}(I)\vee\mathcal{R}(J),$ and applying
(2), we get the desired inclusion. ∎

Expanding upon the previously introduced definition of the radical of an ideal
in a quantale, we shall now introduce two additional types of radicals
specific to quantales: nilradicals and Jacobson radicals. Additionally, we
shall explore the relationships that exist among these distinct types of
radicals.

###### Definition 2.2.3. —

Let $\mathcal{Q}$ be a quantale.

  1. (1)

An element $x$ of $\mathcal{Q}$ is called _nilpotent_ if $x^{n}=\bot$ for some
positive integer $n$. The set $\mathcal{N}(\mathcal{Q})$ of all nilpotent
elements of a quantale $\mathcal{Q}$ is called the _nilradical_ of
$\mathcal{Q}$. A quantale $\mathcal{Q}$ is called _reduced_ if
$\mathcal{N}(\mathcal{Q})=0$

  2. (2)

The _Jacobson radical_ $\mathcal{J}(\mathcal{Q})$ of a quantale $\mathcal{Q}$
is defined as the intersection of all maximal ideals in $\mathcal{Q}$.

  3. (3)

A _zero-divisor_ of $\mathcal{Q}$ is an element $x$ of $\mathcal{Q}$ for which
there exists an element $y$ ($\neq\bot$) in $\mathcal{Q}$ such that
$x\mbox{{\small{$\&$}}}y=\bot$. If $\bot\neq\top$ in $\mathcal{Q}$ and if
$\mathcal{Q}$ does not have any nonzero zero-divisor, then $\mathcal{Q}$ is
called a _quantale domain_ (QD). Note that a quantale $\mathcal{Q}$ is a QD if
and only if $0$ is a prime ideal.

###### Proposition 2.2.5. —

Let $\mathcal{Q}$ be a quantale. Then the following hold.

  1. (1)

$\mathcal{N}(\mathcal{Q})\in\mathcal{I}_{\mathcal{Q}}$ and it is the
intersection of prime ideals in $\mathcal{Q}$. Moreover,
$\mathcal{N}(\mathcal{Q})\preccurlyeq\mathcal{J}(\mathcal{Q}).$

  2. (2)

$\mathcal{Q}$ is reduced and has only one minimal prime ideal if and only if
$\mathcal{Q}$ is a QD.

###### Proof.

(1) First we show $\mathcal{N}(\mathcal{Q})$ is an ideal in $\mathcal{Q}$.
Suppose $x\in\mathcal{N}(\mathcal{Q})$ and $y\preccurlyeq x$. Then $x^{n}=0$
for some $n\in\mathds{N}$ and by Lemma 2.1.2(3), $y^{n}\preccurlyeq x^{n}=0$,
implying that $y^{n}=0$, and hence, $y\in\mathcal{N}(\mathcal{Q}).$ Now, let
$x,$ $y\in\mathcal{N}(\mathcal{Q})$. This implies $x^{n}=0$ and $y^{m}=0$ for
some $n,$ $m\in\mathds{N}$. Observe that $(x\vee y)^{m+n-1}$ is the join of
integer multiplies of elements $x^{r}\mbox{{\small{$\&$}}}y^{s}$ (see proof of
Lemma 2.2.4(1)), where $r+s=m+n-1$. Since we cannot have both $r<m$ and $s<n$,
each of these terms vanishes, and hence

$(x\vee y)^{m+n-1}=0.$

The fact that $\mathcal{N}(\mathcal{Q})$ is the intersection of prime ideals
in $\mathcal{Q}$ follows from Proposition 2.2.3. Suppose
$x\in\mathcal{N}(\mathcal{Q}).$ Then there exists $n\geqslant 1$ such that
$x^{n}=\bot$. Let $M$ be a maximal ideal in $\mathcal{Q}$. Then $x^{n}\in M$
and $M$ is prime, $x\in M$, proving that
$\mathcal{N}(\mathcal{Q})\preccurlyeq\mathcal{J}(\mathcal{Q})$.

(2) Suppose $\mathcal{Q}$ is reduced. Then $\mathcal{N}(\mathcal{Q})=0$. By
Proposition 2.2.3, this implies

$0=\bigwedge_{P\in\mathrm{Spec}_{\mathcal{Q}}}P.$

Then by Lemma 2.2.1(2) and hypothesis, there exists a prime ideal $P$ such
that $P=0$. Hence $\mathcal{Q}$ is a QD. The converse follows from the fact
that $\mathcal{Q}$ is a QD if and only if $0$ is a prime ideal in
$\mathcal{Q}$. ∎

Our next objective is to establish the equivalence between semiprime ideals
and radical ideals within a quantale. This equivalence is well-known in the
realm of (noncommutative) rings. However, in the noncommutative scenario, the
concepts of $m$-systems and $n$-systems in rings are necessary. On the other
hand, when dealing with ideals in quantale (or commutative rings), the
presence of multiplicatively closed subsets alone suffices. To demonstrate
this equivalence, we shall first proceed by presenting a series of lemmas.

###### Lemma 2.2.6. —

An ideal $P$ in a quantale $\mathcal{Q}$ is prime if and only if the
complement of $P$ in $\mathcal{Q}$ forms a multiplicatively closed subset of
$\mathcal{Q}$. Here, a subset $S$ of $\mathcal{Q}$ is considered
multiplicatively closed if it satisfies two conditions: $\top\in S$, and for
any $x\in S$ and $y\in S$, implies $x\mbox{{\small{$\&$}}}y\in S$.

###### Proof.

Suppose that $P$ is a prime ideal in $\mathcal{Q}$ and $x,$
$y\in\mathcal{Q}\neg P$. Then by Lemma 2.2.1(1),
$x\mbox{{\small{$\&$}}}y\notin P$, and hence
$x\mbox{{\small{$\&$}}}y\in\mathcal{Q}\neg P.$ Conversely, suppose $x,$
$y\notin P$. This implies that $x,$ $y\in\mathcal{Q}\neg P$. Since
$\mathcal{Q}\neg P$ is multiplicatively closed,
$x\mbox{{\small{$\&$}}}y\in\mathcal{Q}\neg P$, and hence
$x\mbox{{\small{$\&$}}}y\notin P$, proving that $P$ is a prime ideal. ∎

###### Lemma 2.2.7. —

Let $S$ be a multiplicatively closed subset in a quantale $\mathcal{Q}$.
Suppose $P\in\mathcal{I}_{\mathcal{Q}}$ such that it is maximal with respect
to the property: $P\cap S=\emptyset$. Then $P\in\mathrm{Spec}_{\mathcal{Q}}$.

###### Proof.

Suppose $x\notin P$ and $y\notin P$ for some $x,$ $y\in\mathcal{Q}$. By the
property on $P$, this implies that the ideals $\langle x,P\rangle$ and
$\langle y,P\rangle$ both intersect with $S$. Hence, there exist $l,$
$l^{\prime}\in\mathcal{Q}$, $p,$ $p^{\prime}\in P$, and $s,$ $s^{\prime}\in S$
such that $(x\mbox{{\small{$\&$}}}l)\vee(p\mbox{{\small{$\&$}}}s)\in S$ and
$(y\mbox{{\small{$\&$}}}l^{\prime})\vee(p^{\prime}\mbox{{\small{$\&$}}}s^{\prime})\in
S$. Since $S$ is a multiplicatively closed set, we must have
$((x\mbox{{\small{$\&$}}}l)\vee(p\mbox{{\small{$\&$}}}s))\mbox{{\small{$\&$}}}((y\mbox{{\small{$\&$}}}l^{\prime})\vee(p^{\prime}\mbox{{\small{$\&$}}}s^{\prime}))\in
S$. On the other hand,

$\displaystyle((x\mbox{{\small{$\&$}}}l)\vee(p\mbox{{\small{$\&$}}}s))((y\mbox{{\small{$\&$}}}l^{\prime})\vee(p^{\prime}\mbox{{\small{$\&$}}}s^{\prime}))$
$\displaystyle=(x\mbox{{\small{$\&$}}}y\mbox{{\small{$\&$}}}l\mbox{{\small{$\&$}}}l^{\prime})$
$\displaystyle\vee(p^{\prime}\mbox{{\small{$\&$}}}(x\mbox{{\small{$\&$}}}l\mbox{{\small{$\&$}}}s^{\prime}))\vee(p\mbox{{\small{$\&$}}}(y\mbox{{\small{$\&$}}}l^{\prime}\mbox{{\small{$\&$}}}s))\vee(p\mbox{{\small{$\&$}}}p^{\prime}\mbox{{\small{$\&$}}}(s\mbox{{\small{$\&$}}}s^{\prime}))\in\langle
x\mbox{{\small{$\&$}}}y,P\rangle,$

which implies $x\mbox{{\small{$\&$}}}y\notin P$. Hence $P$ is a prime ideal. ∎

###### Lemma 2.2.8. —

Let $\mathcal{Q}$ be a quantale. For every $I\in\mathcal{I}_{\mathcal{Q}}$,
$\mathcal{R}(I)$ is equal to the set

$T=\\{l\in\mathcal{Q}\mid\mathrm{every\;multiplicatively\;closed\;subset\;containing}\;l\;\mathrm{intersects}\;I\\}.$

###### Proof.

Suppose that $r\in T$ and $P\in\mathrm{Spec}_{\mathcal{Q}}$ such that
$I\preccurlyeq P$. Then by Lemma 2.2.6, $\mathcal{Q}\neg P$ is a
multiplicatively closed subset of $\mathcal{Q}$ and $l\notin\mathcal{Q}\neg
P.$ Hence $l\in P$. Conversely, let $l\notin T$. This implies that there
exists a multiplicatively closed subset $S$ of $\mathcal{Q}$ such that $l\in
S$ and $S\wedge I=\emptyset.$ By Zorn’s lemma, there exists an ideal $P$
containing $I$ and maximal with respect to the property that $P\wedge
S=\emptyset$. By Lemma 2.2.7, $P$ is a prime ideal with $l\notin P$. ∎

###### Lemma 2.2.9. —

Suppose $I$ is a semiprime ideal in a quantale $\mathcal{Q}$ and suppose
$x\in\mathcal{Q}\neg I$. Then there exists a multiplicatively closed subset
$S$ of $\mathcal{Q}$ such that $x\in S\preccurlyeq\mathcal{Q}\neg I.$

###### Proof.

Define the elements of $S=\\{x_{1},x_{2},\ldots,x_{n},\ldots\\}$ inductively
as follows: $x_{1}:=x;$ $x_{2}:=x_{1}\mbox{{\small{$\&$}}}x_{1};$ $\ldots$;
$x_{n}:=x_{n-1}\mbox{{\small{$\&$}}}x_{n-1};$ $\ldots$. Obviously $x\in S$ and
it is also easy to see that $x_{i},$ $x_{j}\in S$ implies that
$x_{i}\mbox{{\small{$\&$}}}x_{j}\in S$. ∎

###### Theorem 2.2.10. —

For any ideal $I$ in a quantale $\mathcal{Q}$, the following are equivalent.

  1. (1)

$I$ is semiprime.

  2. (2)

$I$ is an intersection of prime ideals in $\mathcal{Q}$.

  3. (3)

$I$ is a radical ideal.

###### Proof.

From Proposition 2.2.3, it follows that (3)$\Rightarrow$(2). Since the
intersection of semiprime ideals is a semiprime ideal, (2)$\Rightarrow$(1)
follows. What remains is to show that (1)$\Rightarrow$(3) and for that, it is
sufficient to show $\mathcal{R}(I)\preccurlyeq I.$ Suppose that $x\notin I$.
Then $x\in\mathcal{Q}\neg I$ and by Lemma 2.2.9, there exists a
multiplicatively closed subset $S$ of $\mathcal{Q}$ such that $x\in
S\preccurlyeq\mathcal{Q}\neg I.$ But $S\wedge I=\emptyset$ and hence by Lemma
2.2.8, $x\notin\mathcal{R}(I).$ ∎

###### Corollary 2.2.11. —

If $I\in\mathcal{I}_{\mathcal{Q}}$, then $\mathcal{R}(I)$ is the smallest
semiprime ideal in $\mathcal{Q}$ containing $I$.

Next we wish to consider saturation of a subset of a quantale and relate with
prime ideals.

###### Definition 2.2.4. —

Let $\mathcal{Q}$ be a quantale and $S$ be a multiplicatively closed subset of
$\mathcal{Q}$. We say $S$ is _saturated_ if for $x$, $y\in\mathcal{Q}$ and
$x\mbox{{\small{$\&$}}}y\in S$ implies that $x,$ $y\in S$. The _saturation_ of
$S$ is defined as

$\overline{S}=\\{x\in\mathcal{Q}\mid\;\text{there
exists}\;y\in\mathcal{Q}\;\text{such that}\;x\mbox{{\small{$\&$}}}y\in S\\}.$

The following proposition shows that $\overline{S}$ behaves as a closure
operation. Moreover, it is related with prime ideals as expected, and extends
Exercise 7 (from Chapter 3) of [6].

###### Proposition 2.2.12. —

Let $\mathcal{Q}$ be a quantale and $S$ be a multiplicatively closed subset of
$\mathcal{Q}$. Then the following hold.

  1. (1)

$\overline{S}$ is the smallest saturated multiplicatively closed subset in
$\mathcal{Q}$ containing $S$.

  2. (2)

$S$ is satuarted if and only if $\mathcal{Q}\neg S$ is a join of prime ideals.

###### Proof.

(1) If $s\in S$, then $\top\mbox{{\small{$\&$}}}s=s\in\overline{S},$ and hence
$S\preccurlyeq\overline{S}.$ Suppose $x,$ $y\in\overline{S}$. Then there exist
$y,$ $y^{\prime}\in\mathcal{Q}$ such that $x\mbox{{\small{$\&$}}}x^{\prime}\in
S$ and $y\mbox{{\small{$\&$}}}y^{\prime}\in S$. Since $S$ is a
multiplicatively closed set,

$(x\mbox{{\small{$\&$}}}x^{\prime})\mbox{{\small{$\&$}}}(y\mbox{{\small{$\&$}}}y^{\prime})=(x\mbox{{\small{$\&$}}}y)\mbox{{\small{$\&$}}}(x^{\prime}\mbox{{\small{$\&$}}}y^{\prime})\in
S,$

which implies that $x\mbox{{\small{$\&$}}}y\in\overline{S}.$ Therefore,
$\overline{S}$ is a multiplicatively closed subset of $\mathcal{Q}$. To show
$\overline{S}$ is saturated, let $x\mbox{{\small{$\&$}}}y\in\overline{S}$ for
some $x,$ $y\in\mathcal{Q}$. Then there exists $z\in\mathcal{Q}$ such that
$x\mbox{{\small{$\&$}}}(y\mbox{{\small{$\&$}}}z)=(x\mbox{{\small{$\&$}}}y)\mbox{{\small{$\&$}}}z\in
S$. This implies $x\in\overline{S}.$ Similarly, $y\in\overline{S}.$ Finally,
let $T$ be a satuaretd multiplicatively closed subset of $\mathcal{Q}$
containing $S$. Suppose $s\in\overline{S}.$ Then there exits $y\in\mathcal{Q}$
such that $x\mbox{{\small{$\&$}}}y\in S\preccurlyeq T$. Since $T$ is
saturated, $x\in T$.

(2) Suppose $\mathcal{Q}\neg
S=\bigvee\limits_{\lambda\in\Lambda}\\{P_{\lambda}\mid
P_{\lambda}\in\mathrm{Spec}_{\mathcal{Q}}\\}.$ Let $x\mbox{{\small{$\&$}}}y\in
S$ for some $x,$ $y\in\mathcal{Q}$ and if possible, let $x\notin S$. This
implies $x\preccurlyeq p_{\lambda}$ for some $p_{\lambda}\in P_{\lambda}$.
Since $P_{\lambda}$ is an ideal and $x\mbox{{\small{$\&$}}}y\preccurlyeq
x\wedge y\preccurlyeq x\preccurlyeq p_{\lambda}\in P_{\lambda},$ we must have
$x\mbox{{\small{$\&$}}}y\in P_{\lambda}.$ This implies
$x\mbox{{\small{$\&$}}}y\notin S$, a contradiction. This proves that $S$ is
saturated. Conversely, let $S$ be a saturated subset of $\mathcal{Q}$. Let
$a\in\mathcal{Q}\neg S.$ It suffices to show that $a\in P_{\lambda}$ for some
$P_{\lambda}\in\mathrm{Spec}_{\mathcal{Q}}$ such that $P_{\lambda}\wedge
S=\emptyset.$ Suppose

$\Omega=\\{I\in\mathcal{I}_{\mathcal{Q}}\mid a\in I\;\text{and}\;I\wedge
S=\emptyset\\}.$

Observe that $\langle a\rangle\wedge S=\emptyset$. Indeed,
$x_{i}\mbox{{\small{$\&$}}}a\preccurlyeq\bigvee\limits_{i=1}^{n}x_{i}\mbox{{\small{$\&$}}}a$
for $x_{i}\in\mathcal{Q},$ implies $x_{i}\mbox{{\small{$\&$}}}a\in S$ and
since $S$ is saturated, this implies $a\in S$, contradiction. Hence $\langle
a\rangle\in\Omega$, and thus $\Omega\neq\emptyset.$ By Zorn’s lemma, $\Omega$
has a maximal element, say $P$. We claim that $P$ is a prime ideal and proof
of this is already described in Proposition 2.2.3. ∎

###  2.3. Primary ideals and primary decompositions

The decomposition of an ideal into primary ideals holds a significant place in
ideal theory of rings and is considered a traditional cornerstone. The
objective of this section is to establish several classical uniqueness
theorems for ideals for quantales, and hence extend the corresponding results
of [6].

###### Definition 2.3.1. —

A proper ideal $P$ in a quantale $\mathcal{Q}$ is called _primary_ if for $x,$
$y\in\mathcal{Q}$ and $x\mbox{{\small{$\&$}}}y\in I$ imply that $x\in I$ or
$y^{n}\in I$ for some $n\in\mathds{N}$. If $P^{\prime}$ is a primary ideal in
$\mathcal{Q}$ and $\mathcal{R}(P^{\prime})=P$, then we say $P^{\prime}$ is
_$P$ -primary_.

###### Proposition 2.3.1. —

Let $\mathcal{Q}$ be a quantale. Then the following hold.

  1. (1)

Every prime ideal in $\mathcal{Q}$ is primary.

  2. (2)

Let $P$ be a primary ideal in $\mathcal{Q}$. Then $\mathcal{R}(P)$ is the
smallest prime ideal in $\mathcal{Q}$ containing $P$.

###### Proof.

(1) Suppose $P$ is a prime ideal in $\mathcal{Q}$ and
$x\mbox{{\small{$\&$}}}y\in P$ for some $x,$ $y\in\mathcal{Q}$. Then $x\in P$
or $y=y^{1}\in P,$

(2) By Proposition 2.2.3, it suffices to show that $\mathcal{R}(P)$ is a prime
ideal. Suppose $x\mbox{{\small{$\&$}}}y\in\mathcal{R}(P)$ for some $x,$
$y\in\mathcal{Q}$. Then $(x\mbox{{\small{$\&$}}}y)^{n}\in P$ for some
$n\in\mathds{N}$. This implies that $x^{n}\in P$ or $y^{mn}\in P$ for some
$m\in\mathds{N}.$ Hence, $x\in\mathcal{R}(P)$ or $y\in\mathcal{R}(P).$ ∎

When considering a finite intersection of primary ideals in a quantale, with
the condition that all the ideals involved are $P$-primary for a given prime
ideal $P$, the resulting intersection is indeed primary. To establish this
claim, we first require a preliminary result.

###### Lemma 2.3.2. —

If $I_{1},\ldots,I_{n}$ are ideals in a quantale $\mathcal{Q}$ and
$I=\bigwedge\limits_{i=1}^{n}I_{i},$ then
$\mathcal{R}(I)=\bigwedge\limits_{i=1}^{n}\mathcal{R}(I_{i}).$

###### Proof.

Suppose $x\in\mathcal{R}(I)$. Then $x^{n}\in I$ for some $n\in\mathds{N}$.
This implies that $x^{n}\in I_{i}$ for all $i\in\\{1,\ldots,n\\},$ and hence
$x\in\mathcal{R}(I_{i})$ for all $i\in\\{1,\ldots,n\\},$ i.e.,
$x\in\bigwedge\limits_{i=1}^{n}\mathcal{R}(I_{i}).$ Now, to obtain the other
inclusion, let $z\in\bigwedge\limits_{i=1}^{n}\mathcal{R}(I_{i}).$ This
implies that $z^{m_{i}}\in\mathcal{R}(I_{i}),$ where $m_{i}\in\mathds{N},$ for
all $i\in\\{1,\ldots,n\\}.$ Choose $m=\max\limits_{1\preccurlyeq i\preccurlyeq
n}\\{m_{i}\\}$. Then $z^{m}\in I_{i}$ for all $i\in\\{1,\ldots,n\\},$ and
hence $z\in\mathcal{R}(I).$ ∎

###### Theorem 2.3.3. —

If $P$ is a prime ideal in a quantale $\mathcal{Q}$ and if
$P_{1},\ldots,P_{n}$ are $P$-primary ideals in $\mathcal{Q}$, then
$P^{\prime}=\bigwedge\limits_{i=1}^{n}P_{i}$ is also a $P$-primary ideal in
$\mathcal{Q}$.

###### Proof.

By Lemma 2.2.4(5) and Lemma 2.3.2, we have

$\mathcal{R}(P^{\prime})=\mathcal{R}\left(\bigwedge_{i=1}^{n}P_{i}\right)=\bigwedge_{i=1}^{n}\mathcal{R}(P_{i})=\bigwedge_{i=1}^{n}P=P.$

What remains for us to prove is that $P^{\prime}$ is a primary ideal. Let
$x\mbox{{\small{$\&$}}}y\in P^{\prime}$. Then $x\mbox{{\small{$\&$}}}y\in
P_{i}$ for all $1\preccurlyeq i\preccurlyeq n$. If $x\notin P_{j}$ for some
$j$, then $y^{m}\in P_{j}$ for some $m\in\mathds{N}$, because $P_{j}$ is a
primary ideal. It follows that $y\in\mathcal{R}(P_{j})=P$. Since
$\mathcal{R}(P^{\prime})=P$, there exists $n\in\mathds{N}$ such that $y^{n}\in
P^{\prime}$. ∎

###### Proposition 2.3.4. —

Suppose $P$ is a prime ideal in a quantale $\mathcal{Q}$, $P^{\prime}$ is a
$P$-primary ideal, and $x$ is an element in $\mathcal{Q}$. Then the following
hold.

  1. (1)

If $x\in P^{\prime}$ then $(P^{\prime}:x)=\mathcal{Q}$.

  2. (2)

If $x\notin P^{\prime}$, then $(P^{\prime}:x)$ is a $P$-primary ideal.

  3. (3)

If $x\notin P$, then $(P^{\prime}:x)=P^{\prime}$.

###### Proof.

(1) If $x\in P^{\prime}$, then $x\mbox{{\small{$\&$}}}1=x\in P^{\prime}$, and
hence $1\in(P^{\prime}:x),$ implying that $(P^{\prime}:x)=\mathcal{Q}$.

(2) Let $x\notin P^{\prime}$. If $y\in(P^{\prime}:x)$, then
$x\mbox{{\small{$\&$}}}y\in P^{\prime}$. Since $P^{\prime}$ is a primary ideal
and $x\notin P^{\prime}$, there exists $n\in\mathds{N}$ such that $y^{n}\in
P^{\prime}$. Hence $y\in\mathcal{R}(P^{\prime})=P$. Therefore, we have
$P^{\prime}\preccurlyeq(P^{\prime}:x)\preccurlyeq P$, implying that

$P=\mathcal{R}(P^{\prime})\preccurlyeq\mathcal{R}(P^{\prime}:x)\preccurlyeq\mathcal{R}(P)=P.$

Hence $\mathcal{R}(P^{\prime}:x)=P$. It remains for us to show that
$(P^{\prime}:x)$ is a primary ideal. Since $x\notin P^{\prime}$, we have
$\top\notin(P^{\prime}:x)$, so $(P^{\prime}:x)\neq\mathcal{Q}$, i.e. is a
proper ideal in $\mathcal{Q}$. Let $a\mbox{{\small{$\&$}}}b\in(P^{\prime}:x)$.
If $b^{m}\notin P^{\prime}$ for all $m\in\mathds{N}$, then
$b\notin\mathcal{R}(P^{\prime}:x)=P$. However,
$a\mbox{{\small{$\&$}}}b\mbox{{\small{$\&$}}}x\in P^{\prime}$ implies that
$a\mbox{{\small{$\&$}}}x\in P^{\prime}$ or $b^{k}\in P^{\prime}$ for some
$k\in\mathds{N}$, because $P^{\prime}$ is a primary ideal. If $b^{k}\in
P^{\prime}$, then $b\in\mathcal{R}(P^{\prime})=P,$ a contradiction. Therefore,
$a\mbox{{\small{$\&$}}}x\in P^{\prime},$ implying that
$a\in(P^{\prime}:x)$whence $(P^{\prime}:x)$ is a primary ideal.

(3) Let $x\notin P$. If $y\notin P^{\prime}$ and $x\mbox{{\small{$\&$}}}y\in
P^{\prime}$, then $x^{n}\in P^{\prime}$ for some $n\in\mathds{N}$, because
$P^{\prime}$ is a primary ideal. Therefore, $x\in\mathcal{R}(P^{\prime})=P$, a
contradiction. Hence $x\mbox{{\small{$\&$}}}y\notin P^{\prime}$. This implies
that $y\notin(P^{\prime}:x)$. By Proposition 2.1.6(12),
$P^{\prime}\preccurlyeq(P:x),$ and this we have $P^{\prime}=(P^{\prime}:x).$ ∎

###### Definition 2.3.2. —

A _primary decomposition_ of an ideal $I$ in a quantale $\mathcal{Q}$ is an
expression:

(2.3) $I=\bigwedge_{i=1}^{n}P_{i},$

where $P_{i}$ are primary ideals in $\mathcal{Q}$. The expression (2.3) is
said to be _minimal_ if

  1. (1)

$\mathcal{R}(P_{1}),\ldots,\mathcal{R}(P_{n})$ are distinct.

  2. (2)

$\bigwedge\limits_{j\neq i}P_{j}\nsubseteq P_{i},$ for all $i.$

If an ideal $I$ in $\mathcal{Q}$ has a primary decomposition, then we say that
$I$ is _decomposible_.

###### Proposition 2.3.5. —

A primary decomposition may be replaced by a minimal primary decomposition.

###### Proof.

Consider a primary decomposition $I=\bigwedge\limits_{i=1}^{n}P_{i},$ where
$P_{i}$ are primary ideal in $\mathcal{Q}$. If

$\mathcal{R}(P_{i_{1}})=\cdots=\mathcal{R}(P_{i_{k}})=P,$

then by Theorem 2.3.3, $P^{\prime}=\bigwedge\limits_{j=1}^{k}P_{i_{j}}$ is a
$P$-primary ideal. Therefore, we can replace $P_{i_{1}},\ldots,P_{i_{k}}$ by
$P^{\prime}$, and continue the process to guarantee that the condition (1) of
Definition 2.3.2 holds. If condition (2) of Definition 2.3.2 does not hold,
then we can eliminate ideals until it does hold, without changing the overall
intersection. ∎

While primary decompositions, or minimal primary decompositions, may not be
unique, they possess certain uniqueness properties, which are outlined in the
following theorem.

###### Theorem 2.3.6. —

Let $I$ be a decomposible ideal in a quantale and
$I=\bigwedge\limits_{i=1}P_{i}$, a minimal primary decomposition. Let
$P_{i}^{\prime}=\mathcal{R}(P_{i}),$ for $i=1,\ldots,n$. Then the set
$\\{P_{1}^{\prime},\ldots,P_{n}^{\prime}\\}$ is composed of the prime ideals
$P$ of $\mathcal{Q}$ such that $P=\mathcal{R}((I:x)),$ for some
$x\in\mathcal{Q}$.

###### Proof.

Suppose $x\in\mathcal{Q}$. By Proposition 2.1.6(16) and Lemma 2.3.2, we have

$\mathcal{R}((I:x))=\mathcal{R}\left(\left(\bigwedge_{i=1}^{n}P_{i}:x\right)\right)=\mathcal{R}\left(\bigwedge_{i=1}^{n}(P_{i}:x)\right)=\bigwedge_{i=1}^{n}\mathcal{R}((P_{i}:x)).$

From (1) and (2) of Proposition 2.3.4, we obtain

$\mathcal{R}((I:x))=\bigwedge_{i=1}^{n}\mathcal{R}((P_{i}:x))=\bigwedge_{i,x\notin
P_{i}}P_{i}^{\prime}.$

If the intersection of a finite set of ideals is a prime ideal, then the
intersection is equal to one of the ideals; thus, if $\mathcal{R}(I:x)$ is a
prime ideal, then

$\mathcal{R}((I:x))\preccurlyeq\\{P_{i}^{\prime}\mid x\notin
P_{i}\\}\preccurlyeq\\{P_{1}^{\prime},\ldots,P_{n}^{\prime}\\}.$

For the converse, let $i\in\\{1,\ldots,n\\}.$ Because the primary
decomposition is minimal, for each $i$, there exists
$x_{i}\in\left(\bigwedge\limits_{j\neq i}P_{j}\right)\neg P_{i}.$ If
$y\in(P_{i}:x_{i}),$ then $y\mbox{{\small{$\&$}}}x_{i}\in P_{i}$. Therefore,

$y\mbox{{\small{$\&$}}}x_{i}\in P_{i}\wedge\left(\bigwedge_{j\neq
i}P_{j}\right)=I,$

which implies that $y\in(I:x_{i}).$ Hence

$(P_{i}:x_{i})\preccurlyeq(I:x_{i})\preccurlyeq(P_{i}^{\prime}:x_{i}),$

where the second inclusion follows from the fact that $I\preccurlyeq P_{i}$.
Therefore, $(P_{i}:x_{i})=(I:x_{i}).$ By Proposition 2.3.4(2), we have

$\mathcal{R}((I:x_{i}))=\mathcal{R}((P_{i}:x_{i}))=P_{i}^{\prime}.$

From this it follows that $P_{i}^{\prime}$ form the set of those ideals
$\mathcal{R}((I:x))$ which are prime ideals in $\mathcal{Q}$. ∎

Suppose $I$ is a decomposable ideal and $I=\bigwedge\limits_{i=1}^{n}P_{i}$, a
minimal primary decomposition. Let us denote
$\mathcal{R}(P_{i})=P_{i}^{\prime}$. We say that the prime ideals
$P_{1}^{\prime},\ldots,P_{n}^{\prime}$ _belong to_ $I$. The minimal elements
of the set $S=\\{P_{1}^{\prime},\ldots,P_{n}^{\prime}\\}$ with respect to
inclusion are said to be _isolated_ prime ideals belonging to $I$, and the
others are called _embedded_ prime ideals. Define

$I^{\uparrow_{\mathrm{Spec}_{\mathcal{Q}}}}=\\{P\in\mathrm{Spec}_{\mathcal{Q}}\mid
P\supseteq I\\}.$

We shall show that the minimal elements of the set S are the minimal elements
of the set $I^{\uparrow_{\mathrm{Spec}_{\mathcal{Q}}}}$. To obtain that, we
need the following result.

###### Proposition 2.3.7. —

Let $I$ be a decomposable ideal in a quantale $\mathcal{Q}$,
$I=\bigwedge\limits_{i=1}^{n}P_{i}$ a minimal primary decomposition, with $S$
the set of prime ideals belonging to $I$ . If $P\in
I^{\uparrow_{\mathrm{Spec}_{\mathcal{Q}}}}$, then $P$ contains an isolated
prime ideal $P_{j}^{\prime}$.

###### Proof.

Set $P_{i}^{\prime}=\mathcal{R}(P_{i}),$ for $i\in\\{1,\ldots,n\\}.$ Applying
Lemma 2.3.2, we obtain

$\bigwedge_{i=1}^{n}P_{i}^{\prime}=\bigwedge_{i=1}^{n}\mathcal{R}(P_{i})=\mathcal{R}(I)\preccurlyeq\mathcal{R}(P)=P.$

However, if a prime ideal contains an intersection of ideals, then at least
one of the ideals in the intersection is contained in the prime ideal,
therefore $P_{j}^{\prime}\preccurlyeq P$, for some $j$. The result now
follows. ∎

###### Corollary 2.3.8. —

The isolated prime ideals are the minimal elements in
$I^{\uparrow_{\mathrm{Spec}_{\mathcal{Q}}}}.$

###### Proof.

In Proposition 2.3.7, we have observed that $S\preccurlyeq
I^{\uparrow_{\mathrm{Spec}_{\mathcal{Q}}}}$, so a fortiori if $P_{j}^{\prime}$
is an isolated prime in $S$, then $P_{j}\in
I^{\uparrow_{\mathrm{Spec}_{\mathcal{Q}}}}$. Suppose now that $P\in
I^{\uparrow_{\mathrm{Spec}_{\mathcal{Q}}}}$, with $P\preccurlyeq
P_{j}^{\prime}$. From Proposition 2.3.7, it follows that there exists an
isolated prime ideal $P_{k}^{\prime}\preccurlyeq P$, so
$P_{k}^{\prime}\preccurlyeq P_{j}^{\prime}$. Since $P_{j}^{\prime}$ is
isolated, $P_{k}^{\prime}=P_{j}^{\prime}$, hence $P=P_{j}^{\prime}$, and it
follows that $P_{j}^{\prime}$ is minimal in
$I^{\uparrow_{\mathrm{Spec}_{\mathcal{Q}}}}$. ∎

Although the primary ideals in different minimal decompositions of an ideal
are not necessarily the same, the primary ideals whose radicals are isolated
are the same. We aim now to establish this.

###### Lemma 2.3.9. —

Let $I$ be a decomposable ideal in a quantale $\mathcal{Q}$ and $P_{j}$ an
ideal in a minimal decomposition of $I$ such that $\mathcal{R}(P_{j})$ is an
isolated prime ideal. Then $P_{j}$ is composed of the elements
$a\in\mathcal{Q}$ for which there exists $b\notin\mathcal{R}(P_{j})$ with
$a\mbox{{\small{$\&$}}}b\in I$.

###### Proof.

Suppose $I=\bigwedge\limits_{i=1}^{n}P_{i}$ is a minimal decomposition, with
$\mathcal{R}(P_{j})$ isolated. We claim that
$P_{i}\nsubseteq\mathcal{R}(P_{j})$, for $i\neq j$. If possible, let
$P_{i}\preccurlyeq\mathcal{R}(P_{j})$ and let $x\in\mathcal{R}(P_{i})$; then
$x^{n}\in P_{i}$, for some $n\in\mathds{N},$ which implies that there exists
$m\in\mathds{N}$ such that $x^{nm}\in P_{j}$, because
$P_{i}\preccurlyeq\mathcal{R}(P_{j})$, hence $x\in\mathcal{R}(P_{j})$. It
follows that $\mathcal{R}(P_{i})\preccurlyeq\mathcal{R}(P_{j})$, which is
impossible, because $\mathcal{R}(P_{j})$ is isolated. Thus
$P_{i}\nleqslant\mathcal{R}(P_{j})$, as claimed.

Now let $a\in P_{j}$. From what we have just seen, for $i\neq j$, there exists
$b_{i}\in P_{i}\neg\mathcal{R}(P_{j}).$ Set $b=\prod\limits_{i\neq j}b_{i}$.
As $\mathcal{R}(P_{j})$ is a prime ideal, we have $b\notin\mathcal{R}(P_{j})$.
However, $a\mbox{{\small{$\&$}}}b\in P_{i}$, for $i\neq j$, because $b_{i}\in
P_{i}$. Also, given that $a\in P_{j}$, we also have
$a\mbox{{\small{$\&$}}}b\in P_{j}$ and so

$a\mbox{{\small{$\&$}}}b\in\bigwedge_{i=1}^{n}P_{i}=I.$

Therefore, $a$ is an element in $\mathcal{Q}$ for which there exists
$b\notin\mathcal{R}(P_{j})$ with $a\mbox{{\small{$\&$}}}b\in I$. We now
consider the converse. Suppose that $a\in\mathcal{Q}$ and
$b\notin\mathcal{R}(P_{j})$ are such that $a\mbox{{\small{$\&$}}}b\in I$.
Since $I\preccurlyeq P_{j}$, we have $a\mbox{{\small{$\&$}}}b\in P_{j}$. As
$P_{j}$ is primary and $b^{n}\notin P_{j}$, for any $n\in\mathds{N}$, we must
have $a\in P_{j}$. This concludes the proof. ∎

###### Corollary 2.3.10. —

If $I$ is a decomposible ideal in a quantale $\mathcal{Q}$, then the primary
ideals in a minimal decomposition of $I$ corresponding to isolated prime
ideals are uniquely determined.

###  2.4. Irreducible and strongly irreducible ideals

The concept of strongly irreducible ideals, originally known as primitive
ideals, was introduced in [19] for commutative rings. In [8, p. 301, Exercise
34], these ideals are referred to as quasi-prime. The term “strongly
irreducible” was initially employed for noncommutative rings in [7]. Strongly
irreducible ideals (of rings) is significant both from algebraic and
topological aspects. For the detail, we refer our reader to [2, 7, 13, 16].

###### Definition 2.4.1. —

Suppose $\mathcal{Q}$ is a quantale.

  1. (1)

An ideal $I$ of $\mathcal{Q}$ is called _irreducible_ if $J\wedge
J^{\prime}=I$ implies that either $J=I$ or $J^{\prime}=I$ for all $J,$
$J^{\prime}\in\mathcal{I}_{\mathcal{Q}}$.

  2. (2)

An ideal $I$ of $\mathcal{Q}$ is called _strongly irreducible_ if $J\wedge
J^{\prime}\preccurlyeq I$ implies that either $J\preccurlyeq I$ or
$J^{\prime}\preccurlyeq I$ for all $J,$
$J^{\prime}\in\mathcal{I}_{\mathcal{Q}}$.

We shall use the notation $\mathrm{Irr}(\mathcal{Q})$ to represent the set of
irreducible ideals in $\mathcal{Q}$ and $\mathrm{Irr}^{+}(\mathcal{Q})$ to
denote the set of strongly irreducible ideals in $\mathcal{Q}$.

The following result gives an equivalent definition of a strongly irreducible
ideal in terms of elements of a quantale, and it extends Proposition 7.33 of
[22] and Theorem 20 of [7].

###### Proposition 2.4.1. —

An ideal $I$ in a quantale $\mathcal{Q}$ is strongly irreducible if and only
if for all $a$, $b\in\mathcal{Q}$ satisfy the condition $:$ $\langle
a\rangle\wedge\langle b\rangle\preccurlyeq I$ implies either $a\in I$ or $b\in
I$.

###### Proof.

Suppose that $I$ is a strongly irreducible ideal in $\mathcal{Q}$ and $\langle
a\rangle\wedge\langle b\rangle\preccurlyeq I$ for some $a,$ $b\in\mathcal{Q}.$
Since $I$ is strongly irreducible, we must have either $a\in\langle
a\rangle\preccurlyeq I$ or $b\in\langle b\rangle\preccurlyeq I$. For the
converse, suppose that $I$ is an ideal in $\mathcal{Q}$ that is not strongly
irreducible. Then there are ideals $J$ and $K$ of $\mathcal{Q}$ satisfy
$J\wedge K\preccurlyeq I,$ but $J\nleqslant I$ and $K\nleqslant I.$ This
implies there exist $j\in J$ and $k\in K$ such that $\langle
j\rangle\wedge\langle k\rangle\preccurlyeq I$, but neither $j\notin I$ nor
$k\in I$, a contradiction. ∎

In Theorem 2.2.10, it was noted that a radical ideal could be presented as the
intersection of prime ideals that contain it. Now, we aim to illustrate in the
next proposition that any proper ideal can be expressed in a comparable
fashion, albeit utilizing irreducible ideals instead. This result extends
Proposition 7.35 of [22]. However, prior to advancing further, we must
establish a lemma.

###### Lemma 2.4.2. —

Consider a quantale $\mathcal{Q}$. Assume that $\bot\neq x\in\mathcal{Q}$ and
$I$ is a proper ideal in $\mathcal{Q}$ such that $x\notin I$. In this case,
there exists an irreducible ideal $J$ of $\mathcal{Q}$ satisfying the
conditions $I\preccurlyeq J$ and $x\notin J.$

###### Proof.

Let $\\{J_{\lambda}\\}_{\lambda\in\Lambda}$ be a chain of ideals in
$\mathcal{Q}$ such that $I\preccurlyeq J_{\lambda}$ and $x\notin J_{\lambda}$
for all $\lambda\in\Lambda$. Then
$I\preccurlyeq\bigvee_{\lambda\in\Lambda}J_{\lambda}$ and $x\notin
J_{\lambda}$, for all $\lambda\in\Lambda$. By Zorn’s lemma, there exists a
maximal element $J$ of this chain. Suppose that $J=J_{1}\wedge J_{2}.$ By the
maximality condition of $J$, we must have $x\in J_{1}$ and $x\in J_{2},$ and
hence $x\in J_{1}\wedge J_{2}=J,$ a contradiction. Therefore, $J$ is the
required irreducible ideal. ∎

###### Proposition 2.4.3. —

If $I$ is a proper ideal in a quantale $\mathcal{Q}$, then
$I=\bigwedge\limits_{I\preccurlyeq J}\\{J\mid
J\in\mathrm{Irr}(\mathcal{Q})\\}$.

###### Proof.

By Lemma 2.4.2, there exists an irreducible ideal $J$ of $\mathcal{Q}$ such
that $I\preccurlyeq J$. Let

$J^{\prime}=\bigwedge_{I\preccurlyeq J}\\{J\mid
J\in\mathrm{Irr}(\mathcal{Q})\\}.$

Then $I\preccurlyeq J^{\prime}$. We claim that $J^{\prime}=I.$ If
$J^{\prime}\neq I$, then there exists an $x\in J^{\prime}\neg I$, and by Lemma
2.4.2, there exists an irreducible ideal $J^{\prime\prime}$ such that $x\notin
J^{\prime\prime}$ and $I\preccurlyeq J^{\prime\prime}$, a contradiction. ∎

In the context of a Noetherian ring, it is widely recognized that radicals can
be expressed as a finite intersection of prime ideals. Expanding upon this
idea, the following proposition extends this result to include any ideal by
utilizing irreducible ideals. This result also extends Proposition 7.3 of
[40]. We say a quantale _Noetherian_ if every ascending chain of ideals
eventually becomes stationary.

###### Proposition 2.4.4. —

If $\mathcal{Q}$ is a Noetherian quantale, then every ideal in $\mathcal{Q}$
can be represented as the intersection of a finite number of irreducible
ideals in $\mathcal{Q}$.

###### Proof.

Suppose that

$\mathcal{F}=\left\\{J\in\mathcal{R}(\mathcal{Q})\mid
J\neq\bigwedge_{j=1}^{n}I_{j},\;I_{j}\in\mathrm{Irr}(\mathcal{Q})\right\\}.$

It is sufficient to show that $\mathcal{F}=\emptyset$. Since $\mathcal{Q}$ is
Noetherian, $\mathcal{F}$ has a maximal element, say $I$. Since
$I\in\mathcal{F}$, it is not a finite intersection of irreducible ideals in
$\mathcal{Q}$. This implies that $I$ is not irreducible. Hence, there are
ideals $J$ and $K$ such that $I\preccurlyeq J$, $I\preccurlyeq K$, $I\neq J$,
$I\neq K$, and $I=J\wedge K.$ Since $I$ is a maximal element of $\mathcal{F}$,
we must have $J,$ $K\notin\mathcal{F}.$ Therefore, $J$ and $K$ are a finite
intersection of irreducible ideals in $\mathcal{Q}$, which followingly implies
that $I$ is also a finite intersection of irreducible ideals in $\mathcal{Q}$,
a contradiction. ∎

The following proposition establishes the relationships between prime,
semiprime, and strongly irreducible ideals, and it extends Proposition 7.36 of
[22] and Theorem 2.1(i) of [2].

###### Proposition 2.4.5. —

If $P$ is a strongly irreducible ideal in $\mathcal{Q}$, then $P$ is prime if
and only if $P$ is radical.

###### Proof.

Notice that if $P$ is prime, then obviously $P$ is a radical ideal. For the
converse, suppose that $P$ is a radical ideal and
$I\mbox{{\small{$\&$}}}J\preccurlyeq P$, for some $I,$
$J\in\mathcal{I}_{\mathcal{Q}}.$ Then

$I\wedge J\preccurlyeq\mathcal{R}(I\wedge
J)=\mathcal{R}(I\mbox{{\small{$\&$}}}J)\preccurlyeq\mathcal{R}(P)=P,$

where the first inclusion and the first equality respectively follow from
Lemma 2.2.4(1) and Lemma 2.2.4(5). Moreover, the second inclusion and the last
equality respectively follow from Lemma 2.2.4(2) and the fact that $P$ is a
prime ideal. Since $P$ is strongly irreducible, either $I\preccurlyeq P$ or
$J\preccurlyeq P.$ ∎

In the context of commutative rings, it has been established (see [2, Theorem
2.1(ii)]) that every proper ideal is contained in a minimal strongly
irreducible ideal. Remarkably, the same principle applies to strongly
irreducible ideals in quantales.

###### Proposition 2.4.6. —

Every proper ideal in a quantale is contained in a minimal strongly
irreducible ideal.

###### Proof.

Let $I$ be a proper ideal in a quantale $\mathcal{Q}$ and let

$\mathcal{E}=\\{J\mid I\preccurlyeq J,\;J\in\mathrm{Irr}^{+}(\mathcal{Q})\\}.$

Note that every maximal ideal in $\mathcal{Q}$ is prime and by Lemma 2.1.6(6),
every prime ideal is strongly irreducible. Also, observe that by Theorem
2.1.9, every proper ideal is contained in a maximal ideal. Therefore, the set
$\mathcal{E}$ is nonempty. By Zorn’s lemma, $\mathcal{E}$ has a minimal
element, which is our desired minimal strongly irreducible ideal. ∎

The following result demonstrates the scenario in which all ideals in a
quantale are strongly irreducible, and its proof is evident. This result
extends Lemma 3.5 of [2].

###### Proposition 2.4.7. —

Every ideal in a quantale $\mathcal{Q}$ is strongly irreducible if and only if
$\mathcal{I}_{\mathcal{Q}}$ is totally ordered.

We bring this subsection to a close with a theorem pertaining to arithmetic
quantales, wherein the notions of irreducibility and strongly irreducibility
align. The proof of this theorem is identical to that of [32, Theorem 3 and
Theorem 7]

###### Theorem 2.4.8. —

In a arithmetic quantale $\mathcal{Q}$, an ideal in $\mathcal{Q}$ is
irreducible if and only if it is strongly irreducible. Conversely, if an
irreducible ideal in a quantale $\mathcal{Q}$ is strongly irreducible, then
$\mathcal{Q}$ is arithmetic.

###### Corollary 2.4.9. —

In an arithmetic quantale, any ideal is the intersection of all strongly
irreducible ideals containing it.

###### Concluding Remarks 2.4.10. —

Building upon the framework of the “ideal theory of quantales” that we have
begun to develop in this paper, we shall now provide a brief outline of our
future work. In [24], our focus be on examining the properties of different
ideals in relation to the quotient of a quantale. We shall also introduce the
concept of localization for a quantale. Additionally, by considering modules
over quantales, we shall extend some of the significant results in commutative
algebra.

On the other hand, in [25], our objective be to establish lower topologies on
distinguished classes of ideals in a quantale, and this extend the work done
in [13] and [16] for rings. We shall investigate the topological properties of
“quantale spaces,” including quasi-compactness, separation properties,
connectedness, continuity, and spectral spaces (as defined in [30]).
Specifically, we shall delve into the detailed examination of quantale spaces
associated with prime ideals, minimal prime ideals, maximal ideals, and
strongly irreducible ideals in a quantale. These investigations respectively
extend the work carried out in [27], [31], [42], and [2] for rings.
Furthermore, we shall demonstrate that the quantale space of proper ideals is
spectral, thereby generalizing the corresponding result presented in [23] for
rings.

While our focus in this paper was on a commutative quantale with identity $1$,
it is worth noting that the developed theory can be further extended to a
complete lattice $(\mathsf{L},\preceq,\bot,\top)$ equipped with an operation
denoted as $\odot$ that retains Axioms (1), (2), and (4) from Definition
2.1.1, as well as the condition:

$x\odot(y\vee z)=(x\odot y)\vee(x\odot z),$

for all $x,$ $y,$ and $z$ in $\mathsf{L}.$

The attentive reader may have observed that we have thus far assumed
commutativity in our quantales. However, by relaxing the commutativity axiom,
we naturally arrive at the concept of primitive ideals (as described in [34])
in quantales. In the forthcoming paper [26], our objective be to explore
Jacobson’s structure theory for quantales. Specifically, we shall formulate
and prove the Jacobson-Cheveley Density Theorem for quantales. In line with
[33], we shall also examine the Jacobson topology on the primitive ideals in a
quantale.

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