# Quantum nonlocality in presence of strong measurement dependence

Ivan Šupić  Département de Physique Appliquée, Université de Genève, 1211
Genève, Switzerland  CNRS, LIP6, Sorbonne Université, 4 place Jussieu, 75005
Paris, France<EMAIL_ADDRESS>Jean-Daniel Bancal  Département de Physique
Appliquée, Université de Genève, 1211 Genève, Switzerland  Université Paris-
Saclay, CEA, CNRS, Institut de Physique Théorique, 91191, Gif-sur-Yvette,
France  Nicolas Brunner  Département de Physique Appliquée, Université de
Genève, 1211 Genève, Switzerland

(August 29, 2024)

###### Abstract

It is well known that the effect of quantum nonlocality, as witnessed by
violation of a Bell inequality, can be observed even when relaxing the
assumption of measurement independence, i.e. allowing for the source to be
partially correlated with the choices of measurement settings. But what is the
minimal amount of measurement independence needed for observing quantum
nonlocality? Here we explore this question, and consider models with strong
measurement-dependent locality, where measurement choices can be perfectly
determined in almost all rounds of the Bell test. Yet, we show that quantum
nonlocality can still be observed in this scenario, which we conjecture is
minimal within the framework we use. We also discuss potential applications in
randomness amplification.

##  I Introduction

Quantum theory allows for strong nonlocal correlations is witnessed by the
violation of Bell inequalities Bell . This phenomenon, known as quantum
nonlocality, has been verified experimentally in a variety of physical
platforms, see e.g. Aspect ; Tittel ; Rowe ; Christensen ; Hensen ; Giustina ;
Shalm , and represent the basic resource for quantum information processing in
the so-called device-independent setting.

A Bell test typically involves a number of distant observers (say Alice and
Bob) performing randomly chosen local measurements on a shared physical
system. Bell inequalities are then usually derived under two assumptions (see
e.g. reviewBell ): (i) the choices of local measurements made by each party
are independent from the source distributing the shared physical resource, and
(ii) the measurement output of each party is determined solely by their input
and a shared local (hidden) variable. Clearly, the first assumption is
important, since if the source could know a priori the choice of measurement
setting for each round, then all possible correlations can be reproduced by a
classical model. Therefore, the observation of (quantum) nonlocality relies on
the assumption that the choice of measurement settings cannot be perfectly
determined at the source. In other words, there must be some level of
measurement independence. But then, how much measurement independence is
required for observing quantum nonlocality?

This question has attracted broad attention in recent years, and has been
discussed following various approaches, see e.g. Hall_2010 ; Hall_2011 ;
Barrett_2011 ; Banik_2012 ; Koh_2012 ; Thinh_2013 ; Tan ; Chaves_2015 ;
Friedman_2019 . A notable approach is that of Pütz et al. Puetz_2014 ;
putz2016measurement  who presented a general framework for addressing these
questions. In particular, they derived Bell inequalities for testing local
models with measurement dependent locality (MDL), i.e. where the above
assumption (i) is relaxed. Remarkably, violation of such inequalities are
possible for any level of measurement dependence, demonstrating that quantum
nonlocality can be observed even if measurement settings can be almost
perfectly determined by the source. Hence quantum theory allows for
measurement-dependent nonlocality. These ideas have been tested experimentally
Aktas_2015 , and shown to be relevant for the task of randomness amplification
Kessler  and network quantum nonlocality Supic

A natural question at this point is whether the above works have identified
the minimal requirements in terms of measurement independence, or if quantum
nonlocality could in fact be demonstrated considering an even stronger form of
measurement dependence.

This is the main motivation behind the present work. Our main result is that
quantum theory allows in fact for a much stronger form of measurement-
dependent nonlocality. Specifically, we consider the approach of Pütz et al.
Puetz_2014 ; putz2016measurement , who showed that the measurement inputs can
be almost perfectly determined in each round of the Bell test. This means that
in every round, the inputs feature some (possibly arbitrarily small) level of
randomness with respect to the source. In contrast, we consider models where,
in almost all rounds, the source can fully determine the measurement inputs.
Hence we only require the presence of rounds where the inputs cannot be
perfectly determined, and that these rounds have a non-vanishing probability
to occur. We then show that quantum correlations are still incompatible with
respect to these models. Our work thus shows that the requirement for
demonstrating quantum nonlocality can still be considerably relaxed compared
to previously known results. We conclude with a discussion of further
questions and applications for randomness amplification.

##  II General setting

We consider first bipartite Bell tests involving two non-communicating agents,
Alice and Bob. Alice can choose between several measurements, labeled with
$x\in\\{0,1,\cdots,m_{\text{A}}-1\\}$. Her outcomes are labelled with
$a\in\\{0,1,\cdots,n_{\text{A}}-1\\}$. For Bob the measurement choice label is
$y\in\\{0,1,\cdots,m_{\text{B}}-1\\}$ and the corresponding outcome
$b\in\\{0,1,\cdots,n_{\text{B}}-1\\}$. The experiment is characterized by the
set of conditional probabilities of obtaining the pair of outcomes $a,b$ when
the chosen inputs are $x$ and $y$:

$\mathscr{P}=\\{p(a,b|x,y)\\}_{a,b,x,y},$ (1)

usually called a behaviour. Without loss of generality the correlation
probabilities can be written as

$\displaystyle p(a,b|x,y)$ $\displaystyle=\int_{\lambda}d\lambda
p(\lambda)p(x,y|\lambda)p(a,b|x,y,\lambda),$ (2)

where $p(\lambda)$, $p(x,y|\lambda)$ and $p(a,b|x,y,\lambda)$ are valid
probability distributions. The classical variable $\lambda$ thus encodes the
correlations between various measurement outputs.

Bell Bell  formalized the concept of an LHV model via the following
requirements. (i) Measurement independence: the variable $\lambda$ is
completely independent from the inputs $x$ and $y$, i.e.

$\displaystyle p(x,y|\lambda)=p(x,y)\quad\forall\lambda$ (3)

(ii) Bell locality: locally, each output is determined solely by the input and
the shared variable $\lambda$, i.e.

$\displaystyle p(a,b|x,y,\lambda)=p(a|x,\lambda)p(b|y,\lambda)\quad\forall
x,y,a,b,\lambda.$ (4)

A behaviour is then termed (Bell) local if it admits a decomposition of the
form:

$p(a,b|x,y)=\int_{\lambda}d\lambda p(\lambda)p(a|x,\lambda)p(b|y,\lambda).$
(5)

We denote such a behaviour as $\bar{\mathscr{P}}_{\textrm{loc}}$ with
corresponding probabilities $p_{\textrm{loc}}(a,b|x,y)$. All such behaviours
satisfy Bell inequalities, which are based on linear functional of the joint
probabilities, and take the form:

$\mathscr{I}\equiv\sum_{a,b,x,y}\omega_{a,b,x,y}p_{\textrm{loc}}(a,b|x,y)\leq\beta_{\mathsf{loc}},$
(6)

where $\omega_{a,b,x,y}$ are real coefficients and $\beta_{\mathsf{loc}}$ is
termed the local bound of the Bell inequality.

Violation of such a Bell inequality in presence of measurement independence
implies nonlocality. This is possible in quantum theory, by performing well-
chosen sets of local measurements on a shared entangled state
$\varrho^{\text{A}\text{B}}$. In general, a quantum behaviour takes the form:

$p(a,b|x,y)=\textrm{Tr}\left[\mathsf{M}_{a|x}\otimes\mathsf{M}_{b|y}\varrho^{\text{A}\text{B}}\right],$
(7)

where the sets of operators $\\{\mathsf{M}_{a|x}\\}$ and
$\\{\mathsf{M}_{b|y}\\}$ represent the local measurements of Alice and Bob
respectively.

Before continuing, let us introduce two additional bounds for a Bell
inequality, relevant for our work. First, the maximal value of a Bell
expression (the functional defined in (6)) in quantum theory (optimized over
all quantum behaviours) is called the quantum bound and denoted by
$\beta_{\textrm{q}}$. Second, the maximal value of a Bell expression
(considering any valid joint probability distribution) is called the algebraic
bound and denoted by $\beta_{\textrm{alg}}$.

Figure 1: We consider a Bell test with relaxed measurement independence. That
is, the source (i.e. the classical variable $\Lambda$) can be correlated to
the choice of measurement inputs (variables $X$ and $Y$) of both parties. We
show that quantum nonlocality can exhibit strong measurement-dependent
nonlocality, i.e. the quantum predictions cannot be explained by a local model
even if the source $\Lambda$ completely determines both inputs in almost all
rounds of the experiment.

Let us now come back to the assumptions behind local models. Our focus here is
on the assumption of measurement independence. From a fundamental perspective,
it is natural to ask whether this assumption could be partially relaxed. That
is, could one still demonstrate quantum nonlocality when considering partial
correlations between the source (i.e. the shared variable $\lambda$) and the
measurement inputs $x$ and $y$. Refs Thinh_2013 ; Puetz_2014 ;
putz2016measurement  introduced a general framework to tackle this question.
In particular, they proposed to relax the usual measurement independence
condition to the following:

$\xi\leq p(x,y|\lambda)\leq\eta,\qquad\forall x,y,\lambda,$ (8)

where $\xi>0$ and consequently $\eta<1$, as $\sum_{x,y}p(x,y|\lambda)=1$.
Hence the measurement inputs are no longer independent from the shared
variable $\lambda$. Importantly the inputs can still not be perfectly
determined from $\lambda$, as ensured by the condition $\xi>0$. More
generally, one can set given values to the parameters $\xi$ and $\eta$ which
then quantify the level of measurement independence. One can then define
classes of local models with relaxed measurement independence, satisfying the
condition of Bell locality (Eq. (4)) and the partial measurement independence
condition in Eq. (8) for given values of $\xi$ and $\eta$. Ref. Puetz_2014
then showed how to construct Bell inequalities for these models. A key idea
consists in considering joint distributions $p(a,b,x,y)$ instead of
conditional ones. Starting from a standard Bell inequality, as in (6), one can
then construct a novel Bell inequality of the form

$\displaystyle\tilde{\mathscr{I}}_{\xi,\eta}\leq\xi\eta\beta_{\textrm{loc}},$

where $\tilde{\mathscr{I}}_{\xi,\eta}$ is a linear functional of the joint
probabilities $p(a,b,x,y)$ (with coefficients obtained from the original Bell
functional).

Violation of such a Bell inequality under the condition Eq. (8) implies so-
called “measurement-dependent nonlocality”. Remarkably, Pütz et al. Puetz_2014
showed that quantum theory allows for measurement-dependent nonlocality even
for an arbitrarily small level of measurement independence, i.e. for any
$\xi>0$. Their example is connected to the quantum correlations arising from
the well-known Hardy paradox Hardy .

##  III Strong measurement-dependent nonlocality

We are now in position to present our main result, namely that quantum
correlations can in fact exhibit an even stronger form of measurement-
dependent nonlocality. To be more precise, let us first notice that in the
approach of Pütz et al. Puetz_2014  reviewed above, the relaxed measurement
independence condition (8) must hold in every round of the Bell experiment.
That is, in every round, the inputs cannot be perfectly determined from the
shared variable $\lambda$.

Below we consider local models where this constraint is almost fully relaxed.
Specifically, we consider two types of rounds. First, we have rounds where the
shared variable perfectly determines the inputs (i.e. full measurement
dependence). These rounds are denoted by the set of shared variables
$\Lambda^{\prime\prime}$. Second, we have rounds with relaxed measurement
independence, i.e. the condition (8) applies for a fixed value of $\xi$ and
$\eta$. This set of rounds is denoted by $\Lambda^{\prime}$. Let us
characterize the relative probabilities of these sets:

$\displaystyle\int_{\Lambda^{\prime}}d\lambda
p(\lambda)=q,\qquad\int_{\Lambda^{\prime\prime}}d\lambda p(\lambda)=1-q.$ (9)

Note that we must require that $q>0$ in order to possibly observe nonlocality.
Below, we show how to construct a relevant Bell inequality, starting from any
standard Bell inequality (6), in order to test the above model, for any values
of $q>0$ and $\xi>0$. Moreover, we exhibit an example of such a Bell
inequality, where quantum correlations lead to a Bell violation for any values
of $q>0$ and $\xi>0$. We say that such correlations feature strong
measurement-dependent nonlocality.

Consider now an arbitrary Bell inequality (6). As in Puetz_2014  we denote
$\omega_{a,b,x,y}^{+}=\omega_{a,b,x,y}$ if $\omega_{a,b,x,y}>0$ and
$\omega_{a,b,x,y}^{-}=\omega_{a,b,x,y}$ if $\omega_{a,b,x,y}>0$. Then the Bell
inequality has the form

$\displaystyle\mathscr{I}$
$\displaystyle\equiv\sum_{a,b,x,y}\left[\omega_{a,b,x,y}^{+}p_{\textrm{loc}}(a,b|x,y)-\omega_{a,b,x,y}^{-}p_{\textrm{loc}}(a,b|x,y)\right]$
(10) $\displaystyle\leq\beta_{\mathsf{loc}}.$

The quantum bound of the inequality, equal to $\beta_{\mathrm{q}}$ can be
decomposed as
$\beta_{\mathrm{q}}=\beta_{\mathrm{q}}^{+}-\beta_{\mathrm{q}}^{-}$, where we
introduced notation

$\displaystyle\beta_{\mathrm{q}}^{+}$
$\displaystyle=\sum_{a,b,x,y}\omega_{a,b,x,y}^{+}p_{\mathrm{q}}(a,b|x,y)$ (11)
$\displaystyle\beta_{\mathrm{q}}^{-}$
$\displaystyle=\sum_{a,b,x,y}\omega_{a,b,x,y}^{-}p_{\mathrm{q}}(a,b|x,y),$
(12)

and $\\{p_{q}(a,b|x,y)\\}$ is a behaviour reaching the quantum bound. The
measurement-dependent Bell inequality from Puetz_2014  reads

$\displaystyle\tilde{\mathscr{I}}_{\xi,\eta}=\sum_{a,b,x,y}\bigg{(}\xi\omega_{a,b,x,y}^{+}p(a,b,x,y)-\eta\omega_{a,b,x,y}^{-}p(a,b,x,y)\bigg{)}.$
(13)

By separating the sets $\Lambda^{\prime}$ and $\Lambda^{\prime\prime}$ and
taking into account the decomposition of local joint probabilities we get

$\displaystyle\tilde{\mathscr{I}}_{\xi,\eta}$
$\displaystyle=\sum_{a,b,x,y}\bigg{(}\xi\omega_{a,b,x,y}^{+}(1-q)p(a,b,x,y)-\eta\omega_{a,b,x,y}^{-}(1-q)p(a,b,x,y)\bigg{)}+$
$\displaystyle\qquad+\sum_{a,b,x,y}\left(\xi\omega_{a,b,x,y}^{+}q\int_{\Lambda^{\prime}}d\lambda\frac{p(\lambda)}{q}p(x,y|\lambda)p(a|x,\lambda)p(b|y,\lambda)-\eta\omega_{a,b,x,y}^{-}q\int_{\Lambda^{\prime}}d\lambda\frac{p(\lambda)}{q}p(x,y|\lambda)p(a|x,\lambda)p(b|y,\lambda)\right).$

The second line is simply the renormalized measurement-dependent inequality
for the local variables satisfying the condition (8). Hence its local bound is
$q\xi\eta\beta_{\mathrm{loc}}$, where $q$ comes from the normalization of the
hidden variable probability distribution. In the first line, as the condition
(8) is not satisfied, the second term can be equal to $0$. Observe now that
the algebraic bound is reached if for all $a,b,x,y$ such that
$\omega_{a,b,x,y}<0$ it holds $p(a,b,x,y)=0$. As the condition (8) is not
satisfied, the optimal value of the first term in the equation is achieved
when $p(x,y|\lambda)=1$ for all $x,y$ and $\lambda\in\Lambda^{\prime\prime}$.
Hence the maximal value of the first line is

$\sum_{a,b,x,y}\xi\omega_{a,b,x,y}^{+}(1-q)p(a,b,x,y)=\xi(1-q)\tilde{\beta}_{\mathrm{alg}},$

where $\tilde{\beta}_{\mathrm{alg}}$ is the algebraic bound of the inequality
$\sum_{a,b,x,y}\left[\omega_{a,b,x,y}^{+}p_{\textrm{loc}}(a,b,x,y)-\omega_{a,b,x,y}^{-}p_{\textrm{loc}}(a,b,x,y)\right]$.
This implies that the local bound for the strong measurement-dependent LHV
models is

$\beta_{\mathrm{loc}}^{smd}\equiv(1-q)\xi\tilde{\beta}_{\mathrm{alg}}+q\xi\eta{\beta}_{\mathrm{loc}}.$
(14)

Thus, for hidden variables satisfying condition (8) when
$\lambda\in\Lambda^{\prime}$ and $\Lambda^{\prime}$ satisfying condition (9)
the bound of the measurement-dependent Bell inequality becomes:

$\mathscr{I}^{md}\leq\beta_{\mathrm{loc}}^{smd}.$ (15)

The next part consists in evaluating the quantum bound of the expression
$\tilde{\mathscr{I}}_{\xi,\eta}$. For this, it is useful to know the
distribution of measurement inputs, which can be observed in the Bell test.
From now on, let us assume that all possible input pairs appear with equal
probabilities, _i.e._ $p(x,y)=1/(m_{A}m_{B})$. From the definitions (11) and
(12), we get the quantum bound

$\beta_{\mathrm{q}}^{smd}=\frac{1}{m_{A}m_{B}}\left(\xi\beta_{\mathrm{q}}^{+}-\eta\beta_{\mathrm{q}}^{-}\right)\,.$
(16)

Strong measurement-dependent nonlocality is then observed when
$\beta_{\mathrm{q}}^{smd}>\beta_{\mathrm{loc}}^{smd}$. Below we discuss when
such a Bell violation is possible.

For most Bell inequalities, we have that
$\beta_{\mathrm{q}}<\beta_{\mathrm{alg}}$. It follows that one can detect
strong measurement-dependent nonlocality for some limited range of parameters
of the model. Notably, one will require that $q>q_{0}$, where $q_{0}$ is a
strictly positive constant. This means that the rounds where the local
variable cannot perfectly determine the inputs have a certain minimal
probability to occur. This is for example the case for the quantum
correlations presented in Ref. Puetz_2014 .

It is however possible to construct quantum correlations that are incompatible
with any local model with strong MDL, i.e. considering any parameter values of
$q>0$ and $\xi>0$. We now present an example of such quantum correlations. Our
starting point are quantum nonlocal correlation arising from the Peres-Mermin
(PM) magic-square game Peres_1991 ; Mermin , which have been proposed in Ref.
Aolita2012 . This work also constructs a Bell inequality to detect these
nonlocal correlations. The key property of this Bell inequality is that the
quantum bound coincides with the algebraic one; a feature sometimes referred
to as quantum “pseudo-telepathy”.

Following Ref. Aolita2012  we consider a bipartite Bell test with three inputs
per party. For each input, each party outputs two bits, denoted
$a=(a_{1},a_{2})$ and for Bob with $b=(b_{1},b_{2})$, with
$a_{i},b_{j}\in\\{-1,1\\}$. For our purpose, we need to construct a variant of
the Bell inequality of Ref. Aolita2012 . Let us first introduce two types of
correlators:

$\displaystyle\tilde{C}_{x,y}^{f(a,b)}$
$\displaystyle=\sum_{a,b}2^{\frac{f(a,b)+1}{2}}(-16)^{-\frac{f(a,b)-1}{2}}p(a,b|x,y),$
(17) $\displaystyle\bar{C}_{x,y}^{f(a,b)}$
$\displaystyle=\sum_{a,b}16^{\frac{f(a,b)+1}{2}}(-2)^{-\frac{f(a,b)-1}{2}}p(a,b|x,y).$
(18)

While it is common to use correlators taking values in the interval $[-1,1]$
for Bell inequalities, note that the above modified correlators take values in
the following intervals: $-16\leq\tilde{C}_{x,y}^{f(a,b)}\leq 2$ and
$-2\leq\bar{C}_{x,y}^{f(a,b)}\leq 16$ (it will become clear below why we need
this). We now construct the following Bell expression

$\displaystyle I$
$\displaystyle=\tilde{C}_{1,1}^{a_{1}b_{1}}+\tilde{C}_{1,2}^{a_{2}b_{1}}+\tilde{C}_{2,1}^{a_{1}b_{2}}+\tilde{C}_{2,2}^{a_{2}b_{2}}+\tilde{C}_{1,3}^{a_{1}a_{2}b_{1}}+$
$\displaystyle+\tilde{C}_{2,3}^{a_{1}a_{2}b_{2}}+\tilde{C}_{3,3}^{a_{1}a_{2}b_{1}b_{2}}+\tilde{C}_{3,1}^{a_{1}b_{1}b_{2}}-\bar{C}_{3,2}^{a_{2}b_{1}b_{2}}\,,$
(19)

with algebraic bound $\beta_{alg}=18$. Using the quantum strategy based on the
PM magic-square game as in Aolita2012 , we get a nonlocal distribution that
reaches $I=18$, hence we have that $\beta_{q}=\beta_{alg}$.

Now the important property for our purpose is that the local bound of $I$ is
$\beta_{loc}=0$ (hence the use of modified correlators), which simply follows
from inspection over all deterministic strategies.

We can now apply the method outlined above to adapt this Bell inequality to
the measurement-dependent scenario, namely

$\displaystyle\tilde{\mathscr{I}}_{\xi,\eta}$
$\displaystyle=\tilde{C}_{1,1,\xi,\eta}^{a_{1}b_{1}}+\tilde{C}_{1,2,\xi,\eta}^{a_{2}b_{1}}+\tilde{C}_{2,1,\xi,\eta}^{a_{1}b_{2}}+\tilde{C}_{2,2,\xi,\eta}^{a_{2}b_{2}}+\tilde{C}_{1,3,\xi,\eta}^{a_{1}a_{2}b_{1}}+$
$\displaystyle+\tilde{C}_{2,3,\xi,\eta}^{a_{1}a_{2}b_{2}}+\tilde{C}_{3,3,\xi,\eta}^{a_{1}a_{2}b_{1}b_{2}}+\tilde{C}_{3,1,\xi,\eta}^{a_{1}b_{1}b_{2}}-\bar{C}_{3,2,\xi,\eta}^{a_{2}b_{1}b_{2}},$
(20)

where $\tilde{C}_{x,y,\xi,\eta}^{f(a,b)}$ and
$\bar{C}_{x,y,\xi,\eta}^{f(a,b)}$ are now weighted correlators of joint
distributions, defined as

$\displaystyle\tilde{C}_{x,y,\xi,\eta}^{f(a,b)}$ $\displaystyle=\sum_{a,b}\xi
2^{\frac{f(a,b)+1}{2}}(-16\eta)^{-\frac{f(a,b)-1}{2}}p(a,b,x,y),$ (21)
$\displaystyle\bar{C}_{x,y,\xi,\eta}^{f(a,b)}$ $\displaystyle=\sum_{a,b}\eta
16^{\frac{f(a,b)+1}{2}}(-2\xi)^{-\frac{f(a,b)-1}{2}}p(a,b,x,y).$ (22)

Notice that for $\tilde{\mathscr{I}}_{\xi,\eta}$ we have
$\tilde{\beta}_{\mathrm{alg}}=2$, hence from eq. (14) we obtain the local
bound $\beta_{\mathrm{loc}}^{smd}=2(1-q)\xi$, while the quantum bound is
$\beta_{\mathrm{q}}^{smd}=2\xi$. Hence we obtain a quantum violation for any
parameter values $q>0$ and $\xi>0$. On Fig. 2 we show for which values of
parameters $\xi$ and $q$ nonlocality with strong measurement dependence can be
observed if noisy states are used to violate the Mermin-Peres magic square
game.

Figure 2: The graph depicts nonlocality in the presence of strong measurement-
dependence attained by the states
${\lvert\phi_{v}\rangle\\!\langle\phi_{v}\rvert}=v{\lvert\phi_{+}\rangle\\!\langle\phi_{+}\rvert}+(1-v)\mathds{1}/16$,
where ${\lvert\phi_{+}\rangle}=\sum_{j=0}^{3}{\lvert jj\rangle}/2$. The
measurements correspond to those used to obtain the maximal violation of the
Mermin-Peres magic square game. For given visibility $v$ nonlocality is
observed if the values of parameters $\xi$ and $q$ belong to the region above
the corresponding curve. The parameter $\eta$ is taken to be equal to
$1/9+\xi$. The red bold curve corresponds to a visibility comparable to the
experiment of Ref. Aolita2012 , based on a hyper-entangled state..

In the Appendix, we also present a similar construction based on the Mermin
inequality, considering a tripartite Bell test.

##  IV Connection to randomness amplification

While the question of observing quantum nonlocality under conditions of
relaxed measurement independence is of rather fundamental nature, these ideas
also have implications from a more applied perspective. Indeed, one prominent
application of quantum nonlocality is the task of generating randomness. This
process is known as “device-independent” quantum random number generation (DI-
QRNG), which explores the possibility to generate certified random numbers
under minimal assumptions on the devices; see e.g. Acin_2016  for a recent
review.

The most relevant aspect of DI-QRNG here is the task of DI randomness
amplification Colbeck_2012 . The idea is to perform a Bell test using
imperfect sources of randomness to generate the inputs, and being able to
certify more randomness in the outputs, hence amplifying randomness via
nonlocal correlations. Previous works in this line have demonstrated
randomness amplification considering so-called Santha-Vazirani sources for
inputs Colbeck_2012 ; Gallego_2013 ; miller2016robust ; Brand_o_2016 ;
Ramanathan_2016 . Interestingly, randomness amplification can be directly
connected to the measurement-dependent nonlocality Kessler , specifically to
the approach of Ref. Puetz_2014 , which leads to protocols with improved
performance.

Given this strong connection, we believe that our approach can have direct
consequences in the context of randomness amplification. In particular, our
results should translate into stronger forms of randomness amplifications,
going beyond Santha-Vazirani sources. Another interesting question is to
understand how our approach connects to the work of Ref. Thinh_2013 , where
min-entropy sources are considered.

##  V Discussion

We explored the question of observing quantum nonlocality considering local
models with relaxed measurement independence. Using the approach developed by
Pütz et al. Puetz_2014 , we showed that the requirements in terms of
measurement independence can still be considerably relaxed, and presented an
example of quantum correlations featuring nonlocality in presence of strong
measurement dependence. This suggests that nonlocality could be demonstrated
with strings of measurement settings having only a very small amount of global
min entropy.

The main open question is arguably whether our work has now reached the
absolute minimal requirements in terms of measurement independence allowing
for quantum nonlocality. Of course, this question is only meaningful within a
certain framework. For the approach developed in Pütz et al. Puetz_2014 , we
believe that our work has identified the minimal requirements. An interesting
question is now to see how this result may impact applications in randomness
amplification.

Finally, our work has also interesting consequences from the point of view of
network nonlocality. Following the construction of Ref. Supic , one can embed
our example of strong measurement-dependent quantum nonlocality in a triangle
network without inputs. In this way, one obtains a quantum distribution that
cannot be reproduced classically even when the three sources of the triangle
network are perfectly correlated in almost all rounds of the Bell test.

_Acknowledgements.—_ We acknowledge financial support from the Swiss National
Science Foundation (project 2000021_192244/1 and NCCR SwissMAP) and ERC
Starting grant QUSCO.

## References

  * (1) J. S. Bell.  On the einstein podolsky rosen paradox.  Physics Physique Fizika, 1:195–200, Nov 1964.  doi:10.1103/PhysicsPhysiqueFizika.1.195.
  * (2) Alain Aspect, Philippe Grangier, and Gérard Roger.  Experimental tests of realistic local theories via bell’s theorem.  Phys. Rev. Lett., 47:460–463, Aug 1981.  doi:10.1103/PhysRevLett.47.460.
  * (3) W. Tittel, J. Brendel, H. Zbinden, and N. Gisin.  Violation of bell inequalities by photons more than 10 km apart.  Phys. Rev. Lett., 81:3563–3566, Oct 1998.  doi:10.1103/PhysRevLett.81.3563.
  * (4) M. A. Rowe, D. Kielpinski, V. Meyer, C. A. Sackett, W. M. Itano, C. Monroe, and D. J. Wineland.  Experimental violation of a bell’s inequality with efficient detection.  Nature, 409(6822):791–794, 2001.
  * (5) B. G. Christensen, K. T. McCusker, J. B. Altepeter, B. Calkins, T. Gerrits, A. E. Lita, A. Miller, L. K. Shalm, Y. Zhang, S. W. Nam, N. Brunner, C. C. W. Lim, N. Gisin, and P. G. Kwiat.  Detection-loophole-free test of quantum nonlocality, and applications.  Phys. Rev. Lett., 111:130406, Sep 2013.  doi:10.1103/PhysRevLett.111.130406.
  * (6) Bas Hensen, Hannes Bernien, Anaïs E Dréau, Andreas Reiserer, Norbert Kalb, Machiel S Blok, Just Ruitenberg, Raymond FL Vermeulen, Raymond N Schouten, Carlos Abellán, et al.  Loophole-free bell inequality violation using electron spins separated by 1.3 kilometres.  Nature, 526(7575):682–686, 2015.  doi:10.1038/nature15759.
  * (7) Marissa Giustina, Marijn A. M. Versteegh, Sören Wengerowsky, Johannes Handsteiner, Armin Hochrainer, Kevin Phelan, Fabian Steinlechner, Johannes Kofler, Jan-Åke Larsson, Carlos Abellán, Waldimar Amaya, Valerio Pruneri, Morgan W. Mitchell, Jörn Beyer, Thomas Gerrits, Adriana E. Lita, Lynden K. Shalm, Sae Woo Nam, Thomas Scheidl, Rupert Ursin, Bernhard Wittmann, and Anton Zeilinger.  Significant-loophole-free test of bell’s theorem with entangled photons.  Phys. Rev. Lett., 115:250401, Dec 2015.  doi:10.1103/PhysRevLett.115.250401.
  * (8) Lynden K. Shalm, Evan Meyer-Scott, Bradley G. Christensen, Peter Bierhorst, Michael A. Wayne, Martin J. Stevens, Thomas Gerrits, Scott Glancy, Deny R. Hamel, Michael S. Allman, Kevin J. Coakley, Shellee D. Dyer, Carson Hodge, Adriana E. Lita, Varun B. Verma, Camilla Lambrocco, Edward Tortorici, Alan L. Migdall, Yanbao Zhang, Daniel R. Kumor, William H. Farr, Francesco Marsili, Matthew D. Shaw, Jeffrey A. Stern, Carlos Abellán, Waldimar Amaya, Valerio Pruneri, Thomas Jennewein, Morgan W. Mitchell, Paul G. Kwiat, Joshua C. Bienfang, Richard P. Mirin, Emanuel Knill, and Sae Woo Nam.  Strong loophole-free test of local realism.  Phys. Rev. Lett., 115:250402, Dec 2015.  doi:10.1103/PhysRevLett.115.250402.
  * (9) Nicolas Brunner, Daniel Cavalcanti, Stefano Pironio, Valerio Scarani, and Stephanie Wehner.  Bell nonlocality.  Rev. Mod. Phys., 86:419–478, Apr 2014.  URL: https://link.aps.org/doi/10.1103/RevModPhys.86.419, doi:10.1103/RevModPhys.86.419.
  * (10) Michael J. W. Hall.  Local deterministic model of singlet state correlations based on relaxing measurement independence.  Phys. Rev. Lett., 105:250404, Dec 2010.  doi:10.1103/PhysRevLett.105.250404.
  * (11) Michael J. W. Hall.  Relaxed bell inequalities and kochen-specker theorems.  Phys. Rev. A, 84:022102, Aug 2011.  doi:10.1103/PhysRevA.84.022102.
  * (12) Jonathan Barrett and Nicolas Gisin.  How much measurement independence is needed to demonstrate nonlocality?  Phys. Rev. Lett., 106:100406, Mar 2011.  doi:10.1103/PhysRevLett.106.100406.
  * (13) Manik Banik, MD. Rajjak Gazi, Subhadipa Das, Ashutosh Rai, and Samir Kunkri.  Optimal free will on one side in reproducing the singlet correlation.  Journal of Physics A: Mathematical and Theoretical, 45(20):205301, May 2012.  doi:10.1088/1751-8113/45/20/205301.
  * (14) Dax Enshan Koh, Michael J. W. Hall, Setiawan, James E. Pope, Chiara Marletto, Alastair Kay, Valerio Scarani, and Artur Ekert.  Effects of reduced measurement independence on bell-based randomness expansion.  Phys. Rev. Lett., 109:160404, Oct 2012.  doi:10.1103/PhysRevLett.109.160404.
  * (15) Le Phuc Thinh, Lana Sheridan, and Valerio Scarani.  Bell tests with min-entropy sources.  Phys. Rev. A, 87:062121, Jun 2013.  doi:10.1103/PhysRevA.87.062121.
  * (16) Ernest Y.-Z. Tan, Yu Cai, and Valerio Scarani.  Measurement-dependent locality beyond independent and identically distributed runs.  Phys. Rev. A, 94:032117, Sep 2016.  doi:10.1103/PhysRevA.94.032117.
  * (17) R. Chaves, R. Kueng, J. B. Brask, and D. Gross.  Unifying framework for relaxations of the causal assumptions in bell’s theorem.  Phys. Rev. Lett., 114:140403, Apr 2015.  doi:10.1103/PhysRevLett.114.140403.
  * (18) Andrew S. Friedman, Alan H. Guth, Michael J. W. Hall, David I. Kaiser, and Jason Gallicchio.  Relaxed bell inequalities with arbitrary measurement dependence for each observer.  Physical Review A, 99(1), Jan 2019.  doi:10.1103/physreva.99.012121.
  * (19) Gilles Pütz, Denis Rosset, Tomer Jack Barnea, Yeong-Cherng Liang, and Nicolas Gisin.  Arbitrarily small amount of measurement independence is sufficient to manifest quantum nonlocality.  Physical Review Letters, 113(19), Nov 2014.  doi:10.1103/physrevlett.113.190402.
  * (20) Gilles Pütz and Nicolas Gisin.  Measurement dependent locality.  New Journal of Physics, 18(5):055006, 2016.  doi:10.1088/1367-2630/18/5/055006.
  * (21) Djeylan Aktas, Sébastien Tanzilli, Anthony Martin, Gilles Pütz, Rob Thew, and Nicolas Gisin.  Demonstration of quantum nonlocality in the presence of measurement dependence.  Physical Review Letters, 114(22), Jun 2015.  doi:10.1103/physrevlett.114.220404.
  * (22) Max Kessler and Rotem Arnon-Friedman.  Device-independent randomness amplification and privatization.  IEEE Journal on Selected Areas in Information Theory, 1(2):568–584, 2020.  doi:10.1109/JSAIT.2020.3012498.
  * (23) Ivan Šupić, Jean-Daniel Bancal, and Nicolas Brunner.  Quantum nonlocality in networks can be demonstrated with an arbitrarily small level of independence between the sources.  Phys. Rev. Lett., 125:240403, Dec 2020.  doi:10.1103/PhysRevLett.125.240403.
  * (24) Lucien Hardy.  Nonlocality for two particles without inequalities for almost all entangled states.  Phys. Rev. Lett., 71:1665–1668, Sep 1993.  URL: https://link.aps.org/doi/10.1103/PhysRevLett.71.1665, doi:10.1103/PhysRevLett.71.1665.
  * (25) A Peres.  Two simple proofs of the kochen-specker theorem.  Journal of Physics A: Mathematical and General, 24(4):L175–L178, feb 1991.  doi:10.1088/0305-4470/24/4/003.
  * (26) N. David Mermin.  Extreme quantum entanglement in a superposition of macroscopically distinct states.  Phys. Rev. Lett., 65:1838–1840, 1990.  doi:10.1103/PhysRevLett.65.1838.
  * (27) Leandro Aolita, Rodrigo Gallego, Antonio Acín, Andrea Chiuri, Giuseppe Vallone, Paolo Mataloni, and Adán Cabello.  Fully nonlocal quantum correlations.  Phys. Rev. A, 85:032107, Mar 2012.  URL: https://link.aps.org/doi/10.1103/PhysRevA.85.032107, doi:10.1103/PhysRevA.85.032107.
  * (28) Antonio Acín and Lluis Masanes.  Certified randomness in quantum physics.  Nature, 540(7632):213–219, Dec 2016.  doi:10.1038/nature20119.
  * (29) Roger Colbeck and Renato Renner.  Free randomness can be amplified.  Nature Physics, 8(6):450–453, May 2012.  doi:10.1038/nphys2300.
  * (30) Rodrigo Gallego, Lluis Masanes, Gonzalo De La Torre, Chirag Dhara, Leandro Aolita, and Antonio Acín.  Full randomness from arbitrarily deterministic events.  Nature Communications, 4(1), Oct 2013.  doi:10.1038/ncomms3654.
  * (31) Carl A Miller and Yaoyun Shi.  Robust protocols for securely expanding randomness and distributing keys using untrusted quantum devices.  Journal of the ACM (JACM), 63(4):1–63, 2016.  doi:10.1145/2885493.
  * (32) Fernando G. S. L. Brandao, Ravishankar Ramanathan, Andrzej Grudka, Karol Horodecki, Michał Horodecki, Paweł Horodecki, Tomasz Szarek, and Hanna Wojewódka.  Realistic noise-tolerant randomness amplification using finite number of devices.  Nature Communications, 7(1), Apr 2016.  doi:10.1038/ncomms11345.
  * (33) Ravishankar Ramanathan, Fernando G. S. L. Brandao, Karol Horodecki, Michał Horodecki, Paweł Horodecki, and Hanna Wojewódka.  Randomness amplification under minimal fundamental assumptions on the devices.  Physical Review Letters, 117(23), Nov 2016.  doi:10.1103/physrevlett.117.230501.

##  VI Appendix

In this Appendix, we show another example of quantum correlations featuring
strong MDL for any parameter values $q>0$ and $\xi>0$. This is based on a
tripartite Bell test, involving a third party, Charlie, whose input is denoted
$z$ and output $c$. The main idea is to start from the Mermin Bell inequality
Mermin  which also has the property that the quantum bound is equal to the
algebraic one. The optimal quantum strategy is that of the famous GHZ paradox.

As our construction in the main text, we must however first modify the Mermin
inequality in a form that is suitable for our purpose. This implies using
suitable definitions for correlators. For a given set of inputs $x,y,z$ we
define the weighted correlator:

$\tilde{C}_{x,y,z}=\sum_{a,b,c}3^{a\oplus b\oplus c\oplus 1}(-1)^{a\oplus
b\oplus c}p(a,b,c|x,y,z),$ (23)

with $a,b,c\in\\{0,1\\}$. We observe that $-1\leq\tilde{C}_{x,y,z}\leq 3$.
From this, we now construct an alternative version of the Mermin Bell
inequality, namely

$\tilde{I}=\tilde{C}_{0,0,0}-\tilde{C}_{1,1,0}-\tilde{C}_{1,0,1}-\tilde{C}_{0,1,1}.$
(24)

The quantum bound for this inequality is $\tilde{I}=4$, which can be attained
by considering a three-qubit GHZ state and each party performing local
measurements in the Pauli $X$ and $Y$ basis (similarly as in the well-known
GHZ paradox, giving maximal violation of the Mermin Bell inequality). It turns
out that this quantum bound coincides with the algebraic bound of the
inequality, i.e. $\beta_{q}=\beta_{alg}$, similarly to the standard Mermin
inequality. The notable difference, which is convenient for our purpose, is
that the local bound is now $\beta_{loc}=0$.

We can now apply the method outlined above to adapt this inequality to the
measurement-dependent scenario, namely

$\tilde{I}^{md}=\tilde{C}^{\eta,\xi}_{0,0,0}-\tilde{C}^{\eta,\xi}_{1,1,0}-\tilde{C}^{\eta,\xi}_{1,0,1}-\tilde{C}^{\eta,\xi}_{0,1,1},$
(25)

where $\tilde{C}^{\eta,\xi}_{x,y,z}$ is now a weighted correlator of joint
distributions, defined as

$\tilde{C}^{\eta,\xi}_{x,y,z}=\sum_{a,b,c}(3\eta)^{a\oplus b\oplus c\oplus
1}(-\xi)^{a\oplus b\oplus c}p(a,b,c,x,y,z).$ (26)

Finally, we find the local and quantum bounds for this new inequality:
$\beta_{\mathrm{loc}}^{smd}=(1-q)\xi$ and $\beta_{\mathrm{q}}^{smd}=\xi$.
Hence we obtain a quantum violation for any parameter values $q>0$ and
$\xi>0$.


#  $\mathfrak{X}$-elements in multiplicative lattices - A generalization of
$J$-ideals, $n$-ideals and $r$-ideals in rings

Sachin Sarode* and Vinayak Joshi  *Department of Mathematics, Shri Muktanand
College
Gangapur, Dist. Aurangabad - 431 109, India<EMAIL_ADDRESS>**Department of Mathematics, Savitribai Phule Pune University, Pune-411 007,
India<EMAIL_ADDRESS>
<EMAIL_ADDRESS>
(Date: January 17, 2021)

###### Abstract.

In this paper, we introduce a concept of $\mathfrak{X}$-element with respect
to an $M$-closed set $\mathfrak{X}$ in multiplicative lattices and study
properties of $\mathfrak{X}$-elements. For a particular $M$-closed subset
$\mathfrak{X}$, we define the concept of $r$-element, $n$-element and
$J$-element. These elements generalize the notion of $r$-ideals, $n$-ideals
and $J$-ideals of a commutative ring with unity to multiplicative lattices. In
fact, we prove that an ideal $I$ of a commutative ring $R$ with unity is a
$n$-ideal ($J$-ideal) of $R$ if and only if it is an $n$-element ($J$-element)
of $Id(R)$, the ideal lattice of $R$.

###### 2020 Mathematics Subject Classification:

Primary 13A15, 13C05, 06F10 Secondary 06A11

Keywords: Multiplicative lattice, prime element, $\mathfrak{X}$-element,
$n$-element, $J$-element, $r$-element, commutative ring, $n$-ideal, $r$-ideal,
$J$-ideal.

##  1\. Introduction

The ideal theory of commutative rings with unity is very rich. Many
researchers defined different ideals ranging from prime ideals, maximal
ideals, primary ideals to recently introduced $r$-ideals, $n$-ideals and
$J$-ideals. More details about $r$-ideals, $n$-ideals, and $J$-ideals can be
found in Mohamadian [11], Tekir et al. [12] and, Khashan and Bani-Ata [9]
respectively.

Ward and Dilworth [13] introduced the concept of multiplicative lattices to
generalize the ideal theory of commutative rings with unity. Analogously, the
concepts of a prime element, maximal element, primary elements are defined.

The study of prime elements and its generalization in a multiplicative lattice
is the main focus of many researchers. Different classes of elements and
generalization of a prime element in multiplicative lattices were studied; see
Burton [1], Joshi and Ballal [4], Jayaram [5], Jayaram and Johnson [6, 7],
Jayaram et al. [8], Manjarekar and Bingi [10].

We observed a unifying pattern in the results of $J$-ideals, $n$-ideals and
$r$-ideals of rings. This motivates us to introduce a new class of elements,
namely $\mathfrak{X}$-element in multiplicative lattices. Hence this study
will unify many of the results proved for these ideals and generalize it to
multiplicative lattice settings. Further, for a particular $M$-closed subset
$\mathfrak{X}$, we define the concept of $r$-element, $n$-element and
$J$-element. Hence this justifies the name $\mathfrak{X}$-element. These
elements are the generalizations of $r$-ideals, $n$-ideals and $J$-ideals of a
commutative ring with unity. In fact, we prove that an ideal $I$ of a
commutative ring $R$ with unity is a $n$-ideal($J$-ideal) of $R$ if and only
if it is an $n$-element ($J$-element) of $Id(R)$, the ideal lattice of $R$.

Now, we begin with the necessary concepts and terminology.

######  Definition 1.1.

A nonempty subset $I$ of a lattice $L$ is a semi-ideal if $x\leq a\in I$
implies $x\in I$. A semi-ideal $I$ of $L$ is an ideal if $a\vee b\in I$
whenever $a,b\in I$. An ideal (semi-ideal) $I$ of a lattice $L$ is a proper
ideal (semi-ideal) of $L$ if $I\neq L$. A proper ideal (semi-ideal) $I$ is
prime if $a\wedge b\in I$ implies $a\in I$ or $b\in I$, and it is minimal if
it does not properly contain another prime ideal (prime semi-ideal). For $a\in
L$, let $(a]=\\{x\in L\colon\,x\leq a\\}$. The set $(a]$ is the principal
ideal generated by a.

A lattice $L$ is complete if for any subset $S$ of $L$, we have $\bigvee
S,\bigwedge S\in L$. The smallest element and the greatest element of a
lattice $L$ is denoted by 0 and 1 respectively.

The concept of multiplicative lattices was introduced by Ward and Dilworth
[13] to study the abstract commutative ideal theory of commutative rings.

A complete lattice $L$ is a multiplicative lattice if there exists a binary
operation $``\cdot"$ called the multiplication on $L$ satisfying the following
conditions:

  1. (1)

$a\cdot b=b\cdot a$, for all $a,b\in L$.

  2. (2)

$a\cdot(b\cdot c)=(a\cdot b)\cdot c$, for all $a,b,c\in L$.

  3. (3)

$a\cdot(\bigvee_{\alpha}b_{\alpha})=\bigvee_{\alpha}(a\cdot b_{\alpha})$, for
all $a,b_{\alpha}\in L$, $\alpha\in\Lambda$(an index set).

  4. (4)

$a\cdot 1=a$, for all $a\in L$.

Note that in a multiplicative lattice $L$, $a\cdot b\leq a\wedge b$ for
$a,b\in L$. For this, let $a=a\cdot 1=a\cdot(b\vee 1)=a\cdot b\vee a$. Thus
$a\cdot b\leq a$. Similarly, $a\cdot b\leq b$. This proves that $a\cdot b\leq
a\wedge b$. Moreover, if $a\leq b$ in $L$, then $a\cdot c\leq b\cdot c$ for
every $c\in L$. Also, if $a\leq b$ and $c\leq d$ then $a\cdot c\leq b\cdot d$.

An element $c$ of a complete lattice $L$ is compact if
$c\leq\bigvee_{\alpha}a_{\alpha}$, $\alpha\in\Lambda$ ($\Lambda$ is an index
set) implies $c\leq\bigvee_{i=1}^{n}a_{\alpha_{i}}$, where
$n\in\mathbb{Z}^{+}$. The set of all compact elements of a lattice $L$ is
denoted by $L_{*}$.

A lattice $L$ is compactly generated or algebraic if for every $x\in L$, there
exist $x_{\alpha}\in L_{*}$ for $\alpha\in\Lambda$(an index set) such that
$x=\bigvee_{\alpha}x_{\alpha}$, that is, every element is a join of compact
elements. Equivalently, if $L$ is a compactly generated lattice and if
$a\not\leq b$ for $a,b\in L$, then there exists a nonzero compact element
$c\in L_{*}$ such that $c\leq a$ and $c\not\leq b$.

A multiplicative lattice $L$ is $1$-compact if $1$ is a compact element of
$L$. A multiplicative lattice $L$ is compact if every element of $L$ is a
compact element.

A multiplicative lattice $L$ is a $c$-lattice if $L$ is 1-compact, compactly
generated multiplicative lattice in which the product of two compact elements
is compact. Note that the ideal lattice of a commutative ring $R$ with unity
is always a $c$-lattice.

An element $p$ of a multiplicative lattice $L$ with $p\neq 1$ is prime if
$a\cdot b\leq p$ implies $a\leq p$ or $b\leq p$. It is not difficult to prove
that an element $p$ (with $p\neq 1$) of a $c$-lattice $L$ is prime if $a\cdot
b\leq p$ for $a,b\in L_{*}$ implies $a\leq p$ or $b\leq p$. An element $p$ is
said to be a minimal prime element if there is no prime element $q$ such that
$q<p$. An ideal $P$ of a commutative ring $R$ with unity is prime if and only
if it is a prime element of $Id(R)$, the ideal lattice of $R$.

Let $L$ be a $c$-lattice and $a\in L$. Then the radical of $a$ is denoted by
$\sqrt{a}$ and given by $\sqrt{a}=\bigvee\\{x\in L_{*}\;|\;\ x^{n}\leq a$ for
some $n\in\mathbb{N}\\}$. Note that if any compact element $c\leq\sqrt{a}$,
then $c^{m}\leq a$ for some $m\in\mathbb{N}$. An element $a$ of a $c$-lattice
is a radical element, if $a=\sqrt{a}$. A $c$-lattice is called a domain if $0$
is a prime element of $L$.

A proper element $i$ of a $c$-lattice $L$ is called  primary element if
whenever $a\cdot b\leq i$ for some $a,b\in L$ then either $a\leq i$ or
$b\leq\sqrt{i}$.

A proper element $m$ of a multiplicative lattice is said to be maximal, if
$m\leq n<1$, then $m=n$. The set of all maximal elements of $L$ is denoted by
Max$(L)$. The Jacobson radical of $L$ is the set
$J(L)=\displaystyle\bigwedge\\{m\ |\ m\in\text{Max}(L)\\}$. It is easy to
observe that a maximal element of a $c$-lattice $L$ is prime. A $c$-lattice
$L$ is said to be local, if $L$ has the unique maximal element $m$. In this
case, we write $(L;m)$.

A non-empty subset $\mathfrak{X}$ of $L_{*}$(set of all compact elements) in a
$c$-lattice $L$ is multiplicatively closed if $s_{1}\cdot
s_{2}\in\mathfrak{X}$, whenever $s_{1},s_{2}\in\mathfrak{X}$.

A non-empty subset $\mathfrak{X}$ of a multiplicative lattice $L$ is called
$M$-closed if $a,b\in\mathfrak{X}$, then $a\cdot b\in\mathfrak{X}$.

From the definitions, it is clear that every $M$-closed subset $A$ of a
$c$-lattice is a multiplicatively closed subset of $L$. The converse is not
true. However, if $L$ is a compact lattice or finite, then $L=L_{*}$ and hence
both definitions coincide with each other. Further, if $p$ is a prime element
of a $c$-lattice $L$, then $L\setminus(p]$ is an $M$-closed subset of $L$.

In a multiplicative lattice $L$, an element $a\in L$ is nilpotent if $a^{n}=0$
for some $n\in\mathbb{Z}^{+}$, and $L$ is reduced if the only nilpotent
element is 0. The set of all nilpotent elements of a multiplicative lattice
$L$ is denoted by Nil$(L)$.

We denote the set $Z(L)$ of zero-divisors in $L$ by the set
$Z(L)=\bigl{\\{}x\in L\ |\ x\cdot y=0\text{ for some }y\in
L\setminus\\{0\\}\bigr{\\}}$. Clearly, Nil$(L)\subseteq Z(L)$.

Let $L$ be a multiplicative lattice and $a,b\in L$. Then
$(a:b)=\bigvee\\{x\;|\;x\cdot b\leq a\\}$. Note that $x\cdot b\leq
a\Leftrightarrow x\leq(a:b)$. Clearly, $a\leq(a:b)$ and $(a:b)\cdot b\leq a$
for $a,b\in L$. If $a\in L$, then $ann_{L}(a)=\bigvee\\{x\in L\;|\>a\cdot
x=0\\}$.

For undefined concepts in lattices, see Grätzer [3].

##  2\. $\mathfrak{X}$-Elements in multiplicative Lattices

We introduce the concept of an $\mathfrak{X}$-element in multiplicative
lattices.

######  Definition 2.1.

Let $L$ be a multiplicative lattice and $\mathfrak{X}$ be an $M$-closed subset
of $L$. A proper element $i$ of a multiplicative lattice $L$ is called an
$\mathfrak{X}$-element, if $a\cdot b~{}\leq i$ with $a~{}\notin\mathfrak{X}$
implies $b~{}\leq i$ for $a,~{}b~{}\in L$.

######  Example 2.2.

Consider a lattice $K$ whose Hasse diagram is shown in Figure 1. On $K$,
define the trivial multiplication $x\cdot y=0=y\cdot x$ for every
$x,~{}y\not\in\\{1\\}$ and $x\cdot 1=x=1\cdot x$ for every $x\in K$. It is
easy to see that $K$ is a multiplicative lattice. Moreover, $K$ is non-
reduced. If we take $\mathfrak{X}=\\{0,~{}a,~{}b,~{}c,~{}d\\}$, then every
proper element of $K$ is an $\mathfrak{X}$-element of $K$.

$0$$c$$a$$b$$d$$1$$K$

Figure 1. A multiplicative lattice in which every proper element is an
$\mathfrak{X}$-element

######  Remark 2.3.

Note that a proper element of a multiplicative lattice $L$ is an
$\mathfrak{X}$-element or not, depends on an $M$-closed subset $\mathfrak{X}$
under consideration. If $x$ is an $\mathfrak{X}_{1}$-element with respect to
an $M$-closed subset $\mathfrak{X}_{1}$, then $x$ may or may not be an
$\mathfrak{X}_{2}$-element with respect to an $M$-closed subset
$\mathfrak{X}_{2}$ different from $\mathfrak{X}_{1}$.

Also, note that if $L$ is a multiplicative lattice and $\mathfrak{X}=\\{1\\}$
is an $M$-closed subset of $L$, then $L$ does not contain an
$\mathfrak{X}$-element.

######  Lemma 2.4.

Let $L$ be a multiplicative lattice and $\mathfrak{X}$ be an $M$-closed subset
of $L$. If $i$ is an $\mathfrak{X}$-element of $L$, then
$(i]~{}\subseteq\mathfrak{X}$. In particular, if $(i]=\mathfrak{X}$, then $i$
is an $\mathfrak{X}$-element of $L$ if and only if $i$ is a prime element of
$L$.

###### Proof.

Suppose $i$ is an $\mathfrak{X}$-element of a multiplicative lattice $L$ and
let $x\in(i]$. Suppose on the contrary that $x\notin\mathfrak{X}$. Clearly,
$x\cdot 1\leq i$ with $x\notin\mathfrak{X}$. Since $i$ is an
$\mathfrak{X}$-element, we get $1\leq i$, a contradiction to the fact that $i$
is a proper element of $L$. Therefore $x\in\mathfrak{X}$ and hence
$(i]~{}\subseteq\mathfrak{X}$.

Now, we prove “in particular” part. Suppose that $(i]=\mathfrak{X}$ and $i$ is
an $\mathfrak{X}$-element of $L$. Let $a,b\in L$ such that $a\cdot b\leq i$
and $a\nleq i$, i.e., $a\notin\mathfrak{X}$. As $i$ is an
$\mathfrak{X}$-element, $b\leq i$. So $i$ is a prime element of $L$.

Conversely, suppose that $(i]=\mathfrak{X}$ and $i$ is a prime element of $L$.
Let $a,b\in L$ such that $a\cdot b\leq i$ with $a\not\in\mathfrak{X}=(i]$. By
primeness of $i$, $b\leq i$. Thus $i$ is an $\mathfrak{X}$-element of $L$. ∎

######  Remark 2.5.

The converse of the Lemma 2.4 need not be true in general, i.e., if $i$ is a
proper element of a multiplicative lattice $L$ such that $i\in\mathfrak{X}$,
then $i$ need not be an $\mathfrak{X}$-element of $L$. Consider the ideal
lattice $L$ of the ring $\mathbb{Z}_{12}$. Clearly, $L$ is a non-reduced
lattice. Put $\mathfrak{X}=\\{(0),~{}(6)\\}$. Then $(0),(6)\in\mathfrak{X}$
but $(0)$ and $(6)$ are not $\mathfrak{X}$-elements of $L$.

######  Lemma 2.6.

Let $L$ be a multiplicative lattice and $\mathfrak{X}$ and
$\mathfrak{X^{\prime}}$ be $M$-closed subsets of $L$ such that
$\mathfrak{X}\subseteq\mathfrak{X^{\prime}}$. If $i$ is an
$\mathfrak{X}$-element of $L$, then $i$ is an $\mathfrak{X^{\prime}}$-element
of $L$.

###### Proof.

follows from the definition of an $\mathfrak{X}$-element. ∎

######  Lemma 2.7.

Let $(L;m)$ be a local lattice. Then every proper element of $L$ is an
$\mathfrak{X}$-element for $\mathfrak{X}=(m]$.

###### Proof.

Let $a\cdot b\leq i$ and $a\notin\mathfrak{X}=(m]$. Since $L$ is local, $a=1$.
Hence in this case $b\leq i$. This proves that $i$ is an
$\mathfrak{X}$-element. ∎

######  Lemma 2.8.

Assume that every proper element of a $c$-lattice $L$ is an
$\mathfrak{X}$-element, where $\mathfrak{X}=(m]$ and $m\in L\setminus\\{1\\}$.
Then $m$ is a unique maximal element of $L$.

###### Proof.

Let $i$ be a proper element of $L$ which is an $\mathfrak{X}$-element. Then by
Lemma 2.4, $i\leq m$. This is true for all proper elements $i$ of $L$. In
particular, it is true for all maximal elements $m^{\prime}$ too. This proves
that $L$ has unique maximal element. Therefore it is a meet-irreducible
element of $L$. ∎

######  Lemma 2.9.

Let $L$ be a multiplicative lattice and $\mathfrak{X}$ be an $M$-closed subset
of $L$. If $\\{i_{j}\\}$, where $j\in\Lambda$ (an index set), is a non-empty
set of $\mathfrak{X}$-elements of $L$, then $\bigwedge_{j}i_{j}$ is also an
$\mathfrak{X}$-element.

###### Proof.

Obvious.∎

######  Remark 2.10.

The join of two $\mathfrak{X}$-elements is not necessarily an
$\mathfrak{X}$-element. Consider the ideal lattice $L$ of $\mathbb{Z}_{15}$
with $\mathfrak{X}=\\{(0),(3),(5)\\}$. Then $(3),(5)$ are
$\mathfrak{X}$-elements of $L$, but $(3)\vee(5)=(1)$ is not
an$\mathfrak{X}$-element.

######  Lemma 2.11.

Let $i$ be a proper element of a $c$-lattice $L$ and $\mathfrak{X}$ be an
$M$-closed subset of $L$. Then $i$ is an $\mathfrak{X}$-element of $L$ if and
only if $i=(i:a)$ for all $a\notin\mathfrak{X}$. In particular, if $i$ is an
$\mathfrak{X}$-element of $L$, then $(i:a)$ is an $\mathfrak{X}$-element of
$L$ for all $a\notin\mathfrak{X}$.

###### Proof.

Suppose $i$ is an $\mathfrak{X}$-element of $L$ and let $a\notin\mathfrak{X}$.
We always have $i\leq(i:a)$. Let $x$ be any compact element such that
$x\leq(i:a)$. Therefore $x\cdot a\leq i$. Since $i$ is an
$\mathfrak{X}$-element and $a\notin\mathfrak{X}$, we get $x\leq i$. Hence
$(i:a)\leq i$, as $L$ is a $c$-lattice. Therefore $i=(i:a)$ for all
$a\notin\mathfrak{X}$.

Conversely, suppose that $i=(i:a)$ for all $a\notin\mathfrak{X}$. Let $c,d\in
L$ such that $c\cdot d~{}\leq i$ with $c~{}\notin\mathfrak{X}$. We claim that
$d\leq i$. Since $c\cdot d~{}\leq i$, we have $d\leq(i:c)$. As
$c~{}\notin\mathfrak{X}$, by the assumption $(i:c)=i$, we have $d\leq i$.
Therefore, $i$ is an $\mathfrak{X}$-element of $L$. Further, “in particular
part” is easy to observe. ∎

######  Lemma 2.12.

Let $i$ be a proper element of a multiplicative lattice $L$ and $\mathfrak{X}$
be an $M$-closed subset of $L$. Then the following statements are equivalent.

  1. (1)

$i$ is an $\mathfrak{X}$-element of $L$.

  2. (2)

$(i:a)$ is an $\mathfrak{X}$-element of $L$ for every $a\nleq i$.

  3. (3)

$\Bigl{(}(i:a)\Bigr{]}\subseteq\mathfrak{X}$ for all $a\nleq i$.

###### Proof.

$(1)\implies(2)$: Suppose that $i$ is an $\mathfrak{X}$-element of $L$ with
$j\nleq i$. Clearly, $(i:j)\not=1$. Let $a,b\in L$ such that $a\cdot
b\leq(i:j)$ with $a\notin\mathfrak{X}$. So $a\cdot b\cdot j\leq i$. As $i$ is
an $\mathfrak{X}$-element and $a\notin\mathfrak{X}$, we get $b\cdot j\leq i$,
i.e., $b\leq(i:j)$. Therefore $(i:j)$ is an $\mathfrak{X}$-element of $L$.

$(2)\implies(3)$: follows from Lemma 2.4.

$(3)\implies(1)$: Suppose that $((i:a)]\subseteq\mathfrak{X}$ for all $a\nleq
i$. Let $c,d\in L$ such that $c\cdot d\leq i$ with $c\notin\mathfrak{X}$. We
claim that $d\leq i$. Suppose $d\nleq i$. So by the assumption and
$c\leq(i:d)$, we have $c\in\mathfrak{X}$, a contradiction. Therefore $d\leq
i$. Hence $i$ is an $\mathfrak{X}$-element of $L$. ∎

######  Lemma 2.13.

Let $L$ be a multiplicative lattice and $\mathfrak{X}$ be an $M$-closed subset
of $L$. If $i$ is a maximal $\mathfrak{X}$-element of $L$, then $i$ is a prime
element of $L$.

###### Proof.

Suppose $i$ is a maximal $\mathfrak{X}$-element of $L$. Let $a,b\in L$ such
that $a\cdot b\leq i$ and $a\nleq i$. Since $i$ is an $\mathfrak{X}$-element
and $a\nleq i$, by Lemma 2.12, $(i:a)$ is an $\mathfrak{X}$-element of $L$. As
$i$ is a maximal $\mathfrak{X}$-element of $L$ and $i\leq(i:a)$, we get
$(i:a)=i$. Therefore $b\leq i$. ∎

######  Lemma 2.14.

Let $j$ be a proper element of a $c$-lattice $L$ and $\mathfrak{X}=(j]$ be an
$M$-closed subset of $L$. Then a proper element $i$ is an
$\mathfrak{X}$-element if and only if the condition $(*)$:

$(*)$: for all $a,~{}b~{}\in L_{*}$ (set of all compact elements), $a\cdot
b~{}\leq i$ with $a~{}\notin\mathfrak{X}$ implies $b~{}\leq i$.

###### Proof.

Assume that the condition $(*)$ holds. Let $a,b\in L$ such that $a\cdot
b~{}\leq i$ with $a~{}\notin\mathfrak{X}$. As $L$ is a $c$-lattice and
$a~{}\nleq j$, there exists $(0\not=)x\in L_{*}$ such that $x\leq a$ and
$x~{}\nleq j$. Now, let $y$ be a compact element such that $y\leq b$. As
$x\cdot y\leq a\cdot b\leq i$ with $x~{}\notin\mathfrak{X}=(j]$, by the
condition $(*)$, $y\leq i$. Thus every compact element $\leq b$ is $\leq i$
and $L$ is a $c$-lattice, we get $b\leq i$. Hence $i$ is an
$\mathfrak{X}$-element. The converse is obvious. ∎

######  Lemma 2.15.

Let $L$ be a $c$-lattice and $\mathfrak{X}=(j]$ be an $M$-closed subset of
$L$. Then for a prime element $i$ of $L$ with $j\leq i$, $i$ is an
$\mathfrak{X}$-element if and only if $i=j$.

###### Proof.

Assume that $i$ is a prime element which is also an $\mathfrak{X}$-element of
$L$. By Lemma 2.4, we have $i\leq j$. This together with $j\leq i$, we have
$i=j$. Conversely, assume that $i=j$ and $i$ is prime. To prove $i$ is an
$\mathfrak{X}$-element, assume that $a\cdot b\leq i$ and
$a\notin\mathfrak{X}$. Then by primeness of $i$ and $i=j$, we have $b\leq i$.
∎

######  Corollary 2.16.

Let $L$ be a $c$-lattice and $\mathfrak{X}=(j]$ be an $M$-closed subset of
$L$. Then for a maximal element $i$ of $L$, $i$ is an $\mathfrak{X}$-element
if and only if $i=j$.

###### Proof.

Assume that $i$ is a maximal element which is also an $\mathfrak{X}$-element
of $L$. By Lemma 2.4, we have $i\leq j$. Thus by the maximality of $i$, we
have $i=j$. Conversely, assume that $i=j$. Since $i$ is maximal, it is prime.
Thus the result follows from Lemma 2.15. ∎

######  Theorem 2.17.

Let L be a $c$-lattice and $\mathfrak{X}=(j]$ be an $M$-closed subset of $L$,
where
$j=\bigwedge P$ where $P=\bigl{\\{}i_{k}\ |\ i_{k}\text{ is a prime elements
of }L\bigr{\\}}$. Then the following statements are equivalent.

  1. (1)

There exists an $\mathfrak{X}$-element in $L$.

  2. (2)

$j$ is a prime element of $L$.

Moreover, if the set $Min(L)$ of all minimal prime elements in $L$ is finite,
then all the above conditions are equivalent to $(*)$: $|Min(L)|=1$.

###### Proof.

(1) $\Rightarrow$ (2): Suppose there exists an $\mathfrak{X}$-element $i$ in
$L$. Let $\beta=\\{x\ |\ x$ is an $\mathfrak{X}$-element in $L\\}$. As
$i\in\beta$, $\beta$ is a poset under induced partial order of $L$. Let
$j_{1}\leq j_{2}\leq\cdots\leq j_{n}\leq\cdots$ be a chain $\mathcal{C}$ in
$\beta$. We claim that $j=\bigvee^{\infty}_{\alpha=1}j_{\alpha}$ is in
$\beta$, i.e., $j=\bigvee^{\infty}_{\alpha=1}j_{\alpha}$ is an
$\mathfrak{X}$-element of $L$. Let $a,b\in L_{*}$ (set of all compact element
of $L$) such that $a\cdot b\leq j=\bigvee^{\infty}_{\alpha=1}j_{\alpha}$ and
$a\not\leq j$. As $L$ is a $c$-lattice and $a,b\in L_{*}$, we get $a\cdot b\in
L_{*}$. Therefore $a\cdot b\leq j_{1}\vee j_{2}\vee\cdots\vee j_{n}$ for some
$j_{1},j_{2},\cdots,j_{n}\in\mathcal{C}$. Since $\mathcal{C}$ is a chain, we
must have $j_{1}\vee j_{2}\vee\cdots\vee j_{n}=j_{\gamma}$ for some $\gamma$,
where $1\leq\gamma\leq n$. Thus $a\cdot b\leq j_{\gamma}$ with $a\nleq
j_{\gamma}$. Since $j_{\gamma}$ is an $\mathfrak{X}$-element of $L$, we have
$b\leq j_{\gamma}$. Thus $b\leq\bigvee^{\infty}_{\alpha=1}j_{\alpha}=j$.
Therefore $j$ is an $\mathfrak{X}$-element of $L$. By Zorn’s Lemma $\beta$ has
a maximal element $w$, that is, $w$ is a maximal $\mathfrak{X}$-element. By
Lemma 2.13, $w$ is a prime element of $L$, that is, $w\in P$. Hence $j\leq w$.
Also, $w$ is an $\mathfrak{X}$-element, we have by Lemma 2.4, $w\leq j$. Thus
$j=w$. Hence $j$ is prime.

(2) $\Rightarrow$ (1): Let $j$ be a prime element. By Lemma 2.15, $j$ is an
$\mathfrak{X}$-element.

We, now, prove $(2)\iff(*)$.

Assume that $j$ is a prime element. Since $j=\bigwedge\bigl{\\{}i_{k}\ |\
i_{k}\text{ is a prime elements of }L\bigr{\\}}$ and the set $Min(L)$ is
finite, without loss of generality, we assume that $\displaystyle
j=\bigwedge_{k=1}^{n}\bigl{\\{}i_{k}\ |\ i_{k}\in Min(L)\bigr{\\}}$. By
primeness of $j$ and $j\leq i_{k}$ for all $k$ with $i_{k}\in Min(L)$, we have
$j=i_{k}$ for all $k$. Thus $|Min(L)|=1$.

Conversely, assume that $|Min(L)|=1$. Let $p$ be the only minimal prime
element in $L$. Then $p\leq i_{k}$ for every $k$. Hence $j=p$. This proves
that $j$ is prime. ∎

######  Lemma 2.18 (L. Fuchs and R. Reis [2, Lemma 2.5]).

Let $L$ be a $c$-lattice and $a\in L$. Then radical of $a$ is given by
$\sqrt{a}=\bigwedge\\{p\in L\colon\ $p is a minimal prime element over $a\\}$.

######  Lemma 2.19.

Let L be a $c$-lattice and $\mathfrak{X}=(j]$ be an $M$-closed subset of $L$,
where
$j=\bigwedge\bigl{\\{}i_{k}\ |\ i_{k}\text{ is a prime element of
}L\bigr{\\}}$. Then a proper element $i$ is an $\mathfrak{X}$-element of $L$
if and only if $i$ is a primary element of $L$ and $\sqrt{i}=j$.

###### Proof.

Suppose $i$ is an $\mathfrak{X}$-element of $L$. By Lemma 2.4,
$(i]\subseteq\mathfrak{X}$. Hence $i\leq\sqrt{i}\leq\sqrt{j}=j$. Clearly, by
Lemma 2.18, $j\leq\sqrt{i}$. Hence $j=\sqrt{i}$. Let $a,b\in L$ such that
$a\cdot b\leq i$. By Theorem 2.17, $j$ is a prime element of $L$. Hence either
$a\leq j$ or $b\leq j$. So either $a\leq\sqrt{i}$ or $b\leq\sqrt{i}$. Hence
$i$ is a primary element of $L$. Conversely, suppose that $i$ is a primary
element of $L$ and $\sqrt{i}=j$. Let $a,b\in L$ such that $a\cdot b\leq i$ and
$a\nleq j$. Since $i$ is a primary element and $a\nleq j=\sqrt{i}$, we get
$b\leq i$. Thus $i$ is an $\mathfrak{X}$-element of $L$. ∎

######  Lemma 2.20.

Let L be a $c$-lattice and $\mathfrak{X}=(j]$ be an $M$-closed subset of $L$,
where
$j=\bigwedge\bigl{\\{}i_{k}\ |\ i_{k}\text{ is a maximal element of
}L\bigr{\\}}$. Then a proper element $i$ is an $\mathfrak{X}$-element of $L$
if and only if $i$ satisfies the following condition

$(*)$: If $a\cdot b\leq i$, then $a\leq i$ or $b\leq m$, where
$m=\bigwedge\bigl{\\{}i_{k}\ |\ i_{k}\text{ is a maximal element\;}\geq
i~{}\bigr{\\}}$ and $m=j$.

###### Proof.

follows on similar lines as that of Lemma 2.19. ∎

######  Lemma 2.21.

Let $L$ be a $c$-lattice and $\mathfrak{X}$ be an $M$-closed subset of $L$.
Let $k$ be an element of $L$ such that $k\notin\mathfrak{X}$. If $i_{1}$ and
$i_{2}$ are $\mathfrak{X}$-elements with $i_{1}k=i_{2}k$, then $i_{1}=i_{2}$.
Further, if $i$ is an element such that $ik$ is an $\mathfrak{X}$-element,
then $ik=i$.

###### Proof.

Clearly, $i_{1}k\leq i_{2}$ with $k\notin\mathfrak{X}$. Since $i_{2}$ is an
$\mathfrak{X}$-element, we have $i_{1}\leq i_{2}$. On similar lines we can
prove that $i_{2}\leq i_{1}$. Thus $i_{1}=i_{2}$. Now, we prove “further”
part. Since $ik$ is an $\mathfrak{X}$-element and $ik\leq ik$ with
$k\notin\mathfrak{X}$, we have $i\leq ik$. The reverse inequality is always
true. Hence $i=ik$. ∎

It is well-known that if a proper ideal $P$ of a commutative ring $R$ with
unity is prime if and only if $R\setminus P$ is a multiplicatively closed
subset of $R$. Analogously, a proper element $p$ of a $c$-lattice $L$ is prime
if and only if $L\setminus(p]$ is an $M$-closed subset of $L$. To characterize
$\mathfrak{X}$-element, we define $\mathfrak{X}$-multiplicatively closed
subset of $L$ as follows.

######  Definition 2.22.

Let $\mathfrak{X}$ be an $M$-closed subset of a $c$-lattice $L$. A non-empty
subset $A$ of $L_{*}$ with $(L_{*}\setminus\mathfrak{X})\subseteq A$ is called
a $\mathfrak{X}$-multiplicatively closed subset of $L$, if
$a_{1}\in(L_{*}\setminus\mathfrak{X})$ and $a_{2}\in A$, then $a_{1}\cdot
a_{2}\in A$.

######  Remark 2.23.

If $A$ is a $\mathfrak{X}$-multiplicatively closed subset of $L$, then $A$
need not be a multiplicatively closed subset of $L$. Consider a multiplicative
lattice $K$ given in Example 2.2. If we take
$\mathfrak{X}=\\{0,~{}a,~{}b,~{}c,~{}d\\}$, then $A=\\{1,~{}c,~{}d\\}$ is an
$\mathfrak{X}$-multiplicatively closed subset of $K$ but $A=\\{1,~{}c,~{}d\\}$
is not a multiplicatively closed subset of $K$, as $c,d\in A$ and $c\cdot
d=0\notin A$.

Also, if $A$ is a multiplicatively closed subset of $L$, then $A$ need not be
a $\mathfrak{X}$-multiplicatively closed subset of $L$ for some $M$-closed
subset $\mathfrak{X}$. Consider the ideal lattice $L$ of the ring
$\mathbb{Z}_{12}$. Then $\\{(1)\\}$ is a multiplicatively closed subset of
$L$, but $\\{(1)\\}$ is not a $\mathfrak{X}$-multiplicatively closed subset of
$L$ for $\mathfrak{X}=\\{(0),~{}(6)\\}$.

######  Lemma 2.24.

Let $i$ be a proper element of a $c$-lattice $L$ and $\mathfrak{X}$ be an
$M$-closed subset of $L$. If $i$ is an $\mathfrak{X}$-element of $L$, then
$L_{*}\setminus(i]$ is an $\mathfrak{X}$-multiplicatively closed subset of
$L$. The converse is true if either $\mathfrak{X}=(j]$ is an $M$-closed subset
of a $c$-lattice $L$ or $L$ is a compact lattice.

###### Proof.

Suppose that $i$ is an $\mathfrak{X}$-element of $L$. By Lemma 2.4,
$(L_{*}\setminus\mathfrak{X})\subseteq(L_{*}\setminus(i])$. Let
$a\in(L_{*}\setminus\mathfrak{X})$ and $b\in(L_{*}\setminus(i])$. We claim
that $a\cdot b\in(L_{*}\setminus(i])$. Suppose on the contrary that $a\cdot
b\notin(L_{*}\setminus(i])$. So $a\cdot b\leq i$. Since $i$ is an
$\mathfrak{X}$-element of $L$ and $a\notin\mathfrak{X}$, we get $b\leq i$, a
contradiction to $b\in(L_{*}\setminus(i])$. Thus $a\cdot
b\in(L_{*}\setminus(i])$. Consequently, $L_{*}\setminus(i]$ is an
$\mathfrak{X}$-multiplicatively closed subset of $L$.

Conversely, suppose that $\mathfrak{X}=(j]$ is an $M$-closed subset of a
$c$-lattice $L$ and $L_{*}\setminus(i]$ is an $\mathfrak{X}$-multiplicatively
closed subset of $L$. Therefore
$\bigl{(}L_{*}\setminus(j]\bigr{)}\subseteq\bigl{(}L_{*}\setminus(i]\bigr{)}$.
In view of Lemma 2.14, to show that $i$ is an $\mathfrak{X}$-element of $L$,
it is enough to show that for $a,b\in L_{*}$ such that $a\cdot b\leq i$ with
$a\not\in\mathfrak{X}=(j]$, we have $b\leq i$. If $b\in(L_{*}\setminus(i])$,
then as $(L_{*}\setminus(i])$ is an $\mathfrak{X}$-multiplicatively closed
subset of $L$, we get $a\cdot b\in(L_{*}\setminus(i])$, a contradiction to
$a\cdot b\leq i$. Therefore $i$ is an $\mathfrak{X}$-element of $L$.

If $L$ is a compact lattice and $\mathfrak{X}$ be an $M$-closed subset of $L$,
then the converse follows similarly. ∎

######  Theorem 2.25.

Let $j$ be a proper element of $c$-lattice $L$ and $\mathfrak{X}=(j]$ be an
$M$-closed subset of $L$. Suppose $a\in L$ and $t\nleq a$ for all $t\in A$,
where $A$ is an $\mathfrak{X}$-multiplicatively closed subset. Then there is
an $\mathfrak{X}$-element $i$ of $L$ such that $a\leq i$ and $i$ is maximal
with respect to $t\nleq i$ for all $t\in A$.

###### Proof.

Let $R=\\{c\in L\;|\ a\leq c\text{ and }\;t\nleq c$ for all $t\in A\\}$.
Clearly, $a\in R$ and hence $R$ is a poset under the induced partial order of
$L$. Let $\mathcal{C}$ be a chain in $R$ and $w=\bigvee\\{d\ |\
d\in\mathcal{C}\\}$. We claim that $w\in R$. Suppose on the contrary that
$w\notin R$, that is, $t\leq w$ for some $t\in A$. Since $t\in A\subseteq
L_{*}$ is compact, we have $t\leq d_{1}\vee d_{2}\vee\dots\vee d_{n}$ for some
$d_{1},d_{2},\cdots,d_{n}\in\mathcal{C}$. As $\mathcal{C}$ is a chain we must
have $d_{1}\vee d_{2}\vee\dots\vee d_{n}=d_{i}$ for some $i$, where $1\leq
i\leq n$. Thus $t\leq d_{i}$, a contradiction. Thus $w\in R$. Hence by Zorn’s
lemma, there is a maximal element $i$ of $R$. Hence $a\leq i$ and $t\not\leq
a$ for all $t\in A$.

In view of Lemma 2.14, to prove $i$ is an $\mathfrak{X}$-element of $L$,
assume that $x,y\in L_{*}$ such that $x\cdot y\leq i$ with
$x\notin\mathfrak{X}=(j]$. Suppose that $y\nleq j$. Clearly, $y\leq(i:x)$ and,
if $i=(i:x)$, then $y\leq i$ and we are done. Hence assume that $i<(i:x)$.
Since $i$ is a maximal element of $R$, $(i:x)\not\in R$. Hence, there exists a
compact element $t_{1}\in A$ such that $t_{1}\leq(i:x)$, that is, $x\cdot
t_{1}\leq i$. But $x\cdot t_{1}\in A$, as $A$ is an
$\mathfrak{X}$-multiplicatively closed subset, $x\in(L_{*}\setminus(j])$ and
$t_{1}\in A$. Thus there exist an element $t_{2}=x\cdot t_{1}\in A$ such that
$t_{2}\leq i$, a contradiction to $i\in R$. Hence $i$ is an
$\mathfrak{X}$-element of $L$.∎

######  Theorem 2.26.

Let $L$ be a compact lattice and $\mathfrak{X}$ be an $M$-closed subset of
$L$. Suppose $a\in L$ and $t\nleq a$ for all $t\in A$, where $A$ is an
$\mathfrak{X}$-multiplicatively closed subset. Then there is an
$\mathfrak{X}$-element $i$ of $L$ such that $a\leq i$ and $i$ is maximal with
respect to $t\nleq i$ for all $t\in A$.

###### Proof.

Proof follows on similar lines as that of Theorem 2.25. ∎

##  3\. Applications of $\mathfrak{X}$-Elements

As already mentioned in the introduction, there is a unifying pattern in the
results of $J$-ideals, $n$-ideals and $r$-ideals of a commutative ring with
unity.

In this section, we prove these results of $J$-ideals, $n$-ideals and
$r$-ideals by suitably replacing the set $\mathfrak{X}$ in multiplicative
lattices. Hence most of the results of the papers [9], [11] and [12] becomes
the corollaries of our results.

First, we quote the definitions of $J$-ideals, $n$-ideals and $r$-ideals using
the sets $Z(R)$, $N(R)$ and $J(R)$ ($Z(L)$, $N(L)$ and $J(L)$), the set of
zero-divisors, the nil-radical and the Jacobson radical of a commutative ring
$R$ (multiplicative lattice $L$) respectively.

######  Definition 3.1.

A proper ideal $I$ of a commutative ring $R$ with unity is called:

  * •

$r$-ideal, if $ab\in I$ with $ann_{R}(a)=(0)$ implies $b~{}\in I$ for all
$a,~{}b~{}\in R$ (see Mohamadian [11]).

  * •

$n$-ideal, if $ab\in I$ with $a~{}\not\in\sqrt{0}$ implies $b~{}\in I$ for all
$a,~{}b~{}\in R$ (see U. Tekir et al. [12]).

  * •

$J$-ideal, if $ab\in I$ with $a~{}\not\in J(R)$ implies $b~{}\in I$ for all
$a,~{}b~{}\in R$ (see Khashan and Bani-Ata [9]).

Analogously, we define the concepts of $r$-element, $n$-element and
$J$-element in multiplicative lattices.

######  Definition 3.2.

A proper element $i$ of a multiplicative lattice $L$ is called:

  * •

$r$-element, if $a\cdot b~{}\leq i$ with $a~{}\not\in Z(L)$ implies $b~{}\leq
i$ for all $a,~{}b~{}\in L_{*}$.

  * •

$n$-element, if $a\cdot b~{}\leq i$ with $a~{}\notin\bigl{(}\sqrt{0}\
\\!\bigr{]}$ implies $b~{}\leq i$ for all $a,~{}b~{}\in L_{*}$.

  * •

$J$-element, if $a\cdot b~{}\leq i$ with $a~{}\not\in\bigl{(}J(L)\bigr{]}$
implies $b~{}\leq i$ for all $a,~{}b~{}\in L_{*}$.

We quote the following three results to prove that a proper ideal $I$ of a
commutative ring $R$ with unity is an $r$-ideal, $n$-ideal and $J$-ideal if
and only if it is an $r$-element, $n$-element and $J$-element of the
multiplicative lattice $Id(R)$, the set of all ideals of $R$, respectively.

######  Theorem 3.3 (Mohamadian [11, Lemma 2.5]).

Let $R$ be a commutative ring with unity and $I$ be a proper ideal of $R$.
Then $I$ is an $r$-ideal if and only if whenever $J$ and $K$ are ideals of $R$
with $J\not\subseteq Z(R)$ and $JK\subseteq I$, then $K\subseteq I$.

######  Theorem 3.4 (U. Tekir et al. [12, Theorem 2.7]).

Let $R$ be a commutative ring with unity and $I$ a proper ideal of $R$. Then
the following are equivalent:

  1. (1)

$I$ is an $n$-ideal of $R$.

  2. (2)

$I=(I:a)$ for every $a\notin\sqrt{0}$.

  3. (3)

For ideals $J$ and $K$ of $R$, $JK\subseteq I$ with
$J\cap(R-\sqrt{0})\not=\emptyset$ implies $K\subseteq I$.

######  Theorem 3.5 (H. A. Khashan et al. [9, Proposition 2.10]).

Let $R$ be a commutative ring with unity and $I$ a proper ideal of $R$. Then
the following are equivalent:

  1. (1)

$I$ is a $J$-ideal of $R$.

  2. (2)

$I=(I:a)$ for every $a\not\in J(R)$.

  3. (3)

For ideals $A$ and $B$ of $R$, $AB\subseteq I$ with $A\nsubseteq J(R)$ implies
$B\subseteq I$.

######  Theorem 3.6.

Let $R$ be a Noetherian ring with unity. Then $I$ is an $r$-ideal of $R$ if
and only if $I$ is an $r$-element of the multiplicative lattice $L=Id(R)$,
where $Id(R)$ is the ideal lattice of $R$.

###### Proof.

Suppose that $I$ is an $r$-ideal of $R$. Let $J,K$ be any ideals of $R$ such
that $J\cdot K\leq I$ in $L$ with $J\notin Z(L)$, that is, $ann_{L}(J)=0_{L}$,
where $(0_{R})$ is the least element of $L$, denoted by $0_{L}$. We claim that
$ann_{R}(J)=(0_{R})$. Suppose on the contrary that $(0_{R}\not=)x\in
ann_{R}(J)$. Hence $(x)J=(0_{R})$, a contradiction to the $ann_{L}(J)=0_{L}$.
Hence $ann_{R}(J)=(0_{R})$.

Now, we prove that $J\not\subseteq Z(R)$. Suppose on the contrary that
$J\subseteq Z(R)$. Since $R$ is Noetherian, $J\subseteq
Z(R)=\bigcup_{i=1}^{n}P_{i}$, where $P_{i}$’s are associate primes. By Prime
Avoidance Theorem $J\subseteq P_{k}$ for some $k$. Since $P_{k}$ is an
associated prime, we have $P_{k}=0:x$ for some $x\in R$. But this will
contradicts the fact that $ann_{R}(J)=(0)$. Hence $J\not\subseteq Z(R)$. By
Theorem 3.3, $K\subseteq I$, i.e., $K\leq I$. Thus $I$ is an $r$-element of
$L$.

Conversely, suppose that $I$ is a $r$-element of $L$. Let $a,b\in R$ such that
$a\cdot b\in I$ with $ann_{R}(a)=(0_{R})$. We claim that $b\in I$. Since
$(a\cdot b)=(a)\cdot(b)\subseteq I$, we have $a^{\prime}\cdot b^{\prime}\leq
I$ in $L$, where $a^{\prime}=(a),\ b^{\prime}=(b)$. Clearly, $a^{\prime}\notin
Z(L)$. Hence $b^{\prime}\leq I$, i.e., $b\in I$. Thus $I$ is an $r$-ideal of
$R$. ∎

######  Remark 3.7.

From the proof of Theorem 3.6, it is clear that every $r$-element of $Id(R)$
is an $r$-ideal of $R$. However, for the converse we need the assumption that
a ring is Noetherian. It should be noted that if we replace “Noetherian ring”
by “ring satisfies strongly annihilator condition” still the result is true.
By strong annihilator condition, we mean, for given ideal $I$ of $R$, there
exists $a\in I$ such that $ann_{R}(I)=ann_{R}(a)$.

Further, we are unable to find an example to show that the condition that the
ring is Noetherian or satisfies strongly annihilator condition is necessary to
prove the above Theorem 3.6. Hence we raise the following question.

######  Question 3.8.

Let $I$ be an $r$-ideal of a commutative ring $R$ with unity. Is $I$ an
$r$-element of $Id(R)$?

######  Theorem 3.9.

Let $R$ be a commutative ring with unity. Then $I$ is an $n$-ideal of $R$ with
unity if and only if $I$ is a $n$-element of multiplicative lattice $L=Id(R)$,
where $Id(R)$ is the ideal lattice of $R$.

###### Proof.

Suppose that $I$ is an $n$-ideal of $R$. Let $J,K$ be any finitely generated
ideals of $R$ such that $J\cdot K\leq I$ with $J\nleq\sqrt{0_{L}}$ in $L$. It
is known that finitely generated ideals of $R$ are compact elements of
$Id(R)$. Since $J\nleq\sqrt{0_{L}}$, we get $J^{n}\not=0_{L}=(0_{R})$ for
every $n\in\mathbb{N}$. Hence $J\cap(R\setminus\sqrt{0_{R}})\not=\emptyset$.
By Theorem 3.4, $K\subseteq I$, i.e., $K\leq I$. Therefore $I$ is a
$n$-element of $L$.

Conversely, suppose that $I$ is a $n$-element of $L$. Let $a,b\in R$ such that
$a\cdot b\in I$ with $a\notin\sqrt{0_{R}}$. We claim that $b\in I$. Since
$(a\cdot b)=(a)\cdot(b)\subseteq I$, we have $a^{\prime}\cdot b^{\prime}\leq
I$ in $L$, where $a^{\prime}=(a),\ b^{\prime}=(b)\in L_{*}$. Clearly,
$a^{\prime}\nleq\sqrt{0_{L}}$. Hence $b^{\prime}\leq I$, i.e., $b\in I$. Thus
$I$ is an $n$-ideal of $R$. ∎

######  Theorem 3.10.

Let $R$ be a commutative ring with unity. Then $I$ is a $J$-ideal of $R$ if
and only if $I$ is a $J$-element of multiplicative lattice $L=Id(R)$, where
$Id(R)$ is the ideal lattice of $R$.

###### Proof.

Suppose that $I$ is a $J$-ideal of $R$. Let $A,B$ be finitely generated ideals
of $R$ (which are compact elements of $Id(R)$) such that $A\cdot B\leq I$ with
$A\nleq J(L)$ in $L=Id(R)$. Since $A\nleq J(L)$, we get $A\nsubseteq J(R)$. By
Theorem 3.5, $B\subseteq I$, i.e., $B\leq I$ in $L$. Hence $I$ is a
$J$-element of $L$.

Conversely, suppose that $I$ is a $J$-element of $L$. Let $a,b\in R$ such that
$a\cdot b\in I$ with $a\notin J(R)$. We claim that $b\in I$. Since $(a\cdot
b)=(a)\cdot(b)\subseteq I$, we have $a^{\prime}\cdot b^{\prime}\leq I$ in $L$,
where $a^{\prime}=(a),\ b^{\prime}=(b)\in L_{*}$. Clearly, $a^{\prime}\nleq
J(L)$. Hence $b^{\prime}\leq I$, i.e., $b\in I$. Hence $I$ is a $J$-ideal of
$R$. ∎

Let $L$ be a multiplicative lattice. Then one can see the each of the sets
$Z(L)$, $\bigl{(}\sqrt{0}\ \\!\bigr{]}$ and $\bigl{(}J(L)\\!\bigr{]}$ are
multiplicatively $M$-closed subsets of $L$. So if we replace $\mathfrak{X}$ by
these sets, then we get the results of $r$-element, $n$-element and
$J$-element respectively.

We quote some of these results for ready reference.

One can see that in a $c$-lattice $L$, $\bigl{(}L_{*}\setminus
Z(L)\bigr{)}\subseteq\bigl{(}L_{*}\setminus\bigl{(}\sqrt{0}\
\\!\bigr{]}\bigr{)}$ and $\bigl{(}\sqrt{0}\
\\!\bigr{]}\subseteq\bigl{(}J(L)\\!\bigr{]}$. For this, let
$x\in\bigl{(}L_{*}\setminus Z(L)\bigr{)}$ and $x\in\bigl{(}\sqrt{0}\
\\!\bigr{]}$. Then $x^{n}=0$ for some $n\in\mathbb{N}$. Thus $x\in Z(L)$, a
contradiction. This proves the inclusion $\bigl{(}L_{*}\setminus
Z(L)\bigr{)}\subseteq\bigl{(}L_{*}\setminus\bigl{(}\sqrt{0}\
\\!\bigr{]}\bigr{)}$. Now, for the second inclusion, let $y$ be any compact
element such that $y\in\bigl{(}\sqrt{0}\ \\!\bigr{]}$. Then $y^{k}=0$ for some
$k\in\mathbb{N}$. Let $m$ be a maximal element of $L$. Then it is prime. This
together with $y^{k}=0\leq m$ implies that $y\leq m$. This further yields that
$y\in J(L)=\bigwedge_{k\in\Lambda}m_{k}$. Since $L$ is a $c$-lattice and every
compact element below $\sqrt{0}$ is below $J(L)$, we have $\sqrt{0}\leq J(L)$.

Hence by Lemma 2.6, we have the following result.

######  Proposition 3.11.

Let $L$ be a $c$-lattice. Then every $n$-element of $L$ is a $r$-element as
well as it is a $J$-element of $L$.

By Proposition 3.11, Theorems 3.6, 3.9, and Theorem 3.10, we have:

######  Proposition 3.12.

Let $R$ be a commutative ring with unity. Then every $n$-ideal of $R$ is a
$r$-ideal as well as it is a $J$-ideal.

From Lemma 2.4, we get Proposition 2.2 of [9] and Proposition 2.3 of [12].
Also, Proposition 2.4 of [12] follows from Lemma 2.9. It is easy to observe
that Proposition 2.10 of [9] and Theorem 2.7 of [12] follows from Lemma 2.11.
We observe that Proposition 2.13 of [9] follows from Lemmas 2.4 and 2.13. Note
that Theorem 2.17 strengthens Theorem 2.12 of [12]. Lemmas 2.19 and 2.20
generalizes the equivalence of $(i)$ and $(ii)$ in Corollary 2.13 of [12] and
Proposition 2.20 of [9] respectively. One can see that Lemma 2.21 extends
Proposition 2.16 of [12] and Proposition 2.21 of [9]. Lastly, Theorem 2.23 of
[12] and Proposition 2.29 of [9] follows from Theorem 2.25.

For the following result, we need a little more explanation.

######  Proposition 3.13 ([9, Proposition 2.3]).

Let $R$ be a commutative ring with unity. Then the following are equivalent.

  1. (1)

$R$ is a local ring.

  2. (2)

Every proper ideal of $R$ is a $J$-ideal.

###### Proof.

$(1)\implies(2):$ It is clear that the ideal lattice $Id(R)$ of $R$ is a local
lattice. Further, $J(L)=m$, where $m$ is the unique maximal element of
$Id(R)$. Hence by Lemma 2.7, every proper element of $L$ is an
$\mathfrak{X}$-element, where $\mathfrak{X}=(m]=\bigl{(}J(L)\\!\bigr{]}$. That
is, every proper element of $L$ is a $J$-element. By Theorem 3.10, every
proper ideal of $R$ is a $J$-ideal.

$(2)\implies(1):$ follows from Lemma 2.8 ∎

Finally, the results of $n$-multiplicatively closed subset and
$J$-multiplicatively closed subset can be obtained by using Lemma 2.24.
Further, the results of $r$-ideals can be deduced from our results for
Noetherian rings, since Theorem 3.6 is available for Noetherian lattice
settings. If Question 3.8 has an affirmative answer, then our results will
extend most of the results of $r$-ideals of a commutative ring with unity.

## References

  * [1] R. G. Burton, Fractional elements in multiplicative lattices, Pacific J. Math. 56 (1) (1975), 35-49.
  * [2] L. Fuchs and R. Reis, On lattice-ordered commutative semigroups, Algebra Universalis 50 (3) (2003), 341-357.
  * [3] G. Grätzer, General Lattice Theory, Birkhäuser Basel (1998).
  * [4] Vinayak Joshi and S. B. Ballal, A note on n-Baer multiplicative lattices, Southeast Asian Bull. Math. 39 (2015), 67-76.
  * [5] C. Jayaram, Primary elements in Prüfer lattices, Czech. Math. J. 52 (127) (2002), 585-593.
  * [6] C. Jayaram and E. W. Johnson, s-prime elements in multiplicative lattices, Period. Math. Hungar. 31 (1995), 201-208.
  * [7] C. Jayaram and E. W. Johnson, Strong compact elements in multiplicative lattices, Czech. Math. J. 47 (122) (1997), 105-112.
  * [8] C. Jayaram, U. Tekir and E. Yetkin, 2-absorbing and weakly 2-absorbing elements in multiplicative lattices, Communications in Algebra 42 (2014), 1-16.
  * [9] H. A. Khashan and A. B. Bani-Ata, $J$-ideals of commutative rings, Int. Electron. J. Algebra 29 (2021), 148-164.
  * [10] C. S. Manjarekar and A. V. Bingi, $\phi$-prime and $\phi$-primary elements in multiplicative lattices, Algebra, Volume 2014, Article ID 890312, 7 pages http://dx.doi.org/10.1155/2014/890312.
  * [11] R. Mohamadian, $r$-ideals in commutative rings, Turk. J. Math. 39 (2015), 733-749.
  * [12] U. Tekir, S. Koc and K. H. Oral, $n$-ideals of commutative rings, Filomat 31 (10) (2017), 2933-2941.
  * [13] M. Ward and R. P. Dilworth, Residuated lattices, Trans. Amer. Math. Soc. 45 (1939), 335-354.