11institutetext: FEEC, UNICAMP, Brazil 22institutetext: NeuralMind, Brazil

# ExaRanker-Open:
Synthetic Explanation for IR using Open-Source LLMs

Fernando Ferraretto 11 Thiago Laitz 11 Roberto Lotufo 1122
Rodrigo Nogueira 11

###### Abstract

ExaRanker recently introduced an approach to training information retrieval
(IR) models, incorporating natural language explanations as additional labels.
The method addresses the challenge of limited labeled examples, leading to
improvements in the effectiveness of IR models. However, the initial results
were based on proprietary language models such as GPT-3.5, which posed
constraints on dataset size due to its cost and data privacy. In this paper,
we introduce ExaRanker-Open, where we adapt and explore the use of open-source
language models to generate explanations. The method has been tested using
different LLMs and datasets sizes to better comprehend the effective
contribution of data augmentation. Our findings reveal that incorporating
explanations consistently enhances neural rankers, with benefits escalating as
the LLM size increases. Notably, the data augmentation method proves
advantageous even with large datasets, as evidenced by ExaRanker surpassing
the target baseline by 0.6 nDCG@10 points in our study. To encourage further
advancements by the research community, we have open-sourced both the code and
datasets at https://github.com/unicamp-dl/ExaRanker.

##  1 Introduction and Background

Previous research by Ferraretto et al. [4] introduced a method for training
retrieval models, incorporating natural language explanations as additional
labels. This approach mitigates the need for large amounts of training
examples.

The method is built on the observation that a categorical label (e.g.,
true/false) provides limited information about the task to be learned by the
AI model. This limitation makes it more challenging for the model to
comprehend the subtleties of the task. It has been demonstrated that the
learning process becomes more efficient when natural language explanations are
provided. These explanations elucidate, for example, why a passage is relevant
or not to a given query.

The ExaRanker model used GPT-3.5 [9] in a few-shot manner to autonomaticaly
augment training examples with explanations. The implementation of this method
has demonstrated a reduction in the dataset size dependency, performing on par
with a categorical dataset three times larger. However, the previous results
suggest that the effectiveness of incorporating explanations decreases as the
number of training examples increases. Additionally, the study faced
limitations in the number of examples that were augmented, as it relied on a
LLM exposed only via a paid API to generate explanations.

This work extends upon the ExaRanker method by leveraging open-source language
models (LLMs) for generating explanations. This extension enables the method
to be efficiently utilized for automatically augmenting training examples with
explanations at a low cost and guaranteed data privacy. Furthermore, we
present new results using the original method on a larger dataset, confirming
that explanations consistently yield benefits regardless of dataset size. We
refer to the model from Ferraretto et al. [4] as ExaRanker-v1, and the model
introduced in this paper as ExaRanker-Open.

##  2 Methodology

In this section, the ExaRanker-Open method is presented and detailed on how it
differs from the original ExaRanker.

We use the same samples randomly chosen in the prior study, consisting of 15k
pairs of (query, relevant passage), and an additional 15k pairs of (query,
non-relevant passage) from the training set of the MS MARCO passage ranking
dataset [1]. Two open-source LLMs have been used for generating explanations
and augmenting the dataset: llama-2-7B-chat-hf and llama-2-70b-chat-hf [14].

In addition to the dataset with 30k samples, we created a larger version
containing 100k samples. This dataset is composed of 50k pairs of query-
passage relevant and another 50k non-relevant pairs. In total, four versions
of the dataset were generated for this experiment, as detailed in the Table 1.

After the dataset generation, a T5-base [10] model has been fine-tuned for up
to 30 epochs on all dataset variations. The fine-tuning process employed the
AdamW [6] optimizer with a learning rate set of $3e-5$, weight decay of
$0.01$, and a batch size of $128$ examples ($64$ positives and $64$
negatives). Both the maximum number of input and output tokens were restricted
to $512$, with sequences surpassing these limits truncated during training and
inference. For comparative analysis, another T5-base model was also fine-tuned
using identical hyperparameters and dataset specifications, excluding the
explanations in the target text, i.e. using categorial labels only. The
finetuning method for this baseline model is similar to one used by monoT5
[8], except that here we use smaller versions of the MS MARCO dataset and a
different optimizer.

LLM model |  Relevant Samples |  Total Samples
---|---|---
Llama-2-7B |  15k |  30k
Llama-2-7B |  50k |  100k
Llama-2-70B |  15k |  30k
Llama-2-70B |  50k |  100k
Table 1: Datasets generated with the open-source LLMs.

Besides the ExaRanker-Open method as described before, the prior ExaRanker-v1
experiment was expanded to encompass a 20 times larger dataset of 300k
relevant pairs of query-passage and 300k non-relevant pairs. Following the
same methodology, samples were randomly selected, and utilizing the few-shot
prompt, explanations were generated by GPT-3.5-turbo. We used greedy decoding
and restricting the output to 256 tokens.

The next step is to fine-tuned a T5-base model over 30 epochs with
explanations. The experiment was conducted across eight different dataset
sizes: 2.5k, 5k, 10k, 15k, 50k, 100k, 150k, and 300k relevant pairs of query-
passage. In all cases, an equal number of non-relevant passages were employed,
resulting in the smallest dataset comprising 5k samples and the largest
reaching 600k. As done before, another monoT5 was fine-tuned within the same
dataset but without explanations. This model serves as a benchmark for
comparison, enabling an evaluation of the benefits derived from using
explanations to augment the dataset and fine-tune the neural ranker in eight
different dataset sizes.

To evaluate the advantages of incorporating explanations, each fine-tuned
model was evaluated in a zero-shot manner across six datasets from the BEIR
benchmark [13]: Robust04 [15], TREC-COVID [11], DBPedia [5], FiQA [7], TREC-
NEWS [12], and NFCorpus [2]. TREC-DL 2020 [3] serves as a validation set for
selecting the best checkpoint. Different attempts were made for each
experiment, with this study reporting the average results obtained.

##  3 Results

The results obtained with open-source LLMs are shown in the Table 2. The
initial row represents the T5-base model fine-tuned without data augmentation,
solely relying on categorical labels. The second row reflects the earlier
results of ExaRanker-v1 using GPT-3.5-turbo. The last two rows present the
outcomes of this study, employing Llama-2-70B and Llama-2-7B, respectively.

As evident, the zero-shot effectiveness is enhanced when employing a larger
LLM model. Illustrated in Figure 1, the performance of Llama-2-7B surpasses
that of the monoT5 without explanations in both evaluated dataset sizes, 15k
and 50k relevant pairs of query-passage. However, Llama-2-70B outperforms
Llama-2-7B, specially when using the larger dataset with 50k query-relevant
passage pairs. Ultimately, ExaRanker-v1, using GPT-3.5, remains the top-
performing model.

These results strongly suggest the quality of data augmentation produced by
each Llama model size. As expected, larger models exhibit superior natural
language processing capabilities, leading to a more substantial extraction of
signals that can be effectively utilized during the fine-tuning phase.

Table 2: Results (nDCG@10) of open-source LLMs. Average zero-shot (all except
DL 20). The column “Ft Pos.” is the number of positive training examples on
which the model was finetuned.

Model | LLM | Ft Pos. | DL 20 | Robust | Covid | Dbp | FiQA | News | NFC | Avg ZS
---|---|---|---|---|---|---|---|---|---|---
monoT5 | n/a | 15k | 0.656 | 0.523 | 0.746 | 0.392 | 0.382 | 0.409 | 0.344 | 0.466
ExaRanker-v1 | GPT-3.5 | 15k | 0.683 | 0.531 | 0.752 | 0.403 | 0.408 | 0.415 | 0.352 | 0.477
ExaRanker-Open | Llama-2-70B | 15k | 0.653 | 0.551 | 0.730 | 0.398 | 0.393 | 0.425 | 0.341 | 0.473
ExaRanker-Open | Llama-2-7B | 15k | 0.662 | 0.523 | 0.737 | 0.407 | 0.392 | 0.421 | 0.344 | 0.471
monoT5 | n/a | 50k | 0.653 | 0.534 | 0.757 | 0.384 | 0.396 | 0.426 | 0.350 | 0.475
ExaRanker-v1 | GPT-3.5 | 50k | 0.673 | 0.540 | 0.778 | 0.423 | 0.413 | 0.431 | 0.349 | 0.489
ExaRanker-Open | Llama-2-70B | 50k | 0.670 | 0.563 | 0.757 | 0.414 | 0.403 | 0.440 | 0.345 | 0.487
ExaRanker-Open | Llama-2-7B | 50k | 0.670 | 0.529 | 0.741 | 0.419 | 0.398 | 0.439 | 0.349 | 0.479

Figure 1: Average zero-shot results on 6 datasets of the BEIR benchmark with
respect to training dataset size, comparing the 4 models evaluated.

The results of the second experiment, an extension of ExaRanker-v1, are shown
in the Table 3. Contrary to the results presented in the original ExaRanker
paper, here we observe that data augmentation consistently yields benefits,
even with large datasets. On average, the effectiveness of ExaRanker-v1 is 1.4
nDCG@10 points higher than that of the monoT5 trained without explanations.

Indeed, the model introduced by our method outperforms the strong monoT5-400k
baseline, by 0.8 NDCG@10 points, as illustrated in Figure 2. This baseline is
the monoT5-base model, which was trained over the entire MS MARCO dataset,
consisting of 400k relevant pairs. In contrast, our model is trained on a
dataset with 300k relevant pairs but still achieves superior performance.

Table 3: Results (nDCG@10) of larger datasets. Average zero-shot (all except DL 20). The column “Ft Pos.” is the number of positive training examples on which the model was finetuned. The "*" marker denotes the results from Ferraretto et al. [4]. Model | Ft Pos. | DL 20 | Robust | Covid | Dbp | FiQA | News | NFC | Avg ZS
---|---|---|---|---|---|---|---|---|---
monoT5 | 300k | 0.662 | 0.532 | 0.780 | 0.412 | 0.403 | 0.446 | 0.350 | 0.487
ExaRanker-v1 | 300k | 0.682 | 0.558 | 0.784 | 0.427 | 0.416 | 0.451 | 0.349 | 0.497
monoT5 | 150k | 0.663 | 0.537 | 0.790 | 0.395 | 0.396 | 0.443 | 0.349 | 0.485
ExaRanker-v1 | 150k | 0.684 | 0.559 | 0.781 | 0.426 | 0.413 | 0.447 | 0.348 | 0.496
monoT5 | 100k | 0.658 | 0.528 | 0.774 | 0.400 | 0.396 | 0.434 | 0.350 | 0.480
ExaRanker-v1 | 100k | 0.677 | 0.552 | 0.776 | 0.419 | 0.412 | 0.440 | 0.350 | 0.492
monoT5 | 50k | 0.653 | 0.534 | 0.757 | 0.384 | 0.396 | 0.426 | 0.350 | 0.475
ExaRanker-v1 | 50k | 0.673 | 0.540 | 0.778 | 0.423 | 0.413 | 0.431 | 0.349 | 0.489
monoT5 | 15k | 0.656 | 0.523 | 0.746 | 0.392 | 0.382 | 0.409 | 0.344 | 0.466
ExaRanker-v1 | 15k | 0.683 | 0.531 | 0.752 | 0.403 | 0.408 | 0.415 | 0.352 | 0.477
monoT5∗ | 10k | 0.643 | 0.510 | 0.749 | 0.379 | 0.374 | 0.426 | 0.341 | 0.463
ExaRanker-v1∗ | 10k | 0.667 | 0.527 | 0.752 | 0.409 | 0.393 | 0.418 | 0.347 | 0.474
monoT5∗ | 5k | 0.625 | 0.488 | 0.693 | 0.364 | 0.337 | 0.417 | 0.328 | 0.438
ExaRanker-v1∗ | 5k | 0.665 | 0.505 | 0.750 | 0.389 | 0.380 | 0.414 | 0.345 | 0.464
monoT5∗ | 2.5k | 0.611 | 0.486 | 0.666 | 0.334 | 0.328 | 0.370 | 0.325 | 0.418
ExaRanker-v1∗ | 2.5k | 0.650 | 0.496 | 0.686 | 0.393 | 0.306 | 0.398 | 0.335 | 0.436
Figure 2: Average zero-shot results on 6 datasets of the BEIR benchmark.
monoT5-400k is finetuned on the 400k relevant query-passage pairs from MS
MARCO without explanations. Note the log scale in horizontal axis.

##  4 Conclusion

This study introduces ExaRanker-Open, an enhanced iteration of ExaRanker-v1
[4] that uses open-source LLMs to augment categorical datasets with natural
language explanations. The findings demonstrate the advantages of this
approach within IR domain, revealing enhancement in the effectiveness of the
models fine-tuned through this methodology. Furthermore, experiments on larger
training datasets validates the advantages of data augmentation when employing
a neural ranker for IR tasks. ExaRanker outperforms a strong baseline trained
on a larger dataset but without natural language explanations associated in
the target.

## Acknowledgments

This research was partially funded by grant 2022/01640-2 from Fundação de
Amparo à Pesquisa do Estado de São Paulo (FAPESP). Roberto Lotufo is partially
supported by CNPq (The Brazilian National Council for Scientific and
Technological Development) under grant 313047/2022-7.

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#  multiplication $(m,n)$-hypermodules

Mahdi Anbarloei  Department of Mathematics, Faculty of Sciences, Imam Khomeini
International University, Qazvin, Iran<EMAIL_ADDRESS>
###### Abstract.

The concept of multiplication $(m,n)$-hypermodules was introduced by Ameri and
Norouzi in [2]. Here we intend to investigate extensively the multiplication
$(m,n)$-hypermodules. Let $(M,f,g)$ be a $(m,n)$-hypermodule (with canonical
$(m,n)$-hypergroups) over a commutative Krasner $(m,n)$-hyperring $(R,h,k)$. A
$(m,n)$-hypermodule $(M,f,g)$ over $(R,h,k)$ is called a multiplication
$(m,n)$-hypermodule if for each subhypermodule $N$ of $M$, there exists a
hyperideal $I$ of $R$ such that $N=g(I,1^{(n-2)},M)$.

###### Key words and phrases:

$n$-ary prime hyperideal, maximal hyperideal, multiplication
$(m,n)$-hypermodule.

###### 2010 Mathematics Subject Classification:

16Y99, 20N20

##  1\. Introduction

Algebraic hyperstructures represent a natural extension of classical algebraic
structures. Introduction of the structures was first done by Marty at the
$8^{th}$ Congress of Scandinavian Mathematicians in 1934 [15]. Later
on,various authors studied it such as by Mittas [19],[20] and Corsini [8],[9].
In 1904, Kasner introduce the notion of n-ary algebras in a lecture in a
annual meeting [13]. It is worth noting that Dorente wrote the first paper on
the theory of n-ary groups in 1928 [11]. The notion of Krasner hyperrings was
introduced by Krasner for the first time in [14]. Also, we can see some
properties on Krasner hyperrings in [18] and [21]. In [10], Davvaz and
Vougiouklis defined the notion of $n$-ary hypergroups which is a
generalization of hypergroups in the sense of Marty. The concept of
$(m,n)$-hyperrings was introduced in [16]. Davvaz et al. introduced Krasner
$(m,n)$\- hyperrings as a generalization of $(m,n)$-rings and studied some
results in this context in [17]. The notions of maximal hyperideal, $n$-ary
prime hyperideal, $n$-ary primary hyperideal and the radical of a hyperideal
in a Krasner $(m,n)$-hyperring were introduced in [1]. Also, Ostadhadi and
Davvaz studied the isomorphism theorems of ring theory and Krasner hyperring
theory which are derived in the context of Krasner $(m,n)$-hyperrings in [23].
Anvariyeh et al defined the notion of $(m,n)$\- hypermodules over
$(m,n)$-hyperrings in [6]. After, the concepts of free and canonical
$(m,n)$-hypermodules were definded and studied in [5] and [7]. In [2], Ameri
and Norouzi introduced a class of the $(m,n)$-hypermodules, which they defined
as multiplication $(m,n)$-hypermodules. The $(m,n)$-hypermodule $(M,f,g)$ over
$(R,h,k)$ is called a multiplication $(m,n)$\- hypermodule if for each
subhypermodule $N$ of $M$, there exists a hyperideal $I$ of $R$ such that
$N=g(I,1^{(n-2)},M)$.

The paper is orgnized as follows. In section 2, we have given some basic
definitions and results of $n$-ary hyperstructures which we need to develop
our paper. In section 3, we have proved some important results of the
multiplications $(m,n)$-hypermodule. In section 4, we have studied the primary
subhypermodules in a multiplication $(m,n)$-hyperodule. In section 5, we have
investigated intersections and sums of the multiplication
$(m,n)$-hypermodules.

##  2\. Preliminaries

In this section, we recall some basic terms and definitions concerning $n$-ary
hyperstructures which we need to develop our paper.
A mapping $f:H^{n}\longrightarrow P^{*}(H)$ is called an $n$-ary
hyperoperation, where $P^{*}(H)$ is the set of all the non-empty subsets of
$H$. An algebraic system $(H,f)$, where $f$ is an $n$-ary hyperoperation
defined on $H$, is called an $n$-ary hypergroupoid.
We shall use the following abbreviated notation:
The sequence $x_{i},x_{i+1},...,x_{j}$ will be denoted by $x^{j}_{i}$. For
$j<i$, $x^{j}_{i}$ is the empty symbol. In this convention

$f(x_{1},...,x_{i},y_{i+1},...,y_{j},z_{j+1},...,z_{n})$
will be written as $f(x^{i}_{1},y^{j}_{i+1},z^{n}_{j+1})$. In the case when
$y_{i+1}=...=y_{j}=y$ the last expression will be written in the form
$f(x^{i}_{1},y^{(j-i)},z^{n}_{j+1})$.

For non-empty subsets $A_{1},...,A_{n}$ of $H$ we define

$f(A^{n}_{1})=f(A_{1},...,A_{n})=\bigcup\\{f(x^{n}_{1})\ |\ x_{i}\in
A_{i},i=1,...,n\\}$.
An n-ary hyperoperation $f$ is called associative if

$f(x^{i-1}_{1},f(x_{i}^{n+i-1}),x^{2n-1}_{n+i})=f(x^{j-1}_{1},f(x_{j}^{n+j-1}),x_{n+j}^{2n-1}),$
hold for every $1\leq i<j\leq n$ and all $x_{1},x_{2},...,x_{2n-1}\in H$. An
$n$-ary hypergroupoid with the associative $n$-ary hyperoperation is called an
$n$-ary semihypergroup.

An $n$-ary hypergroupoid $(H,f)$ in which the equation $b\in
f(a_{1}^{i-1},x_{i},a_{i+1}^{n})$ has a solution $x_{i}\in H$ for every
$a_{1}^{i-1},a_{i+1}^{n},b\in H$ and $1\leq i\leq n$, is called an $n$-ary
quasihypergroup, when $(H,f)$ is an $n$-ary semihypergroup, $(H,f)$ is called
an $n$-ary hypergroup.

An $n$-ary hypergroupoid $(H,f)$ is commutative if for all
$\sigma\in\mathbb{S}_{n}$, the group of all permutations of
$\\{1,2,3,...,n\\}$, and for every $a_{1}^{n}\in H$ we have
$f(a_{1},...,a_{n})=f(a_{\sigma(1)},...,a_{\sigma(n)})$. If an $a_{1}^{n}\in
H$ we denote $a_{\sigma(1)}^{\sigma(n)}$ as the
$(a_{\sigma(1)},...,a_{\sigma(n)})$. We assume throughout this paper that all
Krasner $(m,n)$-hyperrings are commutative.

If $f$ is an $n$-ary hyperoperation and $t=l(n-1)+1$, then $t$-ary
hyperoperation $f_{(l)}$ is given by

$f_{(l)}(x_{1}^{l(n-1)+1})=f(f(...,f(f(x^{n}_{1}),x_{n+1}^{2n-1}),...),x_{(l-1)(n-1)+1}^{l(n-1)+1})$.

######  Definition 2.1.

([17]). Let $(H,f)$ be an $n$-ary hypergroup and $B$ be a non-empty subset of
$H$. $B$ is called an $n$-ary subhypergroup of $(H,f)$, if
$f(x^{n}_{1})\subseteq B$ for $x^{n}_{1}\in B$, and the equation $b\in
f(b^{i-1}_{1},x_{i},b^{n}_{i+1})$ has a solution $x_{i}\in B$ for every
$b^{i-1}_{1},b^{n}_{i+1},b\in B$ and $1\leq i\leq n$.
An element $e\in H$ is called a scalar neutral element if
$x=f(e^{(i-1)},x,e^{(n-i)})$, for every $1\leq i\leq n$ and for every $x\in
H$.

An element $0$ of an $n$-ary semihypergroup $(H,g)$ is called a zero element
if for every $x^{n}_{2}\in H$ we have
$g(0,x^{n}_{2})=g(x_{2},0,x^{n}_{3})=...=g(x^{n}_{2},0)=0$. If $0$ and
$0^{\prime}$are two zero elements, then $0=g(0^{\prime},0^{(n-1)})=0^{\prime}$
and so the zero element is unique.

######  Definition 2.2.

([12]). Let $(H,f)$ be a $n$-ary hypergroup. $(H,f)$ is called a canonical
$n$-ary hypergroup if
(1) there exists a unique $e\in H$, such that for every $x\in
H,f(x,e^{(n-1)})=x$;
(2) for all $x\in H$ there exists a unique $x^{-1}\in H$, such that $e\in
f(x,x^{-1},e^{(n-2)})$;
(3) if $x\in f(x^{n}_{1})$, then for all $i$, we have $x_{i}\in
f(x,x^{-1},...,x^{-1}_{i-1},x^{-1}_{i+1},...,x^{-1}_{n})$.

We say that e is the scalar identity of $(H,f)$ and $x^{-1}$ is the inverse of
$x$. Notice that the inverse of $e$ is $e$.

######  Definition 2.3.

([17])A Krasner $(m,n)$-hyperring is an algebraic hyperstructure $(R,f,g)$
which satisfies the following axioms:
(1) $(R,f$) is a canonical $m$-ary hypergroup;
(2) $(R,g)$ is a $n$-ary semigroup;
(3) the $n$-ary operation $g$ is distributive with respect to the $m$-ary
hyperoperation $f$ , i.e., for every $a^{i-1}_{1},a^{n}_{i+1},x^{m}_{1}\in R$,
and $1\leq i\leq n$,

$g(a^{i-1}_{1},f(x^{m}_{1}),a^{n}_{i+1})=f(g(a^{i-1}_{1},x_{1},a^{n}_{i+1}),...,g(a^{i-1}_{1},x_{m},a^{n}_{i+1}))$;
(4) $0$ is a zero element (absorbing element) of the $n$-ary operation $g$,
i.e., for every $x^{n}_{2}\in R$ we have

$g(0,x^{n}_{2})=g(x_{2},0,x^{n}_{3})=...=g(x^{n}_{2},0)=0$.

A non-empty subset $S$ of $R$ is called a subhyperring of $R$ if $(S,f,g)$ is
a Krasner $(m,n)$-hyperring. Let $I$ be a non-empty subset of $R$, we say that
$I$ is a hyperideal of $(R,f,g)$ if $(I,f)$ is an $m$-ary subhypergroup of
$(R,f)$ and $g(x^{i-1}_{1},I,x_{i+1}^{n})\subseteq I$, for every $x^{n}_{1}\in
R$ and $1\leq i\leq n$.

######  Definition 2.4.

([6]) Let $M$ be a nonempty set. Then $(M,f,g)$ is an $(m,n)$-hypermodule over
an $(m,n)$\- hyperring $(R,h,k)$, if $(M,f)$ is an $m$-ary hypergroup and the
map

$g:\underbrace{R\times...\times R}_{n-1}\times M\longrightarrow P^{*}(M)$
statisfied the following conditions:

$(i)\
g(r_{1}^{n-1},f(x_{1}^{m}))=f(g(r_{1}^{n-1},x_{1}),...,g(r_{1}^{n-1},x_{m}))$

$(ii)\
g(r_{1}^{i-1},h(s_{1}^{m}),r_{i+1}^{n-1},x)=f(g(r_{1}^{i-1},s_{1},r_{i+1}^{n-1},x),...,g(r_{1}^{i-1}s_{m},r_{i+1}^{n-1},x))$

$(ii)\
g(r_{1}^{i-1},k(r_{i}^{i+n-1}),r_{i+m}^{n+m-2},x)=g(r_{1}^{n-1},g(r_{m}^{n+m-2},x))$

$(iv)\ \\{0\\}=g(r_{1}^{i-1},0,r_{i+1}^{n-1},x)$.

If $g$ is an $n$-ary hyperoperation, $T_{1},...,T_{n-1}$ are subsets of $R$
and $M^{\prime}\subseteq M$, we set

$g(T_{1}^{n-1},M^{\prime})=\bigcup\\{g(r_{1}^{n-1},m)\ |\ r_{i}\in T_{i},1\leq
i\leq n-1,m\in M^{\prime}\\}.$
A canonical $(m,n)$-hypermodule $(M,f,g)$ is an $(m,n)$-ary hypermodule with a
canonical m-ary hypergroup $(M,f)$ over a Krasner $(m,n)$-hyperring $(R,h,k)$.

Let $(M,f,g)$ be an $(m,n)$-hypermodule over $(R,h,k)$. A non-empty subset $N$
of $M$ is said to be an $(m,n)$-subhypermodule of $M$ if $(N,f)$ is a $m$-ary
subhypergroup of $(M,f)$ and $g(R^{(n-1)},N)\in P^{*}(N)$.

Let $(M,f,g)$ be an $(m,n)$-hypermodule and $N$ be a subhypermodule of $M$.
Then the hyperideal $S_{N}$ is considered as follows:

$S_{N}=\\{r\in R\ |\ g(r,1^{(n-2)},M)\subseteq N\\}$

######  Definition 2.5.

([4]) Let $M$ be an $(m,n)$-hypermodule over $(R,h,k)$. A proper
subhypermodule $K$ of $M$ is said to be maximal, if for $N\leq M$ with
$K\subseteq N\subseteq M$, we have either $K=N$ or $N=M$.

######  Definition 2.6.

([4]) Let $M$ be an $(m,n)$-hypermodule over $(R,h,k)$. A proper
subhypermodule $N$ of $M$ is said to be $n$-ary primary, if
$g(r_{1}^{n-1},x)\subseteq Q$ with $r_{1}^{n-1}\in R$ and $x\in M\backslash
Q$, implies that

$g(k(r_{1}^{(t)},1^{(n-t)}),...,k(r_{n-1}^{(t)},1^{(n-t)}),M)\subseteq N$
for $t\leq n$, and for $t>n$ such that $t=l(n-1)+1$

$g(k_{(l)}(r_{1}^{(t)}),...,k_{(l)}(r_{n-1}^{(t)}),M)\subseteq N$
for some $t\in\mathbb{N}$ and $l>0$.

######  Definition 2.7.

([4]) Let $N$ be a subhypermodule of an $(m,n)$-hypermodule $(M,f,g)$ over
$(R,h,k)$. Then the set

$R/N=\\{f(x^{i-1}_{1},N,x^{m}_{i+1})\ |\ x^{i-1}_{1},x^{m}_{i+1}\in M\\}$
endowed with m-ary hyperoperation $f$ which for all
$x_{11}^{1m},...,x_{m1}^{mm}\in R$

$F(f(x_{11}^{1(i-1)},N,x^{1m}_{1(i+1)}),...,f(x_{m1}^{m(i-1)},N,x^{mm}_{m(i+1)}))$

$\hskip 8.5359pt=\\{(f(t_{1}^{i-1},N,t_{i+1}^{m})\ |\ t_{1}\in
f(x_{11}^{m1}),...,t_{m}\in f(x_{1m}^{mm})\\}$
and with $n$-ary hyperoperation $G:\underbrace{R\times...\times R}_{n-1}\times
M/N\longrightarrow P^{*}(M/N)$ which for all $x_{1}^{i-1},x_{i+1}^{m}\in M$
and $r_{1}^{n-1}\in R$

$G(r_{1}^{n-1},f(x_{1}^{i-1},N,x_{i+1}^{m}))$

$\hskip 8.5359pt=\\{f(z_{1}^{i-1},N,z_{i+1}^{m}\ |\ z_{1}\in
g(r_{1}^{n-1},x_{1}),...,z_{m}\in g(r_{1}^{n-1},x_{m})\\}$
is an $(m,n)$-hypermodule over $(R,h,k)$, and $(R/N,F,G)$ is called the
quotient $(m,n)$-hypermodule of $M$ by $N$.

######  Definition 2.8.

([3]) Let $(M,f,g)$ be an $(m,n)$-hypermodule over $(R,h,k)$ and $P$ be a
maximal hyperideal of $R$. Then

(1) we define

$\mathcal{X}_{P}(M)=\\{m\in M\ |\ \exists p\in P;0\in
g(h(1,-p,0^{(m-2)}),1^{(n-2)},m)\\}$ $\mathcal{X}^{P}(M)=\\{m\in M\ |\ \exists
p\in P;\\{0\\}=g(h(1,-p,0^{(m-2)}),1^{(n-2)},m)\\}.$

In this paper we let $\mathcal{X}_{P}(M)=\mathcal{X}^{P}(M)$ for every maximal
hyperideal $P$ of $R$.

(2) $M$ is called an n-ary $P$-cyclic if there exist $q\in P$ and $m\in M$
such that $g(h(1,-q,0^{(m-2)}),1^{(n-2)},M)\subseteq g(R,1^{(n-2)},m)$.

Also, we let $x\in g(I,1^{(n-2)},m)$ implies that there exists $a\in I$ such
that $\\{x\\}=g(a,1^{(n-2)},m)$, and $\\{0\\}=h(r,-r,0^{(m-2)})$, for all
$x,m\in M$, $r\in R$ and $I\subseteq R$.

######  Definition 2.9.

([2]) For every nonzero element $m$ of $(m,n$)-hypermodule $(M,f,g)$ over
$(R,h,k)$, we define

$F_{m}=\\{r\in R\ |\ 0\in g(r,1^{(n-2)},m);r\neq 0\\}.$
It is clear that $F_{m}$ is a hyperideal of $(R,h,k)$. The $(m,n)$-hypermodule
$(M,f,g)$ is said to be faithful, if $F_{m}=\\{0\\}$ for all nonzero elements
$m\in M$, that is $0\in g(r,1^{(n-2)},m)$ implies that $r=0$, for $r\in R$.

##  3\. the multiplication $(m,n)$-hypermodules

The notion of multiplication $(m,n)$-hypermodules was defined in [2]. The
$(m,n)$-hypermodule $(M,f,g)$ over $(R,h,k)$ is called a multiplication
$(m,n)$\- hypermodule if for each subhypermodule $N$ of $M$, there exists a
hyperideal $I$ of $R$ such that $N=g(I,1^{(n-2)},M)$.

######  Example 3.1.

$(m,n)$-hypermodule $(\mathbb{Z},f,g)$ over $(m,n)$-hyperring
$(\mathbb{Z},h,k)$ is multiplication. (see Example 3.5 in [2])

######  Lemma 3.2.

Let $(M,f,g)$ be a multiplication $(m,n)$-hypermodule over $(R,h,k)$ and let
$J_{(m,n)}(R)$ be intersection of all maximal hyperideals of $R$. If
$M=g(I,1^{(n-2)},M)$ for some hyperideal $I\subseteq J_{(m,n)}(R)$, then
$M=0$.

###### Proof.

If $m\in M$ then $g(R,1^{(n-2)},m)=g(A,1^{(n-2)},M)$ for some hyperideal $A$
of $R$. Then we get

$\hskip 28.45274ptm\in g(1^{(n-1)},m)\subseteq g(R,1^{(n-2)},m)$

$\hskip 105.2751pt=g(A,1^{(n-2)},M)$

$\hskip 105.2751pt=g(A,1^{(n-2)},g(I,1^{(n-2)},M))$

$\hskip 105.2751pt=g(k(A,I,1^{(n-2)}),1^{(n-2)},M)$

$\hskip 105.2751pt=g(I,1^{(n-2)},g(A,1^{(n-2)},M))$

$\hskip 105.2751pt=g(I,1^{(n-2)},g(R,1^{(n-2)},m)$

$\hskip 105.2751pt=g(k(I,R,1^{(n-2)}),1^{(n-2)},m)$

$\hskip 105.2751pt=g(I,1^{(n-2)},m)$.
It imlies that there exists some $r\in I$ such that $m\in
f(g(r,1^{(n-2)},m),0^{(m-1)})$. Then we get
$0\in f(-g(r,1^{(n-2)},m),m,0^{(m-2)})\subseteq
f(g(-r,1^{(n-2)},m),g(1^{(n-1)},m),g(0,1^{(m-2)},m))$

$\hskip 139.41832pt=g(h(-r,1,0^{(m-2)}),1^{(n-2)},m)$.
By Corollary 4.15 in [1], Since $r\in J_{(m,n)}(R)$ then $h(-r,1,0^{(m-2)})$
is invertible which means there exists $s\in R$ such that
$k(s,h(-r,1,0^{(m-2)}),1^{(n-2)})=1$. Thus

$\hskip 34.14322pt\\{0\\}=g(s,1^{(n-2)},0)\subseteq
g(s,1^{(n-2)},g(h(-r,1,0^{(m-2)}),1^{(n-2)},m))$

$\hskip 113.81102pt=g(k(s,h(-r,1,0^{(m-2)}),1^{(n-2)}),1^{(n-2)},m)$

$\hskip 113.81102pt=g(1,1^{(n-2)},m)$

$\hskip 113.81102pt=g(1^{(n-1)},m)$

$\hskip 113.81102pt=\\{m\\}$.
Hence $m=0$ which implies $M=0$. ∎

######  Theorem 3.3.

Let $(M,f,g)$ be a nonzero multiplication $(m,n)$-hypermodule over $(R,h,k)$
such that $R$ has the only one maximal hyperideal. Then $M$ is cyclic.

###### Proof.

Let $(M,f,g)$ be a nonzero multiplication $(m,n)$-hypermodule over $(R,h,k)$.
Assume that $I$ is the only maximal hyperideal of $R$. The Lemma 3.2 shows
that $g(I,1^{(n-2)},M)\neq M$. Then there exists some $m\in M$ such that
$m\notin g(I,1^{(n-2)},M)$. Thus we have $g(R,1^{(n-2)},m)=g(J,1^{(n-2)},M)$
for some hyperideal $J$ of $R$. Let $J\neq R$. It follows that $J\subseteq I$.
Hence we get $g(R,1^{(n-2)},m)=g(J,1^{(n-2)},M)\subseteq g(I,1^{(n-2)},M)$.
Since $m\in g(1^{(n-1)},m)\subseteq g(R,1^{(n-2)},m)$, then we conclude that
$m\in g(I,1^{(n-2)},M)$ which is contradiction. Thus $J=R$ which implies
$g(R,1^{(n-2)},m)=g(J,1^{(n-2)},M)=g(R,1^{(n-2)},M)=M$ and the proof is
completed. ∎

######  Theorem 3.4.

If $(M,f,g)$ is a multiplication hypermodule over $(R,h,k)$ then $M/N_{1}$
with $N_{1}=g(J_{(m,n)}(R),1^{(n-2)},M)$ is a multiplication hypermodule over
$(R,h,k)$ such that $J_{(m,n)}(R)$ is intersection of all maximal hyperideals
of $R$.

###### Proof.

Let $N_{1}=g(J_{(m,n)}(R),1^{(n-2)},M)$ in which $J_{(m,n)}(R)$ is
intersection of all maximal hyperideals of $R$. Let $N$ be a subhypermodule of
$M$ such that $N_{1}\subseteq N$ and $N/N_{1}$ be a subhypermodule of
$M/N_{1}$. Since $M$ is a multiplication hypermodule over $R$ then there
exists some hyperideal $I$ of $R$ such that $N=g(I,1^{(n-2)},M)$. Since
$N/N_{1}=\\{f(a_{1}^{m-1},N_{1})\ \ \ |a_{1}^{m-1}\in N\\}$

$\hskip
14.22636pt=\\{f(g(r_{1},1^{(n-2)},x_{1}),...,g(r_{m-1},1^{(n-2)},x_{m-1}),N_{1})|r_{1}^{m-1}\in
I,x_{1}^{m-1}\in N\\}$

$\hskip
14.22636pt=\\{f(g(r_{1},1^{(n-2)},x_{1}),...,g(r_{m-1},1^{(n-2)},x_{m-1}),f(N_{1}^{(m-1)},0)|r_{1}^{m-1}\in
I,$

$\hskip 22.76228ptx_{1}^{m-1}\in N\\}$

$\hskip
14.22636pt=\\{f(f(g(r_{1},1^{(n-2)},x_{1}),N_{1},0^{(m-2)}),...,f(g(r_{m-1},1^{(n-2)},x_{m-1}),N_{1},0^{(m-2)})$

$\hskip 22.76228pt,0)|r_{1}^{m-1}\in I,x_{1}^{m-1}\in N\\}$

$\hskip 14.22636pt=\\{f(\\{f(t_{1},N_{1},0^{(m-2)})\ |\ t_{1}\in
g(r_{1},1^{(n-2)},x_{1})\\},...,\\{f(t_{m-1},N_{1},0^{(m-2)})\ |\ t_{m-1}$

$\hskip 22.76228pt\in g(r_{m-1},1^{(n-2)},x_{m-1})\\},0)|r_{1}^{m-1}\in
I,x_{1}^{m-1}\in N\\}$

$\hskip 14.22636pt=F(\\{f(t_{1},N_{1},0^{(m-2)})\ |\ t_{1}\in
g(r_{1},1^{(n-2)},x_{1})\\},...,\\{f(t_{m-1},N_{1},0^{(m-2)})\ |\ t_{m-1}$

$\hskip 22.76228pt\in g(r_{m-1},1^{(n-2)},x_{m-1})\\},0))$

$\hskip
14.22636pt=F(G(r_{1},1^{(n-2)},f(x_{1},N_{1},0^{(m-2)})),...,G(r_{m-1},1^{(n-2)},f(x_{m-1},N_{1},0^{(m-2)})),0)$

$\hskip
14.22636pt=G(r_{1}^{m-1},F(f(x_{1},N_{1},0^{(m-2)}),...,f(x_{m-1},N_{1},0^{(m-2)}),0))$

$\hskip 14.22636pt=\bigcup\\{G(r,1^{(n-2)},f(x,N_{1},0^{(m-2)}))\ |\
r=k(r_{1}^{m-1},1^{n-m+1}),r_{1}^{m-1}\in I,x\in$

$\hskip 22.76228ptf(x_{1}^{m-1},0),x_{1}^{m-1}\in M\\}$

$\hskip 14.22636pt=G(I,1^{(n-2)},M/N_{1})$.
Hence $M/N_{1}$ with $N_{1}=g(J_{(m,n)}(R),1^{(n-2)},M)$ is a multiplication
hypermodule over $(R,h,k)$. ∎

######  Theorem 3.5.

Let $(M,f,g)$ be an $(m,n)$-hypermodule over $(R,h,k)$. $(M,f,g)$ is
multiplication if and only if

1)
$\bigcap_{\lambda\in\Lambda}g(I_{\lambda},1^{(n-2)},M)=g(\bigcap_{\lambda\in\Lambda}[f(I_{\lambda},S_{0},0^{(m-2)}],1^{(n-2)},M)$
for all nonempty family of hyperideals $I_{\lambda}$ of $R$.

2) For every hyperideal $A$ of $R$ and the subhypermodule $N$ of $M$ with
$N\subset g(A,1^{(n-2)},M)$, there exists a hyperideal $B\subset A$ such that
$N\subseteq g(B,1^{(n-2)},M)$.

###### Proof.

Since $M$ over $R/S_{0}$ is faithful, then we are done, by Theorem 3.11 in
[2]. ∎

######  Definition 3.6.

An $(m,n)$-hypermodule $(M,f,g)$ over $(R,h,k)$ is called cofinitely generated
if for each nonempty family of subhypermodules $N_{i}$ of $M$, $\bigcap_{i\in
I}N_{i}=0$ then there exists a finite subset $\Lambda^{\prime}of\Lambda$ such
that $\bigcap_{i\in I^{\prime}}N_{i}=0$.

######  Theorem 3.7.

Let $(M,f,g)$ be a faithful multiplication $(m,n)$-hypermodule over $(R,h,k)$.
Then $M$ is cofinitely generated if and only if $R$ is cofinitely generated.

###### Proof.

$\Longrightarrow$ Let $(M,f,g)$ be a faithful multiplication
$(m,n)$-hypermodule over $(R,h,k)$ and $\bigcap_{i\in I}J_{i}=0$ for a
nonempty family of hyperideals $J_{i}$ of $R$. Then we have

$\bigcap_{i\in I}g(J_{i},1^{(n-2)},M)=g(\bigcap_{i\in
I}J_{i},1^{(n-2)},M)=g(0,1^{(n-2)},M)=0$
by Theorem 3.11 in [2]. Since $M$ is cofinitely generated, then there exists a
finite subset $I^{\prime}$ of $I$ with

$\bigcap_{i\in I^{\prime}}g(J_{i},1^{(n-2)},M)=g(\bigcap_{i\in
I^{\prime}}J_{i},1^{(n-2)},M)=0$.
This implies that $\bigcap_{i\in I^{\prime}}J_{i}=0$, since $M$ is faithful.
Then $(R,h,k)$ is cofinitely generated.
$\Longleftarrow$ Let $(R,h,k)$ be cofinitely generated and $\bigcap_{i\in
I}T_{i}=0$ for a nonempty family of subhypermodules $T_{i}$ of $M$. Since $M$
is a multiplication $(m,n)$-hypermodule, then for each $i\in I$ there exists a
hyperideal $J_{i}$ of $R$ such that $T_{i}=g(J_{i},1^{(n-2)},M)$. Now we have

$g(\bigcap_{i\in I}J_{i},1^{(n-2)},M)\subseteq\bigcap_{i\in
I}g(J_{i},1^{(n-2)},M)=\bigcap_{i\in I}T_{i}=0$.
Since $M$ is faithful then $\bigcap_{i\in I}J_{i}=0$. Since $R$ is cofinitely
generated, then there exists a finite subset $I^{\prime}$ of $I$ such that
$\bigcap_{i\in I^{\prime}}J_{i}=0$. Hence

$\bigcap_{i\in I^{\prime}}T_{i}=\bigcap_{i\in I^{\prime}}g(J_{i},1^{(n-2)},M)$

$\hskip 36.98866pt=g(\bigcap_{i\in I^{\prime}}J_{i},1^{(n-2)},M)$

$\hskip 36.98866pt=g(0,1^{(n-2)},M)$

$\hskip 36.98866pt=0$.
Thus $(m,n)$-hypermodule $(M,f,g)$ is cofinitely generated. ∎

Let $(M,f,g)$ be an $(m,n)$-hypermodule over $(R,h,k)$. We denote the family
of all hyperideals $A$ of $R$ such that $g(A,1^{(n-2)},M)=M$ by $\omega$ and
denote intersection of all members of $\omega$ by $\omega(M)$.

######  Theorem 3.8.

Let $(M,f,g)$ be an faithful multiplication $(m,n)$-hypermodule over $(R,h,k)$
and $\omega(M)=B$. Then:

(1) $x\in g(B,1^{(n-2)},m)$, for each $x\in M$.

(2) $B=B.B$.

(3) $B\subseteq P$ or $h(B,P,0^{(n-2)})=R$, for each $n$-ary prime hyperideal
$P$ of $R$.

(4) $M$ is a multiplication $(m,n)$-hypermodule over $B$.

(5) $g(C,1^{(n-2)},M)\neq M$, for each proper hyperideal $C$ of $R$

###### Proof.

(1) Let $B=\omega(M)=\bigcap_{A\in\omega}A$. Theorem 3.11 in [2] shows that
$g(B,1^{(n-2)},M)=g(\bigcap_{A\in\omega}A,1^{(n-2)},M)=\bigcap_{A\in\omega}(g(A,1^{(n-2)},M)=\bigcap_{A\in\omega}M=M$.
Let $x\in R$. Since $M$ is multiplication, then
$g(R,1^{(n-2)},x)=g(D,1^{(n-2)},M)$ for some hyperideal $D$ of $R$. Then we
get

$x\in g(1^{(n-1)},x)\subseteq g(R,1^{(n-2)},x)$

$\hskip 68.28644pt=g(D,1^{(n-2)},M)$

$\hskip 68.28644pt=g(D,1^{(n-2)},g(B,1^{(n-2)},M))$

$\hskip 68.28644pt=g(k(D,1^{(n-2)},B),1^{(n-2)},M)$

$\hskip 68.28644pt=g(k(B,1^{(n-2)},D),1^{(n-2)},M)$

$\hskip 68.28644pt=g(B,1^{(n-2)},g(D,1^{(n-2)},M))$

$\hskip 68.28644pt=g(B,1^{(n-2)},g(R,1^{(n-2)},x))$

$\hskip 68.28644pt=g(k(B,R,1^{(n-2)}),1^{(n-2)},x)$

$\hskip 68.28644pt=g(B,1^{(n-2)},x)$
(2) Since $M=g(B,1^{(n-2)},M)$, then

$\hskip 56.9055ptM=g(B,1^{(n-2)},M)$

$\hskip 71.13188pt=g(B,1^{(n-2)},g(B,1^{(n-2)},M))$

$\hskip 71.13188pt=g(k(B^{(2)},1^{(n-2)}),1^{(n-2)},M)$

$\hskip 71.13188pt\subseteq g(B.B,1^{(n-2)},M)$
Since $g(B.B,1^{(n-2)},M)\subseteq M$ then $M=g(B.B,1^{(n-2)},M)$ which
implies $B.B\in\omega$ then $B\subseteq B.B$. Since $B.B\subseteq B$ then we
get $B=B.B$.
(3) If $M=g(P,1^{(n-2)},M)$ for some $n$-ary prime hyperideal $P$ of $R$ then
$P\in\omega$ and so $B\subseteq P$. If $M\neq g(P,1^{(n-2)},M)$ then there
exists some $m\in M$ such that $m\notin g(P,1^{(n-2)},M)$. Since $M$ is a
multiplication $(m,n)$-hypermodule, then we have
$g(R,1^{(n-2)},m)=g(C,1^{(n-2)},M)$ for some hyperideal $C$ of $R$. If
$C\subseteq P$ then $g(C,1^{(n-2)},M)\subseteq g(P,1^{(n-2)},M)$ which implies
$m\in g(P,1^{(n-2)},M)$ which is a contradiction. Therefore $C\nsubseteq P$.
Since for each $m\in M$, $m\in g(B,1^{(n-2)},m)$. Then there exists $b\in B$
such that $m\in f(g(b,1^{(n-2)},m),0^{(n-1)})$. Thus

$\hskip 28.45274pt0\in f(-g(b,1^{(n-2)},m),m,0^{(m-2)})$

$\hskip 36.98866pt\subseteq
f(g(-b,1^{(n-2)},m),g(1^{(n-1)},m),g(0,1^{(n-2)},m))$

$\hskip 36.98866pt=g(h(-b,1,0^{(m-2)}),1^{(n-2)},m)$

$\hskip 36.98866pt\subseteq g(h(-b,1,0^{(m-2)}),1^{(n-2)},g(1^{(n-1)},m))$

$\hskip 36.98866pt\subseteq g(h(-b,1,0^{(m-2)}),1^{(n-2)},g(R,1^{(n-2)},m))$

$\hskip 36.98866pt=g(h(-b,1,0^{(m-2)}),1^{(n-2)},g(C,1^{(n-2)},M))$

$\hskip 36.98866pt=g(k(h(-b,1,0^{(m-2)}),C,1^{(n-2)}),1^{(n-2)},M)$
Since $M$ is faithful then $k(h(-b,1,0^{(m-2)}),C,1^{(n-2)})=0$ then Since $P$
is an $n$-ary hyperideal of $R$ and
$k(h(-b,1,0^{(m-2)}),C,1^{(n-2)})=\\{0\\}\subseteq P$ with $C\nsubseteq P$
then we conclude that $h(-b,1,0^{(m-2)})\subseteq P$. Thus
$h(b,h(-b,1,0^{(m-2)}),0^{(n-2)})\subseteq f(B,P,0^{(n-2)})$ which means
$h(b,-b,1,0^{(n-3)})\subseteq h(B,P,0^{(n-2)})$.
Since $0\in h(b,-b,0^{(n-2)})$ then

$\hskip 19.91684pt1=h(0^{(n-1)},1)$

$\hskip 28.45274pt=h(0,1,0^{(n-2)})$

$\hskip 28.45274pt\subseteq h(h(b,-b,1,0^{(n-3)}),1,0^{(n-2)})$

$\hskip 28.45274pt=h(b,-b,1,0^{(n-3)})$

$\hskip 28.45274pt\subseteq h(B,P,0^{(n-2)})$.
Consequently, $R=h(B,P,0^{(n-2)})$.
(4) Let $N$ be an subhypermodule of $(m,n)$-hypermodule $M$ over $B$. Since
$N=g(B,1^{(n-2)},N)$, by (1), then $N$ is an subhypermodule of
$(m,n)$-hypermodule $M$ over $R$. Thus $N=g(D,1^{(n-2)},M)$ for some
hyperideal $D$ of $R$. It follows that

$\hskip 25.6073ptN=g(D,1^{(n-2)},M)$

$\hskip 36.98866pt=g(B,D,1^{(n-2)},M)$

$\hskip 36.98866pt=g(k(B,D,1^{(n-2)}),1^{(n-2)},M)$

$\hskip 36.98866pt\subseteq g(B.D,1^{(n-2)},M)$
Since $g(B.D,1^{(n-2)},M)\subseteq N$ then $g(B.D,1^{(n-2)},M)=N$. Since $B.D$
is a hyperideal of $B$, then $M$ is a multiplication $(m,n)$-hypermodule over
$B$.
(5) Let $g(C,1^{(n-2)},M)=M$, for some proper hyperideal $C$ of $R$. Therefore
we get

$\hskip 28.45274ptM=g(C,1^{(n-2)},M)$

$\hskip 42.67912pt=g(C,1^{(n-2)},g(B,1^{(n-2)},M))$

$\hskip 42.67912pt=g(k(C,B,1^{(n-2)}),1^{(n-2)},M)$

$\hskip 42.67912pt\subseteq g(C.B,1^{(n-2)},M)$.
It follows that $C.B\in\omega$. Since $b=\omega(M)$ then $B\subseteq
C.B\subseteq C$. since $C\subseteq B$ then we get $C=B$. ∎

######  Theorem 3.9.

Let $(M,f,g)$ be a nonzero $(m,n)$\- hypermodule over $(R,h,k)$. Let
$A_{1}=\\{P\in Max(R)\ |\ M\neq g(P,1^{(n-2)},M)\\}$ and $A_{2}=\\{P\in
Max(R)\ |\ S_{0}\subseteq P\\}$ such that $Max(R)$ is the family of all
maximal subhypermodules of $R$. Then

$J_{(m,n)}(M)=g(\bigcap\\{P\ |\ P\in A_{1}\\},1^{(n-2)},M)=g(\bigcap\\{P\ |\
P\in A_{2}\\},1^{(n-2)},M)$

###### Proof.

Suppose that $B_{1}=\bigcap\\{P\ |\ P\in A_{1}\\}$ and $B_{2}=\bigcap\\{P\ |\
P\in A_{2}\\}$. Let $K$ is a maximal subhypermodule of $M$. Theorem 3.16 in
[2] shows that $K=g(P,1^{(n-2)},M)\neq M$, for some maximal hyperideal $P$ of
$R$. By Theorem 3.11 (i) in [2],

$\hskip 14.22636ptg(\bigcap\\{P\ |\ P\in
A_{2}\\},1^{(n-2)},M)=\bigcap\\{g(P,1^{(n-2)},M)\ |\ P\in A_{2}\\}$

$\hskip 142.26378pt\subseteq\bigcap\\{g(P,1^{(n-2)},M)\ |\ P\in A_{1}\\}$

$\hskip 142.26378pt=g(\bigcap\\{P\ |\ P\in A_{1}\\},1^{(n-2)},M)$

$\hskip 142.26378pt=J_{(m,n)}(M)$
Now, let $Q\in A_{2}$. If $g(Q,1^{(n-2)},M)=M$ then we get
$J_{(m,n)}(M)\subseteq g(Q,1^{(n-2)},M)$ which means $J_{(m,n)}(M)\subseteq
g(\bigcap\\{P\ |\ P\in A_{2}\\},1^{(n-2)},M)$. If $g(Q,1^{(n-2)},M)\neq M$
then we conclude that $Q\in A_{1}$ which implies $J_{(m,n)}(M)\subseteq
g(Q,1^{(n-2)},M)$ which means $J_{(m,n)}(M)\subseteq g(\bigcap\\{P\ |\ P\in
A_{2}\\},1^{(n-2)},M)$. ∎

######  Definition 3.10.

[7] Let $M$ is an $(m,n)$-hypermodule over $(R,h,k)$. Let $M_{1}^{t}$ be
subhypermodules of $M$ and $t=l(m-1)+1$, then $M$ is called a (internal)
direct sum $M_{1}\oplus...\oplus M_{t}$ if satisfies the following axioms:
(1) $M=f_{(l)}(M_{1},...,M_{t})$ and
(2) $M_{i}\cap f_{(l)}(M_{1},...,M_{i-1},0,M_{i+1},...M_{t})=\\{0\\}$

######  Theorem 3.11.

Let $M=\bigoplus_{\lambda\in\Lambda}M_{\lambda}$ for some $(m,n)$-hypermodules
$M_{\lambda}$ over $(R,h,k)$ and $\lambda=l(m-1)+1$. Then $M$ is a
multiplication $(m,n)$-hypermodule over $(R,h,k)$ if and only if
$(i)$ $M_{\lambda}$ is a multiplication $(m,n)$-hypermodule for each
$\lambda\in\Lambda$
$(ii)$ for each $\lambda\in\Lambda$ there exists hyperideal $I_{\lambda}$ of
$R$ such that $g(I_{\lambda},1^{(n-2)},M_{\lambda})=M_{\lambda}$ and
$g(I_{\lambda},1^{(n-2)},\hat{M_{\lambda}})=0$ where
$\hat{M_{\lambda}}=\bigoplus_{\mu\neq\lambda}M_{\mu}$.

###### Proof.

$\Longrightarrow$Let $M$ is a multiplication $(m,n)$-hypermodule over
$(R,h,k)$ and $\lambda\in\Lambda$ such that $\lambda=l(m-1)+1$. Then there
exists hyperideal $I_{\lambda}$ of $R$ such that
$g(I_{\lambda},1^{(n-2)},M)=M_{\lambda}$ which implies
$g(I_{\lambda},1^{(n-2)},\hat{M_{\lambda}})\cong
g(I_{\lambda},1^{(n-2)},M/M_{\lambda})=M_{\lambda}/M_{\lambda}=0$. Therefore
we have
$M_{\lambda}=g(I_{\lambda},1^{(n-2)},M)=g(I_{\lambda},1^{(n-2)},M_{\lambda}\bigoplus\hat{M_{\lambda}})=g(I_{\lambda},1^{(n-2)},M_{\lambda})$.
$\Longleftarrow$ Let $P$ be a maximal hyperideal of $R$ such that
$\mathcal{X}_{P}(M_{\lambda})=M_{\lambda}$ for all $\lambda\in\Lambda$. Then
we get
$M=\bigoplus_{\lambda\in\Lambda}M_{\lambda}=\bigoplus_{\lambda\in\Lambda}\mathcal{X}_{P}(M_{\lambda})=\mathcal{X}_{P}(\bigoplus_{\lambda\in\Lambda}M_{\lambda})=\mathcal{X}_{P}(M).$
If $\mathcal{X}_{P}(M_{\gamma})\neq M_{\gamma}$, for some $\gamma\in\Lambda$
then we conclude that $M_{\gamma}$ is $n$-ary $P-$cyclic, by Theorem 3.8 in
[2]. It means there exist $m\in M_{\gamma}$ and $p\in P$ with

$g(h(1,-p,0^{(m-2)}),1^{(n-2)},M_{\gamma})\subseteq g(R,1^{(n-2)},m)$.
Thus we have

$\hskip
28.45274ptg(I_{\gamma},1^{(n-2)},M)=g(I_{\gamma},1^{(n-2)},M_{\gamma}\bigoplus\hat{M_{\gamma}})$

$\hskip 99.58464pt=g(I_{\gamma},1^{(n-2)},M_{\gamma})\bigoplus
g((I_{\gamma},1^{(n-2)},\hat{M_{\gamma}})$

$\hskip 99.58464pt=g(I_{\gamma},1^{(n-2)},M_{\gamma})$

$\hskip 99.58464pt=M_{\gamma}$.
It follows that

$g(h(1,-p,0^{(m-2)}),I_{\gamma},1^{(n-3)},M)=g(h(1,-p,0^{(m-2)}),1^{(n-2)},M_{\gamma})$

$\hskip 145.10922pt\subseteq g(R,1^{(n-2)},m).$
Now let $I_{\gamma}\subseteq P$. It follows that
$M_{\gamma}=g(I_{\gamma},1^{(n-2)},M_{\gamma})\subseteq
g(P,1^{(n-2)},M_{\gamma})$ which means $g(P,1^{(n-2)},M_{\gamma})=M_{\gamma}$.
By Theorem 3.8 in [2], we have $\mathcal{X}_{(}M_{\gamma})=M_{\gamma}$, since
$M_{\gamma}$ is a multiplication $(m,n)$-hypermodule. This is a contradiction.
Thus $I_{\gamma}\nsubseteq P$ and so
$g(h(1,-p,0^{(m-2)}),1^{(n-2)},I_{\gamma})\nsubseteq P$.Hence
$h(1,-q,0^{(m-2)})\subseteq k(h(1,-p,0^{(m-2)}),I_{\gamma},1^{(n-2)})$ for
some $q\in P$. Then

$g(h(1,-q,0^{(m-2)}),1^{(n-2)},M)\subseteq
g(k(h(1,-p,0^{(m-2)}),I_{\gamma},1^{(n-2)}),1^{(n-2)},M)$

$\hskip 130.88284pt=g(h(1,-p,0^{(m-2)}),I_{\gamma},1^{(n-3)},M)$

$\hskip 130.88284pt\subseteq g(R,1^{(n-2)},m)$.
It means $M$ is $n$-ary $P$-cyclic. Consequently, $M$ is a multiplication
$(m,n)$-hypermodule, by Theorem 3.8 in [2]. ∎

##  4\. The primary subhypermodules of multiplication $(m,n)$-hyperodules

######  Lemma 4.1.

Let $(M,f,g)$ be a faithful multiplication $(m,n)$-hypermodule over $(R,h,k)$
such that $Q$ is an $n$-ary primary hyperideal of $R$. If
$g(r,1^{(n-2)},m)\subseteq g(Q,1^{((n-2)},M)$, for some $r\in R$ , $m\in M$
then $r\in{\sqrt{Q}}^{(m,n)}$ or $m\in g(Q,1^{((n-2)},M)$.

###### Proof.

Let $r\notin{\sqrt{Q}}^{(m,n)}$. Suppose that $E=\\{a\in R\ |\
g(a,1^{(n-2)},m)\subseteq g(Q,1^{((n-2)},M)\\}$. Let $E\neq R$. Therefore
there exists a maximal hyperideal $P$ of $R$ such that $E\subseteq P$. If
$m\in\mathcal{X}_{P}(M)$, then for some $p\in P$,
$g(h(1,-p,0^{(m-2)}),1,m)=\\{0\\}$. This means that
$g(h(1,-p,0^{(m-2)}),1,m)\subseteq g(Q,1^{((n-2)},M)$. Hence
$h(1,-p,0^{(m-2)})\subseteq E\subseteq Q$ and so $1\in Q$, which is a
contradiction. Thus $m\notin\mathcal{X}_{P}(M)$. Then $M$ is n-ary $P$-cyclic,
by Theorem 3.8 in [2]. Then there exists $x\in M$, $p\in P$ such that

$g(h(1,-p,0^{(m-2)}),1^{(n-2)},M)\subseteq g(R,1^{(n-2)},x)$.
This means $g(h(1,-p,0^{(m-2)}),1^{(n-2)},m)=g(U,1^{(n-2)},x)$ for some
$U\subseteq R$. We have

$g(r,h(1,-p,0^{(m-2)}),1^{(n-3)},m)=g(h(1,-p,0^{(m-2)}),1^{(n-2)},g(r,1^{(n-2)},m))$

$\hskip 136.5733pt\subseteq g(h(1,-p,0^{(m-2)}),1^{(n-2)},g(Q,1^{(n-2)},M))$

$\hskip 136.5733pt\subseteq g(Q,1^{(n-2)},g(h(1,-p,0^{(m-2)}),1^{(n-2)},M))$

$\hskip 136.5733pt\subseteq g(Q,1^{(n-2)},g(R,1^{(n-2)},x))$

$\hskip 136.5733pt=g(Q,1^{(n-2)},x)$
Thus for some $Q_{1}\subseteq Q$, we get

$g(r,h(1,-p,0^{(m-2)}),1^{(n-3)},m)=g(Q_{1},1^{(n-2)},x)$
Also,

$g(k(r,U,1^{(n-2)}),1^{(n-2)},x)=g(r,1^{(n-2)},g(U,1^{(n-2)},x))$

$\hskip 116.65646pt=g(r,1^{(n-2)},g(h(1,-p,0^{(m-2)}),1^{(n-2)},m))$

$\hskip 116.65646pt=g(r,1^{(n-3)},h(1,-p,0^{(m-2)}),m)$

$\hskip 116.65646pt\subseteq g(Q,1^{(n-2)},x).$
Let $u\in U$. Then for some $Q_{1}\subseteq Q$ we have

$g(k(r,u,1^{(n-2)}),1^{(n-2)},x)=g(Q_{1},1^{(n-2)},x)$

$\hskip 116.65646pt=f(g(Q_{1},1^{(n-2)},x),0^{(m-1)})$
This means $0\in
f(g(k(r,u,1^{(n-2)}),1^{(n-2)},x),-g(Q_{1},1^{(n-2)},x),0^{(m-2)}))$

$\hskip 48.36958pt=g(h(k(r,u,1^{(n-2)}),-Q_{1},0^{(m-2)}),1^{(m-2)},x)$.
Therefore $0\in h(k(r,u,1^{(n-2)}),-Q_{1},0^{(m-2)})$, since $M$ is faithful.
This implies that $k(r,u,1^{(n-2)})\in Q_{1}$. Thus

$k(u,k(h(1,-p,0^{(m-2)}),r,1^{(n-2)}),1^{(n-2)})=k(k(u,r,1^{(n-2)}),h(1,-p,0^{(m-2)}),1^{(n-2)})$

$\hskip 176.407pt\subseteq k(Q_{1},h(1,-p,0^{(m-2)}),1^{(n-2)})$

$\hskip 176.407pt\subseteq k(Q,h(1,-p,0^{(m-2)}),1^{(n-2)})$

$\hskip 176.407pt\subseteq Q$.
Since $Q$ is an n-ary primary hyperideal of $R$, we have

$u\in Q$ or
$k(1,k(h(1,-p,0^{(m-2)}),r,1^{(n-2)}),1^{(n-2)})\in{\sqrt{Q}}^{(m,n)}$.
Let
$k(1,k(h(1,-p,0^{(m-2)}),r,1^{(n-2)}),1^{(n-2)})=k(h(1,-p,0^{(m-2)}),r,1^{(n-2)})\in{\sqrt{Q}}^{(m,n)}$.
Since $Q$ is an n-ary primary hyperideal of $R$, then ${\sqrt{Q}}^{(m,n)}$ is
an n-ary prime hyperideal, by Theorem 4.28 in [1]. Since
$r\notin{\sqrt{Q}}^{(m,n)}$ and $1\notin{\sqrt{Q}}^{(m,n)}$, then
$h(1,-p,0^{(m-2)})\in{\sqrt{Q}}^{(m,n)}\subseteq Q\subseteq E\subseteq P$ and
so $1\in P$, a contradiction. Then we conclude that $u\in Q$. Therefore
$g(h(1,-p,0^{(m-2)}),1^{(n-2)},m)=g(U,1^{(n-2)},m)\subseteq g(Q,1^{(n-2)},M)$.
By the definition of the set $E$, we have $h(1,-p,0^{(m-2)})\subseteq
E\subseteq P$. This is a contradiction. Consequently, $E=R$ and so $m\in
g(1^{(n-1)},m)\subseteq g(Q,1^{(n-2)},M)$. ∎

######  Corollary 4.2.

Let $(M,f,g)$ be a faithful multiplication $(m,n)$-hypermodule over $(R,h,k)$.
Let $N$ be a subhypermodule of $M$ such that $N=g(Q,1^{(n-2)},M)$and $M\neq
g(Q,1^{(n-2)},M)$ for some n-ary primary hyperideal $Q$ of $R$. Then $N$ is an
$n$-ary primary subhypermodule of $M$.

###### Proof.

Suppose that $N$ be a subhypermodule of a faithful multiplication
$(m,n)$-hypermodule $M$ such that $N=g(Q,1^{(n-2)},M)$ for some n-ary primary
hyperideal $Q$ of $R$. Let for $r_{1}^{n-1}\in R$ and $m\in M\backslash N$,
$g(r_{1}^{n-1},m)\subseteq N$. Since $N=g(Q,1^{(n-2)},M)$ and
$g(k(r_{1}^{n-1},1),1^{(n-2)},m)=g(r_{1}^{n-1},m)\subseteq g(Q,1^{(n-2)},M)$,
we have $k(r_{1}^{n-1},1)\in\sqrt{Q}^{(m,n)}$, By Lemma 4.1. Since
$\sqrt{Q}^{(m,n)}\subseteq\sqrt{S_{N}}^{(m,n)}$ with $S_{N}=\\{r\in R\ |\
g(r,1^{(n-2)},M)\subseteq N\\}$, then $N$ is an $n$-ary primary subhypermodule
of $M$. ∎

######  Corollary 4.3.

Let $(M,f,g)$ be a multiplication $(m,n)$-hypermodule over $(R,h,k)$. The
followings are equivalent for a proper subhypermodule $N$ of $M$:
(1) $N$ is an $n$-ary primary subhypermodule of $M$.
(2) $S_{N}=\\{r\in R\ |\ g(r,1^{(n-2)},M)\subseteq N\\}$ is an $n$-ary primary
hyperideal of $R$.
(3) $N=g(Q,1^{(n-2)},M)$ for some $n$-ary primary hyperideal $Q$ of $R$ with
$S_{0}\subseteq Q$.

###### Proof.

$(1)\Longrightarrow(2)$ It follows from Theorem 5.3 in [4].
$(2)\Longrightarrow(3)$ It is clear by Remark 3.2 in [2].
$(3)\Longrightarrow(1)$ Let for $r_{1}^{n-1}\in R$ and $m\in M\backslash N$,
$g(r_{1}^{n-1},m)\subseteq N$. Then,

$g(1^{(n-1)},g(r_{1},1^{(n-2)},g(k(r_{2}^{n-1},1^{(2)}),1^{(n-2)},m)))$

$\hskip
14.22636pt=g(1^{(n-2)},k(r_{1},1^{(n-1)},g(k(r_{2}^{n-1},1^{(2)}),1^{(n-2)},m))$

$\hskip 14.22636pt\vdots$

$\hskip
14.22636pt=g(1,k(r_{1},1^{(n-1)}),...,k(r_{n-2},1^{(n-1)}),g(r_{n-1},1^{(n-2)},m))$

$\hskip
14.22636pt=g(k(r_{1},1^{(n-1)}),...,k(r_{n-2},1^{(n-1)}),k(r_{n-1},1^{(n-1)}),m)$

$\hskip 14.22636pt=(r_{1}^{n-1},m)$

$\hskip 14.22636pt\subseteq N=g(Q,1^{(n-2)},M)$
for n-ary primary hyperideal $Q$ of $R$. Since $1\notin\sqrt{Q}^{(m,n)}$, then

$g(r_{1},1^{(n-2)},g(k(r_{2}^{n-1},1^{(2)}),1^{(n-2)},m))\subseteq
g(Q,1^{(n-2)},M)$
by Lemma 4.1. By using Lemma 4.1 again, we have

$r_{1}\in\sqrt{Q}^{(m,n)}$ or $g(k(r_{2}^{n-1},1^{(2)}),1^{(n-2)},m)\subseteq
g(Q,1^{(n-2)},M)$
In the second case, we have

$g(r_{2},g(r_{3}^{n-1},1^{(2)}),1^{(n-3)},m))\subseteq g(Q,1^{(n-2)},M)$
Therefore we get

$r_{2}\in\sqrt{Q}^{(m,n)}$ or $g(r_{3}^{n-1},1^{(2)}),1^{(n-3)},m)\subseteq
g(Q,1^{(n-2)},M)$,
by Lemma 4.1. In the second case, by using a similar argument we get
$r_{n-1}\in\sqrt{Q}^{(m,n)}$ or $m\in g(Q,1^{(n-1)},M)=N$. Since $m\notin N$,
then we have $r_{i}\in\sqrt{Q}^{(m,n)}$ for some $1\leq i\leq n-1$ and so
$k(r_{1}^{n-1},1)\in\sqrt{Q}^{(m,n)}$. Since
$\sqrt{Q}^{(m,n)}\subseteq\sqrt{S_{N}}^{(m,n)}$, then
$k(r_{1}^{n-1},1)\in\sqrt{S_{N}}^{(m,n)}$. It follows that $N$ is an $n$-ary
primary subhypermodule of $M$. ∎

##  5\.  intersections and sums of the multiplication $(m,n)$-hypermodule

######  Theorem 5.1.

Let $(M,f,g)$ be an $(m,n)$-hypermodule over $(R,h,k)$ such that
$M=f_{(l)}(N_{\lambda}^{(\lambda\in\Lambda)})$ for some multiplication
subhypermodules $N_{\lambda}$ of $M$ and $\lambda=l(m-1)+1$. The followings
are equivalent:
$(1)$ $M$ is a multiplication $(m,n)$-hypermodule.
$(2)$ $N_{\lambda}=g(S_{N_{\lambda}},1^{(n-2)},M)$, for each
$\lambda\in\Lambda$
$(3)$ $R=h(A(m),B,0^{(m-2)})$ for each $m\in M$ such that $A(m)=\\{r\in R\ |\
0\in g(r,1^{(n-2)},m)\\}$ and
$B=f_{(l)}(S_{N_{\lambda}}^{(\lambda\in\Lambda)})$
$(4)$ for every maximal hyperideal $P$ of $R$ either $M=\mathcal{X}_{P}(M)$ or
there exist $a\in\bigcup_{\lambda\in\Lambda}N_{\lambda}$ and $p\in P$ such
that $g(h(1,-p,0^{(m-2)}),1^{(n-2)},M)\subseteq g(R,1^{(n-2)},a)$

###### Proof.

$(1)\Longrightarrow(2)$ It is clear.
$(2)\Longrightarrow(3)$ Let $R\neq h(A(m),B,0^{(m-2)})$ for some $m\in M$.
Then there exists some maximal hyperideal $Q$ of $R$ such that
$h(A(m),B,0^{(m-2)})\subseteq Q$. Then $B\subseteq Q$ which implies
$g(B,1^{(n-2)},M)\subseteq g(Q,1^{(n-2)},M)$ which means
$g(f_{(l)}(S_{N_{\lambda}}^{(\lambda\in\Lambda)}),1^{(n-2)},M)\subseteq
g(Q,1^{(n-2)},M)$. Thus $f_{(l)}(N_{\lambda}^{(\lambda\in\Lambda)})=M\subseteq
g(Q,1^{(n-2)},M)$ and so $M=g(Q,1^{(n-2)},M)$. Since $m\in
M=f_{(l)}(N_{\lambda}^{(\lambda\in\Lambda)})$ then $m\in
f_{(l)}(m_{\lambda}^{(\lambda\in\Lambda)})$ for some elements $m_{\lambda}\in
N_{\lambda}$. Since $N_{\lambda}$ is a multiplication $(m,n)$-hypermodule then
there exists hyperideal $I_{\lambda}$ of $R$ such that
$g(R,1^{(n-2)},m_{\lambda})=g(I_{\lambda},1^{(n-2)},N_{\lambda})$. Thus we get

$\hskip
28.45274ptg(R,1^{(n-2)},m_{\lambda})=g(I_{\lambda},1^{(n-2)},N_{\lambda})$.

$\hskip 102.43008pt=g(I_{\lambda},1^{(n-2)},g(S_{N_{\lambda}},1^{(n-2)},M)$.

$\hskip 102.43008pt=g(I_{\lambda},S_{N_{\lambda}},1^{(n-3)},M)$

$\hskip 102.43008pt=g(I_{\lambda},S_{N_{\lambda}},1^{(n-3)},g(Q,1^{(n-2)},M))$

$\hskip 102.43008pt=g(I_{\lambda},Q,1^{(n-3)},g(S_{N_{\lambda}},1^{(n-2)},M))$

$\hskip 102.43008pt\subseteq g(I_{\lambda},Q,1^{(n-3)},{N_{\lambda}})$

$\hskip 102.43008pt=g(Q,1^{(n-2)},g(I_{\lambda},1^{(n-2)},{N_{\lambda}}))$

$\hskip 102.43008pt=g(Q,1^{(n-2)},g(R,1^{(n-2)},m_{\lambda}))$

$\hskip 102.43008pt=g(Q,R,1^{(n-3)},m_{\lambda})$

$\hskip 102.43008pt=g(Q,1^{(n-2)},m_{\lambda})$.
Then there exists $q_{\lambda}\in Q$ such that $m_{\lambda}\in
f(g(q_{\lambda},1^{(n-2)},m_{\lambda}),0^{(m-1)})$ which implies

$\hskip 28.45274pt0\in
f(-g(q_{\lambda},1^{(n-2)},m_{\lambda}),m_{\lambda},0^{(m-2)})$

$\hskip 36.98866pt\subseteq
f(g(-q_{\lambda},1^{(n-2)},m_{\lambda}),g(1^{(n-1)},m_{\lambda}),g(0,1^{(n-2)},m_{\lambda}))$

$\hskip 36.98866pt=g(h(-q_{\lambda},1,0^{(m-2)}),1^{(n-2)},m_{\lambda}))$.
Thus we get $0\in
g(g(h(-q_{\lambda},1,0^{(m-2)})^{(\lambda\in\Lambda)}),1^{(n-2)},m)$ which
means

$g(h(-q_{\lambda},1,0^{(m-2)})^{(\lambda\in\Lambda)})\in A(m)\subseteq Q$.
Since all $q_{\lambda}$’s belong to $Q$ then we conclude that $1\in Q$ which
is a contradiction. Consequently, $R=h(A(m),B,0^{(m-2)})$.
$(3)\Longrightarrow(4)$ Suppose that for some maximal hyperideal $P$ of $R$,
$M\neq\mathcal{X}_{P}(M)$. Then there exists some $m\in M$ such that $0\notin
g(h(1,-p,0^{(m-2)}),1^{(n-2)},m)$ for all $p\in P$. If $A(m)\nsubseteq P$ then
there exists $p\in P$ such that $h(1,-p,0^{(m-2)})\subseteq A(m)$ which means
$0\in g(h(1,-p,0^{(m-2)}),1^{(n-2)},m)$ which is a contradiction. Then
$A(m)\subseteq P$ and so $B\notin P$ by (3). This follows that
$S_{N_{\lambda}}\nsubseteq P$ for some $\lambda\in\Lambda$. Then there exists
$p\in P$ such that $h(1,-p,0^{(m-2)})\subseteq S_{N_{\lambda}}$. Thus
$g(h(1,-p,0^{(m-2)}),1^{(n-2)},M)\subseteq
g(S_{N_{\lambda}},1^{(n-2)},M)\subseteq N_{\lambda}$. Let
$\mathcal{X}_{P}(N_{\lambda})=N_{\lambda}$ and $x\in
g(S_{N_{\lambda}},1^{(n-2)},m)\subseteq N_{\lambda}$. Then for all $p\in P$ we
have
$0\in g(h(1,-p,0^{(m-2)}),1^{(n-2)},x)\subseteq
g(h(1,-p,0^{(m-2)}),1^{(n-2)},g(S_{N_{\lambda}},1^{(n-2)},m))$

$\hskip
133.72786pt=g(g(h(1,-p,0^{(m-2)}),S_{N_{\lambda}},1^{(n-2)}),1^{(n-3)},m)$
Therefore $g(h(1,-p,0^{(m-2)}),S_{N_{\lambda}},1^{(n-2)})\subseteq
A(m)\subseteq P$. Since $P$ is an n-ary prime hyperideal and
$h(1,-p,0^{(m-2)})\nsubseteq P$ then we get $S_{N_{\lambda}}\subseteq P$ which
is a contradiction. Then $\mathcal{X}_{P}(N_{\lambda})\neq N_{\lambda}$. Hence
$N_{\lambda}$ is n-ary $P$-cyclic, by Theorem 3.8 in [2]. Then there exist
$a\in N_{\lambda}$ and $q\in Q$ such that
$g(h(1,-p,0^{(m-2)}),1^{(n-2)},N_{\lambda})\subseteq g(R,1^{(n-2)},a)$. By
$g(h(1,-p,0^{(m-2)}),1^{(n-2)},M)\subseteq
g(S_{N_{\lambda}},1^{(n-2)},M)\subseteq N_{\lambda}$ it follows that
$g(h(1,-q,0^{(m-2)}),h(1,-p,0^{(m-2)}),1^{(n-3)},M)\subseteq
g(R,1^{(n-2)},a)$. It is clear that
$h(p,q,-k(p,q,1^{(n-2)}),0^{(m-3)})\subseteq P$.
Let $t\in h(p,q,-k(p,q,1^{(n-2)}),0^{(m-3)})$. Then

$g(h(1,-t,0^{(m-2)}),1^{(n-2)},M)$

$\hskip 19.91684pt\subseteq
g(h(1,-(h(p,q,-k(p,q,1^{(n-2)}),0^{(m-3)}),0^{(m-2)}),1^{(n-2)},M)$

$\hskip 19.91684pt=g(h(1,-p,-q,k(p,q,1^{(n-2)}),0^{(m-4)}),1^{(n-2)},M)$

$\hskip
19.91684pt=g(h(1,-p,h(-q,k(p,q,1^{(n-2)}),0^{(m-2)})),0^{(m-3)}),1^{(n-2)},M)$

$\hskip
19.91684pt=g(h(1,-p,h(-q,-(-k(p,q,1^{(n-2)})),0^{(m-2)})),0^{(m-3)}),1^{(n-2)},M)$

$\hskip
19.91684pt=g(h(1,-p,h(-q,k(-p,-q,1^{(n-2)})),0^{(m-2)})),0^{(m-3)}),1^{(n-2)},M)$

$\hskip
19.91684pt=g(h(1,-p,k(-q,h(1,-p,0^{(m-2)}),1^{(n-2)})),0^{(m-3)}),1^{(n-2)},M)$

$\hskip
19.91684pt=g(h(h(1,-p,0^{(m-2)}),k(-q,h(1,-p,0^{(m-2)}),1^{(n-2)})),0^{(m-2)}),1^{(n-2)},M)$

$\hskip
19.91684pt=g(k(h(1,-p,0^{(m-2)}),h(1,-q,0^{(m-2)}),1^{(m-2)}),1^{(n-2)},M)$

$\hskip 19.91684pt\subseteq g(R,1^{(n-2)},a)$.

$(4)\Longrightarrow(1)$ This is obvious by Theorem 3.8 in [2]. ∎

######  Corollary 5.2.

Let $(M,f,g)$ be an $(m,n)$-hypermodule over $(R,h,k)$ such that
$R=h_{(l)}(S_{N_{\lambda}}^{(\lambda\in\Lambda)})$ for some multiplication
subhypermodules $N_{\lambda}$ of $M$ and $\lambda=l(m-1)+1$. Then $M$ is a
multiplication hypermodule.

###### Proof.

Since

$\hskip 28.45274ptM=g(R,1^{(n-2)},M)$

$\hskip
42.67912pt=g(h_{(l)}(S_{N_{\lambda}}^{(\lambda\in\Lambda)}),1^{(n-2)},M)$

$\hskip
42.67912pt=f_{(l)}(g(S_{N_{\lambda}},1^{(n-2)},M)^{(\lambda\in\Lambda)}))$

$\hskip 42.67912pt\subseteq f_{(l)}(N_{\lambda}^{(\lambda\in\Lambda)})$

$\hskip 42.67912pt\subseteq M$,
then $M=f_{(l)}(N_{\lambda}^{(\lambda\in\Lambda)})$. Thus $M$ is a
multiplication hypermodule, by Theorem 5.1. ∎

######  Theorem 5.3.

Let $(M,f,g)$ be an $(m,n)$-hypermodule over $(R,h,k)$ such that
$M=f_{(l)}(N_{\lambda}^{(\lambda\in\Lambda)})$ for some multiplication
subhypermodules $N_{\lambda}$ of $M$ and $\lambda=l(m-1)+1$. Then for each
subhypermodule $N$ of $M$, $N=f_{(l)}((N\cap
N_{\lambda})^{(\lambda\in\Lambda)})$.

###### Proof.

Let $N$ be a arbitrary subhypermodule of $M$. Then

$\hskip 28.45274ptN=g(S_{N},1^{(n-2)},M)$

$\hskip
39.83368pt=g(S_{N},1^{(n-2)},f_{(l)}(N_{\lambda}^{(\lambda\in\Lambda)}))$

$\hskip
39.83368pt=f_{(l)}((g(S_{N},1^{(n-2)},N_{\lambda})^{(\lambda\in\Lambda)})$

$\hskip 39.83368pt\subseteq f_{(l)}((N\cap
N_{\lambda})^{(\lambda\in\Lambda)})$

$\hskip 39.83368pt\subseteq N$.
Consequently, $N=f_{(l)}((N\cap N_{\lambda})^{(\lambda\in\Lambda)})$. ∎

######  Theorem 5.4.

Let $(M,f,g)$ be an $(m,n)$-hypermodule over $(R,h,k)$. Let $H$ and $K$ be two
multiplication subhypermodules of $M$ such that $H$, $K$ and
$f(H,K,1^{(n-2)})$ are multiplication hypermodules .Then $H\cap K$ is a
multiplication hypermodule.

###### Proof.

Let $\mathcal{X}_{P}(H\cap K)\neq H\cap K$ for some maximal hyperideal $P$ of
$R$. Then we conclude that $\mathcal{X}_{P}(H)\neq H$, $\mathcal{X}_{P}(K)\neq
K$ and $\mathcal{X}_{P}(f(H,K,0^{(m-2)}))\neq f(H,K,0^{(m-2)})$. By Theorem
3.8 in [2] and Theorem 5.1, there exist $a\in H$, $b\in K$, $c\in H\cup K$ and
$p,p^{\prime},p^{\prime\prime}\in P$ such that

$\hskip 28.45274ptg(h(1,-p,0^{(m-2)}),1^{(n-2)},H)\subseteq g(R,1^{(n-2)},a)$

$\hskip 28.45274ptg(h(1,-p^{\prime},0^{(m-2)}),1^{(n-2)},K)\subseteq
g(R,1^{(n-2)},b)$
and

$\hskip
28.45274ptg(h(1,-p^{\prime\prime},0^{(m-2)}),1^{(n-2)},f(H,K,0^{(m-2)}))\subseteq
g(R,1^{(n-2)},c)$.
If $c\in H$ then

$\hskip 8.5359ptg(h(1,-p^{\prime\prime},0^{(m-2)}),1^{(n-2)},b)\subseteq
g(h(1,-p^{\prime\prime},0^{(m-2)}),1^{(n-2)},f(H,K,0^{(m-2)}))$

$\hskip 139.41832pt\subseteq g(R,1^{(n-2)},c)$

$\hskip 139.41832pt\subseteq H$
which implies

$\hskip 28.45274ptg(h(1,-p^{\prime\prime},0^{(m-2)}),1^{(n-2)},b)\subseteq
H\cap K$.
Let $s\in
h(p^{\prime},p^{\prime\prime},-k(p^{\prime},p^{\prime\prime},1^{(n-2)}),0^{(m-3)})\subseteq
P$. Then
$g(h(1,-s,0^{(m-2)}),1^{(n-2)},H\cap K)$

$\subseteq
g(h(1,-(h(p^{\prime},p^{\prime\prime},-k(p^{\prime},p^{\prime\prime},1^{(n-2)}),0^{(m-3)}),0^{(m-2)}),1^{(n-2)},H\cap
K)$

$=g(h(1,-p^{\prime},-p^{\prime\prime},k(p^{\prime},p^{\prime\prime},1^{(n-2)}),0^{(m-4)}),1^{(n-2)},H\cap
K)$

$=g(h(1,-p^{\prime},h(-p^{\prime\prime},k(p^{\prime},p^{\prime\prime},1^{(n-2)}),0^{(m-2)})),0^{(m-3)}),1^{(n-2)},H\cap
K)$

$=g(h(1,-p^{\prime},h(-p^{\prime\prime},-(-k(p^{\prime},p^{\prime\prime},1^{(n-2)})),0^{(m-2)})),0^{(m-3)}),1^{(n-2)},H\cap
K)$

$=g(h(1,-p^{\prime},h(-p^{\prime\prime},k(-p^{\prime},-p^{\prime\prime},1^{(n-2)})),0^{(m-2)})),0^{(m-3)}),1^{(n-2)},H\cap
K)$

$=g(h(1,-p^{\prime},k(-p^{\prime\prime},h(1,-p^{\prime},0^{(m-2)}),1^{(n-2)})),0^{(m-3)}),1^{(n-2)},H\cap
K)$

$=g(h(h(1,-p^{\prime},0^{(m-2)}),k(-p^{\prime\prime},h(1,-p^{\prime},0^{(m-2)}),1^{(n-2)})),0^{(m-2)}),1^{(n-2)},H\cap
K)$

$=g(k(h(1,-p^{\prime},0^{(m-2)}),h(1,-p^{\prime\prime},0^{(m-2)}),1^{(n-2)}),1^{(n-2)},H\cap
K)$

$\subseteq
g(h(1,-p^{\prime\prime},0^{(m-2)}),1^{(n-2)},g(h(1,-p^{\prime},0^{(m-2)}),1^{(n-2)},K))$

$\subseteq g(h(1,-p^{\prime\prime},0^{(m-2)}),1^{(n-2)},g(R,1^{(n-2)},b))$

$=g(R,1^{(n-2)},g(h(1,-p^{\prime\prime},0^{(m-2)}),1^{(n-2)},b))$.
Then there exists $m\in
g(h(1,-p^{\prime\prime},0^{(m-2)}),1^{(n-2)},b)\subseteq H\cap K$ such that

$g(h(1,-s,0^{(m-2)}),1^{(n-2)},H\cap K)\subseteq g(R,1^{(n-2)},m)$
which means $H\cap K$ is n-ary $P$-cyclic. Similarly, by $c\in K$ it follows
that $H\cap K$ is n-ary $P$-cyclic. Thus $H\cap K$ is a multiplication
hypermodule, by Theorem 3.8 in [2]. ∎

######  Lemma 5.5.

Let $(M,f,g)$ be an $(m,n)$-hypermodule over $(R,h,k)$. Let subhypermodules
$N_{1}$ and $N_{2}$ of $M$ be multiplication. If $R=h(A,B,0^{(m-2)})$ such
that $A=\\{r\in R\ |\ g(r,1^{(n-2)},N_{2})\subseteq N_{1}\\}$ and $B=\\{r\in
R\ |\ g(r,1^{(n-2)},N_{1})\subseteq N_{2}\\}$, then $f(N_{1},N_{2},0^{(m-2)})$
is a multiplication $(m,n)$-hypermodules.

###### Proof.

This follows from Corollary 5.2. ∎

######  Theorem 5.6.

Let $(M,f,g)$ be an $(m,n)$-hypermodule over $(R,h,k)$. Let
$L,N_{1},...,N_{t}$ be subhypermodules of $M$ such that $L$, $N_{i}$,
$f(L,N_{i},0^{(m-2)})$ $(1\leq i\leq t)$ and $N=\bigcap_{i=1}^{t}N_{i}$ are
multiplication $(m,n)$-hypermodules. Then $f(L,N,0^{(m-2)})$ is a
multiplication $(m,n)$-hypermodule.

###### Proof.

Let $\mathcal{X}_{P}(f(L,N,0^{(m-2)}))\neq f(L,N,0^{(m-2)})$ for some maximal
hyperideal $P$ of $R$. Then $\mathcal{X}_{P}(f(L,N_{i},0^{(m-2)})\neq
f(L,N_{i},0^{(m-2)})$ for each $1\leq i\leq t$, since
$f(L,N,0^{(m-2)})\subseteq f(L,N_{i},0^{(m-2)})$. By Theorem 5.1 we conclude
that $h(A_{i},B_{i},0^{(m-2)})\nsubseteq P$ with $A_{i}=\\{r\in R\ |\
g(r,1^{(n-2)},N_{i})\subseteq L\\}$ and $B_{i}=\\{r\in R\ |\
g(r,1^{(n-2)},L)\subseteq N_{i}\\}$. Let $C=\\{r\in R\ |\
g(r,1^{(n-2)},N)\subseteq L\\}$. Then $A_{i}\subseteq C$. Therefore we have
$h(B_{i},C,0^{(m-2)})\nsubseteq P$ for each $1\leq i\leq t$. Let $B=\\{r\in R\
|\ g(r,1^{(n-2)},L)\subseteq N\\}$. Then we get
$h(B,C,0^{(m-2)})=h(\bigcap_{i=1}^{t}B_{i},C,0^{(m-2)})\nsubseteq P$ which
means $B\nsubseteq P$ or $C\nsubseteq P$. Hence there exists $p\in P$ such
that $h(1,-p,0^{(m-2)})\subseteq B\cup C=B^{\prime}\cup C^{\prime}$ with
$B^{\prime}=\\{r\in R\ |\ g(r,1^{(n-2)},f(L,N,0^{(m-2)}))\subseteq N\\}$ and
$C^{\prime}=\\{r\in R\ |\ g(r,1^{(n-2)},f(L,N,0^{(m-2)}))\subseteq L\\}$. Then
we have

$\hskip 14.22636ptg(h(1,-p,0^{(m-2)}),1^{(n-2)},f(L,N,0^{(m-2)}))\subseteq L$
or

$\hskip 14.22636ptg(h(1,-p,0^{(m-2)}),1^{(n-2)},f(L,N,0^{(m-2)}))\subseteq N$.
Since $L$ and $N$ are multiplication $(m,n)$-hypermodules, then by Lemma 5.5,
we conclude that $f(L,\bigcap_{i=1}^{t}N_{i},0^{(m-2)})=f(L,N,0^{(m-2)})$ is a
multiplication $(m,n)$-hypermodule. ∎

######  Theorem 5.7.

Let $(M,f,g)$ be an $(m,n)$-hypermodule over $(R,h,k)$. Let $N_{1},...,N_{n}$
be subhypermodules of $M$ such that for all $1\leq i<j\leq n$,
$f(N_{i},N_{j},0^{(m-2)})$ is a multiplication $(m,n)$-hypermodule. Then
$N_{1},...,N_{n}$ are multiplication $(m,n)$-hypermodules if and only if
$N=\bigcap_{i=1}^{n}N_{i}$ is a multiplication $(m,n)$-hypermodule.

###### Proof.

$\Longrightarrow$ Let $N_{1},...,N_{n}$ are multiplication
$(m,n)$-hypermodules. By induction on $n$, $N_{2}\cap...\cap N_{n}$ is a
multiplication $(m,n)$-hypermodule. Now, by using Theorem 5.6 and Theorem 5.4,
the proof is completed.

$\Longleftarrow$ Let $N=\bigcap_{i=1}^{n}N_{i}$ be a multiplication
$(m,n)$-hypermodule. Assume that for some maximal hyperideal $P$ of $R$,
$\mathcal{X}_{P}(N_{1})\neq N_{1}$. This implies that for all $2\leq i\leq n$,
$\mathcal{X}_{P}(f(N_{1},N_{i},0^{(m-2)}))\neq f(N_{1},N_{i},0^{(m-2)})$. Then
there exist $x_{i}\in N_{1}\cup N_{i}$ and $p_{i}\in P$ such that
$g(h(1,-p_{i},1^{(m-2)}),1^{(n-2)},f(N_{1},N_{i},0^{(m-2)}))\subseteq
g(R,1^{(n-2)},x_{i})$, by Theorem 5.1. If $x_{i}\in N_{1}$ then $N_{1}$ is an
n-ary $P$-cyclic. If $x_{i}\in N_{i}$ $(2\leq i\leq n)$, then

$g(h(1,-{p_{2}},0^{(m-2)}),...,h(1,-p_{n},0^{(m-2)}),N_{1})\subseteq\bigcap_{i=1}^{n}N_{i}$.
By Theorem 3.8 in [2] and using an argument similar to that in the proof of
Theorem 5.1 $((3)\Longrightarrow(4))$, we conclude that $N_{1}$ is
multiplication. Similarly, we can show that for all $2\leq i\leq n$, $N_{i}$
is multiplication. ∎

######  Corollary 5.8.

Let $(M,f,g)$ be an $(m,n)$-hypermodule over $(R,h,k)$ such that
subhypermodules $N_{1},...,N_{n},N_{n+1},...,N_{k}$ of $M$ are multiplication.
If for all $1\leq i<j\leq k$, $f(N_{i},N_{j},0^{(m-2)})$ is a multiplication
$(m,n)$-hypermodule. Then $f(N,L,0^{(m-2)})$ is a multiplication
$(m,n)$-hypermodule with $N=N_{1}\cap...\cap N_{n}$ and $L=N_{n+1}\cap...\cap
N_{k}$.

###### Proof.

By Theorem 5.7, $N$ is a multiplication $(m,n)$-hypermodule. Theorem 5.6 shows
that for each $n+1\leq i\leq k$, $f(N,N_{i},0^{(m-2)})$ is a multiplication
$(m,n)$-hypermodule. Thus $f(N,N_{n+1}\cap...\cap N_{k},0^{(m-2)})$ is a
multiplication $(m,n)$-hypermodule, by Theorem 5.6. ∎

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