On $\alpha$-Tall Modules

Abstract

In this article, we introduce and study $\alpha$-tall modules. We show that an $\alpha$-tall module, where $\alpha \ge 0$, is a tall module, i.e. M contains a submodule N such that N and $\frac{M}{N}$ are both non-Noetherian. We observe that every submodule of $\alpha$-tall modules is countably generated, where $\alpha$ is countable. It is shown that if M is a $\beta$-atomic module, where $\beta =\alpha +2$, for some ordinal $\alpha$, then M is $\alpha$-tall. It is also proved that if M is an $\alpha$-atomic module, where $\alpha$ is a limit ordinal, then M is both an $\alpha$-tall and $\alpha$-short module.

1 Introduction

Lemonnier [20] introduced the concept of deviation and codeviation of an arbitrary poset, which in particular, when applied to the lattice of all submodules of a module $M_{R}$ give the concepts of Krull dimension (in the sense of Rentschler and Gabriel, see [6, 10, 19]) and dual Krull dimension of M, respectively. Later, Chambless in [8] studied dual Krull dimension and called it N-dimension. Karamzadeh also extensively studied the latter dimension in his Ph.D. thesis [12] and called it Noetherian dimension. Roberts in [23] calls this dual dimension again Krull dimension. The latter dimension is also called dual Krull dimension in some other articles, see for example [1–5]. The Noetherian dimension of an R-module M is denoted by ${{ n}\hbox {-}\mathrm{dim}}\, M$ and by ${{ k}\hbox {-}\mathrm{dim}}\, M$; we denote the Krull dimension of M, see [10–16, 18, 19, 21, 22] for more details. It is convenient, when we are dealing with the latter dimensions, to begin our list of ordinals with $-1$. If an R-module M has Noetherian dimension and $\alpha$ is an ordinal number, then M is called $\alpha$-atomic if ${{ n}\hbox {-}\mathrm{dim}}\, M=\alpha$ and ${{ n}\hbox {-}\mathrm{dim}}\, N<\alpha$, for all proper submodule N of M. An R-module M is called atomic if it is $\alpha$-atomic for some ordinal $\alpha$ (note, atomic modules are also called conotable, critical and N-critical in some other articles for example [3, 8, 20]). In [24], Sarath defines an R-module M to be tall if M contains a submodule N such that N and $\frac{M}{N}$ are both non-Noetherian. Bilhan and Smith [7] introduced short modules. In [9], Davoudian, Karamzadeh and Shirali introduced $\alpha$-short modules, (a 0-short module is just a short module, i.e. for each submodule N of M either N or $\frac{M}{N}$ is Noetherian). They show that if M is an $\alpha$-short module, then the Noetherian dimension of M is either $\alpha$ or $\alpha +1$. In this article, we introduce and study $\alpha$-tall modules. We show that an $\alpha$-tall module where $\alpha \ge 0$ is tall. Our aim of studying the concept of $\alpha$-tall modules, in this article, is twofold. We aim to extend the concept of tall modules and at the same time provide a dual to the concept of $\alpha$-short modules. Tall modules are not necessarily Noetherian and one dose not know how far they are from being Noetherian. Although $\alpha$-tall modules for $\alpha >-1$ are similarly non-Noetherian, but the ordinal $\alpha$ measures how these modules deviate from being Noetherian (note, it is observed that if M is $\alpha$-tall, then $\alpha <{{ n}\hbox {-}\mathrm{dim}}\, M\le \alpha +2$). This is an advantage for $\alpha$-tall modules over tall modules. We observe that if M is an $\alpha$-tall module, then each non-zero submodule (each non-zero factor module) of M which is not simple is $\beta$-tall for some $\beta \le \alpha$. All modules in this paper are assumed to be unital modules over an associative ring with unit.

2 Preliminaries

In this section, we recall some useful facts on modules with Noetherian dimension [8, 10, 12, 20].

We recall that the Noetherian dimension of an R-module M is defined by transfinite recursion as follows: If $M=0$, then ${{ n}\hbox {-}\mathrm{dim}}\, M=-1$. If $\alpha$ is an ordinal and ${{ n}\hbox {-}\mathrm{dim}}\, \not < \alpha$, then ${{ n}\hbox {-}\mathrm{dim}}\, M=\alpha$, provided there is no infinite ascending chain $M_{1}\subseteq M_{2}\subseteq \ldots$ of submodules of M such that ${{ n}\hbox {-}\mathrm{dim}}\, \frac{M_{i+1}}{M_{i}}\not <\alpha$ for each $i\ge 1$. It is possible that there is no ordinal $\alpha$ such that ${{ n}\hbox {-}\mathrm{dim}}\, M=\alpha$. In that case, we say M dose not have Noetherian dimension

Lemma 2.1

If M is an R-module and for each submodule N of M, either N or $\frac{M}{N}$ has Noetherian dimension, then so does M.

Proposition 2.2

If M is an R-module, then we have ${{ n}\hbox {-}\mathrm{dim}}\, M=\sup \{{{ n}\hbox {-}\mathrm{dim}}\, \frac{M}{N}:N\ is\ a\ non$-$zero\ proper \ submodule\ of\ M\}$, if either side exists.

Proposition 2.3

If M is an R-module, then we have ${{ n}\hbox {-}\mathrm{dim}}\, M\le \sup \{{{ n}\hbox {-}\mathrm{dim}}\, N:N\ is\ a\ proper\ submodule\ of\ M\}$+1, if either side exists.

Lemma 2.4

If N is a submodule of an R-module M, then ${{ n}\hbox {-}\mathrm{dim}}\, M=\sup \{{{ n}\hbox {-}\mathrm{dim}}\, \frac{M}{N}, {{ n}\hbox {-}\mathrm{dim}}\, N\}$, if either side exists.

Remark 2.5

An R-module M is called $\alpha$-short, if for each submodule N of M, either ${{ n}\hbox {-}\mathrm{dim}}\, N\le \alpha$ or ${{ n}\hbox {-}\mathrm{dim}}\, \frac{M}{N}\le \alpha$ and $\alpha$ is the least ordinal number with this property, see [9]. Clearly each 0-short module is just a short module, we recall that an R-module M is called a short module if for each submodule N of M, either N or $\frac{M}{N}$ is Noetherian, see [7].

We cite the following facts from [9].

Remark 2.6

If M is an R-module with ${{ n}\hbox {-}\mathrm{dim}}\, M=\alpha$, then M is $\beta$-short for some $\beta \le \alpha$.

In view of Remark 2.5 and Lemma 2.1, the following corollary is now evident.

Corollary 2.7

Let M be an $\alpha$-short module. Then M has Noetherian dimension and ${{ n}\hbox {-}\mathrm{dim}}\, M\ge \alpha$.

The following is also in [9, Proposition 1.12]. We give the proof for completeness.

Proposition 2.8

If M is an $\alpha$-short R-module, then either ${{ n}\hbox {-}\mathrm{dim}}\, M=\alpha$ or ${{ n}\hbox {-}\mathrm{dim}}\, M=\alpha +1$.

Proof

Clearly in view of Remark 2.6 and Corollary 2.7, we have ${{ n}\hbox {-}\mathrm{dim}}\, M\ge \alpha$. If ${{ n}\hbox {-}\mathrm{dim}}\, M\ne \alpha$, then ${{ n}\hbox {-}\mathrm{dim}}\, M\ge \alpha +1$. Now let $M_{1}\subseteq M_{2}\subseteq M_{3}\subseteq \ldots$ be any ascending chain of submodules of M. If there exists some k such that ${{ n}\hbox {-}\mathrm{dim}}\, \frac{M}{M_{k}}\le \alpha$, then ${{ n}\hbox {-}\mathrm{dim}}\, \frac{M_{i+1}}{M_{i}}\le {{ n}\hbox {-}\mathrm{dim}}\, \frac{M}{M_{i}}={{ n}\hbox {-}\mathrm{dim}}\, \frac{M/M_{k}}{M_{i}/M_{k}}\le {{ n}\hbox {-}\mathrm{dim}}\, \frac{M}{M_{k}}\le \alpha$ for each $i\ge k$. Otherwise ${{ n}\hbox {-}\mathrm{dim}}\, M_{i}\le \alpha$ (note, M is $\alpha$-short) for each i, hence ${{ n}\hbox {-}\mathrm{dim}}\, \frac{M_{i}}{M_{i+1}}\le \alpha$ for each i. Thus in any case, there exists an integer k such that for each $i\ge k$, ${{ n}\hbox {-}\mathrm{dim}}\, \frac{M_{i+1}}{M_{i}}\le \alpha$. This shows that ${{ n}\hbox {-}\mathrm{dim}}\, M\le \alpha +1$, i.e. ${{ n}\hbox {-}\mathrm{dim}}\, M=\alpha +1$. $\square$

3 $\alpha$-Tall Modules

In [24], Sarath defines an R-module M to be tall if M contains a submodule N such that N and $\frac{M}{N}$ are both non-Noetherian. In this section, we introduce and study $\alpha$-tall modules. We observe that an $\alpha$-tall module, where $\alpha \ge 0$, is tall. We show that if M is an $\alpha$-tall module, then each non-zero submodule (non-zero factor module) of M which is not simple is $\beta$-tall for some $\beta \le \alpha$. In particular, we show that each $\alpha$-tall module has Noetherian dimension and its Noetherian dimension is either $\alpha +1$ or $\alpha +2$. We observe that if M is an $\alpha$-tall module, where $\alpha$ is a countable ordinal, then every submodule of M is countably generated. If an R-module M is not tall, then it is a short module and by Davoudian et al. [9, Proposition 1.12], and M has Noetherian dimension and ${{ n}\hbox {-}\mathrm{dim}}\, M\le 1$.

Next, we give our definition of $\alpha$-tall modules.

Definition 3.1

Let M be an R-module and $\beta$ be an ordinal number. Put $\psi _{M}=\{\beta : \textit{there} \ \textit{exists} \ a \ \textit{submodule} \ N \ \textit{of} \ M \ \textit{such} \ \textit{that} \ {{ n}\hbox {-}\mathrm{dim}}\, N>\beta \ \textit{and}\ {{ n}\hbox {-}\mathrm{dim}}\, \frac{M}{N}>\beta \}$. If $\psi _{M}\ne \emptyset$ and $\alpha =\sup \psi _{M}$, then M is called $\alpha$-tall.

Remark 3.2

A simple module M is not $\alpha$-tall for any ordinal number $\alpha$, and every non-zero Noetherian module M which is not simple, is $-1$-tall.

The following results provide a criterion for non-simple $\alpha$-tall modules.

Lemma 3.3

Let M be a non-zero R-module which is not simple. Then $\psi _{M}\ne \emptyset$ if and only if M has Noetherian dimension.

Proof

If $\psi _{M}\ne \emptyset$, then there exists a proper non-zero submodule N of M such that N and $\frac{M}{N}$ both have Noetherian dimension. Thus M has Noetherian dimension, see Lemma 2.4. Conversely, since M is not simple, it has a non-zero proper submodule, N say. By our assumption M has Noetherian dimension, thus both N and $\frac{M}{N}$ have Noetherian dimension, see Lemma 2.4. This implies that $\psi _{M}\ne \emptyset$. $\square$

In view of Lemma 3.3, we have the following results.

Corollary 3.4

Let M be a non-zero R-module which is not simple. Then $\psi _{M}=\emptyset$ if and only if M does not have Noetherian dimension.

Corollary 3.5

Let M be a non-zero R-module which is not simple. Then M has Noetherian dimension if and only if it is $\alpha$-tall for some ordinal number $\alpha$.

Lemma 3.6

Let M be a non-zero R-module which is not simple. Then M is tall if and only if either $\psi _{M}=\emptyset$ or M is $\alpha$-tall for some ordinal number $\alpha \ge 0$.

Proof

If M is $\alpha$-tall for some ordinal number $\alpha \ge 0$, then there exists a submodule N of M such that ${{ n}\hbox {-}\mathrm{dim}}\, N>0$ and ${{ n}\hbox {-}\mathrm{dim}}\, \frac{M}{N}>0$, therefore N and $\frac{M}{N}$ both are non-Noetherian and M is a tall module. Now let $\psi _{M}=\emptyset$, then by Corollary 3.4, M dose not have Noetherian dimension, hence M is a tall module (note, if M is not tall, then M is a short module and by Corollary 2.7, M has Noetherian dimension, which is a contradiction). Conversely, let M be a tall module, it is clear that M is not simple. We consider two cases. If M does not have Noetherian dimension, then by Corollary 3.4, $\psi _{M}=\emptyset$. Otherwise, M has Noetherian dimension. Since M is a tall module, there exists a submodule N of M, such that N and $\frac{M}{N}$ are both non-Noetherian. Thus ${{ n}\hbox {-}\mathrm{dim}}\, N>0$ and ${{ n}\hbox {-}\mathrm{dim}}\, \frac{M}{N}>0$, see Lemma 2.4. Therefore $0\in \psi _{M}$ and hence $\psi _{M}\ne \emptyset$ and M is $\alpha$-tall for some ordinal number $\alpha \ge 0$. $\square$

Proposition 3.7

If M is an $\alpha$-tall R-module, then either ${{ n}\hbox {-}\mathrm{dim}}\, M=\alpha +1$ or ${{ n}\hbox {-}\mathrm{dim}}\, M=\alpha +2$.

Proof

For each submodule N of M, we get either ${{ n}\hbox {-}\mathrm{dim}}\, N\le \alpha +1$ or ${{ n}\hbox {-}\mathrm{dim}}\, \frac{M}{N}\le \alpha +1$ (note, if there exists a submodule N of M such that ${{ n}\hbox {-}\mathrm{dim}}\, N>\alpha +1$ and ${{ n}\hbox {-}\mathrm{dim}}\, \frac{M}{N}>\alpha +1$, then $\alpha +1\in \psi _{M}$ and M is $\gamma$-tall for some $\gamma \ge \alpha +1$, which is a contradiction). Thus M is $\alpha +1$-short and by Proposition 2.8, M has Noetherian dimension and ${{ n}\hbox {-}\mathrm{dim}}\, M=\alpha +1$ or ${{ n}\hbox {-}\mathrm{dim}}\, M=\alpha +2$. $\square$

Here are some elementary properties of $\alpha$-tall modules.

Corollary 3.8

If M is a $-1$-tall module, then either ${{ n}\hbox {-}\mathrm{dim}}\, M=1$ or M is a non-zero Noetherian module which is not simple.

Corollary 3.9

Every $-1$-tall module is a short module and if M is a non-zero short module which is not simple, then it is $-1$-tall.

It is well known that any module with Noetherian dimension has finite uniform dimension. In view of Proposition 3.7, the following corollary is now evident.

Corollary 3.10

Every $\alpha$-tall module has finite uniform dimension.

The following corollary shows that there exists a tall module which is not $\alpha$-tall.

Corollary 3.11

Let M be an R-module. If M has infinite uniform dimension, then it is tall.

Proof

Since M has infinite uniform dimension, we infer that M contains an infinite direct sum such as $X=\sum _{i\in \mathbb N}\oplus M_{i}$. Now we put $N=\sum _{i=2k}\oplus M_{i}$. It is clear that N and $\frac{M}{N}$ both are non-Noetherian. Hence M is tall. $\square$

In view of Corollaries 3.10, 3.11 and Lemma 3.6, we have the following results.

Corollary 3.12

Let M be an R-module. If M has infinite uniform dimension, then it is a tall module which is not $\alpha$-tall, i.e. $\psi _{M}=\emptyset$.

Corollary 3.13

If M is an $\alpha$-tall module, where $\alpha$ is a countable ordinal number, then every submodule of M is countably generated.

Proof

By Proposition 3.7, ${{ n}\hbox {-}\mathrm{dim}}\, M=\alpha +1$ or ${{ n}\hbox {-}\mathrm{dim}}\, M=\alpha +2$. We know that every submodule of a module with countable Noetherian dimension is countably generated, see [17, Corollary 1.8]. Hence we are through. $\square$

For the atomic modules, the following facts are important.

Lemma 3.14

Let M be an $\alpha +2$-atomic module. Then M is an $\alpha$-tall module.

Proof

Since M is $\alpha +2$-atomic, for each proper submodule N of M, we get ${{ n}\hbox {-}\mathrm{dim}}\, N\le \alpha +1$ and ${{ n}\hbox {-}\mathrm{dim}}\, \frac{M}{N}=\alpha +2$. If for each proper submodule N of M we have ${{ n}\hbox {-}\mathrm{dim}}\, N\le \alpha$, then ${{ n}\hbox {-}\mathrm{dim}}\, M\le \sup \{{{ n}\hbox {-}\mathrm{dim}}\, N:N\subsetneq M\}+1\le \alpha +1$ and this is a contradiction. This shows that there exists a proper submodule N of M such that ${{ n}\hbox {-}\mathrm{dim}}\, N>\alpha$. But ${{ n}\hbox {-}\mathrm{dim}}\, \frac{M}{N}=\alpha +2>\alpha$, therefore $\alpha \in \psi _{M}$. For each proper submodule N of M, ${{ n}\hbox {-}\mathrm{dim}}\, N\le \alpha +1$, therefore $\alpha +1\not \in \psi _{M}$. Hence M is an $\alpha$-tall module. $\square$

Lemma 3.15

Let M be an $\alpha$-atomic R-module, where $\alpha$ is a limit ordinal number. Then M is an $\alpha$-tall module.

Proof

Let $\beta <\alpha$ be an ordinal number. If for each proper submodule N of M we have ${{ n}\hbox {-}\mathrm{dim}}\, N\le \beta$, then ${{ n}\hbox {-}\mathrm{dim}}\, M\le \sup \{{{ n}\hbox {-}\mathrm{dim}}\, N:N\subseteq M\}+1\le \beta +1<\alpha$ and it is a contradiction. Thus there exists a proper submodule N of M such that ${{ n}\hbox {-}\mathrm{dim}}\, N>\beta$. Since M is $\alpha$-atomic, we must have ${{ n}\hbox {-}\mathrm{dim}}\, \frac{M}{N}=\alpha >\beta$. This shows that $\beta \in \psi _{M}$. If $\gamma \ge \alpha$ is an ordinal number, then for each submodule N of M we have ${{ n}\hbox {-}\mathrm{dim}}\, N<\gamma$, thus $\gamma \notin \psi _{M}$. Therefore $\sup \psi _{M}=\alpha$, i.e. M is $\alpha$-tall. $\square$

In view of previous Lemma and [9, Proposition 1.16], we have the following corollary.

Corollary 3.16

Let M be an $\alpha$-atomic R-module, where $\alpha$ is a limit ordinal number. Then M is both $\alpha$-tall and $\alpha$-short.

The previous corollary will rise the natural question, namely, what are $\alpha$-tall modules which are also $\alpha$-short, in general?

The following result is evident.

Corollary 3.17

Let M be a tall module. If M has Noetherian dimension, then ${{ n}\hbox {-}\mathrm{dim}}\, M\ge 1$.

Proof

There exists a submodule N of M such that N and $\frac{M}{N}$ both are non-Noetherian. Therefore ${{ n}\hbox {-}\mathrm{dim}}\, N>0$ and ${{ n}\hbox {-}\mathrm{dim}}\, \frac{M}{N}>0$, see Lemma 2.4. Now by Lemma 2.4, we have ${{ n}\hbox {-}\mathrm{dim}}\, M >0$, i.e. ${{ n}\hbox {-}\mathrm{dim}}\, M\ge 1$ and we are done. $\square$

Lemma 3.18

If M is an $\alpha$-tall module, then each non-zero submodule (non-zero factor module) of M which is not simple is $\beta$-tall for some $\beta \le \alpha$.

Proof

By Proposition 3.7, ${{ n}\hbox {-}\mathrm{dim}}\, M=\alpha +1$ or ${{ n}\hbox {-}\mathrm{dim}}\, M=\alpha +2$. Thus for each submodule N of M, N and $\frac{M}{N}$ both have Noetherian dimension, see Lemma 2.4. Now let N and $\frac{M}{N}$ are non-zero R-modules which are not simple, therefore N and $\frac{M}{N}$ both are $\beta$-tall for some ordinal number $\beta$, see Corollary 3.5. Hence in view of Lemma 2.4 and Proposition 3.7, we infer that $\beta \le \alpha +1$. We claim that $\beta \ne \alpha +1$. Since if $\beta =\alpha +1$, then there exists a non-zero proper submodule $N_{1}$ of N $\big ($similarly there exists a non-zero proper submodule $\frac{N_{1}}{N}$ of $\frac{M}{N}\big )$ such that ${{ n}\hbox {-}\mathrm{dim}}\, N_{1}>\alpha +1$ and ${{ n}\hbox {-}\mathrm{dim}}\, \frac{N}{N_{1}}>\alpha +1$ $\big ($similarly ${{ n}\hbox {-}\mathrm{dim}}\, \frac{N_{1}}{N}>\alpha +1$ and ${{ n}\hbox {-}\mathrm{dim}}\, \frac{M}{N_{1}}>\alpha +1\big )$. Therefore ${{ n}\hbox {-}\mathrm{dim}}\, N_{1}>\alpha +1$ and ${{ n}\hbox {-}\mathrm{dim}}\, \frac{M}{N_{1}}>\alpha +1$ $\big ($similarly ${{ n}\hbox {-}\mathrm{dim}}\, \frac{M}{N_{1}}>\alpha +1$ and ${{ n}\hbox {-}\mathrm{dim}}\, N_{1}>\alpha +1\big )$. Thus M is $\gamma$-tall for some $\gamma \ge \alpha +1$, which is a contradiction. $\square$

Proposition 3.19

Let N be a submodule of an R-module M such that N is $\alpha$-tall and $\frac{M}{N}$ is $\beta$-tall. Let $\mu =\sup \{\alpha , \beta \}$, then M is $\gamma$-tall such that $\mu \le \gamma \le \mu +1$.

Proof

Since N is $\alpha$-tall, by Proposition 3.7, ${{ n}\hbox {-}\mathrm{dim}}\, N=\alpha +1$ or ${{ n}\hbox {-}\mathrm{dim}}\, N=\alpha +2$. Similarly since $\frac{M}{N}$ is $\beta$-tall, either ${{ n}\hbox {-}\mathrm{dim}}\, \frac{M}{N}=\beta +1$ or ${{ n}\hbox {-}\mathrm{dim}}\, \frac{M}{N}=\beta +2$. We infer that M has Noetherian dimension and ${{ n}\hbox {-}\mathrm{dim}}\, M=\sup \left\{ {{ n}\hbox {-}\mathrm{dim}}\, N, {{ n}\hbox {-}\mathrm{dim}}\, \frac{M}{N} \right\}$, see Lemma 2.4. Therefore $\mu +1\le {{ n}\hbox {-}\mathrm{dim}}\, M\le \mu +2$. By Corollary 3.5, M is $\gamma$-tall for some ordinal number $\gamma$ and by Proposition 3.7, $\gamma +1\le {{ n}\hbox {-}\mathrm{dim}}\, M\le \gamma +2$. This shows that $\gamma =\mu$ or $\gamma =\mu +1$ (note, by Lemma 3.18, we always have $\mu \le \gamma$) and we are done. $\square$

We also have the following two facts.

Lemma 3.20

Let N be a simple submodule of an R-module M. If $\frac{M}{N}$ is $\beta$-tall, then M is a $\gamma$-tall module for some $\beta \le \gamma \le \beta +1$.

Lemma 3.21

Let N be a maximal submodule of an R-module M. If N is an $\alpha$-tall module, then M is $\gamma$-tall for some $\alpha \le \gamma \le \alpha +1$.

Proposition 3.22

Let M be a non-zero R-module which is not simple and $\alpha$ be an ordinal number. If every non-zero proper factor module of M which is not simple, is $\gamma$-tall for some ordinal number $\gamma \le \alpha$, then M is $\mu$-tall where $\mu \le \alpha$.

Proof

It is clear that for each simple factor module $\frac{M}{N}$ of M, ${{ n}\hbox {-}\mathrm{dim}}\, \frac{M}{N}=0\le \alpha +2$. Now let $0\ne N\subsetneq M$ be any submodule such that $\frac{M}{N}$ is $\gamma$-tall for some ordinal number $\gamma$ with $\gamma \le \alpha$. We infer that ${{ n}\hbox {-}\mathrm{dim}}\, \frac{M}{N}\le \gamma +2\le \alpha +2$, by Proposition 3.7. Thus we have ${{ n}\hbox {-}\mathrm{dim}}\, M=\sup \{{{ n}\hbox {-}\mathrm{dim}}\, \frac{M}{N}:N\ne 0\}$, by Proposition 2.2. This shows that ${{ n}\hbox {-}\mathrm{dim}}\, M\le \alpha +2$. If ${{ n}\hbox {-}\mathrm{dim}}\, M\le \alpha +1$, then it is clear that M is $\mu$-tall for some $\mu \le \alpha$. Hence we may suppose that ${{ n}\hbox {-}\mathrm{dim}}\, M=\alpha +2$. If $0\ne N\subsetneq M$ is a submodule of M, then we are to show that either ${{ n}\hbox {-}\mathrm{dim}}\, \frac{M}{N}\le \alpha +1$ or ${{ n}\hbox {-}\mathrm{dim}}\, N\le \alpha +1$ (note, this implies that $\alpha +1\notin \psi _{M}$ and hence $\sup \psi _{M}\le \alpha$). To this end, let us suppose that ${{ n}\hbox {-}\mathrm{dim}}\, \frac{M}{N}=\alpha +2$ and show that ${{ n}\hbox {-}\mathrm{dim}}\, N\le \alpha +1$. Now let $0\ne N'\subsetneq N\subsetneq M$. Since ${{ n}\hbox {-}\mathrm{dim}}\, \frac{M/N'}{N/N'}={{ n}\hbox {-}\mathrm{dim}}\, \frac{M}{N}=\alpha +2$, we must have ${{ n}\hbox {-}\mathrm{dim}}\, \frac{N}{N'}\le \alpha +1$ (note, $\frac{M}{N'}$ is $\gamma$-tall for some $\gamma \le \alpha$). But ${{ n}\hbox {-}\mathrm{dim}}\, N=\sup \{{{ n}\hbox {-}\mathrm{dim}}\, \frac{N}{N'}:0\ne N'\subseteq N\}\le \alpha +1$ and we are through. $\square$

Proposition 3.23

Let $\alpha$ be an ordinal number and M be a non-zero R-module which is not simple . If every non-zero proper submodule of M which is not simple is $\gamma$-tall for some ordinal number $\gamma \le \alpha$. Then either ${{ n}\hbox {-}\mathrm{dim}}\, M=\alpha +2$ or M is $\mu$-tall for some ordinal number $\mu \le \alpha$. In particular, M is $\mu$-tall for some ordinal number $\mu \le \alpha +1$.

Proof

If M has no non-simple proper submodule, than M is $-1$-tall. Let N be a simple proper submodule of M, then ${{ n}\hbox {-}\mathrm{dim}}\, N=0\le \alpha +2$. Now let $0\ne N\subsetneq M$ be any non-simple submodule of M. Since N is $\gamma$-tall for some ordinal number $\gamma \le \alpha$, we infer that ${{ n}\hbox {-}\mathrm{dim}}\, N\le \gamma +2\le \alpha +2$, by Proposition 3.7. In view of Proposition 2.3, we infer that ${{ n}\hbox {-}\mathrm{dim}}\, M\le \alpha +3$. If ${{ n}\hbox {-}\mathrm{dim}}\, M\le \alpha +2$, then we are through. Hence we may suppose that ${{ n}\hbox {-}\mathrm{dim}}\, M=\alpha +3$ and M is not $\mu$-tall for any $\mu \le \alpha$ and seek a contradiction. Since M is not $\mu$-tall for any $\mu \le \alpha$, we infer that there must exists a proper submodule K of M such that ${{ n}\hbox {-}\mathrm{dim}}\, K\ge \alpha +2$. But we have already observed that ${{ n}\hbox {-}\mathrm{dim}}\, K\le \alpha +2$, therefore ${{ n}\hbox {-}\mathrm{dim}}\, K=\alpha +2$. We now claim that ${{ n}\hbox {-}\mathrm{dim}}\, \frac{M}{K}\le \alpha +2$ which trivially implies that ${{ n}\hbox {-}\mathrm{dim}}\, M=\alpha +2$ and this is a contradiction (note, ${{ n}\hbox {-}\mathrm{dim}}\, M=\alpha +3$). Let $K\subset N'\subset M$. Since ${{ n}\hbox {-}\mathrm{dim}}\, K=\alpha +2$ and $N'$ is $\gamma$-tall for some $\gamma \le \alpha$, we get ${{ n}\hbox {-}\mathrm{dim}}\, \frac{N'}{K}\le \alpha +1$. But ${{ n}\hbox {-}\mathrm{dim}}\, \frac{M}{K}\le \sup \left\{ {{ n}\hbox {-}\mathrm{dim}}\, \frac{N'}{K}:\frac{N'}{K}\subset \frac{M}{K}\right\} +1\le \alpha +2$ and we are done. The final part is now evident. $\square$

Now we have the following definition.

Definition 3.24

Let M be an R-module with Noetherian dimension. For each ordinal $\alpha$, we put:

$$\begin{aligned} G_{\alpha }(M)= & {} \cap {\left\{ N:{{ n}\hbox {-}\mathrm{dim}}\, \frac{M}{N}\le \alpha , N\subseteq M\right\} }\\ H_{\alpha }(M)= & {} \cap \{N:{{ n}\hbox {-}\mathrm{dim}}\, N>\alpha , N\subseteq M\}. \end{aligned}$$

We recall that a submodule N of M is called $\alpha$-coatomic, where $\alpha$ is an ordinal number, if $\frac{M}{N}$ is $\alpha$ atomic. An R-module N is called coatomic if it is $\alpha$-coatomic for some ordinal $\alpha$.

Remark 3.25

If ${{ n}\hbox {-}\mathrm{dim}}\, M>\alpha$ and N be a submodule of M such that ${{ n}\hbox {-}\mathrm{dim}}\, \frac{M}{N}\le \alpha$, then clearly ${{ n}\hbox {-}\mathrm{dim}}\, N>\alpha$. This shows that $H_{\alpha }(M)\subseteq G_{\alpha }(M)$, where ${{ n}\hbox {-}\mathrm{dim}}\, M>\alpha$. If N is an $\alpha$-coatomic submodule of M, then ${{ n}\hbox {-}\mathrm{dim}}\, \frac{M}{N}= \alpha$, thus $G_{\alpha }(M)\subseteq N$.

The following lemma is now immediate.

Lemma 3.26

Let M be an R-module with Noetherian dimension and $\alpha$ be an ordinal number. If ${{ n}\hbox {-}\mathrm{dim}}\, \frac{M}{G_{\alpha }(M)}\le \alpha$ and $H_{\alpha }(G_{\alpha }(M))\ne G_{\alpha }(M)$, then M is $\gamma$-tall for some $\gamma \ge \alpha$.

Proof

Since $H_{\alpha }(G_{\alpha } (M))\ne G_{\alpha }(M)$, we infer that $G_{\alpha }(M)\ne 0$ and ${{ n}\hbox {-}\mathrm{dim}}\, M\ne \alpha$. Thus there exists $P\subsetneq G_{\alpha }(M)$ such that ${{ n}\hbox {-}\mathrm{dim}}\, P>\alpha$. Since ${{ n}\hbox {-}\mathrm{dim}}\, \frac{M}{G_{\alpha }(M)}\le \alpha$, we get ${{ n}\hbox {-}\mathrm{dim}}\, \frac{M/P}{G_{\alpha }(M)/P}={{ n}\hbox {-}\mathrm{dim}}\, \frac{M}{G_{\alpha }(M)}\le \alpha$. If ${{ n}\hbox {-}\mathrm{dim}}\, \frac{G_{\alpha }(M)}{P}\le \alpha$, then ${{ n}\hbox {-}\mathrm{dim}}\, \frac{M}{P}\le \alpha$, see Proposition 2.4. This shows that $G_{\alpha }(M)=P$ and this is a contradiction. Thus ${{ n}\hbox {-}\mathrm{dim}}\, \frac{G_{\alpha }(M)}{P}>\alpha$. This shows that ${{ n}\hbox {-}\mathrm{dim}}\, \frac{M}{P}>\alpha$, see Proposition 2.4. Hence M is $\gamma$-tall for some ordinal number $\gamma \ge \alpha$. $\square$

Now in view of [9, Proposition 2.21], we observe the following result.

Proposition 3.27

The following statements are equivalent for a commutative ring R:

• (1) Every Artinian R-module is Noetherian.

• (2) Every m-short module is both Artinian and Noetherian for all integers $m\ge -1$.

• (3) Every $\alpha$-short module is both Artinian and Noetherian for all ordinal $\alpha$.

• (4) Every m-tall module is both Artinian and Noetherian for all integers $m\ge -1$.

• (5) Every $\alpha$-tall module is both Artinian and Noetherian for all ordinal $\alpha$.

• (6) No homomorphic image of R can be isomorphic to a dense subring of a complete local domain of Krull dimension 1.

Proof

By Proposition 3.7 and [9, Proposition 2.21], we are through. $\square$