# Proof of all submodules of $M$ are f.g. then $M$ is Noetherian

I've been thinking about the following proof that if all submodules of $M$ are finitely generated then $M$ is Noetherian:

Assume $M$ is such that all submodules are finitely generated. Then in particular, $M$ is finitely generated. Now let $N_1 \subset N_2 \subset \dots$ be an increasing chain of submodules in $M$. If the chain is non-stabilising, then for $k$ large enough $N_k$ will eventually contain all generators of $M$. Hence $N_k = N_{k+1} = \dots$.

I think the idea is correct but on the other hand, it sort of doesn't feel right. Is this proof correct? Or if not: can it be fixed? Thank you.

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The basic idea is right, but you are confusing yourself by not making the argument very clear; are you trying to prove it by contradiction? By contrapositive? Also, it's not true that the chain will necessarily contain all generators of $M$. What if the chain stabilizes at a proper submodule of $M$? Rather, it seems like you are trying to combine an argument by contradiction and an argument by contrapositive. What you really need to do is write this out carefully and explicitly. It's also a good idea to pick a type of proof (contradiction, contrapositive), and stick to it. No detours. –  Arturo Magidin Jul 21 '12 at 21:04

The standard three properties are:

Theorem. Let $R$ be a ring, and let $M$ be an $R$-module. The following are equivalent:

1. $M$ has ACC on submodules; that is, if $$N_1\subseteq N_2\subseteq\cdots\subseteq N_m\subseteq\cdots$$ is an ascending chain of submodules of $M$, then the chain stabilizes; that is, there exists $m_0$ such that $N_{m_0}=N_{m_0+k}$ for all $k\geq 0$.

2. Every submodule of $M$ is finitely generated.

3. Every nonempty collection of submodules of $M$ has maximal elements (under inclusion).

Some of the implications require the Axiom of Choice. In the absence of AC, the strongest condition is 3, which implies the other 2 without having to invoke AC.

Proof. 1$\implies$2 (This implication requires at least some Choice). We prove it by contrapositive. Assume there is submodule $N$ of $M$ that is not finitely generated. Then $N\neq 0$, so let $n_1\in N$, $n_1\neq 0$. Let $N_1=\langle n_1\rangle$; then $N_1\neq N$, since $N$ is not finitely generated, so there exists $n_2\in N\setminus N_1$. Let $N_2 = \langle n_1,n_2\rangle$. Then $N_1\subsetneq N_2$.

Assume we have chosen $n_1,\ldots,n_k$, with $N_i = \langle n_1,\ldots,n_i\rangle$, and with $N_i\subsetneq N_{i+1}$, $i=1,\ldots,k-1$. Since $N$ is not finitely generated, $N\setminus N_{k}$ is nonempty; let $n_{k+1}\in N\setminus N_{k}$. Then $N_k\subsetneq \langle n_1,\ldots,n_{k},n_{k+1}\rangle = N_{k+1}$.

Inductively (this requires at least dependent countable choice), we obtain an infinite ascending chain that does not stabilizer, $$N_1\subsetneq N_2\subsetneq \cdots\subsetneq N_m\subsetneq N_{m+1}\subsetneq \cdots.$$

2$\implies$1 This implication holds without the Axiom of Choice, and is the one you are working on:

Let $N_1\subseteq N_2\subseteq\cdots\subseteq N_m\subseteq$ be an ascending chain of submodules. It is easy to verify that $$N = \bigcup_{i=1}^{\infty}N_i$$ is a submodule of $M$. By assumption, $N$ is finitely generated; let $n_1,\ldots,n_k$ be a generating set. For each $i$, there exists $m_i$ such that $n_i\in N_{m_i}$. Let $m=\max\{m_1,\ldots,m_k\}$. Then $N_{m_i}\subseteq N_m$, hence $n_i\in N_{m_i}\subseteq N_m$. Thus, $N = \langle n_1,\ldots,n_k\rangle\subseteq N_m\subseteq N_{m+r}\subseteq N$ for all $r\geq 0$, hence $N_m=N_{m+k}=N$ for all $k\geq 0$. Thus, the chain stabilizes.

2$\implies$3. This follows from Zorn's Lemma. Consider a nonempty collection $$\mathcal{S} of submodules of M, ordered by inclusion. We prove that it satisfies the hypothesis of Zorn's Lemma, and hence has maximal elements. Let \mathcal{C} be a chain of elements of \mathcal{S}; then \cup\mathcal{C} is a submodule of M, and hence is finitely generated; therefore, there exist m_1,\ldots,m_n that generate \cup\mathcal{C}; hence there exist M_1,\ldots,M_n\in \mathcal{C} such that m_i\in M_i. Since \mathcal{C} is a chain, one of M_1,\ldots,M_n contains all the others (finite subsets of a chain always have a maximum); say it is M_0. Then \cup \mathcal{C}=M. Hence, if N is an element of \mathcal{C}, then N\subseteq\cup\mathcal{C}=M_0\in\mathcal{S}. Thus, \mathcal{C} has an upper bound in \mathcal{S}. Since every chain in \mathcal{S} is bounded above in \mathcal{S}, by Zorn's Lemma \mathcal{S} has maximal elements, as desired. 3\implies1 This implication holds without AC. Given an ascending chain of submodules of M, let \mathcal{S} be the collection of these modules. By hypothesis, \mathcal{S} has a maximal element, N_m. Since \mathcal{S} is a chain, a maximal element must be a maximum, so N_m contains all other elements of the chain. In particular, if k\geq 0, then N_{m+k}\subseteq N_m\subseteq N_{m+k}, hence N_m=N_{m+k}, as desired. 3\implies2 This implication holds without AC. Let N be a submodule of M. Let \mathcal{S} be the collection of all finitely generated submodules of N, ordered by inclusion. Since 0\in\mathcal{S}, the latter is nonempty. By hypothesis, \mathcal{S} has maximal elements. Let N_0 be a maximal element. Now, N_0\subseteq N by assumption. Let x\in N. Then \langle N_0,x\rangle is a finitely generated submodule of N, and clearly N_0\subseteq \langle N_0,x\rangle. By the maximality of N_0, we conclude that N_0=\langle N_0,x\rangle, hence x\in N_0; thus, N\subseteq N_0, proving equality, and hence that N is finitely generated, as desired. \Box I don't know a direct proof of 1\implies 3 that does not go through 2, but in any case the implication requires the Axiom of Choice, as far as I know. - I think you can formalise this a bit more. For example, you can start of by saying suppose we have an increasing chain of submodules N_1 \subseteq N_2 \subseteq N_3 \subseteq \ldots  Then you can check that$$N = \bigcup_{i \geq 1} N_i

is a submodule of $M$. Now by assumption all submodules of $M$ are finitely generated so that $N$ is finitely generated. Now how does this tell me that the chain is eventually constant?

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Are you saying yes it's correct here is a way to put it more formally? –  Rudy the Reindeer Jul 21 '12 at 10:32
@ClarkKent I don't think your argument is right. Why for large enough $k$ should $N_k$ contain all the generators of $M$? –  user38268 Jul 21 '12 at 10:34
I thought about that, too. Because in each step $k$ we add at least one element to $N_k$ (otherwise we'd already have a stable chain). We start with a finite number of generators. Hence there is a $k$ such that all the generators have been added. –  Rudy the Reindeer Jul 21 '12 at 10:41
@ClarkKent I think that works too, but I like the way it's put out in my answer above. I learned this a while ago though. –  user38268 Jul 21 '12 at 10:42