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I'm learning about Noetherian and Artinian modules. I came across an example that $\mathbb{Z}$ is a Noetherian module. My question is this; The submodules of $\mathbb{Z}$ as a $\mathbb{Z}$-module are the ideals of $\mathbb{Z}$. So stuff of the form $n\mathbb{Z}, n\in\mathbb{N}$. Now an ideal $n\mathbb{Z}$ is contained in $m\mathbb{Z}$ if and only if $m\mid n$. Since 1 divides every natural number every ascending chain of submodules of $\mathbb{Z}$ will eventually become stationary in $\mathbb{Z}$.

But in the definition of the ascending chain condition, we're told that "for every chain of submodules, $M_{0}\subset M_{1}\subset ...$, of an $R$-module, $\exists k\in \mathbb{N}$ such that $\forall n\geq k, M_{n} = M_{k}$". But in the case of $\mathbb{Z}$ above, there are no other submodules after $\mathbb{Z}$ itself. So I want to know if my thinking is correct. If not, I'd be grateful to be pointed in the right direction to think through this

sandoh
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    The chain $M_0 \subset M_1 \subset M_2 \subset \cdots$ where each $M_i = 5\mathbb Z$ is also stationary but it stabilizies in $5\mathbb Z$. Not every chain has to stabilize in $\mathbb Z$. Besides that, I am not entirely sure what your question is. – Arkady Sep 09 '22 at 22:56
  • From my understanding of the question, you seem to either be suggesting that every chain must eventually stabilize to be $\mathbb{Z}$, which as Arkday points out does not have to be true, or that you are suggesting a chain for which one of the $M_i$ is $\mathbb{Z}$ has 'nowhere to go.' In this case, you are exactly right. But then $M_j= \mathbb{Z}$ for $j \geq i$ and the chain has hence stabilized. – mathematics2x2life Sep 10 '22 at 06:11
  • Perhaps you mean that chains must look something like $12\mathbb{Z} \subset 6\mathbb{Z} \subset 2\mathbb{Z} \subset \cdots$, i.e. always growing (untrue). But you are right in that if they 'look like' this, eventually you 'run out' of divisors and the chain could only 'grow' to $\mathbb{Z}$. But then after that, all the other submodules are $\mathbb{Z}$, and the chain stabilizes. This is a possible proof approach - there are finitely many possible larger submodules containing the first, less for the second, etc., unless it stabilizes. Either way, it stabilizes. Imprecise, but I hope helpful. – mathematics2x2life Sep 10 '22 at 06:20

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Perhaps your confusion stems from the use of the word "chain" in the statement "for every chain of submodules, $M_0 \subset M_1 \subset \dots,$ for an $R$-module, there exists $k \in \mathbb{N}$ such that for all $n \geq k$, $M_n = M_k$."

Given the set $S$ whose elements are the submodules of a given $R$-module $M$, one can partially order $S$ by inclusion. In this context, a chain is a subset $T$ of $S$ such that for $M_i, M_j \in T$, either $M_i \subset M_j$ or $M_j \subset M_i$. Such a subset of $S$, however, does not contain multiple copies of the same element (indeed, it is a general property of sets that no subset $S$ of a set $M$ contains multiple copies of an element of $M$). So, perhaps you would be more comfortable avoiding using the word "chain" and instead stick to the phrasing of ACC as can be found in Atiyah--MacDonald (among other places):

"Every increasing sequence $M_0 \subset M_1 \subset \dots$ in $S$ is stationary (i.e., there exists $n$ such that $M_n = M_{n + 1} = \dots$)."

user1090793
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