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Let $M$ be a Noetherian $R$-module. Show that every non-empty set of proper submodules of $M$ has a maximal element.

My proof:

Suppose for a contradiction that we have set of proper submodules of $M$, let's call it $\mathfrak{M}$, which is non-empty and does not have a maximal element. Choose $N_1 \in \mathfrak{M}$, then it can't be maximal so there's $N_2 \in \mathfrak{M}$ with $N_1 \subsetneq N_2$. But $N_2$ is not maximal, so $\ \exists \ N_3 \in \mathfrak{M}$ such that $N_2 \subsetneq N_3$. Thus we can construct a chain $N_1 \subsetneq N_2 \subsetneq N_3 \subsetneq \dots$, but $M$ is Noetherian, hence this is a contradiction. Thus $\mathfrak{M}$ has a maximal element with respect to inclusion.

Is my proof correct? I'm having some doubts because I recall needing Zorn's lemma to prove this kind of statements.

mrtaurho
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Alessandro
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  • @rschwieb fixed, thanks – Alessandro Apr 21 '21 at 12:56
  • You really ought to eliminate $M$ from being in $\mathfrak M$ explicitly. Then it'll look good. – rschwieb Apr 21 '21 at 13:06
  • @rschwieb What's the issue with $M$ possibly being in $\mathfrak{M}$? – Alessandro Apr 21 '21 at 14:30
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    $M$ is the unique maximal element of the poset of ALL submodules. A maximal submodule is a maximal element of the PROPER submodules of M. I think you should exclude it ahead of time, not say "clearly $M\notin \mathfrak M$", because $M$ is not counted as a maximal submodule ever: maximal submodules are by convention proper. – rschwieb Apr 21 '21 at 14:52
  • @rschwieb Oh I see now, thanks! – Alessandro Apr 21 '21 at 14:56

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Two comments.

First, there is no need to formulate the proof by contradiction; you simply show the contrapositive! That is, if there is some non-empty collection of submodules without a maximal element then there is some non-stabilizing properly ascending chain of submodules. This is just a general note on style. The proof is correct, though (the part rschwieb mentioned in the comments is important but essentially taking care of by requiring $\mathfrak M$ to not have maximal elements; and $M$ is maximal...).

Second, you do not need Zorn's Lemma (i.e. the full Axiom of Choice) but some related, strictly weaker choice principle called the axiom of dependent choice. However, to the best of my knowledge there is no way of avoiding the usage of any choice principle at all in showing this implication. So it matters which definition of Noetherian you are using when you are simultaneously interested in these kinds of things.
This is by the way an interesting elementary application of a choice principle within algebra which is not "every vector space has a basis" or "every field has an algebraic closure".

mrtaurho
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    Wow, I hadn't ever read DC was equivalent to the Baire category theorem and the Löwenheim–Skolem theorem. Those seem like important parallels to the equivalences of full choice with other things! – rschwieb Apr 21 '21 at 13:10
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    @rschwieb From what I've seen so far DC seems to be a very, very natural choice principle but also a powerful, or rather, useful one. It also implies the axiom of countable choice which is very useful in basic topology. – mrtaurho Apr 21 '21 at 13:15
  • @rschwieb BTW: https://math.stackexchange.com/questions/2311187/what-follows-from-axiom-of-dependent-choice-dc-and-what-doesnt – mrtaurho Apr 21 '21 at 13:19