I was wondering wether the proof of the follwing statement I found online was correct.
Let $M$ be an $R$-Module ($R$ commutative and unital).
(a) Every ascending sequence $N_0 \subset N_1 \subset \cdots$ of submodules of $M$ terminates.
(b) Every non-empty set $\mathfrak{M}$ of submodules of $M$ has a maximal element.
Then (a) implies (b).
proof. We can apply Zorn's lemma to $(\mathfrak{M},\subset)$ since any ascending chain terminates and hence is bounded in $\mathfrak{M}$. Therefore there is a maximal element.
My objections is that in order to use Zorn's lemma we need to show that any totally ordered subset of $\mathfrak{M}$ has an upper bound in $\mathfrak{M}$. But in the above proof this fact is only estabished for countable totally ordered subsets. Is there a way to bypass this problem? I know a proof of the above using the axiom of dependent choice. So I'm not asking for a general proof, rather for a proof using Zorn's lemma in this straightforward way.
Best Answer
Your objection to the quoted proof is valid -- the proof is at best leaving out some details of how to deal with chains that are not simple ascending sequences.
We can salvage it by instead doing this:
Lemma. Let $P$ be any partial order. Then $P$ contains either a maximal element or an infinite ascending sequence.
Proof. Apply Zorn's lemma to the set of partial increasing functions $\mathbb N\to P$ (ordered by set inclusion, as it is common in Zorn applications). We get a maximal such function. If its domain has a largest element, then its image must be a maximal element of $P$ (or we could extend the function). Otherwise the image of the function is an infinite ascending sequence.
Now since $\mathfrak M$ is partially ordered but doesn't contain an infinite ascending sequence, it must have a maximal element.
If you want preserve the particular step "apply Zorn's lemma to $(\mathfrak M,\subseteq)$", then I think you'll need something like the above lemma to deal with an arbitrary chain anyway.
Sometimes Zorn's lemma is stated in a variant where the premise is not "every chain has an upper bound", but "every well-ordered subset has an upper bound". This is apparently stronger than the usual statement of the lemma, but is actually what the usual proof from the axiom of choice proves, so the strengthening is actually harmless (since standard Zorn $\Rightarrow$ axiom of choice $\Rightarrow$ strenghtened Zorn).
If you're using this strengthened Zorn's lemma instead, you can use Memerson's suggestion (from a comment) to take the initial $\omega$ elements of the well-ordered subset to get a contradiction with the noetherian property if the well-ordered subset doesn't have a maximum.