Zorn's Lemma in noetherian modules

Question

For noetherian modules, we have in particular the equivalent definitions that the Ascending Chain Condition holds and that every nonempty subset of submodules has a maximal element.

Now I can prove the "ACC $\Rightarrow$ every nonempty set of submodules has a maximal element" implication by the following argument:

Let $S$ be a nonempty set of submodules, take $N_1 \in S$. If $N_1$ is not maximal, there exists a $N_2 \supsetneq N_1$ per definition. By induction we get a chain $N_1 \subsetneq N_2 \subsetneq \dotsb$ and by ACC this induction terminates after finitely many, say $n$, steps. Then $N_n$ is maximal.

My question is: Did I miss the use of Zorn's Lemma somewhere? This proof doesn't seem to use it, and the other implications in the equivalence work fine without it too, yet I've seen some sources (e.g. Wikipedia, first answer here: Is every Noetherian module finitely generated?) that require it. I'm fine with Zorn's Lemma by the way, just feeling a bit stupid right now.

In this case (as it sometimes is) the answer is buried in the ellipsis. — rschwieb, Mar 10 '14 at 18:31

user2345215 · Answer 1 · 2014-03-11T10:49:13.913

4

I must say the argument seems fine to me.

Assuming no module in $S$ is maximal, we can construct a sequence $N_0\subsetneq N_1\subsetneq\ldots$ which contradicts the ACC condition.

But note you need the axiom of choice to construct the sequence, as it's not obvious at all how to choose the modules.

In fact, only the dependent choice is required.

edited Mar 11 '14 at 10:49

answered Mar 10 '14 at 18:01

user2345215

16,803

Seems to me that you’re exactly right here: if you have ACC, a direct appeal to Zorn is usually avoidable. – Lubin Mar 10 '14 at 18:06
It looks more like dependent choice than countable choice. Each step depends on the previous one. – Asaf Karagila Mar 10 '14 at 19:15
@Asaf Karagila: True, but I just wanted to illustrate you don't need the full power of zorn's lemma. – user2345215 Mar 10 '14 at 19:17
That's fine, but countable choice is weaker than dependent choice. – Asaf Karagila Mar 10 '14 at 19:19
I don't see how/if countable choice suffices. I'd guess in countable choice you need to know the sets from which to choose beforehand, but I'm not sure. – Andre Knispel Mar 11 '14 at 10:40
@Andre: You are right, I messed up. For some reason I thought dependent is weaker and even misread Asaf's comment. – user2345215 Mar 11 '14 at 10:48

score 4 · Accepted Answer · answered Mar 10 '14 at 19:18

This is a particularly delicate use of the axiom of choice. Historically the principle of dependent choice, which is used exactly for these sort of inductive constructions, was overlooked by some of the greatest people that opposed to the axiom (e.g. Lebesgue).

When we construct a sequence by induction, we can only assure that for every finite number $n$, we can construct a sequence of length $n$. The axiom of choice, or rather the much weaker principle of dependent choice, is needed to prove the existence of an infinite sequence from that.

Zorn's lemma is used in the following manner. Every chain is finite, therefore it has an upper bound. So there is a maximal element.

Thank you very much, I didn't know a about dependent choice before! — Andre Knispel, Mar 11 '14 at 10:35
Andre, yes, it is not common to mention that when one can just talk about the axiom of choice and move along. — Asaf Karagila, Mar 11 '14 at 10:38

Zorn's Lemma in noetherian modules

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