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In the book commutative algebra by Atiyah-Macdonald,

On page $75$, it says

Let $N$ be a submodule of $M$, and let $\sum$ be the set of all finitely generated submodules of $N$. Then $\sum$ is not empty and therefore has a maximal element.

On page $78$, in exercise $2$, it says

Let $M$ be an $A$-module. If every non-empty set of finitely generated submodules of $M$ has a maximal element, then $M$ is Noetherian.

Can someone explain what's the difference between two statements and what is going on here? Why is the first statement something trivial but the second statement something that has to be proved?

mrtaurho
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jk001
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  • In using Zorn's lemma, I learned that I can use the union of every element in the chain as an upper bound. Why can't I use this in the second case, while I can use it in the first statement? – jk001 Sep 17 '21 at 20:26
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    Ah, my first comment was not so good, and does not actually address the heart of the issue here. (I will delete it shortly). You've left out crucial context from your post, however: in the first portion of your quoted text, $M$ is assumed to be Noetherian! It is not Zorn's lemma being used, but rather this Noetherian property of $M$, which says that every non-empty set of submodules of $M$ has a maximal element. Anyway, the point of the exercise is this: at the beginning of the chapter, it is proved that $M$ is Noetherian iff every non-empty set of submodules of $M$ has a maximal element. – Alex Wertheim Sep 17 '21 at 21:20
  • The point of the exercise is to show that in the "only if" portion of that statement, it is sufficient that this is the case for any non-empty set of finitely generated sub-modules of $M$. – Alex Wertheim Sep 17 '21 at 21:21
  • Oh, I see. Can't believe I missed this crucial part in this chapter. Thanks so much! – jk001 Sep 17 '21 at 22:09
  • You should give some context. The first statement is not true without any hypotheses. If you consider $M=N=\mathbf Q$ (as a $\mathbf Z$-module), then $N$ has finitely generated submodules, but no maximal ones. – tomasz Sep 18 '21 at 00:16

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Well, as a matter of perspective, both statements are non-trivial and have to be proved.

The first statement is actually deduced from Proposition $6.1$, which establishes a general correspodance between chain and maximal conditions. We just happen to be in a situation (given Noetherianess!) such that this proposition applies. So, the chain condition implies the maximal condition always, and the second statement shows that the maximal condition on finitely-generated submodules only implies the converse already.

What you noticed is very important, in my opinion, as it highlights very well one aspect making Noetherian modules (and rings) special: they come equipped with a version of choice! Let me expand on this.

You know Zorn's Lemma which allows you to construct - whenever certain technical conditions are met - maximal elements of various partially ordered set. One such partially ordered set is the set of finitely-generated submodules of a module. (EDIT: The finitely-generated part is (crucially) needed to satisfy the technical conditions and, in fact, needed for all submodules a priori; as noted by Alex Wertheim there arise problems if not (for example, that supposedly all modules would be Noetherian)).

You do not need Zorn to accomplish the same thing in Noetherian modules! This, of course, boils down to your definition of Noetherian. However, proving that the prominent versions (finitely-generated submodules, ascending chain conditon, maximal condition) are equivalent actually requires some form of choice (as far as I know dependent choice is sufficient but also, more or less, the least you need).

So, while in general you can deduce the first statement using Zorn, you can deduce the second using a different definition of Noetherianess (combined with some less powerful choice) alone.

You might also find my answer to a related post interesting.

mrtaurho
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    Are you sure you only need Zorn for the first statement? I wrote a similar comment initially, but then I deleted it because I think that it's not actually true. What if you take $F$ a field, $A = F[X_{1}, X_{2}, \ldots]$ and $M = A$, $N = \langle X_{1}, X_{2}, \ldots \rangle$? The chain of f.g. submodules $\langle X_{1} \rangle \subset \langle X_{1}, X_{2} \rangle \subset \cdots$ of $N$ doesn't have an upper bound in the set of f.g. submodules of $N$, so I don't see how we can use Zorn. But I'd be happy to hear where my error is, if I'm mistaken. – Alex Wertheim Sep 17 '21 at 22:21
  • @AlexWertheim You're right, of course (I just checked Atiyah-MacDonald (should've read your comment more carefully though :D))! Some assumption is necessary as the maximal condition and the chain condition are both equivalent to Notherianess, but obviously not every module is Noetherian. – mrtaurho Sep 17 '21 at 22:25
  • @AlexWertheim I think my general point on intuition regarding these statements, especially on the Zorn-Notherian-relation, still stands although has to interpreted more carefully too. – mrtaurho Sep 17 '21 at 22:28
  • @AlexWertheim I've edited my answer. Do you think it's fine like this or does it still convey a wrong message? – mrtaurho Sep 17 '21 at 22:35