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Let $R$ be a commutative ring with $1$ and let $M$ be a left $R$-module. On page 458 of Dummit and Foote's Algebra, 3rd edition, they show that $M$ is Noetherian (i.e. satisfies A.C.C. on submodules) if and only if every nonempty set of submodules of $M$ has a maximal element under inclusion. I don't understand why the axiom of choice comes into their proof, outlined below.

Their proof is as follows: Let $\Sigma$ be a nonempty set of submodules of $M$, and assume for a contradiction that $\Sigma$ does not have a maximal element under incusion. and take $M_1 \in \Sigma$. Then as $M_1$ is not maximal, there is some $M_2 \in \Sigma \setminus \{M_1\}$ such that $M_1 \subsetneq M_2$. We continue this way to get $M_1 \subsetneq M_2 \subsetneq M_3 \subsetneq \cdots$, contradicting the A.C.C. on submodules.

Where do they use the axiom of choice?

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  • $\begingroup$ The proof as quoted uses dependent choice. $\endgroup$ – Zhen Lin Sep 18 '14 at 16:01
  • $\begingroup$ I just want to comment for future reference that this equivalence can be phrased more easily in terms of Zorn's Lemma, which is known to be equivalent to the Axiom of Choice. $\endgroup$ – Chill2Macht Oct 9 '16 at 17:15
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Index the elements of $\Sigma$ with elements of $I$. By assuming no element of $\Sigma $ is maximal, you assert that each set $\Sigma_i:=\{N\subset M\mid N\supset M_i\}$ is nonempty, and then you need a choice function $C:\{\Sigma_i\mid i\in I\}\to \Sigma$ to legitimately furnish infinitely many witnesses to the nonmaximality of each thing in $\Sigma$ all at once.

Once you have those witnesses, you inductively construct the countable ascending chain you mentioned. Without the choice function, it's unclear "how to choose a sock" in each $\Sigma_i$.

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  • $\begingroup$ Gotcha, thanks. I guess this shows that one can end up using the axiom of choice inadvertently if one is not careful. $\endgroup$ – nigel Sep 18 '14 at 23:25
  • $\begingroup$ @user71815 Somewhat similarly to the above, you'll need a choice function to prove that the "all submodules f.g." is equivalent to the ascending chain condition. Suppose $N$ is a submodule of $M$ which isn't finitely generated. Then $N\setminus N'$ is nonempty for any f.g. submodule $N'<N$ and you need to pick some element out of this nonempty set to make a second f.g. module $N''$ over $N'$. A choice function will enable the existence of these elements, and then you can lift a strictly ascending chain out of the poset of finitely generated submodules of $N$. $\endgroup$ – rschwieb Oct 20 '14 at 15:59
  • $\begingroup$ @user71815 Is it the case that even a simple line like this requires AC? No, of course not. The AoC only becomes necessary when you are attempting to select an element from infinitely many nonempty sets at once. Picking an element from a single nonempty set or finitely many nonempty sets is possible without the AoC. $\endgroup$ – rschwieb Oct 21 '14 at 12:45
  • $\begingroup$ @user71815 the statement of AC seems to concern big Cartesian products... Well, yes, that sounds like part of an equivalent statement. For me, the canonical version of the AoC is "Given a set of infinitely many nonempty sets, there is a choice function for the collection, (a choice function is like this: the set of sets is the domain of the function, and the value of the function evaluated at one of these sets is a point in that set.) There is no difficulty for defining such a function in the finite case: only the infinite case is interesting. $\endgroup$ – rschwieb Oct 21 '14 at 12:49
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    $\begingroup$ @user71815 No, I totally know the feeling you're having. Most people (myself included) run across this during their math education. The issue is that in the process you're describing, the act of picking one element from infinitely many nonempty sets is "swept under the rug" by induction. The fact is that you need to bring those choices into existence all at once with the AoC, and then you can use induction to build a suitable chain of them. $\endgroup$ – rschwieb Oct 21 '14 at 13:13

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