# Why is axiom of choice needed? (Equivalent conditions for Noetherian)

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?

• The proof as quoted uses dependent choice. – Zhen Lin Sep 18 '14 at 16:01
• 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. – Chill2Macht Oct 9 '16 at 17:15

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 $\Sigma_i$.
• @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 $N$. – rschwieb Oct 20 '14 at 15:59