There are two definitions of Noetherian R-module: That all ascending chains of submodules stop, and that all submodules are finitely generated. Typically people prove them equivalent with Axiom of Dependent Choice, like in Why is axiom of choice needed? (Equivalent conditions for Noetherian)

Is it true, and if so how we prove it, that the equivalence of these two statements (or rather, that ACC implies all submodules are f.g.) is itself equivalent to Axiom of Dependence Choice?


1 Answer 1


Normally what you do is you show how a counterexample to Dependent Choice can be used to produce a counterexample to the equivalence. So starting with a tree of height $\omega$ with no maximal nodes and no infinite branches, we define an $R$-module whose ideals are not all finitely generated, and yet every sequence of ideals is finite.

Note that this will also prove the equivalence you're after, since Dependent Choice easily proves the two definitions are equivalent.

So. How do we actually do it? I'm not entirely sure, to be honest. It's a hard question, and I couldn't find it in any of the "usual places" (Howard–Rubin; Herrlich; etc.), and it is often mentioned that the equivalence is a consequence of Dependent Choice.

I imagine this would require a clever construction to prove. But the general gist would probably be to start with a tree $T$ which has neither maximal elements, nor infinite branches, and use it to construct some ring $R$, perhaps somehow a partially ordered ring $\Bbb Z^{(T)}$ as an order exponentiation. Then prove that the ideals have some natural ordering which is inherited from $T$, therefore there is a collection without a maximal element, but every chain of ideals is finite. And devise a way to use this fact to construct an ideal which is not finitely generated.


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