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Do I need axiom of choice in this proof here?

I think not: at each step we choose one element from a set $N - \langle g_1, \dots, g_k \rangle $. So while there is indeed a countable number of sets involved from which we choose elements, I could also think of the process as follows:

Assume $N$ is generated uncountably. Let $C = \{g_1 , g_2, \dots \}$ be a countable subset of the generators of $N$. (So far we have not used choice, right?) We may write $C$ as a countable union of singleton sets $\bigcup_n \{g_n\}$. But now we can write down an explicit choice function: Let $c(\{x\}) = x$. Since the union is countable, we may consider the choice function $\tilde{c}: \mathbb N \to C, n \mapsto g_n$.

From here, we can finish the argument as follows: Then the following is an increasing chain of submodules: $\langle \tilde{c}(1) \rangle \subset \langle \tilde{c}(1), \tilde{c}(2) \rangle \subset \dots$, avoiding the axiom of choice.

Would someone tell me where my argument is flawed? Thanks.

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Rudy the Reindeer
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    Choosing a countable subset does use choice. – Cheerful Parsnip Jul 21 '12 at 11:09
  • You should think about how you would prove a countable subset of an uncountable exists. The proof will boil down to: first I pick $g_1$. Okay that didn't exhaust the set. So now I can pick $g_2$... You are making infinitely many arbitrary choices, so A.C. kicks in. – Cheerful Parsnip Jul 21 '12 at 11:11
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    Hodges [Six impossible rings] showed there exists (a model of ZF minus choice in which there exists) a ring satisfying the ascending chain condition but also having a non-finitely-generated ideal. – Zhen Lin Jul 21 '12 at 11:11
  • @JimConant Right, thanks Jim! – Rudy the Reindeer Jul 21 '12 at 11:24

1 Answers1

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There is a delicate point about the definition of Noetherian which requires the axiom of choice.

Let $X$ be an amorphous set, namely an infinite set that cannot be partitioned into two disjoint infinite subsets. Such $X$ has the interesting property that if ${\cal A\subseteq P}(X)$ is a chain, then $\cal A$ is finite.

Now let $M=\bigoplus_X\mathbb Z$ as a module over $\mathbb Z$. Every chain of modules is finite, simply since it defines a chain of subsets of $X$ and every such chain is finite.

On the other hand, it is clear that $M$ is not finitely generated and therefore not Noetherian in the definition that "every submodule is finitely generated", simply take $M$ itself to be that submodule. We also have that the family of finitely generated submodules does not have a maximal element.

It should be remarked that the equivalence between "every module is finitely generated" and "every non-empty family of submodules has a maximal element" requires the axiom of choice (specifically it requires Dependent Choice, which amongst other things implies that every infinite set has a countably infinite subset).

More:

  1. Can one avoid AC in the proof that in Noetherian rings there is a maximal element for each set?
  2. Where is the Axiom of choice used?
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Asaf Karagila
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    To be precise, it is the implication from "every submodule is finitely generated" to "every nonempty family of submodules has a maximal element" that uses some choice. Going from the latter to the former does not require AC. – Arturo Magidin Jul 21 '12 at 21:09