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I use the definition of a Noetherian ring given by Qiaochu in this: A commutative ring is Noetherian if, for any nonempty collection of ideals $\mathcal{I}$, there is some $I \in \mathcal{I}$ which is not properly contained in any $J \in \mathcal{I}$.

Can we prove the following theorem without using Axiom of Choice?

Theorem Let $A$ be a Noetherian commutative ring. Let $A[[X]]$ be the ring of formal power series over $A$. Then $A[[X]]$ is Noetherian.

As for why I think this question is interesting, please see(particularly Pete Clark's answer): Why worry about the axiom of choice?

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The same question was asked in MathOverflow. – Makoto Kato Jan 14 '14 at 8:22

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