# Being Noetherian Is Not Local…Or Is It?

A commutative ring $R$ can be non-Noetherian and have all of its localisations at prime ideals Noetherian, such as the infamous $\prod_{i=1}^\infty \mathbb{Z}/2$. So being Noetherian is not a local property. However, there is an interesting variant of ‘local’ that does work, which I learnt from Yves Lequain’s paper [1]. It goes like this:

Theorem. Let $R$ be a ring and fix a left maximal ideal $M$ of $R$. Then $R$ is left Noetherian if and only if every left ideal contained in $M$ is finitely generated.

The nice thing about this statement is that it avoids localisation so it’s easy to state for noncommutative rings.
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