## 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. […]