# Atypical definition of "regular ring"?

In my adventures I've finally come by a copy of Examples of Commutative Rings by Hutchins, and almost immediately read something that surprised me.

On page 10, a regular ring is defined as Noetherian ring whose every localization at a maximal ideal is a regular local ring.

Up until now, I thought the standard definition was that it was rather a Noetherian ring whose every localization at a prime ideal is a regular local ring.

I'm not handy with commutative algebra... is the definition using maximal ideals a thing? Or are they equivalent and I just haven't run across the theorem?

Am I apparently going to have to re-interpret everything he says about regular rings?

• The two variants are equivalent. Feb 22, 2018 at 23:43

The definitions are equivalent. The key fact is that the localization of a regular local ring at a prime ideal is still regular (this follows immediately from the characterization of regular local rings as local rings of finite global dimension, for instance). So, suppose you have a ring $R$ which is regular by Hutchins's definition, and let $P$ be any prime ideal in $R$. Let $M$ be a maximal ideal containing $P$; then $R_M$ is regular local. But $R_P$ is just the localization of $R_M$ at the prime $P_M$, so $R_P$ is also regular local.
• Just randomly came across this -- here's a silly example of a property which does not always hold at all prime ideals, even if it does hold at all maximal ideals: "has positive dimension". For example, $\dim \mathbb{Z}_{\mathfrak{m}} = 1$ whenever $\mathfrak{m}$ is maximal, but $\dim \mathbb{Z}_0 = 0$.What's more is that $$\dim R = \sup_{\mathfrak{m}~\text{maximal}} \dim R_{\mathfrak{m}}$$ always holds, so "$R$ has positive dimension" does follow from "$R_{\mathfrak{m}}$ has positive dimension for all maximal ideals $\mathfrak{m}$", and ofc these statements are equivalent when $R$ is local. Feb 7, 2021 at 20:24