# When one defines the localized ring $A_f$, is it tacitly assumed that $f$ is not nilpotent?

Let $$A$$ be a commutative ring, let $$f\in A$$. When one defines the localized ring $$A_f$$, is it tacitly assumed that $$f$$ is not nilpotent?

In Atiyah-McDonald, $$A_f$$ is defined as the localization $$A_f=S^{-1}A$$ around the multiplicatively closed set $$S=\{f^n\}_{n\geq 0}$$. But if $$f$$ is nilpotent, then $$0\in S$$, so $$S^{-1}A$$ includes formal expressions of the sort $$a/0$$. I suppose one could include such expressions and still have a resonable ring $$A_f$$, but is this case intentionally included in the definition?

No. The localization is defined for every multiplicative subset. Just because fractions of the form a/0 may appear does not invalidate the general construction. It works without any case distinctions! When the multiplicative subset contains a zero, you can easily verify that the localization is the zero ring, though. A must-read on this general topic is MO/45951.

• If this example appears on that MO page, it would help if you included a tip on how to find it in order to replace a search of all the answers. (I am reminded of citing a theorem in a book without mentioning the page number.) Searching on "nilpotent" or "localization" or "ring" at that MO link didn't give any related result.
– KCd
Dec 26, 2022 at 6:01
• I linked the MO thread because it gives a better understanding of everything related to the zero, empty set, etc. I am not saying that localizations are mentioned there. This is what I meant with "general topic". Regarding localizations, I think everything is said already here. Dec 26, 2022 at 7:02
• Ah, okay. Maybe they should be mentioned there...
– KCd
Dec 26, 2022 at 7:09
• Yes, I agree :) Dec 26, 2022 at 7:10

Nope!! It's fine for $$f$$ to be nilpotent just a bit boring, what you are noticing is that if $$f$$ is nilpotent then $$A_f$$ is the zero ring.

• Oh, right. Any element $a/f^k$ of $A_f$ is equal to $0/1$, since if $f^N=0$, then $f^N(a\cdot 1-0\cdot f^k)=0\implies a/f^k=0/1$ Dec 25, 2022 at 22:40