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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?

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2 Answers 2

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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.

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  • $\begingroup$ 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. $\endgroup$
    – KCd
    Dec 26, 2022 at 6:01
  • $\begingroup$ 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. $\endgroup$ Dec 26, 2022 at 7:02
  • $\begingroup$ Ah, okay. Maybe they should be mentioned there... $\endgroup$
    – KCd
    Dec 26, 2022 at 7:09
  • $\begingroup$ Yes, I agree :) $\endgroup$ Dec 26, 2022 at 7:10
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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.

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

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