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

Question

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?

Answer 1

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.

Answer 2

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.