Lemma 10.70.8. Let R be a ring, I \subset R an ideal, a \in I, and f \in R. Set R' = R[\frac{I}{a}] and R'' = R[\frac{fI}{fa}]. Then there is a surjective R-algebra map R' \to R'' whose kernel is the set of f-power torsion elements of R'.
Proof. The map is given by sending x/a^ n for x \in I^ n to f^ nx/(fa)^ n. It is straightforward to check this map is well defined and surjective. Since af is a nonzero divisor in R'' (Lemma 10.70.2) we see that the set of f-power torsion elements are mapped to zero. Conversely, if x \in R' and f^ n x \not= 0 for all n > 0, then (af)^ n x \not= 0 for all n as a is a nonzero divisor in R'. It follows that the image of x in R'' is not zero by the description of R'' following Definition 10.70.1. \square
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