Lemma 10.70.7 (080U)—The Stacks project

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Table of contents
Part 1: Preliminaries
Chapter 10: Commutative Algebra
Section 10.70: Blow up algebras
Lemma 10.70.7 (cite)

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Lemma 10.70.7. Let $R$ be a ring, $I \subset R$ an ideal, and $a \in I$. Set $R' = R[\frac{I}{a}]$. If $f \in R$ is such that $V(f) = V(I)$, then $f$ maps to a nonzerodivisor in $R'$ and $R'_ f = R'_ a = R_ a$.

Proof. We will use the results of Lemma 10.70.2 without further mention. The assumption $V(f) = V(I)$ implies $V(fR') = V(IR') = V(aR')$. Hence $a^ n = fb$ and $f^ m = ac$ for some $b, c \in R'$. The lemma follows. $\square$

Comments (2)

Comment #2472 by Dario Weißmann on April 02, 2017 at 22:35

Should it not be $R'_f = R'_a = R_a$ instead of $... = R_f$ ?

Comment #2505 by Johan on April 14, 2017 at 00:05

Thanks, fixed here.

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