
Lemma 10.69.4. 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 10.69.2 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

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

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• 1 comment(s) on Section 10.69: Blow up algebras

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