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Tag 080U

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. $\square$

The code snippet corresponding to this tag is a part of the file algebra.tex and is located in lines 16374–16379 (see updates for more information).

\begin{lemma}
\label{lemma-blowup-in-principal}
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 $R'_f = R'_a = R_a$.
\end{lemma}

\begin{proof}
We will use the results of Lemma \ref{lemma-affine-blowup}
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.
\end{proof}

Comment #2472 by Dario Weißmann on April 2, 2017 a 10:35 pm UTC

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

Comment #2505 by Johan (site) on April 14, 2017 a 12:05 am UTC

Thanks, fixed here.

There is also 1 comment on Section 10.69: Commutative Algebra.

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