# The Stacks Project

## 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 16403–16408 (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}

## Comments (2)

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.

## Add a comment on tag 080U

Your email address will not be published. Required fields are marked.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the lower-right corner).

All contributions are licensed under the GNU Free Documentation License.

In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following box. So in case this where tag 0321 you just have to write 0321. Beware of the difference between the letter 'O' and the digit 0.

This captcha seems more appropriate than the usual illegible gibberish, right?