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.
$\square$

## Post a comment

Your email address will not be published.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 toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.

## Comments (2)

Comment #2472 by Dario Weißmann on

Comment #2505 by Johan on

There are also: