The Stacks project

Lemma 15.9.8. Let $R \to S$ be a ring map. Let $I \subset R$ be an ideal of $R$ and let $J \subset S$ be an ideal of $S$. If the closure of the image of $V(J)$ in $\mathop{\mathrm{Spec}}(R)$ is disjoint from $V(I)$, then there exists an element $f \in R$ which maps to $1$ in $R/I$ and to an element of $J$ in $S$.

Proof. Let $I' \subset R$ be an ideal such that $V(I')$ is the closure of the image of $V(J)$. Then $V(I) \cap V(I') = \emptyset $ by assumption and hence $I + I' = R$ by Algebra, Lemma 10.16.2. Write $1 = g + f$ with $g \in I$ and $f \in I'$. We have $V(f') \supset V(J)$ where $f'$ is the image of $f$ in $S$. Hence $(f')^ n \in J$ for some $n$, see Algebra, Lemma 10.16.2. Replacing $f$ by $f^ n$ we win. $\square$


Comments (0)


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Your email address will not be published.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.




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 input field. As a reminder, this is tag 07M3. Beware of the difference between the letter 'O' and the digit '0'.