The Stacks project

Lemma 15.90.3. Let $R$ be a ring, let $f \in R$, and let $R \to R'$ be a ring map which induces an isomorphism $R/fR \to R'/fR'$. The map $\mathop{\mathrm{Spec}}(R') \amalg \mathop{\mathrm{Spec}}(R_ f) \to \mathop{\mathrm{Spec}}(R)$ is surjective. For example, the map $\mathop{\mathrm{Spec}}(R^\wedge ) \amalg \mathop{\mathrm{Spec}}(R_ f) \to \mathop{\mathrm{Spec}}(R)$ is surjective.

Proof. Recall that $\mathop{\mathrm{Spec}}(R) = V(f) \amalg D(f)$ where $V(f) = \mathop{\mathrm{Spec}}(R/fR)$ and $D(f) = \mathop{\mathrm{Spec}}(R_ f)$, see Algebra, Section 10.17 and especially Lemmas 10.17.7 and 10.17.6. Thus the lemma follows as the map $R \to R/fR$ factors through $R'$. The last statement is a special case of the first statement by Lemma 15.90.1. $\square$

Comments (0)

There are also:

  • 4 comment(s) on Section 15.90: The Beauville-Laszlo theorem

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Your email address will not be published.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 0BNM. Beware of the difference between the letter 'O' and the digit '0'.



View Lemma 15.90.3 as pdf