The Stacks project

Lemma 27.3.2. In Situation 27.3.1. Suppose $U \subset U' \subset S$ are affine opens. Let $A = \mathcal{A}(U)$ and $A' = \mathcal{A}(U')$. The map of rings $A' \to A$ induces a morphism $\mathop{\mathrm{Spec}}(A) \to \mathop{\mathrm{Spec}}(A')$, and the diagram

\[ \xymatrix{ \mathop{\mathrm{Spec}}(A) \ar[r] \ar[d] & \mathop{\mathrm{Spec}}(A') \ar[d] \\ U \ar[r] & U' } \]

is cartesian.

Proof. Let $R = \mathcal{O}_ S(U)$ and $R' = \mathcal{O}_ S(U')$. Note that the map $R \otimes _{R'} A' \to A$ is an isomorphism as $\mathcal{A}$ is quasi-coherent (see Schemes, Lemma 26.7.3 for example). The result follows from the description of the fibre product of affine schemes in Schemes, Lemma 26.6.7. $\square$

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Your email address will not be published.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 01LN. Beware of the difference between the letter 'O' and the digit '0'.



View Lemma 27.3.2 as pdf