The Stacks project

Lemma 26.11.5. Let $X$ be a scheme. Let $U, V$ be affine opens of $X$, and let $x \in U \cap V$. There exists an affine open neighbourhood $W$ of $x$ such that $W$ is a standard open of both $U$ and $V$.

Proof. Write $U = \mathop{\mathrm{Spec}}(A)$ and $V = \mathop{\mathrm{Spec}}(B)$. Say $x$ corresponds to the prime $\mathfrak p \subset A$ and the prime $\mathfrak q \subset B$. We may choose an $f \in A$, $f \not\in \mathfrak p$ such that $D(f) \subset U \cap V$. Note that any standard open of $D(f)$ is a standard open of $\mathop{\mathrm{Spec}}(A) = U$. Hence we may assume that $U \subset V$. In other words, now we may think of $U$ as an affine open of $V$. Next we choose a $g \in B$, $g \not\in \mathfrak q$ such that $D(g) \subset U$. In this case we see that $D(g) = D(g_ A)$ where $g_ A \in A$ denotes the image of $g$ by the map $B \to A$. Thus the lemma is proved. $\square$

Comments (4)

Comment #5536 by Zeyn Sahilliogullari on

g∈A in the second to last sentence should be g∈B

Comment #6840 by Shota Inoue on

A slight typo in the proof: "We may choose a , ..." should be replaced by "We may choose an , ..."

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Your email address will not be published.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