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. $\square$

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 $f\in A$, ..." should be replaced by "We may choose an $f\in A$, ..."

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