The Stacks project

Proof. Let $\mathfrak p \in U$. Pick a presentation $N = R^{\oplus n}/\sum _{j = 1, \ldots , m} R k_ j$. Denote $e_ i$ the image in $N$ of the $i$th basis vector of $R^{\oplus n}$. For each $i \in \{ 1, \ldots , n\}$ choose an element $m_ i \in M_{\mathfrak p}$ such that $\varphi (m_ i) = f_ i e_ i$ for some $f_ i \in R$, $f_ i \not\in \mathfrak p$. This is possible as $\varphi _{\mathfrak p}$ is an isomorphism. Set $f = f_1 \ldots f_ n$ and let $\psi : R_ f^{\oplus n} \to M_ f$ be the map which maps the $i$th basis vector to $m_ i/f_ i$. Note that $\varphi _ f \circ \psi$ is the localization at $f$ of the given map $R^{\oplus n} \to N$. As $\varphi _{\mathfrak p}$ is an isomorphism we see that $\psi (k_ j)$ is an element of $M$ which maps to zero in $M_{\mathfrak p}$. Hence we see that there exist $g_ j \in R$, $g_ j \not\in \mathfrak p$ such that $g_ j \psi (k_ j) = 0$. Setting $g = g_1 \ldots g_ m$, we see that $\psi _ g$ factors through $N_{fg}$ to give a map $\chi : N_{fg} \to M_{fg}$. By construction $\chi$ is a right inverse to $\varphi _{fg}$. It follows that $\chi _\mathfrak p$ is an isomorphism. By Lemma 10.79.1 there is an $h \in R$, $h \not\in \mathfrak p$ such that $\chi _ h : N_{fgh} \to M_{fgh}$ is surjective. Hence $\varphi _{fgh}$ and $\chi _ h$ are mutually inverse maps, which implies that $D(fgh) \subset U$ as desired. $\square$