Lemma 10.79.2. Let $R$ be a ring. Let $\varphi : M \to N$ be a map of $R$-modules with $M$ finite and $N$ finitely presented. Then

is an open subset of $\mathop{\mathrm{Spec}}(R)$.

Lemma 10.79.2. Let $R$ be a ring. Let $\varphi : M \to N$ be a map of $R$-modules with $M$ finite and $N$ finitely presented. Then

\[ U = \{ \mathfrak p \subset R \mid \varphi _{\mathfrak p} : M_{\mathfrak p} \to N_{\mathfrak p} \text{ is an isomorphism}\} \]

is an open subset of $\mathop{\mathrm{Spec}}(R)$.

**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.
$\square$

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## Comments (2)

Comment #2909 by Dario Weißmann on

Comment #2941 by Johan on