The Stacks project

Proof. The equality in the displayed formula follows from Nakayama's lemma. Nakayama's lemma also implies that $U$ is open. See Lemma 10.20.1 especially part (3). If $D(f) \subset U$, then $M_ f \to N_ f$ is surjective on all localizations at primes of $R_ f$, and hence it is surjective by Lemma 10.23.1. $\square$

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. $\square$

Proof. Choose a basis $x_1, \ldots , x_ n \in M_\mathfrak p$. We can choose an $f \in R$, $f \not\in \mathfrak p$ such that $x_ i$ is the image of some $y_ i \in M_ f$. After replacing $y_ i$ by $f^ m y_ i$ for $m \gg 0$ we may assume $y_ i \in M$. Namely, this replaces $x_1, \ldots , x_ n$ by $f^ mx_1, \ldots , f^ mx_ n$ which is still a basis as $f$ maps to a unit in $R_\mathfrak p$. Hence we obtain a homomorphism $\varphi = (y_1, \ldots , y_ n) : R^{\oplus n} \to M$ of $R$-modules whose localization at $\mathfrak p$ is an isomorphism. By Lemma 10.79.2 we can find an $f \in R$, $f \not\in \mathfrak p$ such that $\varphi _\mathfrak q$ is an isomorphism for all primes $\mathfrak q \subset R$ with $f \not\in \mathfrak q$. Then it follows from Lemma 10.23.1 that $\varphi _ f$ is an isomorphism and the proof is complete. $\square$

Proof. To prove the set $U$ is open we may work locally on $\mathop{\mathrm{Spec}}(R)$. Thus we may replace $R$ by a suitable localization and assume that $P_1 = R^{n_1}$ and $P_2 = R^{n_2}$, see Lemma 10.78.2. In this case injectivity of $\varphi \otimes \kappa (\mathfrak p)$ is equivalent to $n_1 \leq n_2$ and some $n_1 \times n_1$ minor $f$ of the matrix of $\varphi$ being invertible in $\kappa (\mathfrak p)$. Thus $D(f) \subset U$. This argument also shows that $P_{1, \mathfrak p} \to P_{2, \mathfrak p}$ is injective for $\mathfrak p \in U$.

Now suppose $D(f) \subset U$. By the remark in the previous paragraph and Lemma 10.23.1 we see that $P_{1, f} \to P_{2, f}$ is injective, i.e., (1)(a) holds. By Lemma 10.78.2 to prove (1)(b) it suffices to prove that $\mathop{\mathrm{Coker}}(\varphi )$ is finite projective locally on $D(f)$. Thus, as we saw above, we may assume that $P_1 = R^{n_1}$ and $P_2 = R^{n_2}$ and that some minor of the matrix of $\varphi$ is invertible in $R$. If the minor in question corresponds to the first $n_1$ basis vectors of $R^{n_2}$, then using the last $n_2 - n_1$ basis vectors we get a map $R^{n_2 - n_1} \to R^{n_2} \to \mathop{\mathrm{Coker}}(\varphi )$ which is easily seen to be an isomorphism.

Openness of $W$ and (2)(a) for $D(f) \subset W$ follow from Lemma 10.79.1. Since $P_{2, f}$ is projective over $R_ f$ we see that $\varphi _ f : P_{1, f} \to P_{2, f}$ has a section and it follows that $\mathop{\mathrm{Ker}}(\varphi )_ f$ is a direct summand of $P_{2, f}$. Therefore $\mathop{\mathrm{Ker}}(\varphi )_ f$ is finite projective. Thus (2)(b) holds as well.

It is clear that $V = U \cap W$ is open and the other statement in (3) follows from (1)(a) and (2)(a). $\square$