The Stacks project

Proof. We first write $(R \to S, M)$ as the directed colimit of a system $(R_\lambda \to S_\lambda , M_\lambda )$ as in as in Lemma 10.127.18. Let $\mathfrak q \subset S$ be a prime. Let $\mathfrak p \subset R$, $\mathfrak q_\lambda \subset S_\lambda$, and $\mathfrak p_\lambda \subset R_\lambda$ the corresponding primes. As seen in the proof of Theorem 10.129.4

is a system as in Lemma 10.127.13, and hence by Lemma 10.128.3 we see that for some $\lambda _{\mathfrak q} \in \Lambda$ for all $\lambda \geq \lambda _{\mathfrak q}$ the module $M_\lambda$ is flat over $R_\lambda$ at the prime $\mathfrak q_{\lambda }$.

By Theorem 10.129.4 we get an open subset $U_\lambda \subset \mathop{\mathrm{Spec}}(S_\lambda )$ such that $M_\lambda$ flat over $R_\lambda$ at all the primes of $U_\lambda$. Denote $V_\lambda \subset \mathop{\mathrm{Spec}}(S)$ the inverse image of $U_\lambda$ under the map $\mathop{\mathrm{Spec}}(S) \to \mathop{\mathrm{Spec}}(S_\lambda )$. The argument above shows that for every $\mathfrak q \in \mathop{\mathrm{Spec}}(S)$ there exists a $\lambda _{\mathfrak q}$ such that $\mathfrak q \in V_\lambda$ for all $\lambda \geq \lambda _{\mathfrak q}$. Since $\mathop{\mathrm{Spec}}(S)$ is quasi-compact we see this implies there exists a single $\lambda _0 \in \Lambda$ such that $V_{\lambda _0} = \mathop{\mathrm{Spec}}(S)$.

The complement $\mathop{\mathrm{Spec}}(S_{\lambda _0}) \setminus U_{\lambda _0}$ is $V(I)$ for some ideal $I \subset S_{\lambda _0}$. As $V_{\lambda _0} = \mathop{\mathrm{Spec}}(S)$ we see that $IS = S$. Choose $f_1, \ldots , f_ r \in I$ and $s_1, \ldots , s_ n \in S$ such that $\sum f_ i s_ i = 1$. Since $\mathop{\mathrm{colim}}\nolimits S_\lambda = S$, after increasing $\lambda _0$ we may assume there exist $s_{i, \lambda _0} \in S_{\lambda _0}$ such that $\sum f_ i s_{i, \lambda _0} = 1$. Hence for this $\lambda _0$ we have $U_{\lambda _0} = \mathop{\mathrm{Spec}}(S_{\lambda _0})$. This proves (1).

Proof of (2). Let $(R_0 \to S_0, M_0)$ be as in (1) and suppose that $R = \mathop{\mathrm{colim}}\nolimits R_\lambda$. Since $R_0$ is a finite type $\mathbf{Z}$ algebra, there exists a $\lambda$ and a map $R_0 \to R_\lambda$ such that $R_0 \to R_\lambda \to R$ is the given map $R_0 \to R$ (see Lemma 10.127.3). Then, part (2) follows by taking $S_\lambda = R_\lambda \otimes _{R_0} S_0$ and $M_\lambda = S_\lambda \otimes _{S_0} M_0$.

Finally, we come to the proof of (3). Let $(R_\lambda \to S_\lambda , M_\lambda )$ be as in (3). Choose $(R_0 \to S_0, M_0)$ and $R_0 \to R$ as in (1). As in the proof of (2), there exists a $\lambda _0$ and a ring map $R_0 \to R_{\lambda _0}$ such that $R_0 \to R_{\lambda _0} \to R$ is the given map $R_0 \to R$. Since $S_0$ is of finite presentation over $R_0$ and since $S = \mathop{\mathrm{colim}}\nolimits S_\lambda$ we see that for some $\lambda _1 \geq \lambda _0$ we get an $R_0$-algebra map $S_0 \to S_{\lambda _1}$ such that the composition $S_0 \to S_{\lambda _1} \to S$ is the given map $S_0 \to S$ (see Lemma 10.127.3). For all $\lambda \geq \lambda _1$ this gives maps

the last isomorphism by assumption. By construction $\mathop{\mathrm{colim}}\nolimits _\lambda \Psi _\lambda$ is an isomorphism. Hence $\Psi _\lambda$ is an isomorphism for all $\lambda$ large enough by Lemma 10.127.8. In the same vein, there exists a $\lambda _2 \geq \lambda _1$ and an $S_0$-module map $M_0 \to M_{\lambda _2}$ such that $M_0 \to M_{\lambda _2} \to M$ is the given map $M_0 \to M$ (see Lemma 10.127.5). For $\lambda \geq \lambda _2$ there is an induced map

and for $\lambda$ large enough this map is an isomorphism by Lemma 10.127.6. This implies (3) because $M_0$ is flat over $R_0$. $\square$