Lemma 10.168.1. Let $R \to S$ be a ring map. Let $M$ be an $S$-module. Assume that

1. $R \to S$ is of finite presentation,

2. $M$ is a finitely presented $S$-module, and

3. $M$ is flat over $R$.

In this case we have the following:

1. There exists a finite type $\mathbf{Z}$-algebra $R_0$ and a finite type ring map $R_0 \to S_0$ and a finite $S_0$-module $M_0$ such that $M_0$ is flat over $R_0$, together with a ring maps $R_0 \to R$ and $S_0 \to S$ and an $S_0$-module map $M_0 \to M$ such that $S \cong R \otimes _{R_0} S_0$ and $M = S \otimes _{S_0} M_0$.

2. If $R = \mathop{\mathrm{colim}}\nolimits _{\lambda \in \Lambda } R_\lambda$ is written as a directed colimit, then there exists a $\lambda$ and a ring map $R_\lambda \to S_\lambda$ of finite presentation, and an $S_\lambda$-module $M_\lambda$ of finite presentation such that $M_\lambda$ is flat over $R_\lambda$ and such that $S = R \otimes _{R_\lambda } S_\lambda$ and $M = S \otimes _{S_{\lambda }} M_\lambda$.

3. If

$(R \to S, M) = \mathop{\mathrm{colim}}\nolimits _{\lambda \in \Lambda } (R_\lambda \to S_\lambda , M_\lambda )$

is written as a directed colimit such that

1. $R_\mu \otimes _{R_\lambda } S_\lambda \to S_\mu$ and $S_\mu \otimes _{S_\lambda } M_\lambda \to M_\mu$ are isomorphisms for $\mu \geq \lambda$,

2. $R_\lambda \to S_\lambda$ is of finite presentation,

3. $M_\lambda$ is a finitely presented $S_\lambda$-module,

then for all sufficiently large $\lambda$ the module $M_\lambda$ is flat over $R_\lambda$.

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

$((R_\lambda )_{\mathfrak p_\lambda }, (S_\lambda )_{\mathfrak q_\lambda }, (M_\lambda )_{\mathfrak q_{\lambda }})$

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

$\Psi _{\lambda } : R_\lambda \otimes _{R_0} S_0 \longrightarrow R_\lambda \otimes _{R_{\lambda _1}} S_{\lambda _1} \cong S_\lambda$

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

$S_\lambda \otimes _{S_0} M_0 \longrightarrow S_\lambda \otimes _{S_{\lambda _2}} M_{\lambda _2} \cong M_\lambda$

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

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