Proof.
As in the proof of Lemma 10.127.11 we may first write $R = \mathop{\mathrm{colim}}\nolimits R_\lambda $ as a directed colimit of local $\mathbf{Z}$-algebras which are essentially of finite type. Denote $\mathfrak p_\lambda $ the maximal ideal of $R_\lambda $. Next, we may assume that for some $\lambda _1 \in \Lambda $ there exist $f_{j, \lambda _1} \in R_{\lambda _1}[x_1, \ldots , x_ n]$ such that
\[ S = \mathop{\mathrm{colim}}\nolimits _{\lambda \geq \lambda _1} S_\lambda , \text{ with } S_\lambda = (R_\lambda [x_1, \ldots , x_ n]/ (f_{1, \lambda }, \ldots , f_{u, \lambda }))_{\mathfrak q_\lambda } \]
For some $\lambda _2 \in \Lambda $, $\lambda _2 \geq \lambda _1$ there exist $g_{j, \lambda _2} \in R_{\lambda _2}[x_1, \ldots , x_ n, y_1, \ldots , y_ m]$ with images $\overline{g}_{j, \lambda _2} \in S_{\lambda _2}[y_1, \ldots , y_ m]$ such that
\[ S' = \mathop{\mathrm{colim}}\nolimits _{\lambda \geq \lambda _2} S'_\lambda , \text{ with } S'_\lambda = (S_\lambda [y_1, \ldots , y_ m]/ (\overline{g}_{1, \lambda }, \ldots , \overline{g}_{v, \lambda }))_{\overline{\mathfrak q}'_\lambda } \]
Note that this also implies that
\[ S'_\lambda = (R_\lambda [x_1, \ldots , x_ n, y_1, \ldots , y_ m]/ (g_{1, \lambda }, \ldots , g_{v, \lambda }))_{\mathfrak q'_\lambda } \]
Choose a presentation
\[ (S')^{\oplus s} \to (S')^{\oplus t} \to M \to 0 \]
of $M$ over $S'$. Let $A \in \text{Mat}(t \times s, S')$ be the matrix of the presentation. For some $\lambda _3 \in \Lambda $, $\lambda _3 \geq \lambda _2$ we can find a matrix $A_{\lambda _3} \in \text{Mat}(t \times s, S_{\lambda _3})$ which maps to $A$. For all $\lambda \geq \lambda _3$ we let $M_\lambda = \mathop{\mathrm{Coker}}((S'_\lambda )^{\oplus s} \xrightarrow {A_\lambda } (S'_\lambda )^{\oplus t})$.
With these choices, we have for each $\lambda _3 \leq \lambda \leq \mu $ that $S_\lambda \otimes _{R_{\lambda }} R_\mu \to S_\mu $ is a localization, $S'_\lambda \otimes _{S_{\lambda }} S_\mu \to S'_\mu $ is a localization, and the map $M_\lambda \otimes _{S'_\lambda } S'_\mu \to M_\mu $ is an isomorphism. This also implies that $S'_\lambda \otimes _{R_{\lambda }} R_\mu \to S'_\mu $ is a localization. Thus, since $M$ is flat over $R$ we see by Lemma 10.128.3 that for all $\lambda $ big enough the module $M_\lambda $ is flat over $R_\lambda $. Moreover, note that $ \mathfrak m = \mathop{\mathrm{colim}}\nolimits \mathfrak p_\lambda $, $ S/\mathfrak mS = \mathop{\mathrm{colim}}\nolimits S_\lambda /\mathfrak p_\lambda S_\lambda $, $ S'/\mathfrak mS' = \mathop{\mathrm{colim}}\nolimits S'_\lambda /\mathfrak p_\lambda S'_\lambda $, and $ M/\mathfrak mM = \mathop{\mathrm{colim}}\nolimits M_\lambda /\mathfrak p_\lambda M_\lambda $. Also, for each $\lambda _3 \leq \lambda \leq \mu $ we see (from the properties listed above) that
\[ S'_\lambda /\mathfrak p_\lambda S'_\lambda \otimes _{S_{\lambda }/\mathfrak p_\lambda S_\lambda } S_\mu /\mathfrak p_\mu S_\mu \longrightarrow S'_\mu /\mathfrak p_\mu S'_\mu \]
is a localization, and the map
\[ M_\lambda / \mathfrak p_\lambda M_\lambda \otimes _{S'_\lambda /\mathfrak p_\lambda S'_\lambda } S'_\mu /\mathfrak p_\mu S'_\mu \longrightarrow M_\mu /\mathfrak p_\mu M_\mu \]
is an isomorphism. Hence the system $(S_\lambda /\mathfrak p_\lambda S_\lambda \to S'_\lambda /\mathfrak p_\lambda S'_\lambda , M_\lambda /\mathfrak p_\lambda M_\lambda )$ is a system as in Lemma 10.127.13 as well. We may apply Lemma 10.128.3 again because $M/\mathfrak m M$ is assumed flat over $S/\mathfrak mS$ and we see that $M_\lambda /\mathfrak p_\lambda M_\lambda $ is flat over $S_\lambda /\mathfrak p_\lambda S_\lambda $ for all $\lambda $ big enough. Thus for $\lambda $ big enough the data $R_\lambda \to S_\lambda \to S'_\lambda , M_\lambda $ satisfies the hypotheses of Lemma 10.99.15. Pick such a $\lambda $. Then $S = S_\lambda \otimes _{R_\lambda } R$ is flat over $R$, and $M = M_\lambda \otimes _{S_\lambda } S$ is flat over $S$ (since the base change of a flat module is flat).
$\square$
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