Lemma 15.20.1. Let $R \to S$ be a ring map. Let $M$ be an $S$-module. Assume

$(R, \mathfrak m)$ is a complete local Noetherian ring,

$S$ is a Noetherian ring, and

$M$ is finite over $S$.

Then there exists an ideal $I \subset \mathfrak m$ such that

$(M/IM)_{\mathfrak q}$ is flat over $R/I$ for all primes $\mathfrak q$ of $S/IS$ lying over $\mathfrak m$, and

if $J \subset R$ is an ideal such that $(M/JM)_{\mathfrak q}$ is flat over $R/J$ for all primes $\mathfrak q$ lying over $\mathfrak m$, then $I \subset J$.

In other words, $I$ is the smallest ideal of $R$ such that (15.18.0.1) holds for $(\overline{R} \to \overline{S}, \overline{\mathfrak m}, \overline{M})$ where $\overline{R} = R/I$, $\overline{S} = S/IS$, $\overline{\mathfrak m} = \mathfrak m/I$ and $\overline{M} = M/IM$.

**Proof.**
Let $J \subset R$ be an ideal. Apply Algebra, Lemma 10.99.11 to the module $M/JM$ over the ring $R/J$. Then we see that $(M/JM)_{\mathfrak q}$ is flat over $R/J$ for all primes $\mathfrak q$ of $S/JS$ if and only if $M/(J + \mathfrak m^ n)M$ is flat over $R/(J + \mathfrak m^ n)$ for all $n \geq 1$. We will use this remark below.

For every $n \geq 1$ the local ring $R/\mathfrak m^ n$ is Artinian. Hence, by Lemma 15.17.1 there exists a smallest ideal $I_ n \supset \mathfrak m^ n$ such that $M/I_ nM$ is flat over $R/I_ n$. It is clear that $I_{n + 1} + \mathfrak m^ n$ is contains $I_ n$ and applying Lemma 15.16.1 we see that $I_ n = I_{n + 1} + \mathfrak m^ n$. Since $R = \mathop{\mathrm{lim}}\nolimits _ n\ R/\mathfrak m^ n$ we see that $I = \mathop{\mathrm{lim}}\nolimits _ n\ I_ n/\mathfrak m^ n$ is an ideal in $R$ such that $I_ n = I + \mathfrak m^ n$ for all $n \geq 1$. By the initial remarks of the proof we see that $I$ verifies (1) and (2). Some details omitted.
$\square$

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