Lemma 15.20.2. With notation $R \to S$, $M$, and $I$ and assumptions as in Lemma 15.20.1. Consider a local homomorphism of local rings $\varphi : (R, \mathfrak m) \to (R', \mathfrak m')$ such that $R'$ is Noetherian. Then the following are equivalent

condition (15.18.0.1) holds for $(R' \to S \otimes _ R R', \mathfrak m', M \otimes _ R R')$, and

$\varphi (I) = 0$.

**Proof.**
The implication (2) $\Rightarrow $ (1) follows from Lemma 15.18.1. Let $\varphi : R \to R'$ be as in the lemma satisfying (1). We have to show that $\varphi (I) = 0$. This is equivalent to the condition that $\varphi (I)R' = 0$. By Artin-Rees in the Noetherian local ring $R'$ (see Algebra, Lemma 10.51.4) this is equivalent to the condition that $\varphi (I)R' + (\mathfrak m')^ n = (\mathfrak m')^ n$ for all $n > 0$. Hence this is equivalent to the condition that the composition $\varphi _ n : R \to R' \to R'/(\mathfrak m')^ n$ annihilates $I$ for each $n$. Now assumption (1) for $\varphi $ implies assumption (1) for $\varphi _ n$ by Lemma 15.18.1. This reduces us to the case where $R'$ is Artinian local.

Assume $R'$ Artinian. Let $J = \mathop{\mathrm{Ker}}(\varphi )$. We have to show that $I \subset J$. By the construction of $I$ in Lemma 15.20.1 it suffices to show that $(M/JM)_{\mathfrak q}$ is flat over $R/J$ for every prime $\mathfrak q$ of $S/JS$ lying over $\mathfrak m$. As $R'$ is Artinian, condition (1) signifies that $M \otimes _ R R'$ is flat over $R'$. As $R'$ is Artinian and $R/J \to R'$ is a local injective ring map, it follows that $R/J$ is Artinian too. Hence the flatness of $M \otimes _ R R' = M/JM \otimes _{R/J} R'$ over $R'$ implies that $M/JM$ is flat over $R/J$ by Algebra, Lemma 10.101.7. This concludes the proof.
$\square$

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