Lemma 15.20.3. With notation $R \to S$, $M$, and $I$ and assumptions as in Lemma 15.20.1. In addition assume that $R \to S$ is of finite type. Then for any local homomorphism of local rings $\varphi : (R, \mathfrak m) \to (R', \mathfrak m')$ the following are equivalent

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

2. $\varphi (I) = 0$.

Proof. The implication (2) $\Rightarrow$ (1) follows from Lemma 15.18.1. Let $\varphi : R \to R'$ be as in the lemma satisfying (1). As $R$ is Noetherian we see that $R \to S$ is of finite presentation and $M$ is an $S$-module of finite presentation. Write $R' = \mathop{\mathrm{colim}}\nolimits _\lambda R_\lambda$ as a directed colimit of local $R$-subalgebras $R_\lambda \subset R'$, with maximal ideals $\mathfrak m_\lambda = R_\lambda \cap \mathfrak m'$ such that each $R_\lambda$ is essentially of finite type over $R$. By Lemma 15.18.3 we see that condition (15.18.0.1) holds for $(R_\lambda \to S \otimes _ R R_\lambda , \mathfrak m_\lambda , M \otimes _ R R_\lambda )$ for some $\lambda$. Hence Lemma 15.20.2 applies to the ring map $R \to R_\lambda$ and we see that $I$ maps to zero in $R_\lambda$, a fortiori it maps to zero in $R'$. $\square$

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