Remark 86.2.4 (Base change by closed immersion). Let $A$ be a Noetherian ring and $I \subset A$ an ideal. Let $\mathfrak a \subset A$ be an ideal. Denote $\bar A = A/\mathfrak a$. Let $\bar I \subset \bar A$ be an ideal such that $I^ c \bar A \subset \bar I$ and $\bar I^ d \subset I\bar A$ for some $c, d \geq 1$. In this case the base change functor (86.2.3.2) for $(A, I)$ to $(\bar A, \bar I)$ is given by $B \mapsto \bar B = B/\mathfrak aB$. Namely, we have

the last equality because any finite $B$-module is $I$-adically complete by Algebra, Lemma 10.96.1 and if annihilated by $\mathfrak a$ also $\bar I$-adically complete by Algebra, Lemma 10.95.9.

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