Lemma 10.40.4. Let $R \to S$ be a flat ring map. Let $M$ be an $R$-module and $m \in M$. Then $\text{Ann}_ R(m) S = \text{Ann}_ S(m \otimes 1)$. If $M$ is a finite $R$-module, then $\text{Ann}_ R(M) S = \text{Ann}_ S(M \otimes _ R S)$.

Proof. Set $I = \text{Ann}_ R(m)$. By definition there is an exact sequence $0 \to I \to R \to M$ where the map $R \to M$ sends $f$ to $fm$. Using flatness we obtain an exact sequence $0 \to I \otimes _ R S \to S \to M \otimes _ R S$ which proves the first assertion. If $m_1, \ldots , m_ n$ is a set of generators of $M$ then $\text{Ann}_ R(M) = \bigcap \text{Ann}_ R(m_ i)$. Similarly $\text{Ann}_ S(M \otimes _ R S) = \bigcap \text{Ann}_ S(m_ i \otimes 1)$. Set $I_ i = \text{Ann}_ R(m_ i)$. Then it suffices to show that $\bigcap _{i = 1, \ldots , n} (I_ i S) = (\bigcap _{i = 1, \ldots , n} I_ i)S$. This is Lemma 10.39.2. $\square$

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