# The Stacks Project

## Tag 07T8

Lemma 10.39.4. 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 10.38.2. $\square$

The code snippet corresponding to this tag is a part of the file algebra.tex and is located in lines 8995–9001 (see updates for more information).

\begin{lemma}
\label{lemma-annihilator-flat-base-change}
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
$\text{Ann}_R(M) S = \text{Ann}_S(M \otimes_R S)$.
\end{lemma}

\begin{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 \ref{lemma-flat-intersect-ideals}.
\end{proof}

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