Lemma 15.24.3. Let $(A, \mathfrak m)$ be a local ring. Let $u : M \to N$ be a map of flat $A$-modules such that $\overline{u} : M/\mathfrak m M \to N/\mathfrak m N$ is injective. If $x \in M$ has content ideal $I$, then $u(x)$ has content ideal $I$ as well.
Proof. It is clear that $u(x) \in IN$. If $u(x) \in I'N$, then $u(x) \in (I' \cap I)N$, see discussion following Definition 15.24.1. Hence it suffices to show: if $x \in I'N$ and $I' \subset I$, $I' \not= I$, then $u(x) \not\in I'N$. Since $I/I'$ is a nonzero finite $A$-module (Lemma 15.24.2) there is a nonzero map $\chi : I/I' \to A/\mathfrak m$ of $A$-modules by Nakayama's lemma (Algebra, Lemma 10.20.1). Since $I$ is the content ideal of $x$ we see that $x \not\in I''M$ where $I'' = \mathop{\mathrm{Ker}}(\chi )$. Hence $x$ is not in the kernel of the map
\[ IM = I \otimes _ A M \xrightarrow {\chi \otimes 1} A/\mathfrak m \otimes M \cong M/\mathfrak m M \]
Applying our hypothesis on $\overline{u}$ we conclude that $u(x)$ does not map to zero under the map
\[ IN = I \otimes _ A N \xrightarrow {\chi \otimes 1} A/\mathfrak m \otimes N \cong N/\mathfrak m N \]
and we conclude. $\square$
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