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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