Definition 15.24.1. Let $A$ be a ring. Let $M$ be a flat $A$-module. Let $x \in M$. If the set of ideals $I$ in $A$ such that $x \in IM$ has a smallest element, we call it the content ideal of $x$.
15.24 Content ideals
The definition may not be what you expect.
Note that since $M$ is flat over $A$, for a pair of ideals $I, I'$ of $A$ we have $IM \cap I'M = (I \cap I')M$ as can be seen by tensoring the exact sequence $0 \to I \cap I' \to I \oplus I' \to I + I' \to 0$ by $M$.
Lemma 15.24.2. Let $A$ be a ring. Let $M$ be a flat $A$-module. Let $x \in M$. The content ideal of $x$, if it exists, is finitely generated.
Proof. Say $x \in IM$. Then we can write $x = \sum _{i = 1, \ldots , n} f_ i x_ i$ with $f_ i \in I$ and $x_ i \in M$. Hence $x \in I'M$ with $I' = (f_1, \ldots , f_ n)$. $\square$
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 10.19.1). 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. $\square$
Lemma 15.24.4. Let $A$ be a ring. Let $M$ be a flat Mittag-Leffler module. Then every element of $M$ has a content ideal.
Proof. This is a special case of Algebra, Lemma 10.90.2. $\square$
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