Content ideal of element of flat module

more-algebra.tex

@@ -3216,6 +3216,87 @@ \section{Torsion and flatness}
 
 
 
+\section{Content ideals}
+\label{section-content}
+
+\noindent
+The definition may not be what you expect.
+
+\begin{definition}
+\label{definition-content-ideal}
+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 {\it content ideal of $x$}.
+\end{definition}
+
+\noindent
+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 $M$.
+
+\begin{lemma}
+\label{lemma-content-finitely-generated}
+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.
+\end{lemma}
+
+\begin{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
+$I' = (f_1, \ldots, f_n)$.
+\end{proof}
+
+\begin{lemma}
+\label{lemma-equal-content}
+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.
+\end{lemma}
+
+\begin{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 \ref{definition-content-ideal}. 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 \ref{lemma-content-finitely-generated}) there is a nonzero map
+$\chi : I/I' \to A/\mathfrak m$ of $A$-modules
+by Nakayama's lemma (Algebra, Lemma \ref{lemma-NAK}).
+Since $I$ is the content ideal of $x$ we see that
+$x \not \in I''M$ where $I'' = \Ker(\chi)$.
+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.
+\end{proof}
+
+\begin{lemma}
+\label{lemma-content-exists-flat-Mittag-Leffler}
+Let $A$ be a ring. Let $M$ be a flat Mittag-Leffler module.
+Then every element of $M$ has a content ideal.
+\end{lemma}
+
+\begin{proof}
+This is a special case of Algebra, Lemma \ref{algebra-lemma-flat-ML-criterion}.
+\end{proof}
+
+
+
+
+
+
+
+
 \section{Flatness and finiteness conditions}
 \label{section-flat-finite}