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 10.20.1). 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. $\square$
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 10.91.2. $\square$
Post a comment
Your email address will not be published. Required fields are marked.
In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).
All contributions are licensed under the GNU Free Documentation License.
Comments (0)