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
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