Loading web-font TeX/Main/Regular

The Stacks project

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


Comments (0)


Your email address will not be published.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).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.