The Stacks project

Lemma 15.124.8. Let $R$ be a Bézout domain.

  1. Every finite submodule of a free module is finite free.

  2. Every finitely presented $R$-module $M$ is a direct sum of a finite free module and a torsion module $M_{tors}$ which is a summand of a module of the form $\bigoplus _{i = 1, \ldots , n} R/f_ iR$ with $f_1, \ldots , f_ n \in R$ nonzero.

Proof. Proof of (1). Let $M \subset F$ be a finite submodule of a free module $F$. Since $M$ is finite, we may assume $F$ is a finite free module (details omitted). Say $F = R^{\oplus n}$. We argue by induction on $n$. If $n = 1$, then $M$ is a finitely generated ideal, hence principal by our assumption that $R$ is Bézout. If $n > 1$, then we consider the image $I$ of $M$ under the projection $R^{\oplus n} \to R$ onto the last summand. If $I = (0)$, then $M \subset R^{\oplus n - 1}$ and we are done by induction. If $I \not= 0$, then $I = (f) \cong R$. Hence $M \cong R \oplus \mathop{\mathrm{Ker}}(M \to I)$ and we are done by induction as well.

Let $M$ be a finitely presented $R$-module. Since the localizations of $R$ are maximal ideals are valuation rings (Lemma 15.124.7) we may apply Lemma 15.124.4. Thus $M$ is a summand of a module of the form $R^{\oplus r} \oplus \bigoplus _{i = 1, \ldots , n} R/f_ iR$ with $f_ i \not= 0$. Since taking the torsion submodule is a functor we see that $M_{tors}$ is a summand of the module $\bigoplus _{i = 1, \ldots , n} R/f_ iR$ and $M/M_{tors}$ is a summand of $R^{\oplus r}$. By the first part of the proof we see that $M/M_{tors}$ is finite free. Hence $M \cong M_{tors} \oplus M/M_{tors}$ as desired. $\square$

Comments (0)

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Your email address will not be published.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).

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.

In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 0ASU. Beware of the difference between the letter 'O' and the digit '0'.