Finitely-generated submodules of a finitely generated module over a PID have a complement?

Question

I found this result about modules:

Let $R$ be a Principal Ideal Domain and $M$ a finitely generated $R$-module. Let $C \subseteq M$ be a cyclic submodule of order $\mu$ such that $\mu \in \text{Ann}(M)$. Then, there is a finitely generated submodule $L \subseteq M$ such that $M = C \oplus L$.

I wonder if this is true for any $C \subseteq M$ finitely generated submodule. That is, suppose that $C = \langle x_1, \ldots, x_k \rangle$ and $\langle x_i \rangle \subseteq M$ has order $\mu_i$ such that $\mu_i \in \text{Ann}(M)$ for each $i$. Is it true that there is a finitely generated submodule $L \subseteq M$ such that $M = C \oplus L$?

Answer 1

$\newcommand{\Set}[1]{\left\{ #1 \right\}}\newcommand{\Span}[1]{\left\langle #1 \right\rangle}$The answer to your rephrased question is no.

Consider $R = \mathbb{Z}$, so that $R$-modules are just abelian groups.

Let $M = \Span{a} \oplus \Span{b}$, where $a$ and $b$ have order $4$. Consider the submodule $C = \Span{a, a + 2 b}$. It satisfies your assumption, but it does not have a complement, as

• $C = \Span{a, 2 b}$;
• $M/C$ has order $2$;
• $C$ contains the set $\Set{2a, 2b, 2a + 2 b}$ of the elements of $M$ of order $2$.

You may want to check pure submodules.