Let $R$ be a principal ideal domain and $M$ a finitely generated $R$-module. Furthermore, let $N$ be a submodule of $M$. Prove or disprove: there exist free submodules $F_N \leq N$, $F_M \leq M$ with $N = F_N \oplus N_{\mathrm{tor}}$, $M =F_M \oplus M_{\mathrm{tor}}$ with $F_N \leq F_M$.
From Hungerford, Theorem 6.6, we know that $M$ is the direct sum of a free submodule of finite rank, call it $F_M$, and $M_{\mathrm{tor}}$. That is, $$M = F_M \oplus M_{\mathrm{tor}}.$$ Note that $F_M$ and $M_{\mathrm{tor}}$ are taken to be submodules of $M$ here.
Now, let $N \leq M$. Recall that every submodule of a finitely generated $R$-module, $R$ a principal ideal domain, is finitely generated. Thus, again by Theorem 6.6 in Hungerford, we know that $N$ is the direct sum of a free submodule of finite rank, call it $F_N$, and $N_{\mathrm{tor}}$. That is, $$N = F_N \oplus N_{\mathrm{tor}}.$$ Note that $F_M \cong M/M_{\mathrm{tor}}$ and $F_N \cong N/N_{\mathrm{tor}}$. It is clear that $N_{\mathrm{tor}} \leq M_{\mathrm{tor}}$. However, it does not seem like we can deduce that $F_N \leq F_M$. Is the statement false? If so, what is a counterexample?
Note: I'm letting "$\leq $" denote submodule here.
Here is the correct solution to this question:
Let $R =\mathbb{Z}$ and $M$ the $R$-module given by $$M = \big\{r(1,0):r \in R\text{ and }(1,0) \in \mathbb{Z} \times \mathbb{Z}/2\mathbb{Z}\big\} \oplus \big\{r(0,1):r \in R\text{ and }(0,1) \in \mathbb{Z}\times \mathbb{Z}/2\mathbb{Z}\big\}$$ For short, we will write $$M = \langle (1,0) \rangle \oplus \langle (0,1) \rangle$$ Note that $R$ is a principal ideal domain and $M$ is finitely generated. Furthermore, notice that $M_{tor} = \langle (0,1)\rangle$. Recall that if $R$ is a principal ideal domain, then for any finitely generated $R$ module $M$, there exists a free module $F_M\leq M$ such that $F \cong M/M_{tor}$ and $M = F_M\oplus M_{tor}$. In our case, $M_{tor} = \langle (0,1) \rangle$. Given this, we know that the free part of $M$ is such that $F_M \cong \mathbb{Z}$. There are two possibilities: $$F_M = \langle (1,1) \rangle \;\;\; or \;\;\; F_M = \langle (1,0) \rangle$$ Indeed, these are the only free modules of rank one in $\mathbb{Z} \times (\mathbb{Z}/2\mathbb{Z})$ which are isomorphic to $\mathbb{Z}$ and $F_M$ cannot be generated by two or more elements in $\mathbb{Z} \times (\mathbb{Z}/2\mathbb{Z})$. Otherwise, $F_M$ would be a free R-module of rank greater than or equal to 2 which is isomorphic to $\mathbb{Z}$, a free $R$ module of rank 1. We conclude that $M$ can be written in two ways: $$M = \langle (1,1) \rangle \oplus \langle (0,1) \rangle \;\;\; or \;\;\; M = \langle (1,0) \rangle \oplus \langle (0,1) \rangle$$ Now, let $N$ be the submodule of $M$ given by $N = \langle (2,1) \rangle$. Notice that $N$ is torsion free. Otherwise, there would exist $a \in \mathbb{Z} \setminus \big\{0\big\}$ and $(2a_1, a_1) \in N \setminus 0$, where $a_1\in R\setminus \big\{0\big\}$, such that $$a(2a_1,a_1) = (2aa_1,aa_1) = (0,0) \Rightarrow 2aa_1 = 0$$ Since $\mathbb{Z}$ is an integral domain, this implies $a= 0$ or $a_1 = 0$, a contradiction. Hence, by Theorem 6.5 of Hungerford, $N$ is free. However, $N$ is not contained in either of the two possibilities for the free part of $M$. For example, $(2,1) \in N$, but $$(2,1) \not \in \langle (1,1) \rangle \;\;\; and \;\;\; (2,1) \not \in \langle (1,0) \rangle$$ We have thus show that if $R$ is a principal ideal domain, M is a finitely generated R-module, and $N \leq M$, then it is not necessarily true that there exists free submodules $F_N \leq N$, $F_M \leq M$ with $N = F_N \oplus N_{tor}$, $M = F_M \oplus M_{tor}$, with $F_N \leq F_M$.