Maximal free submodule over a PID

Question

Let M be a f.g. module over a p.i.d. and T(M) be its torsion submodule. Then M is the direct sum of T(M) and of a free submodule F, unique up to isomorphism, and in addition which is a maximal free submodule. Conversely, let F be a maximal free submodule of M (which is a stronger condition than being of maximal rank). Is it true that F is a direct summand, so that M is the direct sum of T(M) and F?

Answer 1

The claim is false in general, as we shall illustrate below by a counterexample. Before moving on two key observations are needed:

1. as I passingly mentioned in a comment to the original question, there is identity between the collection of submodules of $M$ which intersect $\mathrm{T}(M)$ trivially and the collection of free submodules. Clearly, if $N \leqslant_A M$ is free then $N \cap \mathrm{T}(M)=\mathrm{T}(N)=\{0_M\}$, since free modules over (commutative) integral domains are torsionless. On the other hand, by the same token any submodule $N$ that intersects the torsion submodule trivially must itself be torsionless. $M$ being finitely generated over the Noetherian ring $A$ means that $M$ itself is a Noetherian module, with the consequence that in particular $N$ is finitely generated. Since any torsionless finitely generated $A$-module is free, $N$ is in particular free. The problem can therefore be reformulated as analysing whether any submodule $N$ maximal among those which intersect $\mathrm{T}(M)$ trivially is a direct summand or not.
2. assuming that the submodule $L \leqslant_A M$ is maximal free and at the same time a direct summand, let us remark that it has a unique supplementary, namely the torsion submodule. Indeed, consider an arbitrary supplementary $P$ of $L$. Without proof we mention a very general result:

Lemma. Let $A$ be an arbitrary ring, $S \leqslant_{\mathrm{Mon}} \mathrm{Z}(A)$ a central multiplicative system of $A$, $M$ a left $A$-module, $I$ an arbitrary index set and $N \in \mathscr{S}_{\operatorname{\mathit{A}-\mathrm{Mod}}}(M)^I$ a family of submodules whose (internal) sum is direct. Given any family $P \in \displaystyle\prod_{i \in I}\mathscr{S}_{\operatorname{\mathit{A}-\mathrm{Mod}}}(N_i)$ of submodules, the following relation holds between $S$-saturations: $$\mathrm{Sat}_S\left(\sum_{i \in I}P_i\right)=\sum_{i \in I}\left(\mathrm{Sat}_S(P_i) \cap N_i\right)$$

and apply this result to the particular case of submodules $\{0_M\} \leqslant_A L$ respectively $\{0_M\} \leqslant_A P$ and multiplicative system $A^{\times}\colon=A \setminus \{0_A\}$, yielding the conclusion that $\mathrm{T}(M)=\mathrm{Sat}_{A^{\times}}\left(\{0_M\}\right)=\left(\mathrm{Sat}_{A^{\times}}\left(\{0_M\}\right) \cap L\right)+\left(\mathrm{Sat}_{A^{\times}}\left(\{0_M\}\right) \cap P\right)=\mathrm{T}(L)+\mathrm{T}(P)=\mathrm{T}(P)$. Since $P$ is finitely generated over a principal ideal domain, its torsion submodule must admit a (free) supplementary $Q$ in $P$, and it follows that $L+Q$ and $\mathrm{T}(P)=\mathrm{T}(M)$ are supplementary in $M$. By maximality of $L$, it further follows that $Q \subseteq L+Q=L$, whence $Q \subseteq L \cap P=\{0_M\}$ and thus finally $P=\mathrm{T}(M)$.

Now on to the promised counterexample: consider $A=\mathbb{Z}$ and $M=\mathbb{Z} \times \mathbb{Z}_4$, where I am using the abbreviation $\mathbb{Z}_r\colon=\mathbb{Z}/r\mathbb{Z}$ for any $r \in \mathbb{N}$. It is clear that $\mathrm{T}(M)=\{0\} \times \mathbb{Z}_4$. We argue that the submodule $N\colon=\mathbb{Z}\left(2, \overline{1}\right)$ is maximal free yet not a direct summand (the bar notation $\overline{m}$ denotes the residue class of $m \in \mathbb{Z}$ modulo $4$). The condition $N \cap \mathrm{T}(M)=\{0_M\}$ is easily verified. Consider a submodule $P \leqslant_{\mathbb{Z}} M$ such that $N \leqslant_{\mathbb{Z}} P$ and $P \cap \mathrm{T}(M)=\{0_M\}$. If $p \colon \mathbb{Z} \times \mathbb{Z}_4 \to \mathbb{Z}$ denotes the canonical projection onto the left factor, we have $2\mathbb{Z}=p[N] \leqslant_{\mathbb{Z}} p[P] \leqslant_{\mathbb{Z}} \mathbb{Z}$, which gives the only options $p[P]=\mathbb{Z}$ or $p[P]=2\mathbb{Z}$. The former option leads to a contradiction, since $1 \in p[P]$ meant there existed $a \in \mathbb{Z}_4$ such that $(1, a) \in P$ and therefore $(2, 2a) \in P$. Since we also have $\left(2, \overline{1}\right) \in N \subseteq P$, it follows that $0_M \neq \left(0, 2a-\overline{1}\right) \in P \cap \mathrm{T}(M)$, which constitutes the contradiction (the fact that $\overline{1} \notin 2\mathbb{Z}_4=\left\{\overline{0}, \overline{2}\right\}$ is crucial here). We thus necessarily have $p[P]=2\mathbb{Z}$, so for any $x \in P$ there exist $n \in \mathbb{Z}$ and $t \in \mathbb{Z}_4$ such that $x=(2n, t)$. As we also have $n\left(2, \overline{1}\right)=\left(2n, \overline{n}\right) \in P$, we infer that $\left(0, t-\overline{n}\right) \in P \cap \mathrm{T}(M)=\{0_M\}$ whence $t=\overline{n}$ and thus $x \in N$. This proves the equality $P=N$ and establishes $N$ as maximal among those submodules which intersect $\mathrm{T}(M)$ trivially. However, the second observation tells us that $N$ cannot be a direct summand: indeed, if this were the case it would necessarily follow that $N+\mathrm{T}(M)=M$, whereas in our case $N+\mathrm{T}(M)=2\mathbb{Z} \times \mathbb{Z}_4<_{\mathbb{Z}} M$ is a proper submodule.

Answer 2

Consider $M=\mathbb{Z}$ over $\mathbb{Z}$ and $F=2\mathbb{Z}$, the sequence $0\rightarrow 2\mathbb{Z}\rightarrow\mathbb{Z}\rightarrow\mathbb{Z}/2\mathbb{Z}\rightarrow 0$ does not split and $2\mathbb{Z}$ is maximal.