# Linear Algebra over a Ring/Modules over principal ideal domains

Proposition (torsion-free modules over principal ideal domains are free):

Let $R$ be a principal ideal domain, and let $M$ be a torsion-free module over $R$ . Then $M$ is free.

(On the condition of the axiom of choice.)

Proof: We consider the set of sets $S\subseteq M$ such that $S$ is linearly independent and $M/\langle S\rangle$ is torsion-free. This set may be equipped with the partial order that is given by inclusion. Suppose then that $A$ is a totally ordered set, and $(S_{\alpha })_{\alpha \in A}$ is a family such that $\alpha \leq \beta \Leftrightarrow S_{\alpha }\subseteq S_{\beta }$ . We claim that an upper bound for this chain is given by the union of the sets $S_{\alpha }$ , which we shall denote by $S_{\infty }$ . Indeed, $S_{\infty }$ is linearly independent, since any linear relation within $S_{\infty }$ involves only finitely many elements of $S_{\infty }$ , and we may find a sufficiently large (w.r.t. the order of $A$ ) $\alpha \in A$ such that all these elements are contained within $S_{\alpha }$ , so that by the linear independence of $S_{\alpha }$ the given linear relation must be trivial. Moreover, $S_{\infty }$ has the property that $M/\langle S_{\infty }\rangle$ is torsion-free, since if $m\in M$ and $r\in R$ are given such that $m\notin S_{\infty }$ , but $rm\in S_{\infty }$ (ie. the equivalence class of $m$ in $M/\langle S_{\infty }\rangle$ is torsion), then $rm$ is a linear combination of finitely many elements of $S_{\infty }$ , so that once more we find a sufficiently large $\alpha \in A$ such that $rm\in S_{\alpha }$ , and then the equivalence class of $m$ in $R/\langle S_{\alpha }\rangle$ is torsion, a contradiction.

Thus, Zorn's lemma may be applied, and it yields a maximal linearly independent $S\subseteq M$ such that $M/\langle S\rangle$ is torsion-free. We lead the assumption $\langle S\rangle \neq M$ to a contradiction. Indeed, if we had $\langle S\rangle \neq M$ , then there would be an element $m\in M\setminus \langle S\rangle$ . The set $S\cup \{m\}$ will then be linearly independent, for if there was a linear relation

$0=rm+r_{1}s_{1}+\cdots +r_{n}s_{n}$ (where $r,r_{1},\ldots ,r_{n}\in R$ and $s_{1},\ldots ,s_{n}\in S$ ),

then we would have $(-r)m=r_{1}s_{1}+\cdots +r_{n}s_{n}\in \langle S\rangle$ and the equivalence class of $m$ in $M/\langle S\rangle$ would be torsion. $\Box$ Theorem (Dedekind's theorem):

Let $R$ be a principal ideal domain. Whenever $M$ is a free $R$ -module and $N\leq M$ is a submodule, $N$ is free as well.

(On the condition of the axiom of choice.)

Proof: Since $N$ is a submodule of a torsion-free module, it is itself torsion-free. Thus, the theorem holds, since torsion-free modules over principal ideal domains are free. $\Box$ 