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

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

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

(On the condition of the axiom of choice.)

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

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

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

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

Theorem (Dedekind's theorem):

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

(On the condition of the axiom of choice.)

Proof: Since ${\displaystyle N}$ 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. ${\displaystyle \Box }$