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

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

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

**Proof:** We consider the set of sets such that is linearly independent and is torsion-free. This set may be equipped with the partial order that is given by inclusion. Suppose then that is a totally ordered set, and is a family such that . We claim that an upper bound for this chain is given by the union of the sets , which we shall denote by . Indeed, is linearly independent, since any linear relation within involves only finitely many elements of , and we may find a sufficiently large (w.r.t. the order of ) such that all these elements are contained within , so that by the linear independence of the given linear relation must be trivial. Moreover, has the property that is torsion-free, since if and are given such that , but (ie. the equivalence class of in is torsion), then is a linear combination of finitely many elements of , so that once more we find a sufficiently large such that , and then the equivalence class of in is torsion, a contradiction.

Thus, Zorn's lemma may be applied, and it yields a maximal linearly independent such that is torsion-free. We lead the assumption to a contradiction. Indeed, if we had , then there would be an element . The set will then be linearly independent, for if there was a linear relation

- (where and ),

then we would have and the equivalence class of in would be torsion.

**Theorem (Dedekind's theorem)**:

Let be a principal ideal domain. Whenever is a free -module and is a submodule, is free as well.

**Proof:** Since 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.