Assume $A$ is a nontrival submodule of a free $R $-module, where $R $ is a ring.
Is it true that $A $ is also free?
2026-03-29 10:07:49.1774778869
Submodule of a free module
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All that can really be said in general is that if $R$ is an integral domain then submodules of free modules are torsion-free. On the other hand, every projective module occurs as a submodule (even a direct summand) of a free module, so to find a counterexample it suffices to find any ring $R$ and any projective module over it which is not free. A nice example is a non-principal ideal $I$ of a Dedekind domain with nontrivial class group.
A simpler example of a non-free submodule of a free module is the ideal $I = (x, y)$ of the polynomial ring $k[x, y]$, where $k$ is a field. $I$ cannot be generated by one element (exercise), and two or more elements must be linearly dependent (exercise), so $I$ cannot be free on any number of generators.