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Let $R$ be a [[ring]].
> [!proposition] Proposition. ([[integral domain is a field iff its finitely generated modules are all free]])
> If $R$ is an [[integral domain]], then it is a [[field]] if and only if every [[submodule generated by a subset|finitely generated]] $R$-[[module]] is [[free module|free]].
^proposition
> [!proof]- Proof. ([[integral domain is a field iff its finitely generated modules are all free]])
> ~
>
$\text{field}\implies \text{free}$ is just restating that [[every vector space has a basis|every vector space is free as a module]]. So we just need to show $\text{free} \implies \text{field}$.
>
Okay, suppose every finitely generated $R$-[[module]] is a [[free module]]. Then, in particular, every [[cyclic module]] over $R$ is a [[free module]]. Hence every [[cyclic module]] over $R$ is [[torsion element of a module|torsion-free]]. So we are done, by [[integral domain is a field iff its cyclic modules are torsion-free]].
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#### References
> [!backlink]
> ```dataview
> TABLE rows.file.link as "Further Reading"
> FROM [[]]
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> GROUP BY Tag
> ```
> [!frontlink]
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> ```