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This question already has an answer here:

Let $R$ be a commutative ring with identity. Then $R^n$ ($n$ is some positive integer) is a module over $R$ in a natural way. It is free and finitely generated. Now let $M$ be a finitely generated submodule of $R^n$. Does it follow that $M$ is free? My intuition leads me to believe that this is true, but I am not too sure.

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marked as duplicate by Prism, Dietrich Burde, Siong Thye Goh, Nosrati, Dando18 Nov 8 '17 at 23:23

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This is false in general. For instance, in the polynomial ring $R=K[X,Y]$ ($K$ a field), the ideal $(X,Y)$ is a submodule of the free module $R$ which is not free since $ \{X,Y\}$ is a minimal set of generators, but the satisfy the linear relation $YX-XY=0$.

However this property is true for finitely generated free modules over a P.I.D.

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