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Surely it is true? But somehow I can't think of a way to prove it. Let's assume the ring is commutative with identity.

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Every submodule of a free module of finite rank is finitely generated if and only if the ring is noetherian. – Pierre-Yves Gaillard Sep 15 '11 at 14:13
up vote 10 down vote accepted

Take $R = \mathbb{F_p}[X_i:i\in \omega]$ and $M = (X_i: i\in \omega)$. Then $R$ is free but $M$ is not finitely generated.

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Consider $R=(\mathbb{Z}_p)^{\mathbb{N}}$ the product of ring of $\mathbb{Z}_p$ as many times as the naturals, now $M=(\mathbb{Z}_p)^{(\mathbb{N})}$ the coproduct of $\mathbb{Z}_p$ as many times as the naturals. This counter example works for any $K$ field, but is easy to see when p=2. R is free of rank 1.

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Now I realize that it had better be NOT true, otherwise there would be no need to talk about finitely presented modules.

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