TITLE: Is every submodule of a free module of finite rank finitely generated?
QUESTION [6 upvotes]: 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.

REPLY [5 votes]: 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.