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Are free submodules of finitely generated modules finitely generated?

It feels like this should be true because it seems weird that a finitely generated module will have an infinite linearly independent subset, but I am unable to prove it.

Assume that the ring is commutative with unity.

bedune4
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  • In general a finitely gennerated module can have a submodule which is not finitely generated, in this case the ring must have to be non-Noetherian. For example $A=k[x_1, x_2, \cdots], M=(x_1, x_2, \cdots)$ (infinite variables, $k$ is a field). So I guess your question is whether a f.g. module contains a free submodule of infinite rank or not. Because if it is of finite rank, then it is already generated by finitely many elements. – Krish Sep 12 '17 at 13:27
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    @rschwieb Thank you. I edited the question to make the necessary changes. – bedune4 Sep 12 '17 at 13:29
  • Every non-noetherian ring is an example of a cyclic module with an infinitely generated submodule, so it seems like it might not be too big of a leap for one to exist where the infinitely generated submodule is free. But maybe something rules this out... – rschwieb Sep 12 '17 at 14:20
  • This was not asked, but I think it is nice to point out that this fails in the non-commutative case. If $F$ is a non-finitely generated free group and $\mathbb{Z}F$ is its integral group ring, then the (cyclic) module $\mathbb{Z}F$ contains a non-finitely generated free submodule, namely the augmentation ideal. – SFSH Feb 27 '24 at 17:11

1 Answers1

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If $R$ is a non-trivial commutative ring and $M$ is generated by $n$ elements then $M$ does not contain a free submodule of rank greater than $n$.

To see this, assume we have an injection $R^m \hookrightarrow M$. Since $M$ is finitely generated, there is a surjection $R^n \twoheadrightarrow M$. As $R^m$ is free, the map to $M$ factors through $R^n$ and we get an injection $R^m \hookrightarrow R^n$. As $R$ is nonzero commutative this implies $m \leq n$ (see, for example, here).

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