Let $R$ be a (commutative, otherwise the answer is easy, see the comment below) ring and let $M$ be a finitely generated $R$-module. Is it possible that $M$ admits an infinite linearly independent set? [Cardinality of maximal linearly independent subset](https://mathoverflow.net/questions/30066/cardinality-of-maximal-linearly-independent-subset) seems relevant, but it does not give an answer to my question. Thank you!



Infinite linearly independent set in finitely generated module
In a commutative ring $R$ this does not exist. Better for any $n\ge 0$, if $M$ is an $R$-module generated by $n$ elements, then $R^{n+1}$ doesn't embed into $M$.


Indeed, lifting if necessary, we can suppose $M=R^n$. So we get an $n\times (n+1)$ matrix $u$ over $R$ defining an injective operator $R^{n+1}\to R^n$. Let $B$ be the unital subring generated by entries of $u$. Then $B$ is finitely generated commutative, hence noetherian, and $u$ defines an injective operator $B^{n+1}\to B^n$, contradiction.


