Let $R$ a ring. I firstly thought that all submodule of a finitely generated module where finitely generated, but as I asked here, it's wrong. But is it true if a module is free ? In particular if $\displaystyle M\cong\bigoplus_{i=1}^n R$ and if $N\subsetneq M$, then $\displaystyle N\cong \bigoplus_{i=1}^m R$ for an $m<n$ ?

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    $\begingroup$ Your previous question was marked as a duplicate (in fact it was marked as a duplicate of a duplicate). Did you read the answer to the duplicate question? It answers this question as well (in fact the answer is pretty much the same as Alex's here). There were some votes to close this question as "off-topic" which I disagree with, but you really ought to have read the duplicate post more closely before posting a follow-up. $\endgroup$ – Ken Duna Jan 30 '17 at 15:33

No, this is not true. Take any non-Noetherian ring $R$. Then $R$ is a free $R$-module of rank $1$, but it has an ideal (i.e. submodule) which is not finitely generated.

For a specific example, choose your favorite field $k$ and take $R=k[x_1,x_2,\dots]$, and the ideal $I=(x_1,x_2,\dots)$.


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