$R$ is a commutative ring with unit and $M$ is a finitely generated free $R$-module. Is the submodule of $M$ also finitely generated?


marked as duplicate by user26857, Daniel Fischer Oct 10 '16 at 17:52

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    $\begingroup$ It's always a good idea to look at simple examples first. What about $M=R$? $\endgroup$ – Martin Brandenburg Sep 5 '13 at 16:41

If this were true, we might look at any ring $R$, which is free and generated by $1$ as an $R$ module, and conclude that all its submodules were finitely generated. Thus we would have proven all commutative rings are Noetherian! (If you are unfamiliar with that characterization of Noetherian modules, check out this question)

So you can see that it is not true, since there are non-Noetherian rings.

Andreas's example is a concrete illustration of this: he gave an example of a non-Noetherian ring with an ideal that isn't finitely generated.

  • $\begingroup$ In fact, all f.g. $R$ modules are Noetherian $R$-modules if and only if $R$ is Noetherian. You have shown the only if part, and the other part is (slightly) more involved. $\endgroup$ – Alex Youcis Sep 8 '13 at 3:38

Think of the case when $R$ is the ring of polynomials in countably many variables $x_1, x_2, \dots$ over a field, $M = R$, and take as a submodule the ideal $(x_1, x_2, \dots )$.


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