Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

share|improve this question
2  
It's always a good idea to look at simple examples first. What about $M=R$? –  Martin Brandenburg Sep 5 '13 at 16:41

2 Answers 2

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.

share|improve this answer
    
+1, excellent general explanation. –  Andreas Caranti Sep 5 '13 at 21:20
    
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. –  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 $(x_1, x_2, \dots )$.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.