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Can someone give an example of a ring $R$, a left $R$-module $M$ and a submodule $N$ of $M$ such that $M$ is finitely generated, but $N$ is not finitely generated?

I tried a couple of examples of modules I know and got nothing...

Watson
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Belgi
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1 Answers1

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Consider the simplest possible nontrivial (left) $R$-module: $R$ itself. It's certainly finitely generated, by $\{ 1 \}$. The submodules are exactly the (left) ideals of $R$. So you want a ring which has (left) ideals which are not finitely generated. For example, you could use a non-Noetherian commutative ring, such as $\mathbb{Z}[X_1, X_2, X_3, \ldots ]$.

Chris Eagle
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    Even more is true: A ring $R$ is noetherian if and only if submodules of f.g. modules over $R$ are f.g. – Martin Brandenburg Mar 27 '12 at 10:46
  • But wouldn't the ideal be finitely generates as well? I am sorry I know this is a basic question but I am stuck at this point. So for the given example, $R=\mathbb{Z}[X_1,X_2, \cdots]$, one of the ideals could be the polynomials of $I=\mathbb{Z}[X_1]$. This can also be generated by ${1}$. I know that for all the polynomials in this ideal, we need countably many different powers of $X_1$. Is that why it is not finitely generated? – PythonSage Oct 10 '22 at 09:04
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    @PythonSage Check out my answer here. I have written out almost every detail and it is as clear as crystal. – Sam Wong Dec 29 '22 at 14:27