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Show by an example that an increasing union of finitely generated submodules of M need not be finitely generated.

I was thinking about $R[x_1,x_2,x_3,....]$. Then if we consider the ideal $<x_1>$ , does it form a submodule?

$f(x_1,x_2).g(x_1)\notin<x_1>$, where $f(x_1,x_2)\in R[x_1,x_2,x_3,....]$ and $g(x_1)\in <x_1>$.

So it is not closed under multiplication and hence is not a submodule of $R[x_1,x_2,x_3,....]$

Is this correct?

Can you help me to find an answer for this problem?

Germain
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  • Since $g(x_1)$ is in $<x_1>$, it is of the form $x_1$ times some polynomial in $R$-module $R[x_1, x_2,...]$ so that the $f(x_1,x_2). g(x_1)$ is in $<x_1>$ – Shreya Dec 01 '18 at 21:25

1 Answers1

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You don't quite say what $M$ is. I'm assuming it's a module over a commutative ring $R$. IF $M$ is required to be finitely generated, then in order to get any non-(finitely generated) submodule of $M$ we need $R$ to be non-Noetherian, and your example seems as simple as any.

On the other hand, you don't say that you need $M$ to be finitely generated. If you don't require finite generation of $M$, there are simpler examples: take $R = \mathbb{Z}$, $M = \mathbb{Q}$:

$\mathbb{Z} \subsetneq \frac{1}{2} \mathbb{Z} \subsetneq \frac{1}{6} \mathbb{Z} \subsetneq \ldots \subsetneq \frac{1}{n!} \mathbb{Z} \subsetneq \ldots \subset \mathbb{Q}$.

Pete L. Clark
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