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Let $A$ and $B$ be rings such that $A \subseteq B$ and let $M$ be a finitely generated $A$-module. Is it always the case that $M$ is also finitely generated as a $B$-module?

I ask because there is a characterisation of fractional ideals $F$ of the ring of integers of a number field $K$ which says that they are finitely generated $\mathcal{O}_K$-modules. If we then define the set $\tilde{F} = \lbrace \alpha \in K : \alpha F \subset \mathcal{O}_K\rbrace$, we are able to show that it is a finitely generated $\mathcal{O}_K$-module.

The source I'm reading makes this deduction by first showing that $\tilde{F}$ is finitely generated as a $\Bbb Z$-module, and hence also as an $\mathcal{O}_K$-module, which I presume is just because $\Bbb Z \subseteq \mathcal{O}_K$, and then the generators of $\tilde{F}$ as a $\Bbb Z$-module are just identified with their counterparts in $\mathcal{O}_K$.

Here is the particular quote which I'm having trouble understanding:

Pick a non-zero $y \in F$. Then $y\tilde{F}\subset \mathcal{O}_K$, so that $\tilde{F} \subset (1/y)\mathcal{O}_K$. Therefore, $\tilde{F}$ is a submodule of a finitely generated free $\Bbb Z$-module, so $\tilde{F}$ is a finitely generated $\Bbb Z$-module, hence finitely generated as an $\mathcal{O}_K$-module too.

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    $\begingroup$ Per se, $M $ doesn't have to be a $B $-module at all. But if you assume that it is, and that its $B $-module structure restricts to the given $A $-module structure, then the answer is "yes". $\endgroup$ – darij grinberg Dec 22 '17 at 16:14
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What if $M$ does not even have a $B$-module structure? A classical question in introduction courses is weather the module structure of the subring can be extended to a module structure on the whole ring, the answer to this is no.

For instance consider the finitely generated $\Bbb Z$-module $M=\Bbb Z / 4\Bbb Z$, this module structure can not be extended to a $\Bbb Q$-module structure, construct the element $ a=(1/2) \cdot 1 \in \Bbb Z / 4\Bbb Z$, this element has the property $a + a =1$, but no such element exists in $\Bbb Z / 4\Bbb Z$.

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  • $\begingroup$ How does the author of the quoted example deduce that $\tilde{F}$ must also be finitely generated as an $\mathcal{O}_K$-module then? Does it have to do with the fact that $\mathcal{O}_K$ itself is a free $\Bbb Z$-module? $\endgroup$ – ÍgjøgnumMeg Dec 22 '17 at 20:59

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