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Let $\alpha,\beta\in\mathbb C$ be algebraic integers, so there exist monic $p,q\in\mathbb Z[x]$ such that $p(\alpha)=q(\beta)=0$. It follows that $\mathbb Z[\alpha,\beta]$ is finitely-generated as a $\mathbb Z$-module.

I want to show directly that any submodule of $\mathbb Z[\alpha,\beta]$ is finitely generated.

I'm aware that the result is true in general, since $\mathbb Z$ is a PID and all submodules of a finitely generated module over a PID are finitely generated. But I'm curious if there is a particular direct way in the above case, without going through the general argument.

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    $\begingroup$ When $\beta=0$ this amounts to proving that any submodule of $\mathbb{Z}^n$ is finitely generated (where $n$ is the degree of the minimal polynomial of $\alpha$). If you know that, it follows immediately that a submodule of a finitely generated module is finitely generated, by writing your finitely generated module as a quotient of $\mathbb{Z}^n$ for some $n$. So, it seems that your special case should be just as difficult as the general case. $\endgroup$ – Eric Wofsey Oct 24 '18 at 23:00
  • $\begingroup$ That's a very good observation. Thanks, Eric! $\endgroup$ – Martin Argerami Oct 25 '18 at 0:44
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[Converting my comment into an answer.]

When $\beta=0$ this amounts to proving that any submodule of $\mathbb{Z}^n$ is finitely generated (where $n$ is the degree of the minimal polynomial of $\alpha$). If you know that, it follows immediately that a submodule of a finitely generated $\mathbb{Z}$-module is finitely generated, by writing your finitely generated module as a quotient of $\mathbb{Z}^n$ for some $n$.

So, if you had a simple proof of your special case, you could very easily get a similarly simple proof that every submodule of a finitely generated $\mathbb{Z}$-module is finitely generated. As a result, I would not expect any easier proof to exist in your special case.

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