Finitely generated module over $\mathbb Z$

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.

• 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. – Eric Wofsey Oct 24 '18 at 23:00
• That's a very good observation. Thanks, Eric! – Martin Argerami Oct 25 '18 at 0:44

1 Answer

[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.