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In one of our exercise classes, we had the following task:

Let $K$ be a number field of degree $n$ with ring of integers $O_K$. Let $M$ be a finite $\mathbb{Z}$-submodule of $K$ of rank $n$. Let $\alpha_1, \dotsc, \alpha_n$ be a $\mathbb{Z}$-basis of $M$. Show $d(\alpha_1, \dotsc, \alpha_n) \in \mathbb{Q}$.

In our solution we used the fact that $M$ as a finite $\mathbb{Z}$-submodule of $K$ is free and hence admits an integral basis, i.e. $\alpha_1, \dotsc, \alpha_n$ as a $\mathbb{Z}$-basis of $M$.

In the next step, the claim was that we immediately get that $\alpha_1, \dotsc, \alpha_n$ is a $\mathbb{Q}$-basis of $K$. I do understand that by a looking at the dimension of $M$ and $K$, we know that $\alpha_1, \dotsc, \alpha_n$ can be a generating set of $K$.

However, I do not understand how I can conclude that those $\alpha_i$ are linearly independent over $\mathbb{Q}$? (Of course, from $\alpha_1, \dotsc, \alpha_n$ a $\mathbb{Q}$-basis of $K$, you then get that $d(\alpha_1, \dotsc, \alpha_n) \in \mathbb{Q}$.)

Moreover, we later on in another exercise used that $e_1, \dotsc, e_n$ as a $\mathbb{Z}$-basis of $O_K$ is also a $\mathbb{Q}$-basis of $K$, which seems to follow the same line of argument...

Thank you for your answers!

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    You wrote "...a$\;M\;$ be a **finite** submodule ...." . Did you mean "finitely generated"? Because if you did then it is unnecessary as you also say it is of rank $\;n\;$ . Perhaps you meant *free submodule* ...? – 2017-01-25
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    Let $\sum_i \frac{p_i}{q_i} \alpha_i = 0$. Then, if you multiply by $q_1 \cdots q_n$ you get a linear combination with integer coefficients. Since The $\alpha_i$s are free over $\Bbb{Z}$, all coefficients are $0$. – 2017-01-25
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    @DonAntonio In the task it actually says that it's a finite submodule of rank $n$. But I assume you're right by hinting at finitely generated as we in the following used that $M$ is a free submodule :) – 2017-01-25
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    @Crostul thank you - that was exactly was I was looking for! – 2017-01-25

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