Free $\mathbb{Z}$-submodule of $K$ of rank $n$ gives rise to an order

Question

Let $K$ be a number field and let $M$ be a free $\mathbb{Z}$-submodule of $K$ of rank $n=[K:\mathbb{Q}]$. I want to prove that $$ \mathfrak{o}=\{x\in K : xM\subset M\} $$ defines an order of $K$. That is, it is a subring of $K$ and a free $\mathbb{Z}$-module of rank $n$.

It is easy to see that $\mathfrak{o}$ is a subring of $K$. But how do I see that it is free of rank $n$? I think it would be useful to know that it is contained in the maximal order $\mathcal{O}_K$ because then I know $\mathfrak{o}$ has to be free of rank $\leq n$. Then it would remain to prove that it contains a $\mathbb{Q}$-basis of $K$ for example.

Answer 1

I was being blind. Just take some non-zero $\gamma \in M$. Then $\mathfrak{o}$ is contained in the finitely generated $\mathbb{Z}$-module $\frac{1}{\gamma} M$ and thus is itself finitely generated over $\mathbb{Z}$, which implies that $\mathfrak{o} \subset \mathcal{O}_K$.

If $m_1,\ldots,m_n$ denotes a $\mathbb{Z}$-basis of $M$ and if $\alpha_1,\ldots, \alpha_n$ is a $\mathbb{Q}$-basis of $K$, then, for each $k$, write $\alpha_k m_j$ as a $\mathbb{Q}$-linear combination of the $m_i$'s and multiply by the common denominator $c_k$ of the coefficients, proving that $c_k\alpha_k\in \mathfrak{o}$. Thus, $\mathfrak{o}$ contains the $\mathbb{Q}$-basis $c_1\alpha_1,\ldots, c_n\alpha_n$ of $K$.