The ring of multipliers is an order

$\newcommand{\O}{\mathcal{O}}$ I am trying to solve Exercise 12.3 at page 84 in Neukirch, "Algebraic Number Theory". The exercise is following:

Let $K$ be a number field of degree $n$. Let $\alpha_1, \dots, \alpha_n$ ($\alpha_i \in K$) be a basis of $K/\mathbb{Q}$. \begin{align*} M = \mathbb{Z}\alpha_1 + \dots + \mathbb{Z}\alpha_n, \\ \O = \{ \alpha \in K \mid \alpha M \subset M \}. \end{align*} Show that $\O$ is an order of $K$. ($\O$ is called a ring of multipliers.)

An order is a subring of $\O_K$ which is a free $\mathbb{Z}$-module of rank $n$. I tried to find a basis of $\O$, but I can't.

Hint. Show that there is $d>0$ such that $d\mathcal O_K\subseteq M\subseteq(1/d)\mathcal O_K$.
• $\newcommand{\O}{\mathcal{O}}$ Thanks. I found $d$ such that $dM \subset \O_K$, but I haven't found such that $d\O_K \subset M$ yet. However, it seems to be a expression about not $\O$ but $\O_K$. Can I learn about $\O$ from this expression? – nohm Oct 29 '15 at 9:26
• $a\in\mathcal O_K\implies(ad^2)M\subset (ad^2)(1/d)\mathcal O_K\subset d(a\mathcal O_K)\subset d\mathcal O_K\subset M\implies ad^2\in\mathcal O\implies d^2\mathcal O_K\subset\mathcal O$ – user26857 Oct 29 '15 at 9:55
• $\newcommand{\O}{\mathcal{O}}$ I see, and I found $d$ by using Lemma 2.9 of "ANT". Take $d'$ such that $\alpha_i' := d'\alpha_i \in \O_K$, let $d''$ be the discriminant of $\alpha_1', \dots, \alpha_n'$, then $d = d'd''$. Thanks. – nohm Oct 29 '15 at 12:00