# Elements of the ring of multipliers are integral

For $$K$$ a number field, $$\{\alpha_1, \dots, \alpha_n\}$$ a basis of $$K/\mathbb{Q}$$ and $$M = \mathbb{Z}\alpha_1 + \dots + \mathbb{Z}\alpha_n$$, the corresponding ring of multipliers is defined as $$\mathcal{O} = \{ \alpha \in K : \alpha M \subseteq M \}.$$

I want to prove that this is an order in $$K$$, which requires proving the inclusion $$\mathcal{O} \subseteq \mathcal{O}_K$$. I feel like I am missing something obvious here as I do not understand why this holds. I also have a hard time understanding what $$\alpha$$ are selected by the property $$\alpha M \subseteq M$$, and how $$M$$ and $$\mathcal{O}$$ are related in general, so any help about this is appreciated.

I found this related question, which gives the inclusion $$d \mathcal{O}_K \subseteq \mathcal{O}$$ for some $$d$$ in $$\mathbb{Z}$$ so the proof is complete after that. This is also the only relevant result I found with the term "ring of multipliers" so could it be that this is not the proper terminology?

By assumption, for all $$i$$, $$\alpha\alpha_i=\displaystyle\sum_{j=1}^n a_{ij}\alpha_j,$$ for some $$a_{ij}\in\mathbb{Z}$$.
Set $$A=(a_{ij})$$ and $$v= (\alpha_1 \ \cdots \ \alpha_n)^t$$, so that $$Av=\alpha v$$. In particular, $$\det(\alpha I_n-A)=0$$. Now $$\det(X I_n-A)$$ is a monic polynomial with integer coefficients for which $$\alpha$$ is a root.