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?


1 Answer 1


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.


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