Elements of the ring of multipliers are integral

Question

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?

GreginGre · Answer

By assumption, for all $i$, $alphaalpha_i=displaystylesum_{j=1}^n a_{ij}alpha_j,$ for some $a_{ij}inmathbb{Z}$.
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.

Elements of the ring of multipliers are integral

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