# Eigenvalues of Multiplication in algebraic number field

Finally I have a new question which is puzzling me. I am currently reading a manuscript so there is no reference or link I can provide. I will try to put all the information needed in here. This implies that I hear several definitions for the first time while reading, which on the other hand is assumed and unless I provide references the idea is that everything needed should be obtainable from what is mentioned within the post.

Question 1: (too general)

Let $\mathbb{K}$ be an algebraic number field, i.e. $\mathbb{Q}\hookrightarrow\mathbb{K}$ with $[\mathbb{K}:\mathbb{Q}]=:d<\infty$. Let $b\in\mathbb{K}$ and define $\phi(b)(x):=bx$ for all $x\in\mathbb{K}$. Then $\phi(b)\in\operatorname{End}(\mathbb{Q}^{[\mathbb{K}:\mathbb{Q}]})$. What are the Eigenvalues of $\phi(b)$ (i.e. the roots of the characteristic polynomial)?

What have I got:

As $\mathbb{K}$ is an algebraic number field, we know that $\mathbb{K}\cong\mathbb{Q}(\alpha)$ for some number algebraic over $\mathbb{Q}$ and that for $m\in\mathbb{Q}[z]$ the minimal polynomial of $\alpha$. Letting $\zeta_{1},\ldots,\zeta_{r}$ the real roots of $m$ and $\zeta_{r+1},\overline{\zeta_{r+1}},\ldots,\zeta_{r+s},\overline{\zeta_{r+s}}$ the complex roots of $m$, we know that $\alpha$ is an annihilator of the characteristic polynomial of $\phi(\alpha)$ and hence the minimal polynomial divides the characteristic polynomial and hence $m$ being separable over $\mathbb{Q}$ is a $\mathbb{Q}$-multiple of the charcteristic polynomial for reasons of dimension. This argument holds for all roots of $m$.

What is the issue:

In the manuscript at hand we start off with an element $\zeta\in\mathbb{Q}(\alpha)$ (or $\mathbb{K}$ - as you prefer) satisfying some more restraints:

There is a subring $\mathcal{O}\subseteq \mathbb{Q}(\alpha)$ such that as an additive group $\mathcal{O}\cong\mathbb{Z}^{d}$ (as $d=[\mathbb{Q}(\alpha):\mathbb{Q}]$ and the base field is $\mathbb{Q}$, this implies that $\mathcal{O}$ is a lattice in the $\mathbb{Q}$-vector space $\mathbb{Q}(\alpha)$).

$\mathcal{O}$ is called an order - I am not sure whether this agrees with the standard definition of an order (one sometimes sees the requirement that $1\in\mathcal{O}$ which doesn't follow automatically from this definition). Now the manuscript claims that for $b\in\mathcal{O}$ the eigenvalues of $\phi(b)$ are also given by the roots of $m$. I am not sure but this seems odd to me. If we let $b=0$ then the eigenvalues are all zero. If $1\in\mathcal{O}$, then for $b=1$ the eigenvalues are all 1. Can we say something about the group of units, i.e. if $b\in\mathcal{O}^{\times}=\{o\in\mathcal{O};\exists p\in\mathcal{O}:op=1\}$?

Question 2:

Is there a way to identify the elements $b\in\mathcal{O}$ such that $\phi(b)$ has the roots of $m$ as eigenvalues?

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BTW, a subring is required to contain $1$. –  awllower Aug 23 '13 at 10:49
That depends on whether a ring is required to be unital, no? –  M. Luethi Aug 23 '13 at 10:50
I see. So that terminology might not be standard. And I cannot see why this is an issue: eigen-values are roots of the minimal polynomial, as required? Maybe I mis-understood something. –  awllower Aug 23 '13 at 10:51
The discussion is about "the roots of which minimal polynomial". The statement is that they are the roots of the polynomial of $\alpha$. This is not true for every linear map of course. –  M. Luethi Aug 23 '13 at 10:55
Orders are usually considered as subrings of non-commutative algebras, say matrix algebras $M_{n}(\mathbb{Q})$. I think you are talking about the ring of integers of $\mathbb{Q}(\alpha)$ and its ideals? –  plusepsilon.de Aug 23 '13 at 10:59

If $b\in K$, if $f\in\mathbb Q[x]$ is the minimal polynomial of $b$ and if $n=[K:\mathbb Q(b)]$, then $\phi=f^n$. There is not much to prove: if $n=1$ you can take $1,b,b^2,\dots, b^{k-1}$ as a basis of $K$ ($k=\deg f$) and you get the standard matrix with char. polynomial $f$. For a general $n$, choose a basis $a_1,\dots,a_n$ of $K$ over $\mathbb Q(b)$, so that $b^ta_s$ is a basis of $K$ over $\mathbb Q$, and the matrix of $\phi$ is block-diagonal, with the same block repeating $n$-times. There are nicer proofs (my favorite is: $K\otimes_{\mathbb Q}\mathbb C\cong\mathbb C\oplus\mathbb C\oplus\dots\oplus\mathbb C$ as a $\mathbb C$-algebra, and the inclusion $K\to K\otimes_{\mathbb Q}\mathbb C$ is via all the embeddings of $K$ to $\mathbb C$), but this one will do.