Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
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? – Aug 23 '13 at 10:59

(I'm not sure if I'm aswering the question you actually want to ask, or whether it was already answered in the comments, but anyway)

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.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.