# Positive system of algebraic integers

Let $$\mathbb{A}$$ be the ring of algebraic integers. Consider a sequence $$(d_i)_{i \in I}$$, with $$I$$ a finite set and $$d_i \in \mathbb{A} \cap \mathbb{R}_{\ge 1}$$, such that $$d_i d_j = \sum_{k \in I} n_{i,j,k} d_k,$$ for all $$i,j \in I$$, with $$n_{i,j,k} \in \mathbb{Z}_{\ge 0}$$.

A subset $$J \subset I$$ is called a subsystem if $$\forall i,j \in J$$ and $$\forall k \not \in J$$ then $$n_{i,j,k} = 0$$. Let $$\langle i \rangle$$ be the smallest subsystem containing $$i$$.

Question: Let $$i,j \in I$$ such that $$\langle i \rangle = \langle j \rangle$$. Is it true that $$\mathbb{Z}[d_i] = \mathbb{Z}[d_j]$$?

Observation: The vector $$v = (d_k)_{k \in I}$$ is a common eigenvector for the matrices $$M_i = (n_{i,j,k})_{j,k \in I}$$, with eigenvalue $$d_i$$, because: $$M_i v = (\sum_{k \in I} n_{i,j,k} d_k)_{j \in I} = (d_i d_j)_{j \in I} = d_i v.$$ By Frobenius-Perron Theorem (stated below), the eigenvalue $$d_i$$ of $$M_i$$ is its spectral radius.

Frobenius-Perron Theorem: A square matrix $$M$$ with nonnegative real entries has a non-negative real eigenvalue. If moreover $$M$$ has an eigenvector $$v$$ with strictly positive entries, then the eigenvalue of $$v$$ is the largest non-negative real eigenvalue and is the spectral radius of $$M$$.

No, here are infinitely many counter-examples for $$I = \{1,2\}$$.

Let $$n,m \in \mathbb{Z}_{\ge 1}$$ such that $$m | n^2$$, $$n | 2m^2$$ and $$n \neq 2m$$ (e.g. $$n=m=1$$).

Take $$d_1 = n(1+\sqrt{2})$$, $$d_2 = m(3+2\sqrt{2})$$. Here are the matrices $$M_i = (n_{i,j,k})_{j,k \in I}$$:

$$\left(\begin{matrix}0&\frac{n^2}{m}\\m&2n \end{matrix}\right), \ \left(\begin{matrix}m&2n\\\frac{2m^2}{n}&5m\end{matrix}\right)$$

Now $$\langle 1 \rangle = \langle 2 \rangle = I$$, but $$\mathbb{Z}[d_1] = \mathbb{Z}[n \sqrt{2}] \neq \mathbb{Z}[2m \sqrt{2}] = \mathbb{Z}[d_2]$$, because $$n \neq 2m$$.

Here is a counter-example which is also a fusion ring (the initial motivation), with $$I = \{1,2,3\}$$.

Take $$d_1 = 1$$, $$d_2 = 3+2\sqrt{2}$$, $$d_3 = 4+3\sqrt{2}$$. Here are the matrices $$(M_i)$$:

$$\left(\begin{matrix}1&0&0\\0&1&0\\0&0&1 \end{matrix}\right), \ \left(\begin{matrix}0&1&0\\1&0&4\\0&4&3 \end{matrix}\right), \ \left(\begin{matrix}0&0&1\\0&4&3\\1&3&6 \end{matrix}\right)$$

Now $$\langle 2 \rangle = \langle 3 \rangle = I$$, but $$\mathbb{Z}[d_2] = \mathbb{Z}[2\sqrt{2}] \neq \mathbb{Z}[3\sqrt{2}] = \mathbb{Z}[d_3]$$.

For the notion of fusion ring, see Definition 3.1.7 in the following reference:
P. Etingof, S. Gelaki, D. Nikshych, V. Ostrik; Tensor categories. Mathematical Surveys and Monographs (2015) 205.

Conclusion: We should replace $$\mathbb{Z}[d_i] = \mathbb{Z}[d_j]$$ by $$\mathbb{Q}(d_i) = \mathbb{Q}(d_j)$$ in the question.