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 $M$.