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](https://en.wikipedia.org/wiki/Spectral_radius).


*[Frobenius-Perron Theorem](https://en.wikipedia.org/wiki/Perron%E2%80%93Frobenius_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 $M$.



Positive system of algebraic integers
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 $n \neq 2m$.




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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.




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*Conclusion*: We should replace $\mathbb{Z}[d\_i] = \mathbb{Z}[d\_j]$ by $\mathbb{Q}(d\_i) = \mathbb{Q}(d\_j)$ in the question.


