Suppose that $I, X_1, \ldots, X_{d-1}$ are $n \times n$ matrices with integer entries whose $\mathbb{Z}$-span is a subalgebra of $\mathrm{Mat}_n(\mathbb{Z})$. Suppose that, thought of as a subalgebra of $\mathrm{Mat}_n(\mathbb{C})$, this algebra is semisimple and non-commutative. Thus, by Wedderburn's Theorem, it is isomorphic to a direct product of complete matrix algebras $\mathrm{Mat}_r(\mathbb{C})$, with $r \ge 2$ for at least one factor.
It is possible that there exists a prime $p$ and a field $K$ of characteristic $p$ such that, regarding $I, X_1, \ldots, X_{d-1}$ as elements of $\mathrm{Mat}_n(K)$ by reduction modulo $p$, the subalgebra of $\mathrm{Mat}_n(K)$ spanned by $I, X_1, \ldots, X_{d-1}$ is commutative, and still of dimension $d$?
As a follow-up, in my specific setup, the matrices are the orbital matrices for the action of a finite group $G$ on a set $\Omega$: the orbital matrix with $1$ in position $(\alpha,\beta)$ has $1$s exactly in the positions $(\alpha g, \beta g)$, for $g \in G$. It is known that these matrices span the centralizer algebra of the permutation module (over $\mathbb{Z}$ or any field). Moreover, the algebra is commutative in characteristic zero if and only if the associated permutation character $\pi(g) = |\mathrm{Fix}_\Omega(g)|$ is multiplicity-free.
Is it possible that in this situation the centralizer algebra is non commutative in characteristic zero, but commutative after reduction modulo a prime?
