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.



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> 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 $d$?
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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.



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> Is it possible that in this situation the centralizer algebra is non commutative in characteristic zero, but commutative after reduction modulo a prime?
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Is it possible for the reduction modulo $p$ of an non-commutative semisimple algebra to be commutative?
Example 5.10 of


*Towers, Matthew*, [**Endomorphism algebras of transitive permutation modules for $p$-groups.**](http://dx.doi.org/10.1007/s00013-009-3083-8), Arch. Math. 92, No. 3, 215-227 (2009)


(whose author you might know) gives a positive answer to the second question. The group action involved is of
$$G=\langle x,y,x\mid x^4, y^4, z^4, [x,y]=z\in Z(G)\rangle$$
(which has order $64$) acting on the cosets of the Klein four subgroup generated by $x^2$ and $y^2$,


