Suppose $A,B\in SL(3,F_q)$, where $F_q$ is the finite field of order $q$ and $SL(3,F_q)$, the group of matrices with determinant one and entries from $F_q$ , are such that $A$ has eigenvalues in $F_q$ and $B$ has eigenvalues in $\overline{F_q}\setminus F_q$. Also, $A$ is diagonalizable over $F_q$ and $B$ is diagonalizable over $\overline{F_q}$, the closure of $F_q$. I am trying to show that $A$ and $B$ are not simultaneously diagonalizable by a matrix $P\in SL(3,\overline{F_q})$ (i.e., $\nexists P\in SL(3,\overline{F_q})$ such that $(PAP^{1},PBP^{1})$ are diagonal). I am considering the approach of looking at the $F_q$ algebra generated by $A$ and $B$ and trying to show it is not isomorphic to the algebra generated by $F_q[PAP^{1},PBP^{1}]$. I am looking for some reference which might be helpful in proving the above. Most of the results I saw had been about irreducible representations which is not helpful for my case. I would appreciate it if you could suggest some reference that could be helpful. Thanks in advance for your time.

4$\begingroup$ What about the case where $A$ is the identity matrix? Then they clearly are simultaneously diagonalizable. If $A$ has distinct eigenvalues then you should be OK. $\endgroup$ – Will Sawin Jun 29 '20 at 23:30
I think there is a simpler way to see what is going on: first of all, the hypotheses force the characteristic polynomial of $B$ to be irreducible of degree $3$ over $F_{q}.$ On the other hand, if $PAP^{1}$ and $PBP^{1}$ are both diagonal, then $PAP^{1}$ and $PBP^{1}$ certainly commute. Hence $A$ and $B$ already commute. Since $B$ has irreducible characteristic polynomial, (which is also its minimum polynomial in this situation), and since $B$ leaves every eigenspace of $A$ invariant (as $A$ and $B$ commute), this forces (given the hypotheses) $A$ to have the form $\lambda I$ for some $\lambda \in F_{q}$.

1$\begingroup$ So, if you exclude $A$ being a scalar matrix, what you want to be true indeed is. $\endgroup$ – Geoff Robinson Jun 30 '20 at 12:18
Your intuition is somewhat correct. The characteristic polynomial of $B$ is cubic over $\mathbf F_q$, and has no roots in $\mathbf F_q$ by assumption, hence is irreducible. Thus, the roots all live in $\mathbf F_{q^3}$, and $\mathbf F_q[B] \cong \mathbf F_{q^3}$. If $A$ and $B$ are simultaneously diagonalisable over $\bar {\mathbf F}_q$, then $$\mathbf F_q[A,B] \underset{\mathbf F_q}\otimes \bar{\mathbf F}_q \cong \bar{\mathbf F}_q[PAP^{1},PBP^{1}]$$ is a subalgebra of the diagonal matrices $\bar{\mathbf F}_q^3$, so a dimension count shows that the inclusion $\mathbf F_q[B] \subseteq \mathbf F_q[A,B]$ must be an equality. Since $A$ has eigenvalues in $\mathbf F_q$ and $\mathbf F_q[B] \cong \mathbf F_{q^3}$, this forces $A$ constant (think about which elements of $\mathbf F_{q^3}$ have characteristic polynomial totally split over $\mathbf F_q$).
(Conversely, if $A$ is constant, as Will Sawin noted, then clearly $A$ and $B$ are simultaneously diagonalisable over $\bar{\mathbf F}_q$).

$\begingroup$ Thank you so much for answering. It was very helpful. I was considering $A$ to be the matrices that are not conjugate to the central elements. Sorry I forgot to mention that. $\endgroup$ – C.T. Jun 30 '20 at 0:16