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It is not absolutely irreducible: over the algebraic closure, it is the direct sum of $V^{Fr ^i}$ ($0\leq i \leq e-1$), where $Fr : {\mathbb F}_q\rightarrow {\mathbb F}_q$ is the map $x\mapsto x^p$.

I think $e$ being odd or even does not matter: $V$ is irreducible over ${\mathbb F}_p$. The restriction of scalars may be viewed as follows. First, $V$ viewed as a ${\mathbb F}_p$ vector space has dimension $3e$, and one has an embedding ${\mathbb F}_q\subset End_{{\mathbb F}_p}(V)$. The algebra $M_3({\mathbb F}_q)$ is simply the commutant of this copy of ${\mathbb F}_q$ in $End _{{\mathbb F}_p}(V)$ and is spanned as a ${\mathbb F}_p$ vector space (a small check) by elements of $Ad (SL_2({\mathbb F}_q))$.

Suppose $W\subset V$ is irreducible for the adjoint action of $G=SL_2({\mathbb F}_q)$. Then $W$ is stabilised by $M_3({\mathbb F}_q)$; in particular, $W$ is an ${\mathbb F}_q$ subspace of $V$ (now viewed as a ${\mathbb F}_q$ vector space). But $SL_2({\mathbb F}_q)$ acts irreducibly on $V={\mathbb F}_q ^3$, hence $W=V$. [This irreducibility fails of course when $p=2$].

[Edit] I think I can find a proof that the ${\mathbb F}_p$ algebra $R$ generated by $Ad (SL_2){\mathbb F}_q)$ is all of $M_3({\mathbb F}_q)$. First of all, $V \otimes _{{\mathbb F}_p}{\overline {\mathbb F}_p}$ is the direct sum of $W_i=V^{Fr ^i}$ as $i$ varies from $0$ to $e-1$, and is semi-simple. Therefore, $V$ is semi-simple as an $R$-module. Hence $R$ is semi-simple (the action of $R$ being faithful on $V$). Therefore, $R$ is a product of simple rings $R_i$, each $R_i$ of the form $M_{d_i}({\mathbb F}_{q_i})$ a matrix algebra over a finite field containing ${\mathbb F}_p$.

The module $V_i={\mathbb F}_{q_i}^{d_i}$ is absolutely irreducible for $R_i$. However, $V$ over the algebraic closure, is a sum of irreducible three dimensional representations $W_i$. Therefore, each $V_j$ is some $Wi$ over the algrbraic closure, and is hence three dimensional: $d_i=3$. Moreover, since $R$ commutes with the action of $F={\mathbb F}_{q}$ on $V$, each ${\mathbb F}_{q_i}$ contains $F$. We have thus,

$$M_3({\mathbb F}_q)\supset R\supset R_i=M_3({\mathbb F}_{q_i})$$ which shows that $R=M_3({\mathbb F}_q)$ as needed.

It is not absolutely irreducible: over the algebraic closure, it is the direct sum of $V^{Fr ^i}$ ($0\leq i \leq e-1$), where $\operatorname{Fr}\colon{\mathbb F}_q\rightarrow{\mathbb F}_q$ is the map $x\mapsto x^p$.

I think $e$ being odd or even does not matter: $V$ is irreducible over ${\mathbb F}_p$. The restriction of scalars may be viewed as follows. First, $V$ viewed as a ${\mathbb F}_p$ vector space has dimension $3e$, and one has an embedding ${\mathbb F}_q\subset\operatorname{End}_{{\mathbb F}_p}(V)$. The algebra $M_3({\mathbb F}_q)$ is simply the commutant of this copy of ${\mathbb F}_q$ in $\operatorname{End}_{{\mathbb F}_p}(V)$ and is spanned as a ${\mathbb F}_p$ vector space (a small check) by elements of $\operatorname{Ad}(\operatorname{SL}_2({\mathbb F}_q))$.

Suppose $W\subset V$ is irreducible for the adjoint action of $G=\operatorname{SL}_2({\mathbb F}_q)$. Then $W$ is stabilised by $M_3({\mathbb F}_q)$; in particular, $W$ is an ${\mathbb F}_q$ subspace of $V$ (now viewed as a ${\mathbb F}_q$ vector space). But $\operatorname{SL}_2({\mathbb F}_q)$ acts irreducibly on $V={\mathbb F}_q ^3$, hence $W=V$. [This irreducibility fails of course when $p=2$].

[Edit] I think I can find a proof that the ${\mathbb F}_p$ algebra $R$ generated by $\operatorname{Ad}(\operatorname{SL}_2){\mathbb F}_q)$ is all of $M_3({\mathbb F}_q)$. First of all, $V \otimes _{{\mathbb F}_p}{\overline {\mathbb F}_p}$ is the direct sum of $W_i=V^{Fr ^i}$ as $i$ varies from $0$ to $e-1$, and is semi-simple. Therefore, $V$ is semi-simple as an $R$-module. Hence $R$ is semi-simple (the action of $R$ being faithful on $V$). Therefore, $R$ is a product of simple rings $R_i$, each $R_i$ of the form $M_{d_i}({\mathbb F}_{q_i})$ a matrix algebra over a finite field containing ${\mathbb F}_p$.

The module $V_i={\mathbb F}_{q_i}^{d_i}$ is absolutely irreducible for $R_i$. However, $V$ over the algebraic closure, is a sum of irreducible three dimensional representations $W_i$. Therefore, each $V_j$ is some $Wi$ over the algrbraic closure, and is hence three dimensional: $d_i=3$. Moreover, since $R$ commutes with the action of $F={\mathbb F}_{q}$ on $V$, each ${\mathbb F}_{q_i}$ contains $F$. We have thus,

$$M_3({\mathbb F}_q)\supset R\supset R_i=M_3({\mathbb F}_{q_i})$$ which shows that $R=M_3({\mathbb F}_q)$ as needed.

It is not absolutely irreducible: over the algebraic closure, it is the direct sum of $V^{Fr ^i}$ ($0\leq i \leq e-1$), where $Fr : {\mathbb F}_q\rightarrow {\mathbb F}_q$ is the map $x\mapsto x^p$.

I think $e$ being odd or even does not matter: $V$ is irreducible over ${\mathbb F}_p$. The restriction of scalars may be viewed as follows. First, $V$ viewed as a ${\mathbb F}_p$ vector space has dimension $3e$, and one has an embedding ${\mathbb F}_q\subset End_{{\mathbb F}_p}(V)$. The algebra $M_3({\mathbb F}_q)$ is simply the commutant of this copy of ${\mathbb F}_q$ in $End _{{\mathbb F}_p}(V)$ and is spanned as a ${\mathbb F}_p$ vector space (a small check) by elements of $Ad (SL_2({\mathbb F}_q))$.

Suppose $W\subset V$ is irreducible for the adjoint action of $G=SL_2({\mathbb F}_q)$. Then $W$ is stabilised by $M_3({\mathbb F}_q)$; in particular, $W$ is an ${\mathbb F}_q$ subspace of $V$ (now viewed as a ${\mathbb F}_q$ vector space). But $SL_2({\mathbb F}_q)$ acts irreducibly on $V={\mathbb F}_q ^3$, hence $W=V$. [This irreducibility fails of course when $p=2$].

[Edit] Checking that $M_3({\mathbb F}_q)$ is the ${\mathbb F}_p$ span $R$ of $SL_2({\mathbb F}_q)$ matrices involves the following. Consider the diagonal group $$T=\{g=\begin{pmatrix} a ^2 & 0 & 0 \cr 0 & 1 & 0 \cr 0 & 0 & a^{-2} \end{pmatrix}\}.$$ Since $a\mapsto a^2$ is a nontrivial character of the mutliplicative group (whose order is the element $-1$ in the field ${\mathbb F}_p$, we see that the ${\mathbb F}_p$ span of the matrices in $T$ contains the matrix $$A=\begin{pmatrix} 0 & 0 & 0 \cr 0 & 1 & 0 \cr 0 & 0 & 0 \end{pmatrix}.$$

Next, the algebra $R$ contains the ${\mathbb F}_p$ span of matrices $Ad (u)$ where $u$ is upper triangular unipotent in $SL_2({\mathbb F}_q)$. This contains a matrix of the form $$ B=\begin{pmatrix} 1 & 1 & 1 \cr 0 & 1 & 2 \cr 0 & 0 & 1 \end{pmatrix}$$. By forming the commutator $[A,B]$ which lies in the algebra $R$, we have $C=\begin{pmatrix} 0 & -1 & 0 \cr 0 & 0 & 2 \cr 0 & 0 & 0\end{pmatrix}$ in $R$. The commutator $[C,B]=\begin{pmatrix} 0 & 0 & 1 \cr 0 & 0 & 0 \cr 0 & 0 & 0 \end{pmatrix}$ (or a nonzero multiple of this matrix). Hence $E=[B,C]\in R$. Conjugating $E$ by matrices of the diagonal $T$, we get the ${\mathbb F}_q$ line $L_{13}$ through $E=E_{13}$.

The commutator of $L_{13}$ with the lower triangular matrices corresponding to $B$ and $C$, we get the ${\mathbb F}_q$ line $L_{12}$ through $E_{12}$, etc. We get $L_{ij}$ for all $i\neq j$ and whose ${\mathbb F}_p$ algebra span is all of $M_3({\mathbb F}_q)$.

I am hoping to get a more civilised proof, using that the algebra $R$ is semi-simple and that the only semi-simple algebras over ${\mathbb F}_q$ are matrix algebras.

It is not absolutely irreducible: over the algebraic closure, it is the direct sum of $V^{Fr ^i}$ ($0\leq i \leq e-1$), where $Fr : {\mathbb F}_q\rightarrow {\mathbb F}_q$ is the map $x\mapsto x^p$.

I think $e$ being odd or even does not matter: $V$ is irreducible over ${\mathbb F}_p$. The restriction of scalars may be viewed as follows. First, $V$ viewed as a ${\mathbb F}_p$ vector space has dimension $3e$, and one has an embedding ${\mathbb F}_q\subset End_{{\mathbb F}_p}(V)$. The algebra $M_3({\mathbb F}_q)$ is simply the commutant of this copy of ${\mathbb F}_q$ in $End _{{\mathbb F}_p}(V)$ and is spanned as a ${\mathbb F}_p$ vector space (a small check) by elements of $Ad (SL_2({\mathbb F}_q))$.

Suppose $W\subset V$ is irreducible for the adjoint action of $G=SL_2({\mathbb F}_q)$. Then $W$ is stabilised by $M_3({\mathbb F}_q)$; in particular, $W$ is an ${\mathbb F}_q$ subspace of $V$ (now viewed as a ${\mathbb F}_q$ vector space). But $SL_2({\mathbb F}_q)$ acts irreducibly on $V={\mathbb F}_q ^3$, hence $W=V$. [This irreducibility fails of course when $p=2$].

[Edit] I think I can find a proof that the ${\mathbb F}_p$ algebra $R$ generated by $Ad (SL_2){\mathbb F}_q)$ is all of $M_3({\mathbb F}_q)$. First of all, $V \otimes _{{\mathbb F}_p}{\overline {\mathbb F}_p}$ is the direct sum of $W_i=V^{Fr ^i}$ as $i$ varies from $0$ to $e-1$, and is semi-simple. Therefore, $V$ is semi-simple as an $R$-module. Hence $R$ is semi-simple (the action of $R$ being faithful on $V$). Therefore, $R$ is a product of simple rings $R_i$, each $R_i$ of the form $M_{d_i}({\mathbb F}_{q_i})$ a matrix algebra over a finite field containing ${\mathbb F}_p$.

The module $V_i={\mathbb F}_{q_i}^{d_i}$ is absolutely irreducible for $R_i$. However, $V$ over the algebraic closure, is a sum of irreducible three dimensional representations $W_i$. Therefore, each $V_j$ is some $Wi$ over the algrbraic closure, and is hence three dimensional: $d_i=3$. Moreover, since $R$ commutes with the action of $F={\mathbb F}_{q}$ on $V$, each ${\mathbb F}_{q_i}$ contains $F$. We have thus,

$$M_3({\mathbb F}_q)\supset R\supset R_i=M_3({\mathbb F}_{q_i})$$ which shows that $R=M_3({\mathbb F}_q)$ as needed.

It is not absolutely irreducible: over the algebraic closure, it is the direct sum of $V^{Fr ^i}$ ($0\leq i \leq e-1$), where $Fr : {\mathbb F}_q\rightarrow {\mathbb F}_q$ is the map $x\mapsto x^p$.

I think $e$ being odd or even does not matter: $V$ is irreducible over ${\mathbb F}_p$. The restriction of scalars may be viewed as follows. First, $V$ viewed as a ${\mathbb F}_p$ vector space has dimension $3e$, and one has an embedding ${\mathbb F}_q\subset End_{{\mathbb F}_p}(V)$. The algebra $M_3({\mathbb F}_q)$ is simply the commutant of this copy of ${\mathbb F}_q$ in $End _{{\mathbb F}_p}(V)$ and is spanned as a ${\mathbb F}_p$ vector space (a small check) by elements of $Ad (SL_2({\mathbb F}_q))$.

Suppose $W\subset V$ is irreducible for the adjoint action of $G=SL_2({\mathbb F}_q)$. Then $W$ is stabilised by $M_3({\mathbb F}_q)$; in particular, $W$ is an ${\mathbb F}_q$ subspace of $V$ (now viewed as a ${\mathbb F}_q$ vector space). But $SL_2({\mathbb F}_q)$ acts irreducibly on $V={\mathbb F}_q ^3$, hence $W=V$. [This irreducibility fails of course when $p=2$].

It is not absolutely irreducible: over the algebraic closure, it is the direct sum of $V^{Fr ^i}$ ($0\leq i \leq e-1$), where $Fr : {\mathbb F}_q\rightarrow {\mathbb F}_q$ is the map $x\mapsto x^p$.

I think $e$ being odd or even does not matter: $V$ is irreducible over ${\mathbb F}_p$. The restriction of scalars may be viewed as follows. First, $V$ viewed as a ${\mathbb F}_p$ vector space has dimension $3e$, and one has an embedding ${\mathbb F}_q\subset End_{{\mathbb F}_p}(V)$. The algebra $M_3({\mathbb F}_q)$ is simply the commutant of this copy of ${\mathbb F}_q$ in $End _{{\mathbb F}_p}(V)$ and is spanned as a ${\mathbb F}_p$ vector space (a small check) by elements of $Ad (SL_2({\mathbb F}_q))$.

Suppose $W\subset V$ is irreducible for the adjoint action of $G=SL_2({\mathbb F}_q)$. Then $W$ is stabilised by $M_3({\mathbb F}_q)$; in particular, $W$ is an ${\mathbb F}_q$ subspace of $V$ (now viewed as a ${\mathbb F}_q$ vector space). But $SL_2({\mathbb F}_q)$ acts irreducibly on $V={\mathbb F}_q ^3$, hence $W=V$. [This irreducibility fails of course when $p=2$].

[Edit] Checking that $M_3({\mathbb F}_q)$ is the ${\mathbb F}_p$ span $R$ of $SL_2({\mathbb F}_q)$ matrices involves the following. Consider the diagonal group $$T=\{g=\begin{pmatrix} a ^2 & 0 & 0 \cr 0 & 1 & 0 \cr 0 & 0 & a^{-2} \end{pmatrix}\}.$$ Since $a\mapsto a^2$ is a nontrivial character of the mutliplicative group (whose order is the element $-1$ in the field ${\mathbb F}_p$, we see that the ${\mathbb F}_p$ span of the matrices in $T$ contains the matrix $$A=\begin{pmatrix} 0 & 0 & 0 \cr 0 & 1 & 0 \cr 0 & 0 & 0 \end{pmatrix}.$$

Next, the algebra $R$ contains the ${\mathbb F}_p$ span of matrices $Ad (u)$ where $u$ is upper triangular unipotent in $SL_2({\mathbb F}_q)$. This contains a matrix of the form $$ B=\begin{pmatrix} 1 & 1 & 1 \cr 0 & 1 & 2 \cr 0 & 0 & 1 \end{pmatrix}$$. By forming the commutator $[A,B]$ which lies in the algebra $R$, we have $C=\begin{pmatrix} 0 & -1 & 0 \cr 0 & 0 & 2 \cr 0 & 0 & 0\end{pmatrix}$ in $R$. The commutator $[C,B]=\begin{pmatrix} 0 & 0 & 1 \cr 0 & 0 & 0 \cr 0 & 0 & 0 \end{pmatrix}$ (or a nonzero multiple of this matrix). Hence $E=[B,C]\in R$. Conjugating $E$ by matrices of the diagonal $T$, we get the ${\mathbb F}_q$ line $L_{13}$ through $E=E_{13}$.

The commutator of $L_{13}$ with the lower triangular matrices corresponding to $B$ and $C$, we get the ${\mathbb F}_q$ line $L_{12}$ through $E_{12}$, etc. We get $L_{ij}$ for all $i\neq j$ and whose ${\mathbb F}_p$ algebra span is all of $M_3({\mathbb F}_q)$.

I am hoping to get a more civilised proof, using that the algebra $R$ is semi-simple and that the only semi-simple algebras over ${\mathbb F}_q$ are matrix algebras.