How to prove the adjoint representation of $\text{SL}_2$ is irreducible when $\text{char}(k) \ne 2$? Consider the adjoint representation of $\text{SL}_2$, where it acts by conjugation on its Lie algebra of matrices of the form $\begin{pmatrix} a & b \\ c & -a \\ \end{pmatrix}$, and suppose we look at it over a field $k$ with  $\text{char}(k) \ne 2$. Is there a quick way to prove that this representation is then irreducible? Also, is there a quick way to see why it fails to be irreducible when $\text{char}(k) = 2$?
 A: 
Partial Answer: I'm not sure about semi-simplicity in positive characteristic, so this answer is incomplete. Any proper nontrivial submodule of the adjoint representation has dimension $2$ or $1$. I will show that there are no $1$-dimensional submodules. One also needs to check that there are no $2$-dimensional submodules (unless someone has a slick proof that this doesn't need to be checked).

First, we check that the adjoint representation is not irreducible when $\mathrm{char}(k)=2$. Let 
$$X=\begin{pmatrix}a&b\\c&d\end{pmatrix}\in\mathrm{SL}_2(k),\;\;ad-bc=1.$$
Then, the matrix for $Ad(X)$ in the basis 
$$\left\{\begin{pmatrix}1&0\\0&-1\end{pmatrix},\;\begin{pmatrix}0&1\\0&0\end{pmatrix},\;\begin{pmatrix}0&0\\1&0\end{pmatrix}\right\}$$
is
$$Ad(X)=\begin{pmatrix}ad+bc&-ac&bd\\-2ab&a^2&-b^2\\2cd&-c^2&d^2\end{pmatrix}=\begin{pmatrix}1&-ac&bd\\0&a^2&-b^2\\0&-c^2&d^2\end{pmatrix}$$
so the span of $\begin{pmatrix}1&0\\0&-1\end{pmatrix}$ is a $1$-dimensional submodule. (Note that we used $$ad+bc=(1+bc)+bc=1+2bc=1$$ as $\mathrm{char}(k)=2$.)
Now, assume $\mathrm{char}(k)\neq 2$. Note that since the adjoint representation is $3$-dimensional a proper submodule is either $1$-dimensional or $2$-dimensional. I will show that it contains no $1$-dimensional submodules. Indeed, suppose
$$v=\begin{pmatrix}\alpha&\beta\\\gamma&-\alpha\end{pmatrix}$$
spans a 1-dimensional submodule. Then $v$ is an eigenvector for every $X\in \mathrm{SL}_2(k)$:

$$Ad(X)(v)=\lambda_Xv\;\;\;\mbox{for some }\lambda_X\in k^\times.$$

Take $X=\begin{pmatrix}0&-1\\1&0\end{pmatrix}$ and compute $Ad(X)(v)=-\begin{pmatrix}\alpha&\gamma\\\beta&-\alpha\end{pmatrix}$. Therefore, $\lambda_X=-1$ and $\gamma=\beta$.
Now, $v=\begin{pmatrix}\alpha&\beta\\\beta&-\alpha\end{pmatrix}$. Let $X=\begin{pmatrix}1&1\\0&-1\end{pmatrix}$. Then, $Ad(X)(v)=\begin{pmatrix}\alpha+\beta&-2\alpha\\\beta&-(\alpha+\beta)\end{pmatrix}$. Then, by comparing the $(2,1)$-entries of $Ad(X)(v)$ and $v$, we have $\lambda_X=1$. Therefore, $\alpha=\alpha+\beta$, so $\beta=0$. Finally, we have $\beta=-2\alpha$. Since $\mathrm{char}(k)\neq 2$, $\alpha=0$. Since $v=0$, it does not span a $1$-dimensional submodules.
