= Solution
Here $A_1(K)$ denotes the <adjoint Chevalley group of type A1>; the isogeny form matters, since the simply connected form is $SL_2(K)$ itself. Over $\mathbb C$, take the <sl2 Lie algebra> with <Chevalley basis>
$$
e=\begin{pmatrix}0&1\\0&0\end{pmatrix},\quad
f=\begin{pmatrix}0&0\\1&0\end{pmatrix},\quad
h=\begin{pmatrix}1&0\\0&-1\end{pmatrix}.
$$
Their <Lie brackets> satisfy $[h,e]=2e$, $[h,f]=-2f$, $[e,f]=h$. Conjugation defines a <group homomorphism> $\rho:SL_2(\mathbb C)\to\operatorname{Aut}(\mathfrak{sl}_2(\mathbb C))$. Since $e^2=f^2=0$, $u(t)=\exp(te)$ and $l(s)=\exp(sf)$. Differentiating conjugation, or expanding its finite polynomial on this basis, gives
$$
\rho(u(t))=\exp(t\operatorname{ad}e),\qquad
\rho(l(s))=\exp(s\operatorname{ad}f).
$$
The <adjoint Chevalley group> is generated by these two families. Part (i) therefore proves that it is exactly the image of $\rho$. A matrix commuting with $e$ has the form $\begin{pmatrix}a&b\\0&a\end{pmatrix}$; also commuting with $f$ forces $b=0$. Thus the <kernel of a group homomorphism> $\ker\rho$ consists of scalar matrices of determinant one, namely $\{I,-I\}$. The <first isomorphism theorem> now gives $A_1(\mathbb C)\cong SL_2(\mathbb C)/\{\pm I\}=PSL_2(\mathbb C)$.
For a general <field>, first compute the <Chevalley basis> root operators as integral polynomials, and only then reduce their coefficients to $K$. Explicitly,
$$
\begin{aligned}
x_+(t)e&=e,&x_+(t)h&=h-2te,&x_+(t)f&=f+th-t^2e,\\
x_-(s)f&=f,&x_-(s)h&=h+2sf,&x_-(s)e&=e-sh-s^2f.
\end{aligned}
$$
They are still precisely conjugation by $u(t),l(s)$ on the three-dimensional trace-zero matrix space. These formulas remain valid in characteristic two, where $h=I$ but $e,h,f$ remain <linearly independent>. Part (i) still identifies the generated image with the conjugation image of $SL_2(K)$. The same computation with $e,f$ gives <kernel of a group homomorphism> $\{aI:a^2=1\}$, exactly its scalar center. Consequently
$$
\boxed{A_1(K)\cong SL_2(K)/Z(SL_2(K))=PSL_2(K).}
$$
In characteristic two the scalar center is trivial. The integral formulas avoid the invalid operation of dividing by factorials inside a field of small positive characteristic.
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