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Here $A_1(K)$ denotes the adjoint Chevalley group of type A1; the isogeny form matters, since the simply connected form is $S{L}_2(K)$ itself. Over $\mathbb{C}$, take the sl2 Lie algebra with Chevalley basisTheir Lie brackets satisfy $[h,e]=2e$, $[h,f]=-2f$, $[e,f]=h$. Conjugation defines a group homomorphism $\rho :S{L}_2(\mathbb{C})\to \mathrm{Aut}({\mathfrak{sl}}_2(\mathbb{C}))$. 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, givesThe 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 $\left(\begin{array}{cc}a& b\\ 0& a\end{array}\right)$; 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 $A_1(\mathbb{C})\cong S{L}_2(\mathbb{C})/\{\pm I\}=PS{L}_2(\mathbb{C})$.

For a general field, first compute the Chevalley basis root operators as integral polynomials, and only then reduce their coefficients to $K$. Explicitly,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 $S{L}_2(K)$. The same computation with $e,f$ gives kernel of a group homomorphism $\{aI:a^2=1\}$, exactly its scalar center. ConsequentlyIn 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.