Projective Special Linear Group is an Linear Algebraic Group So I was wondering why the group $PSL(2,K)$ is a linear algebraic group, in the case that the characteristic of $K$ is not equal to $2$. 
Actually there is a description of $PSL(2,K)$, namely:
$PSL(2,K) = SL(2,L)/C,$
whereas $C$ is the center of $SL(2,K)$, namely the subrgoup $x  \cdot I$, with $I$ the identity and $x$ any solution of $x^2 = 1$. 
We can embedd $SL(2,K)$ in the affine space $\mathbb{A}^4$ where the coordinate ring is just $K[x_1,x_2,x_3,x_4]/(x_1x_3 - x_2x_4 - 1)$. We can simply extend the action of $C$ on the affine space $\mathbb{A}^4$. But I dont know houw to embedd $\mathbb{A}^4/C$ in some bigger affine space, s. th. we get a description of $PSL(2,K)$ i.e. s. th. $PSL(2,K)$ is itself an linear algebraic group. 
I would appreciate some hints. 
 A: The general result is that the quotient of a linear algebraic group by a closed normal subgroup is a linear algebraic group.  Surprisingly this doesn't appear to be in Humphries, but it's definitely in Springer.  The proof doesn't work by quotienting all of the ambient affine space, though; IIRC the idea of all these results is to let a group act on a ring of functions, extract a finite-dimensional faithful subrepresentation, and use that to embed the group as a subgroup of some $GL_n$.
A: I haven't checked everything, but I think the following works.
Let $\{x_1,x_2\}$ be a basis for the space that $SL_2(K)$ acts on naturally. That is
$$
x_1=\binom10,\qquad x_2=\binom01,
$$
and 
$$
A=\left(\begin{array}{rr}a&b\\c&d\end{array}\right)
$$
acts as $A.x_1=ax_1+cx_2$, $A.x_2=bx_1+dx_2$. 
The group $SL_2(K)$ then also acts on the 3-dimensional space of homogeneous quadratic polynomials, spanned by $\{x_1^2,x_1x_2,x_2^2\}$. The action goes by the above linear substitutions, so for example $A.x_1^2=(ax_1+cx_2)^2$ et cetera. This gives a homomorphism from $f:SL_2(K)\to GL_3(K)$ - a 3-dimensional representation of $SL_2(K)$ if you wish.
The point is that the matrix $-I_2$ is in the kernel of $f$. Because the center is the only non-trivial normal subgroup of $SL_2$ we have, in fact, $\ker f=\langle -I_2\rangle$.
So the remaining task is to show that $\operatorname{Im} f\cong PSL_2(K)$ is an algebraic subgroup of $GL_3(K)$.
