$GL_2(K)$ and $SL_2(K)\circ_2 T_1$ K is algebraically closed.
Are $GL_2(K)$ and $SL_2(K)\circ_2 T_1$ (central product) isomorphic or equal as algebraic groups? A paper says so but I cannot see it.
I am aware $GL_n(K)$ is isomorphic to the semidirect product of $SL_n(K)$ and $K^*$ as abstract groups.
 A: With $char(K)\ne 2$, let $$G=SL_2(K)\times GL_1(K),\qquad H=\{ (1,1),(-1,-1)\}$$
then $$K[G]=K[x_1,x_2,x_3,x_4,x_5,x_6]/(x_1x_4-x_2x_3-1,x_5x_6-1)$$ and $K[G/H]$ is the subring of $K[G]$ fixed by the automorphism $(g,k)\to (-g,-k)$

that is $K[G/H]$ is the subring generated by the $x_ix_j$, a finitely generated $K$-algebra, which gives our affine variety model for $G/H$.

$$K[GL_2(K)]=K[a,b,c,d,e,f]/(ad-bc-e,ef-1)$$
The group isomorphism $G/H\to GL_2(K)$ is induced by the morphism $G\to GL_2(K),(g,k)\to gk$ given by $$( x_1,x_2,x_3,x_4,x_5,x_6)\to (x_5x_1,x_5x_2,x_5x_3,x_5x_4,x_5^2,x_6^2)$$

Note how the RHS only involves some $x_ix_j$, so it is a regular map $G/H\to GL_2(K)$.

The inverse morphism $GL_2(K)\to G/H$ is given by sending $(a,b,c,d,e,f)$ to the set of $x_ix_j$ with $$x_1=a\sqrt{f},x_2=b\sqrt{f},x_3=c\sqrt{f},x_4=d\sqrt{f},x_5=\sqrt{e},x_6=\sqrt{f}$$ in the $x_ix_j$ there are no square roots anymore, this is a regular map.
Surprisingly there is no need that $K$ is algebraically closed, if $k\in K$ then $\pm (\pmatrix{\sqrt{k}&0\\0&1/\sqrt{k}},\sqrt{k})$ is a $K$-point of $G/H$ whose image in $GL_2$ is $\pmatrix{k&0\\0&1}$.
