TITLE: Is GL2( R ) - > PGL2( R ) surjective?
QUESTION [12 upvotes]: Consider $GL_2$ as the affine group scheme with coordinate ring 
${\mathbb Z}[x_1,x_2,x_3,x_4,y]/(\det\left(\begin{array}{cc}x_1& x_2\\ x_3& x_4\end{array}\right)y-1)$.
The group scheme $PGL_2$ is then given by the subring $S$ of $GL_1$-invariants, which is the subring generated by all monomials of the form $x_ix_iy^2$. 
In this way, we define $PGL_2( R )=Hom(S,R)$ for any ring $R$.
By Hilbert 90 we know that the sequence
$$
0\to GL_1( R )\to GL_2( R )\to PGL_2( R )\to 1
$$
is exact if $R$ is a field. From that one can derive that it stays exact if $R$ is factorial. But what for a general commutative ring with unit? Is it always exact? If not, is there a handy description of all rings for which it is?

REPLY [4 votes]: For a topological perspective, take $R$ to be the tensor product of $\mathbb R$ with the coordinate ring of $PGL_2$. surjectivity would imply that the map of topological spaces $GL_2(\mathbb R)\to PGL_2(\mathbb R)$ has a right inverse, which it does not because the induced map of fundamental groups $\mathbb Z\to \mathbb Z$ is $x\mapsto 2x$. Or you can argue similarly over $\mathbb C$, where the map of fundamental groups is $\mathbb Z\to \mathbb Z/2$.