Consider $GL_2$ as the affine group scheme with coordinate ring ${\mathbb Z}[x_1,x_2,x_3,x_4,y]/(\det\left(\begin{array}{cc}x_1& x_2\\ x_3& x_4\end{array}\right)y1)$. 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?

2$\begingroup$ This boils down to whether it holds when $R$ is the coordinate ring of $PGL_2$, and a positive answer is equivalent to whether there is a setwise section $PGL_2\to GL_2$ defined over $\mathbf{Z}$, if I don't miss anything. $\endgroup$ – YCor May 5 '14 at 12:42

2$\begingroup$ The diagram $1\rightarrow{\rm{GL}}_1\rightarrow {\rm{GL}}_n\rightarrow{\rm{PGL}}_n\rightarrow 1$ of smooth affine $R$group schemes is exact for the etale topology, so the obstruction to surjectivity on $R$points is the triviality of the induced map between the first two etaletopology H$^1$'s, which by descent theory is the map ${\rm{Pic}}(R)\rightarrow {\rm{Vec}}_n(R)$ carrying a line bundle $L$ to $L^{\oplus n}$. So shortexactness on $R$points is equivalent to $L^{\oplus n}\simeq R^n$ iff $L\simeq R$. Considering det, this holds if ${\rm{Pic}}(R)[n]=0$ and conversely for Dedekind $R$. $\endgroup$ – user76758 May 6 '14 at 7:03

$\begingroup$ By the way, since PGL$_n$ represents the automorphism functor of projective $(n1)$space (by deformation theory to bootstrap from the classical case on fieldvalued points), failure of surjectivity is exactly the condition of $\mathbf{P}^{n1}_R$ admitting an $R$automorphism which does not arise from an invertible $n \times n$ matrix over $R$. $\endgroup$ – user76758 May 6 '14 at 7:06
The answer is no. Here is an explicit example. Let $R=\mathbb{Z}[\sqrt{5}]$.
Consider the matrix $$\left(\begin{array}{cc}1+\sqrt{5}& 2\\ 2& 1\sqrt{5}\end{array}\right).$$
It represents an element of $PGL_2(R)$ that is not in the image of $GL_2(R)$.
The motivation for this example is that the ideal $(2,1+\sqrt{5})$ is not principal in $R$ and this should be relevant because the next term in the long exact sequence is $H^1(R,\mathbb{G}_m)$ (and so the sequence written in the question will be exact whenever this $H^1$ vanishes).
I would love to see a more conceptual proof of the failure of $GL_2(R)\to PGL_2(R)$ to be surjective.

1

2$\begingroup$ In general the map will be surjective when Pic(R) is trivial. Brian Conrad's homework on group schemes walks you through this to a certain extent. If R as a Dedekind domain you can think of the elements of $PGL_n(R)$ as being pseudomatrices of fractional ideals all of whose $n$fold products which go into the determinant are principal. If the class group is trivial, then all the entries have to be principal anyway and you're left with a standard matrix in $GL_n(R)$ up to unit multiple in $R$. $\endgroup$ – stankewicz May 5 '14 at 13:53

5$\begingroup$ Slightly better: the pointed set $PGL_2(R)/GL_2(R)$ is identified with the fibre over "0" of $\eta:H^1(X,GL_1) \to H^1(X,GL_2)$, where $X = \mathrm{Spec}(R)$. Interpreting $H^1(X,GL_n)$ as isomorphism classes of rank $n$ bundles, the map $\eta$ sends a line bundle $L$ to the rank $2$ bundle $L \oplus L$. Thus, the question is: can we have a nontrivial line bundle $L$ such that $L \oplus L \simeq \mathcal{O}_X^{\oplus 2}$? The answer is yes: any order $2$ element in $\mathrm{Pic}(X)$ for $X$ a Dedekind domain does the trick (as in the example above). $\endgroup$ – anonymous May 5 '14 at 15:08
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 $\mathbb Z\to \mathbb Z/2$.

1$\begingroup$ alternatively (essentially equivalently, but maybe even more topologically), $R$ can be taken to be taken to be the ring of continuous functions from $PGL_2(\mathbf{R})$ to $\mathbf{R}$. Then $PGL_2(R)$ is the group of continuous selfmaps of $PGL_2(\mathbf{R})$ (with the law on the target). Then the identity selfmap of $PGL_2(\mathbf{R})$ is not in the image, by the same argument. $\endgroup$ – YCor May 5 '14 at 18:53

$\begingroup$ I'd also like to point out that $\pi_1(GL_2(\mathbb R))$ and $\pi_1(GL_2(\mathbb C))$ are two different infinite cyclic groups. The one comes from $SL_2$ and the other does not. $\endgroup$ – Tom Goodwillie May 5 '14 at 22:25