A point in $ PGL(R) $ not in $ GL(R)/R^{\times} $ A bit of notational background first.
Let $k$ be a field and define $ PGL_{n} = Spec(k[x_{ij}]_{(det)}) $, where $i,j = 1,...,n$ and where $k[x_{ij}]_{(det)}$ denote the degree $0$ part of the graded ring $k[x_{ij}]_{det}$.
One has an injective natural map $\phi_{R} : GL_{n}(R)/R^{\times}\hookrightarrow PGL_{n}(R) $, for all $k$-algebras $R$. Now, as the title says, the purpose is to find an $R$ such that $\phi_{R}$ is not surjective.
In particular, I lack a characterization for $x\in PGL_{n}(R)$ to be in the image of $\phi_{R}$. Here is at least a sufficient condition : if $x(\frac f {det})$ is invertible in $R$ (where $f$ is a monomial of degree $n$ in $k[x_{ij}]$), then $x$ is in the image of $\phi_{R}$.
As a remark, this sufficient condition is enough to show that $\phi_{R}$ is surjective whenever $R$ is a local ring. Also note that this condition is not necessary (for example, one could take $n=2$, $R = k[T]$, and the matrix $\left[\begin{array}{c}T&T+1\\T-1&T\end{array}\right]$)
Finally, as stated in the first question of this homework sheet 
http://math.stanford.edu/~conrad/252Page/homework/hmwk3.pdf
one can apparently hope to find an example when $n=2$ and $R$ is a Dedekind ring with a class group having 2-torsion. Brian Conrad gives the (mysterious to me) hint that $I \oplus I \simeq R^{2}$ when $I$ is 2-torsion.
Could someone show how to use this hint to find a point in $ PGL_{2}(R) $ not in $ GL_{2}(R)/R^{\times} $ ? As a bonus, if someone is enthusiastic about explaining why $I \oplus I \simeq R^{2}$ when $I$ is 2-torsion, I would be interested to read it (but take into account that I know basically nothing about class group, appart from what is on Wikipedia).
 A: Let $R=k[x_{ij}]_{(\det)}$, and consider the point $p\in\mathrm{PGL}(R)$ defined by $\mathrm{id}:k[x_{ij}]_{(\det)}\to R$.
If $p$ is in the image of $\phi_R$, then there exists a retraction of the natural immersion $j:k[x_{ij}]_{(\det)}\to k[x_{ij}]_\det$, i.e. a map $\pi:k[x_{ij}]_\det\to k[x_{ij}]_{(\det)}$ such that $\pi\circ j=id_R$. Now, for every $i,j$, fix an integer $k$ and call $\phi(x_{ij})\in k[x_{ij}]$ such that $\pi(x_{ij})=\phi(x_{ij})/\det^k$. We have
$$x_{ij}^n/\det=\pi(x_{ij})^n/\pi(\det)=\phi(x_{ij})^n/\phi(\det)$$
Since $k[x_{ij}]$ is an UFD, $x_{ij}$ divides $\phi(x_{ij})$, we have $\phi(x_{ij})=x_{ij}\psi_{ij}$ and
$$\phi(\det)/\det=\psi_{ij}^n$$
The left hand side doesn't depend on $i,j$: this implies $\psi_{ij}=u_{ij}\psi$ where $u_{ij}\in k$ is a $n$-th root of $1$.
Hence, we have $\pi(x_{ij})=x_{ij}u_{ij}\psi/\det^k$. Now, consider $\det=\det(x)$ as a polynomial, where $x=(x_{ij})$. We have
$$\pi(\det(x))=\det(\pi(x))=\det(ux)\cdot\left(\frac{\psi}{\det(x)^k}\right)^n=\det(ux)\frac{\psi^n}{\det^{nk}(x)}$$
Since $\det(x)$ is invertible, $\pi(\det(x))$ is invertible, too, and this implies that $\psi$ is invertible. But the invertible elements of $k[x_{ij}]_{(\det)}$ are only nonzero constants, hence $\psi$ is constant.
For $n>1$, this is clearly absurd, because $\pi(x_{ij})=x_{ij}u_{ij}\psi/\det^k$ must have degree zero, and $1-\deg(\det^k)=1-nk$ is never zero for $n>1$.
If $n=1$, for every constant $c\in k^*$ we have an homomorphism $k[x]_{x}\to k[x]_{(x)}=k$ sending $x$ to $c$. In particular, for every $k$ algebra $R$ and every point $p:k[x]_{(x)}\to R$ of $PGL_1(R)$, the composition
$$k[x]_{x}\to k[x]_{(x)}\to R$$
defines a point in $GL_1(R)$ over $p$.
