Invertible elements in the ring $K[x,y,z,t]/(xy+zt-1)$ I would like to know how big is the set of invertible elements in the ring $$R=K[x,y,z,t]/(xy+zt-1),$$ where $K$ is any field. In particular whether any invertible element is a (edit: scalar) multiple of $1$, or there is something else. Any help is greatly appreciated.
 A: Here is a geometric proof that $R^*=K^*$.  
Embedd $U=\operatorname{Spec}R$ as an open subvariety into the projective variety 
$$P:=\operatorname{Proj} K[x,y,z,t, w]/(xy+zt-w^2).$$
Note that $xy+zt-w^2$ is irreducible over an algebraic closure of $K$, so $P$ is geometrically integral, hence ${\mathcal O}_P(P)=K$. Using Jacobian criterion, we see that $P$ is smooth, hence normal. 
Let $f\in R^*$ that we consider as a rational function on $P$. As $E:=P\setminus U$ is the zero set of $w$ which is an integral hypersurface in $P$, the divisor of $f$ 
is $\mathrm{div}(f)=nE$ for some integer $n$. If $n\ge 0$, then $f$ has no pole on $P$, so $f\in {\mathcal O}_P(P)=K$ (because $P$ is normal). If $n<0$, then the same reasonning applied to $1/f$ shows that $1/f\in K$. In both cases, $f\in K$. 

EDIT 2. Please forget the wrong answer below.
Edit. Simply the former proof thanks to the comments of the OP.
The only invertible elements are in $K^*$. I don't really like the proof below, but for the moment I don't have an alternative one. 
In $R$, the relation $tz=1-xy$ implies that any element
$f(x,y,z,t)\in R$ can be written as
$$f(x,y,z,t)=g(x,y,z)+th(x,y).$$
Suppose $f$ is invertible in $R$. Let $K^a$ be an algebraic closure of $K$.
Denote by 
$$Z=\{(a,b,c,d)\in (K^a)^4 \mid ab+cd=1\}.$$ 
Then $f(a,b,c,d)\ne 0$ for all $(a,b,c,d)\in Z$.
First observation: there is no common factor $h_1(x,y)$ of $g(x,y,z)$ and
$h(x,y)$ because otherwise $h_1(x,y)$ would be invertible in $R$, but
for any $(a,b)\in (K^a)^2$, there exists $(a,b,c,d)\in Z$, so $h_1(a,b)\ne 0$,
thus $h_1(x,y)\in K^*$.
Now for any $(a,b,c)\in (K^a)^3$ such that $c\ne 0$ and $h(a,b)\ne 0$,
we have $g(a,b,c)+(1-ab)h(a,b)/c \ne 0$ because otherwise
$f(a,b,c,d)=0$ with $d=(1-ab)/c$ and $(a,b,c,d)\in Z$.
This means that in the affine space $\mathbb A^3_K$, we have
$$ V(zg(x,y,z)+(1-xy)h(x,y))\subseteq V(z).$$
Therefore $z \mid f$ and $h=0$.
We are now reduced to the situation $f=g(x,y,z)$. Similar reasoning
show that $g(x,y,z)=\lambda z^n$ with $\lambda \in K^*$ and
$n\ge 0$. Again we see as above that $n>0$ is impossible. So
$f\in K^*$.
A: Let $R=K[X,Y,Z,T]/(XY+ZT-1)$. In the following we denote by $x,y,z,t$ the residue classes of $X,Y,Z,T$ modulo the ideal $(XY+ZT-1)$. Let $f\in R$ invertible. Then its image in $R[x^{-1}]$ is also invertible. But $R[x^{-1}]=K[x,z,t][x^{-1}]$ and $x$, $z$, $t$ are algebraically independent over $K$. Thus $f=cx^n$ with $c\in K$, $c\ne0$, and $n\in\mathbb Z$. Since $R/xR\simeq K[Z,Z^{-1}][Y]$ we get that $x$ is a prime element, and therefore $n=0$. Conclusion: if $f$ is invertible, then $f\in K-\{0\}$.
