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.

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In any ring, every element is a multiple of 1. What did you mean to ask? – Hurkyl Aug 4 '12 at 3:45
@Hurkyl: I assume "multiple of $1$" in this context means scalar multiple of $1$ (that is, element of $K$). – Qiaochu Yuan Aug 4 '12 at 3:47
@Hurkyl: Thank you for your comment. Indeed, I meant to say scalar multiple of 1, that is any element of the form a*1 with a in K\{0}. I will edit the question accordingly. – user26565 Aug 4 '12 at 3:55
In math.stackexchange.com/questions/173021/… it is shown your ring is a UFD. I doubt this helps to find units, but if you continue working with the ring, it could be quite helpful. – Jack Schmidt Aug 4 '12 at 14:42

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. 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 $f\in K^*$.

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Could you please explain, why do you need $h(a,b)\neq 0$ in the paragraph that starts with "Now for any $(a,b,c)\in (K^a)^3$..."? It looks like you only use that $c\neq 0$ when you write d=(1-ab)/c. – user26565 Aug 4 '12 at 16:46
You are right. So the proof is simpler. – Cantlog Aug 5 '12 at 1:29
Your assertion "...any element $f(x,y,z,t)\in R$ can be written as $f(x,y,z,t)=g(x,y,z)+th(x,y).$" is not true. It is false for $t^2$, for example . – Georges Elencwajg Aug 5 '12 at 6:34
I'm trying to learn this sort of approach. I believe I know such things as "the number of poles is the number of zeros" and "all the bad stuff happens on E". I believe "the divisor of f is a multiple of E" is stating that f cannot have both poles and zeros on E. Can someone describe why that is so? – Jack Schmidt Aug 5 '12 at 16:54
@user26565, the proof above is specific to the case when $K$ is a field. In general, the results doesn't hold, e.g. if $K$ is a ring with nilpotent elements, then $1+v\in R^*$ is a unit in $R$ for any nilpotent $v$. But this is probably the only possibility. – Cantlog Aug 6 '12 at 13:39

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 $f\in K-\{0\}$.

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I'm strongly reminded of the reasoning I gave in Bungo's question on associates (now buried in one of the many edits): anyway +1. I wonder just how far this type of reasoning with the units can be extended... – zcn Aug 26 '14 at 19:44