Let $K$ a field not necessarily algebraically closed. I would like to find the coordinate ring of the hyperbola $$ V(XY-1)\subseteq K^{2}. $$ If the field was algebraically closed we could use Hilbert's Nullstellensatz to say that $$ I(V(XY-1))=\sqrt{(XY-1)}=(XY-1)\Rightarrow A(V(XY-1))=K[X,Y]/(XY-1) $$ How could we prove that $I(V(XY-1))=XY-1$ (If this is the case) if $K$ is not algebraically closed?
It would also be easy if the curve was parametrizable, but the hyperbola is not.
Of course we need the field to be infinite. Here are two proofs:
We can do division with remainder w.r.t to $X$ and get $f=h(XY-1)+r$ with $r \in \frac{1}{Y}k[Y]$ and $h \in k(Y)[X].$ For any $x \in K^*$ we evaluate this equation at $(\frac{1}{x},x)$ and get $r(x)=0$, in particular $xr(x)=0$, hence $Yr \in k[Y]$ is the zero-polynomial, hence $r=0$. In particular this implies that $h(XY-1) \in k[Y][X]$. Looking at the coefficients of this polynomial it is easy to deduce that $h \in k[Y][X]$, from which we can conclude that $f \in (XY-1)$.
Note that $K[X,Y]/(XY-1)$ is a one-dimensional domain, i.e. $K[X,Y]/J$ is zero-dimensional if equality does not hold. Then $V(J)$ is finite, but $V(J)=V(XY-1)$ is infinite.
