Field of fractions over $k[X,Y]$

Question

Let $k$ be a algebraicly closed field and $p \in k^2$.

It is to show, that if $z \in k(k^2)=\mathrm{Frac}(k[X,Y])$ is defined in every point $q \in k^2\backslash \{p\}$, then it already holds that $z \in k[X,Y]$.

($z\in k[X,Y]$ defined in $q \in k^2$ means that $\exists u,v \in k[X,Y]$ with $v(q)\neq 0$ and $z=\frac{u}{v}$.)

I started with saying that $k^2 \backslash \{p\}\subseteq \mathrm{Def}(z)=\{q \in k^2 |\,\, \mathrm{z \,\, defined \,\, in \,\,} q\}$ and so it has to hold that $p \in \mathrm{Def}(z)$, because $z \in k[X,Y]$ if and only if $\mathrm{Pol}(z):=k^2 \backslash \mathrm{Def}(z)=\emptyset$ (something that was already proven). But that's as far as I got.

Answer 1

Take any two points $x_1,x_2\in \mathbf A^2$ and the appropriate $u_1,v_1,u_2,v_2$.

Since $k[X,Y]$ is UFD, you can assume without loss of generality that $u_1,v_1$ and $u_2,v_2$ have no common divisors.

Use this to show that $u_1v_2=u_2v_1$ and deduce that $v_1$ divides $v_2$ and vice versa, and conclude that $v_1(x_2)\neq 0$, and, since $x_2$ is arbitrary, by the Nullstellensatz, $v_1$ is constant.

Answer 2

Denote $\Gamma(V,\mathcal{O}_V)$ the ring of regular functions of an affine variety $V$ and $D(f)$ the set of points in $V$ not satisfying $f$. Use the fact that when $\Gamma(V,\mathcal{O}_V)$ is factorial, given $f_1,\dots,f_n \in \Gamma(V,\mathcal{O}_V)$ and their gcd $h$ the natural restriction homomorphism $$ r:\Gamma(D(h),\mathcal{O}_V) \to \Gamma(D(f_1)\cup \dots \cup D(f_n),\mathcal{O}_V) $$ is an isomorphism (you can prove it by induction on $n$). In your case $V=k^2$, $f_1=x$, $f_2=y$.