Let $k$ be a field (possibly not algebraically closed), $F\in k[x,y,z]$ irreducible and homogeneous, $X:=V(F)\subset\mathbb{P}_k^2$. If $z\in k(X)$ is defined at every $P\in V(F)$, prove that $z\in k$. [hint: use Noether's fundamental theorem]
(this is exercise $5.30$ from W. Fulton's Algebraic Curves - an Introduction to Algebraic Geometry)
Here is my attempt: take $H,G\in k[x,y,z]$ homogeneous of same degree and such that $z=\frac{H}{G}$ for all $P\in\mathbb{P}^2$ with $G(P)\neq 0$. In particular, whenever $G(P)\neq 0$ we have: $$H-zG=0\text{ in }\mathcal{O}_{P,X}$$
But since $\mathcal{O}_{P,X}\simeq \mathcal{O}_{P,\mathbb{P}^2}/(F)\mathcal{O}_{P,\mathbb{P}^2}$, the above means $H-zG\in (F)\mathcal{O}_{P,\mathbb{P}^2}$, so there is some $w\in \mathcal{O}_{P,\mathbb{P}^2}$ with $H-zG=wF$, i.e., $H=wF+zG$.
If the construction above works for every $P\in\mathbb{P}^2$, then Noether's fundamental theorem states that there are homogeneous polynomials $A,B$ with degree $\deg(H)-\deg(F)$ and $\deg(H)-\deg(G)=0$ respectively, with $H=AF+BG$, so $\frac{H}{G}=\frac{AF}{G}+B=B\in k$ and we are done.
The problem is that I can't extend the construction for the case $G(P)=0$, so I'm stuck.
How do I solve this?
When $G(P)=0$, then $P\notin X$, hence $F(P)\ne0$ and so $F$ is a unit in $\mathcal O_{P,\Bbb P^2}$. Hence, $(F,G)_P=(1)$ also contains $H$.