For context I'm trying to follow a proof that is linked in the answer to this question, but I'm stuck on one particular point.
Let $k$ be an algebraically closed field, and let $A$ and $B$ be distinct projective planes in $\mathbb{P}_k^3$ intersecting only at a line $L$, and let $X=A\cup B$. Let $D$ be a line on $A$ distinct from $L$, intersecting $L$ only at the point $P$.
Set $U=X\setminus D$. Then any $f\in\Gamma(U)$ is constant on $B\setminus D=B\setminus\{P\}$.
I'm not sure why this should be the case.
I know that a regular function on the projective plane must be constant, and I know that the ring of regular functions of an affine plane with a point removed is isomorphic to the of regular functions on the plane itself.
I thought then that perhaps the same is true for the projective plane. If we say $P=(0:0:1)$, then $B\setminus D=D^+(x)\cup D^+(y)$. Then any regular function on $B\setminus D$ can be written as $\frac{f(x,y,z)}{x^n}$ on $D^+(x)$ and $\frac{g(x,y,z)}{y^m}$ on $D^+(y)$. These must agree on $D^+(x)\cap D^+(y)=D^+(xy)$, but I can't see how to get a contradiction from here.
If non-constant regular functions do exist on $B\setminus D$ then we must somehow use the fact that $f$ is regular on all of $U$, but I can't see how.
Perhaps I've misread the argument in the proof, but everything else follows if I can show this, and I feel like I'm missing something obvious.
Any help would be much appreciated.
You can use the extension theorem for normal varieties to reduce to the projective case.
Here is the statement from Liu's book (Theorem 1.14 Chapter 4), though there should be a version of it in most books of AG (like Shafarevich or Hartshorne).