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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.

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Dave
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1 Answers1

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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).

Theorem: Let $X$ be a normal locally noetherian scheme. Let $F$ be a closed subset of $X$ of codimension at least 2. Then the restriction $$ \mathcal{O}_X(X)\rightarrow\mathcal{O}_X(X\setminus F)$$ is an isomorphism.

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