Problem in proving that $\mathbb{A}^2$ is not homeomorphic to $\mathbb{P}^2$

Question

Let $k$ be an algebraically closed field. All spaces are equipped with the usual Zariski topologies.

All the proofs of this fact that I've seen rely on the fact that two lines in $\mathbb{P}^2$ intersect but this doesn't necessarily hold in $\mathbb{A}^2$. I'm stuck on proving that this property is a "Zariski-topology invariant" (i.e preserved by homeomorphism). All the proofs use this fact without proving it, so I assume it is trivial, but I do not know how to prove it.

Does someone have a hint on how to prove it?

It would be enough for me to prove that lines in $\mathbb{A}^2$ are sent to projective lines to complete the proof, or that the image of an algebraic curve is an algebraic projective curve. But I cannot prove any of these. Any help?

Answer 1

There exist two disjoint irreducible closed subsets both containing more than one point in $\mathbb A^2_k$ but not in $\mathbb P^2_k$ (Bézout).

Answer 2

Well, there are two things.

1. Homeomorphism preserves compactness (can you see why? homeomorphisms being one-to-one preserves unions, inclusions and opennes).

2. $\mathbb{P}^2$ is compact, while $\mathbb{A}^2$ is not.

Answer 3

If you have a little more machinery from algebraic geometry you could notice that $\mathbb{P}^2$ is a complete variety while $\mathbb{A}^2$ is not. To explain a little:

One says that a variety $X$ is complete if for all varieties $Z$ the projection map $X\times Z\rightarrow Z$ is closed, ie sends closed sets to closed sets.

To see that $\mathbb{A}^2$ is not complete, consider $Z=\mathbb{A}^1$. Then $\mathbb{A}^2\times\mathbb{A}^1 = \mathbb{A}^3$, say with co-ordinates $(x,y,z)$. Consider the closed subset $V(xz-1)\subset \mathbb{A}^3$, then the projection map $p:\mathbb{A}^3\rightarrow\mathbb{A}^1$ onto the last factor sends: $$p(V(xz-1)) = \mathbb{A}^1 \setminus 0$$

which is not closed.

It is a (not so easy) theorem that $\mathbb{P}^n$ is a complete variety, see for example in Harris' Intro to Algebraic Geometry book.

This maybe a bit of a high powered answer to your question, but I think it uses some important concept that one should try to become familiar with.

$\textbf{Edit:}$

This does not provide an answer to the OP's question because it is possible for non-complete and complete varieties to be homeomorphic, for example $\mathbb{P}^1$ and $\mathbb{A}^1$. However, they are certainly not isomorphic as varieties.