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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?

JDZ
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Luigi M
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  • Any two curves in $P^2$ have a nonempty intersection. – Elle Najt Dec 01 '14 at 16:02
  • @AreaMan I know that, I do not know how to prove that curves are send to curves (because I think this will generate the absurd you mean) otherwise, how can I prove that this is a topological invariant? – Luigi M Dec 01 '14 at 16:05
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    curves = infinite closed subsets not equal to the entire space – user8268 Dec 01 '14 at 16:14
  • @user8268 does your definition only works in this case? And if I add an external point to a curve will I obtain a curve as well? – Luigi M Dec 01 '14 at 16:23
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    You need to say a curve is a proper closed subset, each of whose components is infinite-a hyperplane with a disjoint point is not a curve. – Kevin Carlson Dec 01 '14 at 16:28
  • This is a very subtle question: the answer has to be expressed in purely topological terms. However $\mathbb P^2_k$ has a certain purely topological property on intersections of irreducible sets whose proof is obtained by superimposing on $\mathbb P^2_k$ a supplementary structure, that of an algebraic variety, and then proving a hard theorem (Bézout) about that structure. – Georges Elencwajg Dec 01 '14 at 20:31
  • @GeorgesElencwajg I've proved Bézout thm during lectures, I'm trying to prove the property of intersections of irreducible sets. The problem is that we did not see lots of algebraic geometry results, we focused on lin. alg. groups instead – Luigi M Dec 01 '14 at 20:48

3 Answers3

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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).

Georges Elencwajg
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  • Thanks for the answer! Actually I think this is the most comprehensible (according to my level of knowledge) answer. I've only one question, I've seen (during lectures) Bezout applied to projective variety of the form $V(f)$, can I use it on a general irreducible closed subset? (or is there a way to reduce the irreducible case to this case?) – Luigi M Dec 01 '14 at 20:09
  • Dear Luigi, yes every irreducible closed subset $C\subsetneq \mathbb P^2_k$ which is not a point is of the form $S=V(f)$ , where $f\in k[X,Y,Z]$ is some homogeneous polynomial of positive degree, unique up to multiplication by a non-zero constant $c\in k^*$: $S=V(f)=V(cf)$. – Georges Elencwajg Dec 01 '14 at 20:18
  • Dear Georges, I've sketch out a proof of this fact, could you please give it a look if it works (I don't feel comfortable enough to rely only on my opinion) Assume V irreducible variety with an infinite number of points in $\mathbb{P}^2$. We can write $V= \cap_{i=1}^r V(f_i)$ thanks to the Noetherianity of $\mathbb{P}^2$. Being $V$ irreducible, we can assume that, for all $i$, $V(f_i)$ are irreducible, otherwise we can restrict to one irreducible component, because $V$ will lie in an irreducible component of each of this varieties. – Luigi M Dec 01 '14 at 21:19
  • So being $V(f_i)$ irreducible we have that $f_i$ is an irreducible homogeneous polynomial. So by Bézout, $V(f_i) \cap V(f_j)$ is a finite number of point, so $V$ is a point being irreducible, absurd. So $V=V(f_i)$ – Luigi M Dec 01 '14 at 21:20
  • Dear Luigi, yes $C$ cannot be included in the intersection of two $V(f_i)$'s but if the course is well organized, you don't need the full force of Bézout to prove that. – Georges Elencwajg Dec 01 '14 at 22:36
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    Is there a method to prove $\mathbf{A}^{2}$ is not homeomorphic to $\mathbf{P}^{2}$ in a more elementary way, without relying on Bézout's Theorem (in case one has not learned it yet)? – JDZ Jun 11 '19 at 20:45
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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.

Moss
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  • But is being complete preserved by mere homeomorphisms of varieties? – Hoot Dec 01 '14 at 18:25
  • Dear gabbering, your argument does not make sense because you are not allowed to use the concept of variety in topology. More concretely , a complete variety can be homeomorphic to a non-complete variety, so that your argument does not prove that $\mathbb A^2_k$ and $\mathbb P^2_k$ are not homeomorphic. – Georges Elencwajg Dec 01 '14 at 19:48
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    Dear @Hoot: you are perfectly right, completeness is not preserved by mere homeomorphisms. For example $\mathbb A^1_k$ and $\mathbb P^1_k$ are homeomorphic. – Georges Elencwajg Dec 01 '14 at 20:00
  • @GeorgesElencwajg: Sorry, I am mistaken. Though I don't understand why a complete variety can be isomorphic to a non-complete variety. – Moss Dec 01 '14 at 21:50
  • Dear gabbering, all irreducible curves over $k$ are homeomomorphic to $\mathbb A^1_k$! They are homeomorphic to $k$ endowed with the cofinite topology, in which the closed sets of $k$ are 1°) $k$ and 2°) the finite subsets of $k$. – Georges Elencwajg Dec 01 '14 at 22:10
  • @GeorgesElencwajg so inparticular $\mathbb{P}^1$ is an irreducible curve over $k$ (is this correct?) hence $\mathbb{P}^1$ and $\mathbb{A}^1$ are homeomorphic?

    They are not however, isomorphic as varieties.

     – Moss Dec 01 '14 at 22:12
  • Yes, $\mathbb P^1_k$ and $\mathbb A^1_k$ are definitely homeomorphic, even if they are not isomorphic at all as varieties. Zariski topology is weird and it took the genius of Zariski and Serre to realize that it could be useful. – Georges Elencwajg Dec 01 '14 at 22:16
  • Indeed, this is strange! Thanks for clearing this up! Should I delete this answer now? – Moss Dec 01 '14 at 22:17
  • Maybe you can modify your answer, emphasizing that while $\mathbb A^2_k$ is not complete and $\mathbb P^2_k$ is, these properties are not purely topological but use the supplementary algebraic variety structure. – Georges Elencwajg Dec 01 '14 at 22:23
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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.

Tom
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