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?

  • $\begingroup$ Any two curves in $P^2$ have a nonempty intersection. $\endgroup$ – Lorenzo Dec 1 '14 at 16:02
  • $\begingroup$ @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? $\endgroup$ – Luigi M Dec 1 '14 at 16:05
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    $\begingroup$ curves = infinite closed subsets not equal to the entire space $\endgroup$ – user8268 Dec 1 '14 at 16:14
  • $\begingroup$ @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? $\endgroup$ – Luigi M Dec 1 '14 at 16:23
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    $\begingroup$ 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. $\endgroup$ – Kevin Carlson Dec 1 '14 at 16:28

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).

  • $\begingroup$ 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?) $\endgroup$ – Luigi M Dec 1 '14 at 20:09
  • $\begingroup$ 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 $c\in k^*$: $S=V(f)=V(cf)$. $\endgroup$ – Georges Elencwajg Dec 1 '14 at 20:18
  • $\begingroup$ 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. $\endgroup$ – Luigi M Dec 1 '14 at 21:19
  • $\begingroup$ 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 $V=V(f_i)$ $\endgroup$ – Luigi M Dec 1 '14 at 21:20
  • $\begingroup$ 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. $\endgroup$ – Georges Elencwajg Dec 1 '14 at 22:36

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.


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.

  • $\begingroup$ But is being complete preserved by mere homeomorphisms of varieties? $\endgroup$ – Hoot Dec 1 '14 at 18:25
  • $\begingroup$ 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. $\endgroup$ – Georges Elencwajg Dec 1 '14 at 19:48
  • $\begingroup$ 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. $\endgroup$ – Georges Elencwajg Dec 1 '14 at 20:00
  • $\begingroup$ @GeorgesElencwajg: Sorry, I am mistaken. Though I don't understand why a complete variety can be isomorphic to a non-complete variety. $\endgroup$ – Moss Dec 1 '14 at 21:50
  • $\begingroup$ 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 $k$. $\endgroup$ – Georges Elencwajg Dec 1 '14 at 22:10

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.

  • $\begingroup$ $\mathbb{A}^2$ is compact in the Zariski topology (quasicompact if for you compact implies Hausdorff.) Otherwise this would be a nice and simple proof. $\endgroup$ – Kevin Carlson Dec 1 '14 at 16:32
  • $\begingroup$ I think that compactness doesn't behave very well with zariski topology (every variety is compact) $\endgroup$ – Luigi M Dec 1 '14 at 16:32
  • $\begingroup$ Replace "compact" by "proper", and the argument works fine. @Luigi $\endgroup$ – Fredrik Meyer Dec 1 '14 at 17:19
  • $\begingroup$ @Frederik: proper over what and in what sense? I don't think any argument involving topological properness will work. $\endgroup$ – Georges Elencwajg Dec 1 '14 at 19:28

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