0
$\begingroup$

I am not sure what is incorrect about the statement.

If the product of closed sets is closed in the product topology shouldn't {(0,0)} be closed in $A^1 \times A^1$ zariski topology, i.e singletons should be closed.

$\endgroup$

1 Answer 1

3
$\begingroup$

Two things :

  • (let us assume the base field is algebraically closed) whenever $a$ and $b$ are closed points in $\mathbb{A}^1$, $(a,b)$ is indeed closed in $\mathbb{A}^2=\mathbb{A}^1\times \mathbb{A}^1$
  • the Zariski topology on $\mathbb{A^2}$ is not the product topology, it is finer. For example $\{y-x^2=0\}$ is not closed under the product topology but it is under the Zariski topology.
$\endgroup$
13
  • 1
    $\begingroup$ Why is it not closed under the product topology? $\endgroup$
    – ben huni
    Apr 18, 2023 at 9:04
  • 2
    $\begingroup$ Proper closed subsets of $\mathbb{A}^1$ are finite, so closed subsets of the product are also finite, or of the form $\mathbb{A^1}\times \mathrm{finite}$. $\{y-x^2\}$ is not. $\endgroup$
    – vanxoo
    Apr 18, 2023 at 9:13
  • 1
    $\begingroup$ @benhuni first of all: two subsets being homeomorphic does not imply that any of them is closed in the ambient space. Secondly what makes you think that the projection is closed/open/quotient map? One of those has to be true for your claim to make sense. But we are not considering product topology here, so standard properties of projection don't hold. And in this case none of them hold, and vanxoo gave you the proof. $\endgroup$
    – freakish
    Apr 18, 2023 at 10:44
  • 1
    $\begingroup$ @benhuni I don't understand what you are saying. What is not true? What do you mean by "topologically not true"? Homeomorphism between what exactly? And why does it matter? Even if the solution set is homeomorphic to $A^1$, so what? This is a different, unrelated question. $\endgroup$
    – freakish
    Apr 18, 2023 at 11:44
  • 2
    $\begingroup$ @benhuni one last time: homeomorphism between what? Existence of some random homeomorphism has nothing to do with the question whether a subset is closed or not. For example $(0,1)$ is homeomorphic to $\mathbb{R}$ but it doesn't make it closed in $\mathbb{R}$. Again: these are unrelated things. There's nothing wrong with the answer. $\endgroup$
    – freakish
    Apr 18, 2023 at 15:31

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .