10
$\begingroup$

Could you please give a hint how to show that the zariski topology on $\mathbb{A}^2$ is not the product topology on $\mathbb{A}^1\times\mathbb{A}^1$

$\endgroup$
  • 1
    $\begingroup$ This is exercise 1.2.2. in Karen Smith's Invitation to Algebraic Geometry. The hint given in the book is to consider the diagonal. $\endgroup$ – student Mar 8 '16 at 0:49
13
$\begingroup$

Assuming $k$ is infinite, consider the topological properties of the line $x=y$.

$\endgroup$
  • 4
    $\begingroup$ Note also that if $k$ is finite, the two topologies do coincide: both are the discrete topology on $k \times k$. $\endgroup$ – Pete L. Clark May 21 '12 at 13:06
17
$\begingroup$

Just to add to Prof Magidin's example: it's a common exercise in point set topology to prove that a topological space $X$ is Hausdorff if and only if the diagonal $\Delta = \{(x, x) : x \in X\}$ is closed in the product topology on $X \times X$. And when $k$ is infinite [in particular, when $k$ is algebraically closed] you know that $\mathbf A^1$ is not Hausdorff: any two non-empty open sets intersect.

By the way, this consideration of the diagonal motivates the correct analogue of the Hausdorff property in algebraic geometry: the notion of a separated morphism of schemes.

$\endgroup$
2
$\begingroup$

Zariski closed sets in $\mathbb{A}^1$ are sets of finitely many points.

So, open sets are compliments of finitely many points.

Let $X,Y$ be topological spaces, then product topology on $X\times Y$ is topology generated by basis elements of the form $U\times V$ where $U$ is open in $X$ and $V$ is open in $Y$.

So, product topology on $\mathbb{A}^1\times \mathbb{A}^1$ is generated by $$\mathbb{A}^1\setminus\{a_1\cdots,a_r\}\times \mathbb{A}^1\setminus\{b_1,\cdots,b_l\}=\mathbb{A}^1\times \mathbb{A}^1\setminus\{(a_i,b_j):1\leq i\leq r;1\leq j\leq l\}$$

Topology generated by basis elements $\{U_{\alpha}\}$ is countable unions of elements of the collection $\{U_{\alpha}\}$

So, any open set in product topology is countable union of sets of the form $$\mathbb{A}^1\times \mathbb{A}^1\setminus\{(a_i,b_j):1\leq i\leq r;1\leq j\leq l\}$$

Countable union of finitely many points is countable.

So, any open set in product topology is complement of countably many points.

So, any closed set in product topology is set of countably many points.

Assuming we are in $\mathbb{C}$, $\{(x,y):x=y\}$ is an infinite uncountable set, closed in Zariski topology, thus not a closed set in product topology.

Thus, product topology is not the same as zariski topology.

$\endgroup$
  • $\begingroup$ [Zariski closed sets in $\mathbb{A}^1=\mathbb{A}^1(k)$ are sets of finitely many points. ]--->Don't you have to suppose that the ground ring $k$ is a domain i.e. a ring without zero divisor ? $\endgroup$ – Duchamp Gérard H. E. Sep 2 '17 at 13:14
  • $\begingroup$ [Topology generated by basis elements $\{U_{\alpha}\}$ is countable unions of elements of the collection $\{U_{\alpha}\}$]-->Unless I missed something in this case, it seems that there is no such countability condition in general, could you elaborate ? $\endgroup$ – Duchamp Gérard H. E. Sep 2 '17 at 13:18
  • $\begingroup$ @DuchampGérardH.E. I assumed that $k$ is a field. Should I add that in answer? $\endgroup$ – user87543 Sep 3 '17 at 3:56
  • 1
    $\begingroup$ [Should I add that in answer?]--->yes, it is safer, and also (a) [compliments]--->[complements] (b) explain why can you restrict to countable unions (I do not see ...). Otherwise the counter example is good but, in my opinion, not for this reason. $\endgroup$ – Duchamp Gérard H. E. Sep 3 '17 at 13:11
  • $\begingroup$ "So, any closed set in product topology is set of countably many points." that is false. Consider a finite set times the whole line, that is uncountable and certainly is not the whole space. $\endgroup$ – Gaston Burrull Jan 18 at 4:19

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.