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I'm pretty new to algebraic geometry. I have not clear what does it mean that the Zariski topology and the product topology over the affine space $\mathbb A^m × \mathbb A^n $ are different.

I'm convinced that if $X\subset \mathbb A^m$ and $Y\subset \mathbb A^n$ are affine variety, then $X×Y$ is an affine variety in $\mathbb A^m × \mathbb A^n$.

So actually, a closed set in the product topology is closed in Zariski topology as well.

I found in the notes that I'm reading an example of a set closed only in Zariski topology, however, I didn't understand it. It says "The subset $V(x-y)=\{(a,a):a∈K\}⊂\mathbb A^2$ is closed in the Zariski topology, but not in the product topology of $\mathbb A^1 × \mathbb A^1$".

However, shouldn't $\{(a,a):a∈K\}$ be closed in the product topology too?

Thanks in advance.

Tereza Tizkova
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Dr. Scotti
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    No, that set is only closed if the space is Hausdorff, but the Zariski topology is definitely not Hausdorff. – Tobias Kildetoft Mar 19 '20 at 17:48
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    The closed subsets of $\mathbb{A}^1$ are exactly the finite sets. What kinds of sets do you get taking the product of a finite set with a finite set? For concreteness, if $W_1 = {1,2,3}$ and $W_2 = {-2, 0}$, what does $W_1 \times W_2$ look like? – Viktor Vaughn Mar 19 '20 at 17:51
  • Yes of corse you're right. Is it true that a closed set in $X×Y $ is a union of set $U_i×V_i $ with $U_i $, $V_i $ closed respectively in $X$, $Y ?$ If yes, the $U_i $ in the example should cover all $\mathbb A^1 $, and the same would be true for the $V_i $. So the unique solution would be that $\mathbb A^1×\mathbb A^1$, and this is not the set of the example. – Dr. Scotti Mar 19 '20 at 19:48
  • @Dorian I'm not sure what you're trying to say in the last part of your comment, but maybe it's easier to see using open sets. Every open subset of $\mathbb{A}^1$ is the complement of a finite set, and every open set in the product topology (for a finite product, anyway) is of the form $U \times V$ where $U$ and $V$ are open subsets of $\mathbb{A}^1$. – Viktor Vaughn Mar 23 '20 at 03:39
  • @RichardD.James Sorry if I am missing something obvious, but "every open set in the product topology (for a finite product, anyway) is of the form $U×V$...." is not true, right? – P-addict Aug 14 '20 at 09:39
  • @P-addict You're right, I should've said every basic open set is of that form, so every open set is a union of such sets. – Viktor Vaughn Aug 14 '20 at 14:49
  • @RichardD.James Ok cool, but then your argument of finiteness doesn't work because we can have arbitrary union of such basic open sets, am I right? – P-addict Aug 14 '20 at 16:13

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