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I'm trying to prove the Zariski topology on $\mathbb A^2$ is not the product topology on $\mathbb A^1\times \mathbb A^1$.

I'm looking for a counter-example based on the fact the closed subsets in $\mathbb A^1$ are the finite ones.

Thanks in advance

user42912
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    I don't have a proof, but I am guessing that in general Zariski topology on $\mathbb{A}^{m+n}$ is not the product topology on $\mathbb{A}^{m}\times\mathbb{A}^{n}$ for any $m, n\in\mathbb{N}$. Could someone verify this? – Prism Aug 12 '14 at 15:10
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    @Prism, I think we can get there from the fact that the product topology on $\mathbb A^2=\mathbb A^1 \times \mathbb A^1$ is not the Zariski topology. Given $k \leq j$, Zariski subspace topology on $\mathbb A^k \subset \mathbb A^{j}$ coincides with the the Zariski topology on $\mathbb A^k$. Considering $\mathbb A^1 \times \mathbb A^1 \subset \mathbb A^m \times \mathbb A^n$ (choose one factor from each), if the two topologies coincide for any $n,m$, then we get a contradiction of the $n=m=1$ case. – vociferous_rutabaga Aug 12 '14 at 15:21
  • @MorganO: Brilliant! :) – Prism Aug 12 '14 at 16:08

3 Answers3

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$(x,x)$ is closed in $\mathbb{A}^2$ being defined by the equation $y-x=0$.

The product topology gives finite sets of points and the horizontal and vertical lines.

user26857
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Rene Schipperus
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A variation on the answer of Rene:

The diagonal $\Delta\subset \mathbb{A}^1\times\mathbb{A}^1$ in the Zariski topology is closed. If it were closed in the product topology, that would imply that $\mathbb{A}^1$ was Hausdorff (with the Zariski topology), which is obviously false. Thus the two topologies cannot coincide.

Daniel Robert-Nicoud
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@Rene has given you a great counterexample. Another way to arrive at an answer is to completely characterize closed sets in $\mathbb A^1 \times \mathbb A^1$ with the product topology, using the fact that closed subsets of $\mathbb A^1$ are finite.

Sets of the form $A \times B$ for $A,B$ sets in $\mathbb A^1$ with finite complements form a basis for the product topology. It's easy to verify that $A\times B = \mathbb A^1 \setminus \{L_1, \dots, L_k\}$ for $L_1, \dots, L_k$ "vertical" or "horizontal" lines (that is, one coordinate or the other is constant). Taking complements, closed subsets of $\mathbb A^1 \times \mathbb A^1$ with the product topology are intersections of finite unions of these lines. Can you show that not all closed sets of $\mathbb A^2$ are of this form?

vociferous_rutabaga
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  • I have to prove this before to use @Rene's argument, no? in another words, I have to characterize the closed subsets of $\mathbb A^1\times \mathbb A^1$ before prove the diagonal is a closed subset of $\mathbb A^2$. – user42912 Aug 12 '14 at 15:29
  • @user42912 You might be able to come up with some other reason why his set cannot be closed, without characterizing all closed sets in the product topology. But you'll have to go through some sort of argument, at which point you might as well hammer out the complete characterization =) – vociferous_rutabaga Aug 12 '14 at 15:31
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    Which other reason? I'm thinking of the following argument: Since $\mathbb A^2$ is not Hausdorff, a diagonal in $\mathbb A^2$ is not closed. – user42912 Aug 12 '14 at 15:39
  • @user42912 That's a great idea! But I think you mean "Since $\mathbb A^1$ is not Hausdorff, the diagonal of $\mathbb A^1 \times \mathbb A^1$ is not closed". You're appealing to this fact, yes? For the record, I did not have a particular reason in mind. – vociferous_rutabaga Aug 12 '14 at 15:41
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    yes, I'm sorry for my mistake. Is it what you thought when you said "you might be able to come up with some other reason why his set cannot be closed..."? – user42912 Aug 12 '14 at 15:44
  • @user42912 no, I was just thinking that there are generally many different ways to show things. – vociferous_rutabaga Aug 12 '14 at 15:46