This is an Exercise in An Invitation to Algebraic Geometry by Karen Smith. I'm not sure what the hint means thus have no clue how to approach. Any thought please? Thanks very much!

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    $\begingroup$ The diagonal in $\mathbb{A}^2$ is closed. Is $\mathbb{A}^1$ Hausdorff? $\endgroup$ – Zhen Lin Jan 10 '16 at 17:23

As already pointed out in the comments one elegant way to show this is by using that $\mathbb{A^1}$ is not Hausdorff, and thus the diagonal is not closed in the product topology. Another, more image-like proof is the following:

The Zariski topology on $\mathbb{A}^1$ is just the cofinite topology. Hence a basis of the product topology is given by the products $U \times V$ where $U, V \subseteq \mathbb{A}^1$ are both only missing finitely many points (if you try to visualize $\mathbb{A}^2$ as $\mathbb{R}^2$ then you can imagine these basis elements as the plane $\mathbb{R}^2$ where we remove finitely many lines parellel to the axes.)

Now take a look at the diagonal $\Delta = \{(x,x) \mid x \in \mathbb{A}^1\}$. Using the above visualization we can already “see” that the complement of $\Delta$ cannot be written as the union of the basis elements described above.

More formally let $U, V \subseteq \mathbb{A}^1$ be open. If $U \neq \emptyset$ or $V \neq \emptyset$, i.e. if $U \times V \neq \emptyset$, then there exist $x \in U \cap V$ because we are dealing with the cofinite topology and $\mathbb{A}^1$ is infinite. Thus $(x,x) \in U \times V$ and $(U \times V) \cap \Delta \neq \emptyset$. Therefore we cannot write $\Delta^C$, which is non-empty, as the union of basis elements $U \times V$ with $U,V \subseteq \mathbb{A}^1$ open.

So $\Delta$ is not closed in the product topology. But $\Delta$ is closed in the Zariski topology of $\mathbb{A}^2$, because it is the zero locus of $P(x,y) = x-y$.

(Notice however, that the product topology is coarser then the Zariski topology.)

PS: For visualization I also find it very useful to think about the Zariski topology on $\mathbb{R}^2$ and the product of the Zariski topology on $\mathbb{R} \times \mathbb{R}$. Instead of the diagonal an even more intuitive subset is then the unit circle $S^1 \subseteq \mathbb{R}^2$. This is certainly closed in the Zariski topology, as it is given as the zero locus of $x^2+y^2-1$. If $S^1$ were closed in the product topology on $\mathbb{R} \times \mathbb{R}$, then we could write $S^1$ by taking a finite number of lines, all parallel to the axes, and then intersecting this by some other lines, all of which are also parallel to the axes. This seems really absurd.

  • $\begingroup$ Or just use the fact that the diagonal being closed in the product topology is equivalent to the original topology being Hausdorff. Not that I'm saying your answer is bad; visualization for intuition is very useful. Just pointing it out to readers/the asker. $\endgroup$ – Dustan Levenstein Jan 10 '16 at 17:40
  • $\begingroup$ This is certainly the more elegant proof, but I like the visual aspect behind this. I added a short nod to this more elegant proof at the beginning. $\endgroup$ – Jendrik Stelzner Jan 10 '16 at 17:44
  • $\begingroup$ @JendrikStelzner Thank you! There is one little spot I'm not clear about. I see that $(U \times V) \cap \Delta \ne \varnothing$. But why "we cannot write $\Delta^C$, which is non-empty, as the union of basis elements $U \times V$ with $U, V \subseteq A^1$ open"? $\endgroup$ – nekodesu Jan 10 '16 at 18:07
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    $\begingroup$ Well, suppose we can write $\Delta^C = \bigcup_{i \in I} (U_i \times V_i)$ where $U_i, V_i \subseteq \mathbb{A}^1$ are open. Then $U_i \times V_i \subseteq \Delta^C$ for all $i \in I$. This is the same as $(U_i \times V_i) \cap \Delta = \emptyset$ for all $i \in I$. But we know that this is only possible if $U_i \times V_i = \emptyset$ for all $i \in I$. So $\Delta^C = \bigcup_{i \in I} (U_i \times V_i) = \emptyset$, which is wrong. $\endgroup$ – Jendrik Stelzner Jan 10 '16 at 18:09
  • $\begingroup$ @JendrikStelzner Thanks! This is very clear now! $\endgroup$ – nekodesu Jan 10 '16 at 23:23

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