# Zariski topology of $\mathbb{C} \times \mathbb{C}$ is finer than the product topology of Zariski topologies on $\mathbb{C}$

I have seen some of the posts on this topic with a slightly algebraic geometric flavour, such as Zariski topology finer than product topology. I try to understand this from a purely topological viewpoint. So the question is, is it true, that the Zariski topology on $$\mathbb{C} \times \mathbb{C}$$ is finer than the product of the Zariski topologies on $$\mathbb{C}$$?

I started with the observation that

$$B = \big\{ U \subseteq \mathbb{C} \times \mathbb{C} \; \vert \; U_1, U_2 \subset \mathcal{O}^{Z}_\mathbb{C}: U = \text{pr}_{1}^{-1}(U_1) \cap \text{pr}^{-1}_{2}(U_2) \big\}$$

defines a basis of the product of the Zariski topologies, which I'll denote as $$\mathcal{O}_{\mathbb{C} \times \mathbb{C}}$$. With $$\mathcal{O}_{\mathbb{C}}^{Z}$$ I denoted the Zariski topology on $$\mathbb{C}$$. $$\text{pr}_i$$ are the canonical surjections for $$\mathbb{C} \times \mathbb{C}$$. Let $$A \subset \mathcal{O}_{\mathbb{C} \times \mathbb{C}}$$ be open, then there are $$U_i \subseteq B$$ such that $$A = \bigcup_{i \in I} U_i$$. Then there are open sets $$V \subseteq \mathcal{O}_{\mathbb{C}\times \mathbb{C}}^{Z}$$ such that $$A \subseteq V$$ is open. Thus, we have that $$\mathcal{O}_{\mathbb{C} \times \mathbb{C}} \subseteq \mathcal{O}^{Z}_{\mathbb{C} \times \mathbb{C}}$$. Therefore, the product of the Zariski topology is finer than the product of the Zariski topologies on $$\mathbb{C}$$.

But I still need to show that this does not work out the other way around, such that $$\mathcal{O}_{\mathbb{C} \times \mathbb{C}} \subset \mathcal{O}^{Z}_{\mathbb{C} \times \mathbb{C}}$$ is a real subset. I tried an argument with the Zariski topology not being Hausdorff, but I stuck with the construction of an open set being in $$\mathcal{O}^{Z}_{\mathbb{C} \times \mathbb{C}}$$ but not in $$\mathcal{O}_{\mathbb{C} \times \mathbb{C}}$$. Thank you for your help in advance.

• On $\mathbb{C}$ a closed subset (in Żarówki topology) is either whole $\mathbb{C}$ or finite. So open subsets are cofinite. And thus in the product open subsets have to be cofinite as well. So all you have to do is to find an open subset in Zariski topology on $\mathbb{C}^2$ which is not cofinite. Or equivalently a (proper) closed subset which is not finite. Or equivalently a polynomial in 2 variables which has infinitely many zeros. May 29, 2022 at 14:22
• Is $\mathbb{C} \times \{0\}$ such a subset? May 29, 2022 at 14:35
• Let's see. What polynomial has this set as zeros? How about $p(x,y)=y$? Yes, that works. May 29, 2022 at 14:39
• Ah okay, that helped a lot. May 29, 2022 at 15:01
• @freakish It is not true that a product of cofinite topologies is a cofinite topology. For instance, $\mathbb C\times\{0\}$ (which Reikiri asked about) is closed in the product topology. May 29, 2022 at 18:56

The comments of @freakish have given a false impression. The product topology makes all projections $$\pi_j:\prod_iX_i\to X_j$$ continuous, and is the coarsest topology with this property.
The result is that there is a sub-basis of the open sets, consisting of the “cylindrical” sets $$\pi_j^{-1}U_j$$ where $$U_j$$ is an open set of one of the factors. So, for instance, $$\Bbb C\times\{w_0\}$$ is closed in the product topology, as is each $$\{z_0\}\times\Bbb C$$. Notice that limiting the closed sets to the cofinite subsets of $$\Bbb C\times\Bbb C$$ does not make either projection continuous. To get the product topology from the subbase, you allow finite intersections of the cylindrical sets, and arbitrary unions of these.
For a Zariski-closed subset, take any obliqilue line $$w=mz+b$$ for $$m\ne0$$, or, much more realistically, a circle $$z^2+w^2=1$$.