Why is the Zariski topology coarser than standard topology

Question

I'm trying to learn about the Zariski topology (without prior knowledge of algebraic sets). I'm asked to prove that if $\tau_1$ is the Zariski topology on $\Bbb{C}^2$ and $\tau_2$ is the standard topology on $\Bbb{C}^2$ then $\tau_1$ is coarser than $\tau_2$. So proving $\tau_1\subseteq\tau_2$ wasn't hard - we can think of every open set in $\tau_1$ basis as $U_f=\{x\in\Bbb{C}^2:f(x)\neq 0\}=\Bbb{C}^2\setminus f^{-1}(\{0\})$ which is obviously open in $\tau_2$ since $f$ is continuous. The problem starts when trying to show $\tau_1\subsetneq\tau_2$. What open set in $\tau_2$ is not open in $\tau_2$?

Answer 1

It's a good first step to understand the version with $\mathbb{C}$ in place of $\mathbb{C}^2$. Here we're looking at single-variable polynomials over $\mathbb{C}$, and these are relatively simple. In particular, we have a good understanding of $\{u: f(u)=0\}$ for such an $f$:

It's either finite or all of $\mathbb{C}$.

This means that any closed in the usual sense subset of $\mathbb{C}$ not satisfying this same "size condition" cannot be Zariski closed. For example:

The unit disc $\{x+yi: x^2+y^2\le 1\}$ is closed in the usual sense and infinite but not all of $\mathbb{C}$.

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OK, now how can we lift this to $\mathbb{C}^2$?

Well, there are various ways to do this, but one I quite like is to consider sections. Suppose $f(u,v)$ is a polynomial over $\mathbb{C}$ in two variables. Fix some $z\in\mathbb{C}$; we then get a single-variable polynomial $$g(u)=f(u,z).$$

• What can we say about $\{u: g(u)=0\}$ (thinking about the previous section of this answer?

• How does that give an example of a subset $A\subseteq\mathbb{C}^2$ which is closed in the usual topology but not in the Zariski topology?

HINT: think about $[0,1]$ ...

Answer 2

Here's a quick proof:

Every Zariski closed subset is quasi-compact (in the Zariski topology). If $\tau_1$ were equal to $\tau_2$, then the same would be true of standard closed sets (in the standard topology), but of course this is not the case, e.g. $\mathbb{C}^2$ itself is closed and not quasi-compact in the standard topology.