Proving that the Zariski topology in $\mathbb{A}^2$ is not the product topology of $\mathbb{A}^1\times \mathbb{A}^1$

Question

I am trying to prove this:

The Zariski topology in $\mathbb{A}^2$ is not the product topology of $\mathbb{A}^1\times \mathbb{A}^1$

Here is the proof:

We will work over a field $k$ which is algebraically closed.

1. We will work with the set $S=V(x-y)$. This set is closed in $\mathbb{A}^2$. We will show that $\mathbb{A}^2-S$ is not open in $\mathbb{A}^1\times \mathbb{A}^1$ (which gives a contradiction).

2. To show that $\mathbb{A}^2-S$ is not open, we will show that no basic set of $\mathbb{A}^1\times \mathbb{A}^1$ is contained inside $\mathbb{A}^2-S$. The basic sets of $\mathbb{A}^1\times \mathbb{A}^1$ are of the form $(\mathbb{A}^1-V(f))\times (\mathbb{A}^1-V(g))$ where $f,g\in k[x]$ ($k[x]$ is a PID as $k$ is a field).

3. Let $V(f)=\{x_1, x_2,\dots, x_n\}$ and $V(g)=\{y_1, y_2,\dots, y_n\}$. We can find a $z\in k$ such that $f(z)\neq 0$ and $g(z)\neq 0$. Then the point $(z,z)\in \mathbb{A}^2$ has the property that $(z,z)\in (\mathbb{A}^1-V(f))\times (\mathbb{A}^1-V(g))$ but $(z,z)\notin \mathbb{A}^2-S$, which proves our claim.

Is the proof fine ?

Answer 1

This is correct ! The geometric argument is that any curve $C \subset \mathbb A^2$ is a closed subset in the usual topology but not in the product topology.