If the product of closed sets is closed in the product topology shouldn't {(0,0)} be closed in $A^1 \times A^1$ zariski topology.

Question

I am not sure what is incorrect about the statement.

If the product of closed sets is closed in the product topology shouldn't {(0,0)} be closed in $A^1 \times A^1$ zariski topology, i.e singletons should be closed.

Answer 1

Two things :

• (let us assume the base field is algebraically closed) whenever $a$ and $b$ are closed points in $\mathbb{A}^1$, $(a,b)$ is indeed closed in $\mathbb{A}^2=\mathbb{A}^1\times \mathbb{A}^1$
• the Zariski topology on $\mathbb{A^2}$ is not the product topology, it is finer. For example $\{y-x^2=0\}$ is not closed under the product topology but it is under the Zariski topology.