Suppose $m,n>0$.Prove that the Zariski topology of $K^n\times K^m$is not equal to the product of the Zariski topologies of $K^n$ and $K^m$.
I have thought:
Take $ K =\mathbb{C}$ and see the shape of the curves forming cerraso in $ \mathbb{C}\times\mathbb{C}$ and $ {\mathbb{C}} ^2$
On
If you are a newcomer, you can first better visualize this for $K=\mathbb{R}$ and $m=n=1$.
The Zariski open sets on $K$ are the complements of finite subsets (and so for every field), in the case of $\mathbb{R}$ note $U_F:=\mathbb{R}\setminus F$ for finite $F$. In the product topology, a fundamental system of neighbourhoods of a point $(x,y)\in \mathbb{R}^2$ is then of the form $U_{F_1}\times U_{F_2}$ with $x\notin F_1$ and $y\notin F_2$ (such a set can be visualized as the complement of finite tartan). By this you see that the line $x=y$ is dense in $\mathbb{R}^2$ (for the product topology) as it always contains an infinite rectangle $]N_1,+\infty[\times ]N_2,+\infty[$ whereas $x=y$ is closed for the Zariski topology of $\mathbb{R}^2$. The (formal) proof is similar (Dieudonné would have written mutatis mutandis) for $K=\mathbb{C}$.
Here's a hint: all closed subsets of $K^m\times K^n$, with the product topology, are (very roughly) of the form $\{(x,y):f(x)=0\text{ and }g(y)=0\}$. The subsets of $K^{m+n}$, with the Zariski topology, are of the form $\{(x,y):f(x,y)=0\}$.