A topological space is Noetherian if there are no infinite descending chains of closed sets. Show that the Zariski topology on $\mathbb{K}^n $ is Noetherian.
I know and understand what is topological space, and Zariski topology.
The Zariski topology on any field $\mathbb{K}^n$ has as closed sets $Z(I)$, where $I \subseteq R = \mathbb{K}[x_1 ,... , x_n]$ an ideal with property $Z(0)= \mathbb{K}^n $ and $Z(1)= \emptyset $.
How I should prove it? Thank you so much !
If $I$ and $J$ are ideal on $R:=k[x_1,\dots,x_n]$, and $I<J$, then $V(J)\subset V(I)$, and viceversa. Then descending chain of closed subsets of $\mathbb{A}^n$ are equivalent to an ascending chain of ideals in the ring $R$. But Hilbert's basis theorem (https://en.wikipedia.org/wiki/Hilbert%27s_basis_theorem) tells us that $R$ is Noetherian. It means that any ascending chain of ideals in $R$ is finite, so any descending chain of closed subsets of $\mathbb{A}^n$ is also finite.
A very naive and easy way to see that $I<J$ implies that $V(J)\subset V(I)$ is that $I$ is generated with fewer equations than $J$, then the set of point vanishing in the generators of $I$ is bigger than the set of points whose vanish in all the generators of $J$.