Is the Zariski topology on a variety $V$ a maximal Noetherian topology?

Question

Let $K$ be an algebraically closed field. By a variety $V$ definable over $K$, I mean a quasi-projective or an algebraic variety in sense of Weil. It is the set of points in an affine or a projective space over some bigger algebraically closed field $L$.

Now consider the Zariski topology $\tau$ on $V$ together with Krull dimension on closed sets. Is it possible to enrich this topology in the naive sense by adding new closed sets, so that the enriched topology $\tau^\prime$ is also a Noetherian topology?

In other words, is the Zariski topology on $V$ a maximal noetherian topology?

Answer 1

Given any Noetherian topological space $(X,\tau)$, if you adjoin one more closed set $C$ the topology always remains Noetherian. That is, let $\tau'$ be the topology generated by $\tau\cup\{X\setminus C\}$; I claim $\tau'$ is Noetherian too. To simplify the argument, I will in fact enlarge $\tau'$ further to the topology $\tau''$ generated by $\tau\cup\{C,X\setminus C\}$. This topology $\tau''$ has a very simple description: it is just the disjoint union topology on $X$ when we decompose $X$ as $C\coprod (X\setminus C)$ and give both $C$ and $X\setminus C$ the subspace topology induced by $\tau$. But $C$ and $X\setminus C$ are both Noetherian in the subspace topology, and a disjoint union of finitely many Noetherian spaces is Noetherian, so $\tau''$ is Noetherian. Explicitly, given a descending sequence of $\tau''$-closed sets, their intersections with $C$ must eventually stabilize and their intersections with $X\setminus C$ must also eventually stabilize, so the entire sequence must eventually stabilize.

So, there do not exist any maximal Noetherian topologies, besides the discrete topologies on finite sets. Any non-discrete Noetherian topology can always be enlarged to another Noetherian topology.