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

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?

• What do you mean by "together with Krull dimension on closed sets"? Nov 14, 2018 at 21:30
• Actually it doesn't really add any more information. I was just emphasising the dimension on sets. If there's such an enrichment, I want it to come with a dimension that extends the original dimension. But that comes naturally, I think, with the noetherian topology. Nov 14, 2018 at 21:34

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.