According to Hartshorne's "Algebraic Geometry", we say that a topological space $X$ is a Zariski space if it is noetherian (i.e. $X$ satisfies the descending chain condition for closed subsets) and every nonempty closed irreducible subset has a unique generic point (Exercise II.3.17.). In this exercise, we show that the underlying topological space of a noetherian scheme is a Zariski space and here arises a question: conversely, every Zariski space occurs as an underlying topological space of some scheme?

I know that if we remove the Zariski assumption and require only to be noetherian, this assertion does not hold, because, for example, $X=\{x,y\}$ with indiscrete topology gives an counterexample. Indeed, this space is obviously noetherian but never appears as an underlying space of scheme because $X$ has no closed points. However, when $X$ is a Zariski space, I can neither find a counterexample nor prove this statement.

Thank you!

  • 2
    $\begingroup$ A note on terminology: the condition that every irreducible closed subset has a unique generic point is called "sober", so in your second paragraph you would be removing the assumption that $X$ is sober, not the "Zariski assumption". As far as actually answering the question, the place to start looking is Mel Hochster's thesis - he characterizes the topological spaces which are the underlying spaces of affine schemes and I bet there's a result in there which points towards what you want. I'll see if I can write an answer later, but I wanted to point you in a hopeful direction, at least. $\endgroup$
    – KReiser
    Jul 23, 2021 at 0:40

1 Answer 1


Every Zariski space is the underlying topological space of an affine scheme. This follows from Hochster's classification of spectral spaces (spaces that are homeomorphic to Spec of a ring) as exactly the sober spaces in which compact open sets are closed under finite intersections and are a basis for the topology. In a Noetherian space, every subset is compact, so every Noetherian sober space is a spectral space.

Note, though, that not every Noetherian sober space is the underlying topological space of a Noetherian scheme. For instance, a Noetherian ring of dimension greater than $1$ always has infinitely many prime ideals (this is a consequence of Krull's principal ideal theorem and prime avoidance; see here for instance). So, a finite sober (or equivalently, just finite $T_0$) space of dimension greater than $1$ cannot be the underlying topological space of a Noetherian scheme, even though it is the spectrum of some (non-Noetherian!) ring.


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