Non-compact subsets of the Zariski space of an integral domain

Abstract

Let V be a minimal valuation overring of an integral domain D and let Zar(D) be the Zariski space of the valuation overrings of D. Starting from a result in the theory of semistar operations, we prove a criterion under which the set Zar(D) \ {V } is not compact. We then use it to prove that, in many cases, Zar(D) is not a Noetherian space, and apply it to the study of the spaces of Kronecker function rings and of Noetherian overrings.

Isolated points of the Zariski space

“…We start from the case L = K(X). Compare the next two results with [28,Corollary 5.5(a)] and [30,Proposition 4.2]. Proposition 7.5.…”

“…In this paper, we want to study the points of Zar(L|D) that are isolated, with respect to the constructible topology. Our starting point is a new interpretation of a result about the compactness of spaces in the form Zar(D) \ {V } [28,Theorem 3.6]: in particular, we show that if V is isolated then V is the integral closure of D[x 1 , . .…”

When the Zariski space is a Noetherian space

“…where X, Y are indeterminates on Q, since in this case Zar(D) can be written as the union of the quotient field of D and two sets homeomorphic to Zar(Q[X]) ≃ Spec(Q[X]), which are Noetherian; from this, it is possible to build examples of non-Prüfer domain whose Zariski spectrum is Noetherian, and having arbitrary finite dimension [22,Example 4.7].…”

“…However, since D is not a field, Zar(D[X]) is not Noetherian by [22,Proposition 5.4]; hence, neither Zar(L|D) can be Noetherian.…”

“…In [22], it was shown that in many cases Zar(D) is not a Noetherian space, i.e., there are subspaces of Zar(D) that are not compact. In particular, it was shown that Zar(D)\{V } (where V is a minimal valuation overring of D) is often non-compact: for example, this happens when dim(V ) > 2 dim(D) [22,Proposition 4.3] or when D is Noetherian and dim(V ) ≥ 2 [22,Corollary 5.2].…”