# Noetherian spectral space comes from noetherian ring?

Let $$X$$ be a spectral space (en.wikipedia.org/wiki/Spectral_space), i.e. a space of the form $$\textrm{Spec}(A)$$ for some commutative ring $$A$$. If $$X$$ is noetherian, does there also exist a noetherian ring $$B$$ such that $$X=\textrm{Spec}(B)$$?

• $\mathrm{Spec}$ is an (anti-)equivalence from commutative rings to affine scheme, so two rings are isomorphic iff their Spec's are. So, if such a noetherian $B$ exists, your $A$ was already isomorphic to it. – Qfwfq Nov 10 '18 at 18:19
• Oh yes, you're totally right, it's the underlying top space of the Spec not the scheme – Qfwfq Nov 10 '18 at 18:21
• I wonder if Hochster's thesis addresses this? Off the top of my head, I don't know how to make the following Noetherian topological space $\{p,q,r\}$, with open sets $\{ \{p,q,r\}, \{p,q\}, \{p\} \}$, as the spectrum of a Noetherian ring (it's the spectrum of the non-discrete valuation ring associated to $\mathbb{Z} \times \mathbb{Z}$ with the lex order). – Karl Schwede Nov 10 '18 at 18:31
• @KarlSchwede - You may want to take look at my comment below. – Pierre-Yves Gaillard Nov 10 '18 at 20:03
• relevant: mathoverflow.net/a/330735/141498 – user141498 Jun 10 '19 at 7:35

Graph $$N_5$$ with poset order topology (i.e. poset $$M=\{p,q,r\}, P_2=\{p,q\}, P_1=\{p\}, Q=\{r\}, N=\phi$$) is not Spec($$A$$) for Noetherian $$A$$ because if $$a \in Q-P_2$$ then 1 = dim$$(A/a)$$ = dim$$(A)-1$$ = 2 by the principal ideal theorem.