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 $X=\textrm{Spec}(B)$?

1$\begingroup$ $\mathrm{Spec}$ 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. $\endgroup$ – Qfwfq Nov 10 '18 at 18:19

1$\begingroup$ Oh yes, you're totally right, it's the underlying top space of the Spec not the scheme $\endgroup$ – Qfwfq Nov 10 '18 at 18:21

7$\begingroup$ 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 nondiscrete valuation ring associated to $\mathbb{Z} \times \mathbb{Z}$ with the lex order). $\endgroup$ – Karl Schwede Nov 10 '18 at 18:31

1$\begingroup$ @KarlSchwede  You may want to take look at my comment below. $\endgroup$ – PierreYves Gaillard Nov 10 '18 at 20:03

$\begingroup$ relevant: mathoverflow.net/a/330735/141498 $\endgroup$ – user141498 Jun 10 '19 at 7:35
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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 $a \in QP_2$ then 1 = dim$(A/a)$ = dim$(A)1$ = 2 by the principal ideal theorem.

6$\begingroup$ See also top of p. 48 in Ring constructions on spectral spaces by Christopher Francis Tedd escholar.manchester.ac.uk/jrul/item/?pid=ukacmanscw:307012  link to the PDF file: escholar.manchester.ac.uk/api/… $\endgroup$ – PierreYves Gaillard Nov 10 '18 at 19:57