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

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    $\begingroup$ $\mathrm{Spec}$ 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
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    $\begingroup$ 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
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    $\begingroup$ 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 non-discrete valuation ring associated to $\mathbb{Z} \times \mathbb{Z}$ with the lex order). $\endgroup$ – Karl Schwede Nov 10 '18 at 18:31
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    $\begingroup$ @KarlSchwede - You may want to take look at my comment below. $\endgroup$ – Pierre-Yves 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 $a \in Q-P_2$ then 1 = dim$(A/a)$ = dim$(A)-1$ = 2 by the principal ideal theorem.

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