Let $X$ be a spectral space ([en.wikipedia.org/wiki/Spectral\_space](https://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 $X=\textrm{Spec}(B)$?



Noetherian spectral space comes from noetherian ring?
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.


