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მამუკა ჯიბლაძე
მამუკა ჯიბლაძე
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Let $X$ be a spectral space (en.wikipedia.org/wiki/Spectral_spaceen.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 $X=\textrm{Spec}(B)$?

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

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