Noetherian spectral space comes from noetherian ring?

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






ag.algebraic-geometry ac.commutative-algebra gn.general-topology







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  • 1




    $mathrmSpec$ 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
    7 hours ago






  • 1




    Oh yes, you're totally right, it's the underlying top space of the Spec not the scheme
    – Qfwfq
    7 hours ago






  • 5




    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 $mathbbZ times mathbbZ$ with the lex order).
    – Karl Schwede
    6 hours ago











  • @KarlSchwede - You may want to take look at my comment below.
    – Pierre-Yves Gaillard
    5 hours ago














up vote
8
down vote

favorite
3












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






ag.algebraic-geometry ac.commutative-algebra gn.general-topology







share|cite|improve this question



















  • 1




    $mathrmSpec$ 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
    7 hours ago






  • 1




    Oh yes, you're totally right, it's the underlying top space of the Spec not the scheme
    – Qfwfq
    7 hours ago






  • 5




    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 $mathbbZ times mathbbZ$ with the lex order).
    – Karl Schwede
    6 hours ago











  • @KarlSchwede - You may want to take look at my comment below.
    – Pierre-Yves Gaillard
    5 hours ago












up vote
8
down vote

favorite
3









up vote
8
down vote

favorite
3






3





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






ag.algebraic-geometry ac.commutative-algebra gn.general-topology







share|cite|improve this question















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






ag.algebraic-geometry ac.commutative-algebra gn.general-topology




ag.algebraic-geometry ac.commutative-algebra gn.general-topology






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share|cite|improve this question













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edited 7 hours ago









მამუკა ჯიბლაძე

7,715242110




7,715242110










asked 7 hours ago









Hans

807411




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  • 1




    $mathrmSpec$ 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
    7 hours ago






  • 1




    Oh yes, you're totally right, it's the underlying top space of the Spec not the scheme
    – Qfwfq
    7 hours ago






  • 5




    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 $mathbbZ times mathbbZ$ with the lex order).
    – Karl Schwede
    6 hours ago











  • @KarlSchwede - You may want to take look at my comment below.
    – Pierre-Yves Gaillard
    5 hours ago












  • 1




    $mathrmSpec$ 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
    7 hours ago






  • 1




    Oh yes, you're totally right, it's the underlying top space of the Spec not the scheme
    – Qfwfq
    7 hours ago






  • 5




    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 $mathbbZ times mathbbZ$ with the lex order).
    – Karl Schwede
    6 hours ago











  • @KarlSchwede - You may want to take look at my comment below.
    – Pierre-Yves Gaillard
    5 hours ago







1




1




$mathrmSpec$ 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
7 hours ago




$mathrmSpec$ 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
7 hours ago




1




1




Oh yes, you're totally right, it's the underlying top space of the Spec not the scheme
– Qfwfq
7 hours ago




Oh yes, you're totally right, it's the underlying top space of the Spec not the scheme
– Qfwfq
7 hours ago




5




5




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 $mathbbZ times mathbbZ$ with the lex order).
– Karl Schwede
6 hours ago





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 $mathbbZ times mathbbZ$ with the lex order).
– Karl Schwede
6 hours ago













@KarlSchwede - You may want to take look at my comment below.
– Pierre-Yves Gaillard
5 hours ago




@KarlSchwede - You may want to take look at my comment below.
– Pierre-Yves Gaillard
5 hours ago










1 Answer
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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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  • 5




    See also top of p. 48 in Ring constructions on spectral spaces by Christopher Francis Tedd escholar.manchester.ac.uk/jrul/item/?pid=uk-ac-man-scw:307012 --- link to the PDF file: escholar.manchester.ac.uk/api/…
    – Pierre-Yves Gaillard
    5 hours ago











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1 Answer
1






active

oldest

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active

oldest

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active

oldest

votes








up vote
5
down vote



accepted










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.






share|cite|improve this answer
















  • 5




    See also top of p. 48 in Ring constructions on spectral spaces by Christopher Francis Tedd escholar.manchester.ac.uk/jrul/item/?pid=uk-ac-man-scw:307012 --- link to the PDF file: escholar.manchester.ac.uk/api/…
    – Pierre-Yves Gaillard
    5 hours ago















up vote
5
down vote



accepted










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.






share|cite|improve this answer
















  • 5




    See also top of p. 48 in Ring constructions on spectral spaces by Christopher Francis Tedd escholar.manchester.ac.uk/jrul/item/?pid=uk-ac-man-scw:307012 --- link to the PDF file: escholar.manchester.ac.uk/api/…
    – Pierre-Yves Gaillard
    5 hours ago













up vote
5
down vote



accepted







up vote
5
down vote



accepted






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.






share|cite|improve this answer












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.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered 5 hours ago









David Lampert

1,654169




1,654169







  • 5




    See also top of p. 48 in Ring constructions on spectral spaces by Christopher Francis Tedd escholar.manchester.ac.uk/jrul/item/?pid=uk-ac-man-scw:307012 --- link to the PDF file: escholar.manchester.ac.uk/api/…
    – Pierre-Yves Gaillard
    5 hours ago













  • 5




    See also top of p. 48 in Ring constructions on spectral spaces by Christopher Francis Tedd escholar.manchester.ac.uk/jrul/item/?pid=uk-ac-man-scw:307012 --- link to the PDF file: escholar.manchester.ac.uk/api/…
    – Pierre-Yves Gaillard
    5 hours ago








5




5




See also top of p. 48 in Ring constructions on spectral spaces by Christopher Francis Tedd escholar.manchester.ac.uk/jrul/item/?pid=uk-ac-man-scw:307012 --- link to the PDF file: escholar.manchester.ac.uk/api/…
– Pierre-Yves Gaillard
5 hours ago





See also top of p. 48 in Ring constructions on spectral spaces by Christopher Francis Tedd escholar.manchester.ac.uk/jrul/item/?pid=uk-ac-man-scw:307012 --- link to the PDF file: escholar.manchester.ac.uk/api/…
– Pierre-Yves Gaillard
5 hours ago


















 

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