Noetherian spectral space comes from noetherian ring?












9














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






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







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  • 1




    $mathrm{Spec}$ 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.
    – Qfwfq
    Nov 10 at 18:19






  • 1




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






  • 7




    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).
    – Karl Schwede
    Nov 10 at 18:31








  • 1




    @KarlSchwede - You may want to take look at my comment below.
    – Pierre-Yves Gaillard
    Nov 10 at 20:03
















9














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






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







share|cite|improve this question




















  • 1




    $mathrm{Spec}$ 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.
    – Qfwfq
    Nov 10 at 18:19






  • 1




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






  • 7




    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).
    – Karl Schwede
    Nov 10 at 18:31








  • 1




    @KarlSchwede - You may want to take look at my comment below.
    – Pierre-Yves Gaillard
    Nov 10 at 20:03














9












9








9


4





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






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 $textrm{Spec}(A)$ for some commutative ring $A$. If $X$ is noetherian, does there also exist a noetherian ring $B$ such that $X=textrm{Spec}(B)$?






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




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






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 10 at 18:17









მამუკა ჯიბლაძე

7,846243110




7,846243110










asked Nov 10 at 18:05









Hans

817512




817512








  • 1




    $mathrm{Spec}$ 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.
    – Qfwfq
    Nov 10 at 18:19






  • 1




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






  • 7




    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).
    – Karl Schwede
    Nov 10 at 18:31








  • 1




    @KarlSchwede - You may want to take look at my comment below.
    – Pierre-Yves Gaillard
    Nov 10 at 20:03














  • 1




    $mathrm{Spec}$ 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.
    – Qfwfq
    Nov 10 at 18:19






  • 1




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






  • 7




    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).
    – Karl Schwede
    Nov 10 at 18:31








  • 1




    @KarlSchwede - You may want to take look at my comment below.
    – Pierre-Yves Gaillard
    Nov 10 at 20:03








1




1




$mathrm{Spec}$ 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.
– Qfwfq
Nov 10 at 18:19




$mathrm{Spec}$ 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.
– Qfwfq
Nov 10 at 18:19




1




1




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




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




7




7




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).
– Karl Schwede
Nov 10 at 18:31






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).
– Karl Schwede
Nov 10 at 18:31






1




1




@KarlSchwede - You may want to take look at my comment below.
– Pierre-Yves Gaillard
Nov 10 at 20:03




@KarlSchwede - You may want to take look at my comment below.
– Pierre-Yves Gaillard
Nov 10 at 20:03










1 Answer
1






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8














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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  • 6




    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
    Nov 10 at 19:57













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8














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

















  • 6




    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
    Nov 10 at 19:57


















8














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

















  • 6




    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
    Nov 10 at 19:57
















8












8








8






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 Nov 10 at 19:55









David Lampert

1,729169




1,729169








  • 6




    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
    Nov 10 at 19:57
















  • 6




    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
    Nov 10 at 19:57










6




6




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
Nov 10 at 19:57






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
Nov 10 at 19:57




















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