Noetherian spectral space comes from noetherian ring?
9
$begingroup$
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
edited Nov 10 '18 at 18:17
მამუკა ჯიბლაძე
8,267345114
asked Nov 10 '18 at 18:05
HansHans
817512
$endgroup$
add a comment |
9
$begingroup$
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
edited Nov 10 '18 at 18:17
მამუკა ჯიბლაძე
8,267345114
asked Nov 10 '18 at 18:05
HansHans
817512
$endgroup$
1
$begingroup$
$mathrmSpec$ 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.
$endgroup$
– Qfwfq
Nov 10 '18 at 18:19
1
$begingroup$
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
7
$begingroup$
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 $mathbbZ times mathbbZ$ with the lex order).
$endgroup$
– Karl Schwede
Nov 10 '18 at 18:31
1
$begingroup$
@KarlSchwede - You may want to take look at my comment below.
$endgroup$
– Pierre-Yves Gaillard
Nov 10 '18 at 20:03
add a comment |
9
9
9
4
$begingroup$
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
edited Nov 10 '18 at 18:17
მამუკა ჯიბლაძე
8,267345114
asked Nov 10 '18 at 18:05
HansHans
817512
$endgroup$
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
edited Nov 10 '18 at 18:17
მამუკა ჯიბლაძე
8,267345114
asked Nov 10 '18 at 18:05
HansHans
817512
edited Nov 10 '18 at 18:17
მამუკა ჯიბლაძე
8,267345114
asked Nov 10 '18 at 18:05
HansHans
817512
edited Nov 10 '18 at 18:17
მამუკა ჯიბლაძე
8,267345114
edited Nov 10 '18 at 18:17
მამუკა ჯიბლაძე
8,267345114
edited Nov 10 '18 at 18:17
მამუკა ჯიბლაძე
8,267345114
8,267345114
asked Nov 10 '18 at 18:05
HansHans
817512
asked Nov 10 '18 at 18:05
HansHans
817512
asked Nov 10 '18 at 18:05
HansHans
817512
817512
1
$begingroup$
$mathrmSpec$ 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.
$endgroup$
– Qfwfq
Nov 10 '18 at 18:19
1
$begingroup$
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
7
$begingroup$
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 $mathbbZ times mathbbZ$ with the lex order).
$endgroup$
– Karl Schwede
Nov 10 '18 at 18:31
1
$begingroup$
@KarlSchwede - You may want to take look at my comment below.
$endgroup$
– Pierre-Yves Gaillard
Nov 10 '18 at 20:03
add a comment |
1
$begingroup$
$mathrmSpec$ 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.
$endgroup$
– Qfwfq
Nov 10 '18 at 18:19
1
$begingroup$
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
7
$begingroup$
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 $mathbbZ times mathbbZ$ with the lex order).
$endgroup$
– Karl Schwede
Nov 10 '18 at 18:31
1
$begingroup$
@KarlSchwede - You may want to take look at my comment below.
$endgroup$
– Pierre-Yves Gaillard
Nov 10 '18 at 20:03
1
1
$begingroup$
$mathrmSpec$ 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.
$endgroup$
– Qfwfq
Nov 10 '18 at 18:19
$begingroup$
$mathrmSpec$ 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.
$endgroup$
– Qfwfq
Nov 10 '18 at 18:19
1
1
$begingroup$
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
$begingroup$
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
7
7
$begingroup$
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 $mathbbZ times mathbbZ$ with the lex order).
$endgroup$
– Karl Schwede
Nov 10 '18 at 18:31
$begingroup$
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 $mathbbZ times mathbbZ$ with the lex order).
$endgroup$
– Karl Schwede
Nov 10 '18 at 18:31
1
1
$begingroup$
@KarlSchwede - You may want to take look at my comment below.
$endgroup$
– Pierre-Yves Gaillard
Nov 10 '18 at 20:03
$begingroup$
@KarlSchwede - You may want to take look at my comment below.
$endgroup$
– Pierre-Yves Gaillard
Nov 10 '18 at 20:03
add a comment |
1 Answer
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oldest
votes
8
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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.
answered Nov 10 '18 at 19:55
David LampertDavid Lampert
1,729169
$endgroup$
6
$begingroup$
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/…
$endgroup$
– Pierre-Yves Gaillard
Nov 10 '18 at 19:57
add a comment |
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1 Answer
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1 Answer
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8
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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.
answered Nov 10 '18 at 19:55
David LampertDavid Lampert
1,729169
$endgroup$
6
$begingroup$
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/…
$endgroup$
– Pierre-Yves Gaillard
Nov 10 '18 at 19:57
add a comment |
8
$begingroup$
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.
answered Nov 10 '18 at 19:55
David LampertDavid Lampert
1,729169
$endgroup$
6
$begingroup$
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/…
$endgroup$
– Pierre-Yves Gaillard
Nov 10 '18 at 19:57
add a comment |
8
8
8
$begingroup$
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.
answered Nov 10 '18 at 19:55
David LampertDavid Lampert
1,729169
$endgroup$
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.
answered Nov 10 '18 at 19:55
David LampertDavid Lampert
1,729169
answered Nov 10 '18 at 19:55
David LampertDavid Lampert
1,729169
answered Nov 10 '18 at 19:55
David LampertDavid Lampert
1,729169
answered Nov 10 '18 at 19:55
David LampertDavid Lampert
1,729169
1,729169
6
$begingroup$
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/…
$endgroup$
– Pierre-Yves Gaillard
Nov 10 '18 at 19:57
add a comment |
6
$begingroup$
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/…
$endgroup$
– Pierre-Yves Gaillard
Nov 10 '18 at 19:57
6
6
$begingroup$
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/…
$endgroup$
– Pierre-Yves Gaillard
Nov 10 '18 at 19:57
$begingroup$
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/…
$endgroup$
– Pierre-Yves Gaillard
Nov 10 '18 at 19:57
add a comment |
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1
$begingroup$
$mathrmSpec$ 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.
$endgroup$
– Qfwfq
Nov 10 '18 at 18:19
1
$begingroup$
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
7
$begingroup$
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 $mathbbZ times mathbbZ$ with the lex order).
$endgroup$
– Karl Schwede
Nov 10 '18 at 18:31
1
$begingroup$
@KarlSchwede - You may want to take look at my comment below.
$endgroup$
– Pierre-Yves Gaillard
Nov 10 '18 at 20:03