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 $textrm{Spec}(A)$ 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
edited Nov 10 at 18:17
მამუკა ჯიბლაძე
7,810242110
asked Nov 10 at 18:05
Hans
812411
add a comment |
up vote
9
down vote
favorite
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
edited Nov 10 at 18:17
მამუკა ჯიბლაძე
7,810242110
asked Nov 10 at 18:05
Hans
812411
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
6
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
@KarlSchwede - You may want to take look at my comment below.
– Pierre-Yves Gaillard
Nov 10 at 20:03
add a comment |
up vote
9
down vote
favorite
up vote
9
down vote
favorite
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
edited Nov 10 at 18:17
მამუკა ჯიბლაძე
7,810242110
asked Nov 10 at 18:05
Hans
812411
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
edited Nov 10 at 18:17
მამუკა ჯიბლაძე
7,810242110
asked Nov 10 at 18:05
Hans
812411
edited Nov 10 at 18:17
მამუკა ჯიბლაძე
7,810242110
asked Nov 10 at 18:05
Hans
812411
edited Nov 10 at 18:17
მამუკა ჯიბლაძე
7,810242110
edited Nov 10 at 18:17
მამუკა ჯიბლაძე
7,810242110
edited Nov 10 at 18:17
მამუკა ჯიბლაძე
7,810242110
7,810242110
asked Nov 10 at 18:05
Hans
812411
asked Nov 10 at 18:05
Hans
812411
asked Nov 10 at 18:05
Hans
812411
812411
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
6
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
@KarlSchwede - You may want to take look at my comment below.
– Pierre-Yves Gaillard
Nov 10 at 20:03
add a comment |
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
6
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
@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
6
6
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
@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
add a comment |
1 Answer
1
active
oldest
votes
up vote
8
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.
answered Nov 10 at 19:55
David Lampert
1,709169
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
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
8
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.
answered Nov 10 at 19:55
David Lampert
1,709169
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
add a comment |
up vote
8
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.
answered Nov 10 at 19:55
David Lampert
1,709169
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
add a comment |
up vote
8
down vote
accepted
up vote
8
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.
answered Nov 10 at 19:55
David Lampert
1,709169
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 at 19:55
David Lampert
1,709169
answered Nov 10 at 19:55
David Lampert
1,709169
answered Nov 10 at 19:55
David Lampert
1,709169
answered Nov 10 at 19:55
David Lampert
1,709169
1,709169
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
add a comment |
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
add a comment |
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$mathrm{Spec}$ 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
6
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
@KarlSchwede - You may want to take look at my comment below.
– Pierre-Yves Gaillard
Nov 10 at 20:03