Noetherian spectral space comes from noetherian ring?
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
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
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
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
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
7,821242110
7,821242110
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
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
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
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
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.
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.
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.
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.
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
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 |
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmathoverflow.net%2fquestions%2f315031%2fnoetherian-spectral-space-comes-from-noetherian-ring%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
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