Rings of Krull dimension one
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I have to write a monograph about commutative rings with Krull dimension $1$, but I can't find results, so I am looking foward for some references, and some results to search. Also, I would appreciate a lot to know if there is some result of the kind:
$$ dim(A)=1 iff ~?$$
Thanks in advance.
abstract-algebra ring-theory commutative-algebra krull-dimension
edited Jun 16 '16 at 21:18
user26857
39.3k124183
asked Jun 16 '16 at 21:17
e.turattie.turatti
811517
$endgroup$
add a comment |
$begingroup$
I have to write a monograph about commutative rings with Krull dimension $1$, but I can't find results, so I am looking foward for some references, and some results to search. Also, I would appreciate a lot to know if there is some result of the kind:
$$ dim(A)=1 iff ~?$$
Thanks in advance.
abstract-algebra ring-theory commutative-algebra krull-dimension
edited Jun 16 '16 at 21:18
user26857
39.3k124183
asked Jun 16 '16 at 21:17
e.turattie.turatti
811517
$endgroup$
$begingroup$
I'm afraid there is no such characterisation. A discrete valuation ring has dimension one, and there's a characterisation of Dedekind domains in terms of dimension (among other properties).
$endgroup$
– Bernard
Jun 16 '16 at 22:20
$begingroup$
This is way too broad. Maybe make some further assumptions like noetherian domains of Krull dimension 1.
$endgroup$
– MooS
Jun 17 '16 at 9:03
add a comment |
$begingroup$
I have to write a monograph about commutative rings with Krull dimension $1$, but I can't find results, so I am looking foward for some references, and some results to search. Also, I would appreciate a lot to know if there is some result of the kind:
$$ dim(A)=1 iff ~?$$
Thanks in advance.
abstract-algebra ring-theory commutative-algebra krull-dimension
edited Jun 16 '16 at 21:18
user26857
39.3k124183
asked Jun 16 '16 at 21:17
e.turattie.turatti
811517
$endgroup$
I have to write a monograph about commutative rings with Krull dimension $1$, but I can't find results, so I am looking foward for some references, and some results to search. Also, I would appreciate a lot to know if there is some result of the kind:
$$ dim(A)=1 iff ~?$$
Thanks in advance.
abstract-algebra ring-theory commutative-algebra krull-dimension
abstract-algebra ring-theory commutative-algebra krull-dimension
edited Jun 16 '16 at 21:18
user26857
39.3k124183
asked Jun 16 '16 at 21:17
e.turattie.turatti
811517
edited Jun 16 '16 at 21:18
user26857
39.3k124183
asked Jun 16 '16 at 21:17
e.turattie.turatti
811517
edited Jun 16 '16 at 21:18
user26857
39.3k124183
edited Jun 16 '16 at 21:18
user26857
39.3k124183
edited Jun 16 '16 at 21:18
user26857
39.3k124183
39.3k124183
asked Jun 16 '16 at 21:17
e.turattie.turatti
811517
asked Jun 16 '16 at 21:17
e.turattie.turatti
811517
asked Jun 16 '16 at 21:17
e.turattie.turatti
811517
811517
$begingroup$
I'm afraid there is no such characterisation. A discrete valuation ring has dimension one, and there's a characterisation of Dedekind domains in terms of dimension (among other properties).
$endgroup$
– Bernard
Jun 16 '16 at 22:20
$begingroup$
This is way too broad. Maybe make some further assumptions like noetherian domains of Krull dimension 1.
$endgroup$
– MooS
Jun 17 '16 at 9:03
add a comment |
$begingroup$
I'm afraid there is no such characterisation. A discrete valuation ring has dimension one, and there's a characterisation of Dedekind domains in terms of dimension (among other properties).
$endgroup$
– Bernard
Jun 16 '16 at 22:20
$begingroup$
This is way too broad. Maybe make some further assumptions like noetherian domains of Krull dimension 1.
$endgroup$
– MooS
Jun 17 '16 at 9:03
$begingroup$
I'm afraid there is no such characterisation. A discrete valuation ring has dimension one, and there's a characterisation of Dedekind domains in terms of dimension (among other properties).
$endgroup$
– Bernard
Jun 16 '16 at 22:20
$begingroup$
I'm afraid there is no such characterisation. A discrete valuation ring has dimension one, and there's a characterisation of Dedekind domains in terms of dimension (among other properties).
$endgroup$
– Bernard
Jun 16 '16 at 22:20
$begingroup$
This is way too broad. Maybe make some further assumptions like noetherian domains of Krull dimension 1.
$endgroup$
– MooS
Jun 17 '16 at 9:03
$begingroup$
This is way too broad. Maybe make some further assumptions like noetherian domains of Krull dimension 1.
$endgroup$
– MooS
Jun 17 '16 at 9:03
add a comment |
1 Answer
1
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While there is no characterization of one-dimensional rings, there are various theorems involving them. Here is a small sample, off the top of my head.
- A UFD is a PID if and only if it is one-dimensional.
- A valuation domain is completely integrally closed if and only if it is a one-dimensional. Therefore a one-dimensional Prufer domain is completely integrally closed. (But the converse is false. For example, the ring of integer-valued polynomials is a 2-dimensional completely integrally closed Prufer domain.)
- One-dimensional domains have almost stable rank 1.
- One-dimensional Bezout domains are elementary divisor rings.
Edited: Accidentally wrote "rings" instead of "domains" in a couple spots. Corrected an error pointed out in the comments.
answered Jun 21 '16 at 1:51
Jason JuettJason Juett
81759
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add a comment |
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$begingroup$
While there is no characterization of one-dimensional rings, there are various theorems involving them. Here is a small sample, off the top of my head.
- A UFD is a PID if and only if it is one-dimensional.
- A valuation domain is completely integrally closed if and only if it is a one-dimensional. Therefore a one-dimensional Prufer domain is completely integrally closed. (But the converse is false. For example, the ring of integer-valued polynomials is a 2-dimensional completely integrally closed Prufer domain.)
- One-dimensional domains have almost stable rank 1.
- One-dimensional Bezout domains are elementary divisor rings.
Edited: Accidentally wrote "rings" instead of "domains" in a couple spots. Corrected an error pointed out in the comments.
answered Jun 21 '16 at 1:51
Jason JuettJason Juett
81759
$endgroup$
add a comment |
$begingroup$
While there is no characterization of one-dimensional rings, there are various theorems involving them. Here is a small sample, off the top of my head.
- A UFD is a PID if and only if it is one-dimensional.
- A valuation domain is completely integrally closed if and only if it is a one-dimensional. Therefore a one-dimensional Prufer domain is completely integrally closed. (But the converse is false. For example, the ring of integer-valued polynomials is a 2-dimensional completely integrally closed Prufer domain.)
- One-dimensional domains have almost stable rank 1.
- One-dimensional Bezout domains are elementary divisor rings.
Edited: Accidentally wrote "rings" instead of "domains" in a couple spots. Corrected an error pointed out in the comments.
answered Jun 21 '16 at 1:51
Jason JuettJason Juett
81759
$endgroup$
add a comment |
$begingroup$
While there is no characterization of one-dimensional rings, there are various theorems involving them. Here is a small sample, off the top of my head.
- A UFD is a PID if and only if it is one-dimensional.
- A valuation domain is completely integrally closed if and only if it is a one-dimensional. Therefore a one-dimensional Prufer domain is completely integrally closed. (But the converse is false. For example, the ring of integer-valued polynomials is a 2-dimensional completely integrally closed Prufer domain.)
- One-dimensional domains have almost stable rank 1.
- One-dimensional Bezout domains are elementary divisor rings.
Edited: Accidentally wrote "rings" instead of "domains" in a couple spots. Corrected an error pointed out in the comments.
answered Jun 21 '16 at 1:51
Jason JuettJason Juett
81759
$endgroup$
While there is no characterization of one-dimensional rings, there are various theorems involving them. Here is a small sample, off the top of my head.
- A UFD is a PID if and only if it is one-dimensional.
- A valuation domain is completely integrally closed if and only if it is a one-dimensional. Therefore a one-dimensional Prufer domain is completely integrally closed. (But the converse is false. For example, the ring of integer-valued polynomials is a 2-dimensional completely integrally closed Prufer domain.)
- One-dimensional domains have almost stable rank 1.
- One-dimensional Bezout domains are elementary divisor rings.
Edited: Accidentally wrote "rings" instead of "domains" in a couple spots. Corrected an error pointed out in the comments.
answered Jun 21 '16 at 1:51
Jason JuettJason Juett
81759
edited Dec 16 '18 at 11:50
answered Jun 21 '16 at 1:51
Jason JuettJason Juett
81759
answered Jun 21 '16 at 1:51
Jason JuettJason Juett
81759
answered Jun 21 '16 at 1:51
Jason JuettJason Juett
81759
81759
add a comment |
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$begingroup$
I'm afraid there is no such characterisation. A discrete valuation ring has dimension one, and there's a characterisation of Dedekind domains in terms of dimension (among other properties).
$endgroup$
– Bernard
Jun 16 '16 at 22:20
$begingroup$
This is way too broad. Maybe make some further assumptions like noetherian domains of Krull dimension 1.
$endgroup$
– MooS
Jun 17 '16 at 9:03