Rings of Krull dimension one












3












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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.










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$endgroup$












  • $begingroup$
    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
















3












$begingroup$


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.










share|cite|improve this question











$endgroup$












  • $begingroup$
    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














3












3








3





$begingroup$


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.










share|cite|improve this question











$endgroup$




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






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share|cite|improve this question













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share|cite|improve this question








edited Jun 16 '16 at 21:18









user26857

39.3k124183




39.3k124183










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


















  • $begingroup$
    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










1 Answer
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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.




  1. A UFD is a PID if and only if it is one-dimensional.

  2. 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.)

  3. One-dimensional domains have almost stable rank 1.

  4. 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.






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    1 Answer
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    1 Answer
    1






    active

    oldest

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    active

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    active

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    2












    $begingroup$

    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.




    1. A UFD is a PID if and only if it is one-dimensional.

    2. 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.)

    3. One-dimensional domains have almost stable rank 1.

    4. 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.






    share|cite|improve this answer











    $endgroup$


















      2












      $begingroup$

      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.




      1. A UFD is a PID if and only if it is one-dimensional.

      2. 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.)

      3. One-dimensional domains have almost stable rank 1.

      4. 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.






      share|cite|improve this answer











      $endgroup$
















        2












        2








        2





        $begingroup$

        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.




        1. A UFD is a PID if and only if it is one-dimensional.

        2. 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.)

        3. One-dimensional domains have almost stable rank 1.

        4. 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.






        share|cite|improve this answer











        $endgroup$



        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.




        1. A UFD is a PID if and only if it is one-dimensional.

        2. 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.)

        3. One-dimensional domains have almost stable rank 1.

        4. 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.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Dec 16 '18 at 11:50

























        answered Jun 21 '16 at 1:51









        Jason JuettJason Juett

        81759




        81759






























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