Krull dimension, how to compute?











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There is something I don't get about Krull's dimension.
If $K$ is a field, then the only ideals are ${ 0 }$ and $K$ so the only prime ideal is ${0}$. So the Krull's dimension is $0$. But if we have an Artinian ring, we have that every prime ideal is maximal. So if we take P a maximal ideal, ${0 } subsetneq P$ is a chain of ideal which seems to be of length one. So why is the Krull's dimension of an Artinian ring $0$?






krull-dimension







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    An Artinian ring in which $(0)$ is a prime ideal is a field. So if $P$ is a nonzero prime ideal, then $(0)$ isn't a prime ideal.
    – Qiaochu Yuan
    18 hours ago












  • Oh, for some reason I thought that $(0)$ was prime in any ring but clearly it is only true when R is integral.
    – roi_saumon
    17 hours ago















up vote
0
down vote

favorite
1












There is something I don't get about Krull's dimension.
If $K$ is a field, then the only ideals are ${ 0 }$ and $K$ so the only prime ideal is ${0}$. So the Krull's dimension is $0$. But if we have an Artinian ring, we have that every prime ideal is maximal. So if we take P a maximal ideal, ${0 } subsetneq P$ is a chain of ideal which seems to be of length one. So why is the Krull's dimension of an Artinian ring $0$?






krull-dimension







share|cite|improve this question


















  • 1




    An Artinian ring in which $(0)$ is a prime ideal is a field. So if $P$ is a nonzero prime ideal, then $(0)$ isn't a prime ideal.
    – Qiaochu Yuan
    18 hours ago












  • Oh, for some reason I thought that $(0)$ was prime in any ring but clearly it is only true when R is integral.
    – roi_saumon
    17 hours ago













up vote
0
down vote

favorite
1









up vote
0
down vote

favorite
1






1





There is something I don't get about Krull's dimension.
If $K$ is a field, then the only ideals are ${ 0 }$ and $K$ so the only prime ideal is ${0}$. So the Krull's dimension is $0$. But if we have an Artinian ring, we have that every prime ideal is maximal. So if we take P a maximal ideal, ${0 } subsetneq P$ is a chain of ideal which seems to be of length one. So why is the Krull's dimension of an Artinian ring $0$?






krull-dimension







share|cite|improve this question













There is something I don't get about Krull's dimension.
If $K$ is a field, then the only ideals are ${ 0 }$ and $K$ so the only prime ideal is ${0}$. So the Krull's dimension is $0$. But if we have an Artinian ring, we have that every prime ideal is maximal. So if we take P a maximal ideal, ${0 } subsetneq P$ is a chain of ideal which seems to be of length one. So why is the Krull's dimension of an Artinian ring $0$?






krull-dimension




krull-dimension






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share|cite|improve this question











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asked 18 hours ago









roi_saumon

1717




1717








  • 1




    An Artinian ring in which $(0)$ is a prime ideal is a field. So if $P$ is a nonzero prime ideal, then $(0)$ isn't a prime ideal.
    – Qiaochu Yuan
    18 hours ago












  • Oh, for some reason I thought that $(0)$ was prime in any ring but clearly it is only true when R is integral.
    – roi_saumon
    17 hours ago














  • 1




    An Artinian ring in which $(0)$ is a prime ideal is a field. So if $P$ is a nonzero prime ideal, then $(0)$ isn't a prime ideal.
    – Qiaochu Yuan
    18 hours ago












  • Oh, for some reason I thought that $(0)$ was prime in any ring but clearly it is only true when R is integral.
    – roi_saumon
    17 hours ago








1




1




An Artinian ring in which $(0)$ is a prime ideal is a field. So if $P$ is a nonzero prime ideal, then $(0)$ isn't a prime ideal.
– Qiaochu Yuan
18 hours ago






An Artinian ring in which $(0)$ is a prime ideal is a field. So if $P$ is a nonzero prime ideal, then $(0)$ isn't a prime ideal.
– Qiaochu Yuan
18 hours ago














Oh, for some reason I thought that $(0)$ was prime in any ring but clearly it is only true when R is integral.
– roi_saumon
17 hours ago




Oh, for some reason I thought that $(0)$ was prime in any ring but clearly it is only true when R is integral.
– roi_saumon
17 hours ago










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So if we take P a maximal ideal, ${0 } subsetneq P$ is a chain of ideal which seems to be of length one.




It is a chain of length one. But it isn't a chain of prime ideals. And for Krull dimension we look only at chains of prime ideals.



For a (commutative) ring $R$ the ideal ${0}$ is prime if and only if $R$ is an integral domain. And an Artinian ring is an integral domain if and only if it is a field. So if $R$ is Artinian but not field then ${0}$ is not prime. On the other hand if $R$ is a field than it has no proper ideal. So the chain ${0}subsetneq P$ is never a chain of prime ideals when $R$ is artinian.



Also note that fields are special Artinian rings. And indeed they all have Krull dimension $0$.






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    So if we take P a maximal ideal, ${0 } subsetneq P$ is a chain of ideal which seems to be of length one.




    It is a chain of length one. But it isn't a chain of prime ideals. And for Krull dimension we look only at chains of prime ideals.



    For a (commutative) ring $R$ the ideal ${0}$ is prime if and only if $R$ is an integral domain. And an Artinian ring is an integral domain if and only if it is a field. So if $R$ is Artinian but not field then ${0}$ is not prime. On the other hand if $R$ is a field than it has no proper ideal. So the chain ${0}subsetneq P$ is never a chain of prime ideals when $R$ is artinian.



    Also note that fields are special Artinian rings. And indeed they all have Krull dimension $0$.






    share|cite|improve this answer



























      up vote
      2
      down vote



      accepted











      So if we take P a maximal ideal, ${0 } subsetneq P$ is a chain of ideal which seems to be of length one.




      It is a chain of length one. But it isn't a chain of prime ideals. And for Krull dimension we look only at chains of prime ideals.



      For a (commutative) ring $R$ the ideal ${0}$ is prime if and only if $R$ is an integral domain. And an Artinian ring is an integral domain if and only if it is a field. So if $R$ is Artinian but not field then ${0}$ is not prime. On the other hand if $R$ is a field than it has no proper ideal. So the chain ${0}subsetneq P$ is never a chain of prime ideals when $R$ is artinian.



      Also note that fields are special Artinian rings. And indeed they all have Krull dimension $0$.






      share|cite|improve this answer

























        up vote
        2
        down vote



        accepted







        up vote
        2
        down vote



        accepted







        So if we take P a maximal ideal, ${0 } subsetneq P$ is a chain of ideal which seems to be of length one.




        It is a chain of length one. But it isn't a chain of prime ideals. And for Krull dimension we look only at chains of prime ideals.



        For a (commutative) ring $R$ the ideal ${0}$ is prime if and only if $R$ is an integral domain. And an Artinian ring is an integral domain if and only if it is a field. So if $R$ is Artinian but not field then ${0}$ is not prime. On the other hand if $R$ is a field than it has no proper ideal. So the chain ${0}subsetneq P$ is never a chain of prime ideals when $R$ is artinian.



        Also note that fields are special Artinian rings. And indeed they all have Krull dimension $0$.






        share|cite|improve this answer















        So if we take P a maximal ideal, ${0 } subsetneq P$ is a chain of ideal which seems to be of length one.




        It is a chain of length one. But it isn't a chain of prime ideals. And for Krull dimension we look only at chains of prime ideals.



        For a (commutative) ring $R$ the ideal ${0}$ is prime if and only if $R$ is an integral domain. And an Artinian ring is an integral domain if and only if it is a field. So if $R$ is Artinian but not field then ${0}$ is not prime. On the other hand if $R$ is a field than it has no proper ideal. So the chain ${0}subsetneq P$ is never a chain of prime ideals when $R$ is artinian.



        Also note that fields are special Artinian rings. And indeed they all have Krull dimension $0$.







        share|cite|improve this answer














        share|cite|improve this answer



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        edited 18 hours ago

























        answered 18 hours ago









        freakish

        10.3k1526




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