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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asked 18 hours ago

roi_saumon

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
  

  
  
  
  

add a comment | 

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

asked 18 hours ago

roi_saumon

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
  

  
  
  
  

add a comment | 

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

asked 18 hours ago

roi_saumon

1717

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

share|cite|improve this question

asked 18 hours ago

roi_saumon

1717

share|cite|improve this question

asked 18 hours ago

roi_saumon

1717

share|cite|improve this question

share|cite|improve this question

asked 18 hours ago

roi_saumon

1717

asked 18 hours ago

roi_saumon

1717

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
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  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

add a comment | 

1 Answer
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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

edited 18 hours ago

answered 18 hours ago

freakish

10.3k1526

add a comment | 

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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

edited 18 hours ago

answered 18 hours ago

freakish

10.3k1526

add a comment | 

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

edited 18 hours ago

answered 18 hours ago

freakish

10.3k1526

add a comment | 

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

edited 18 hours ago

answered 18 hours ago

freakish

10.3k1526

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

edited 18 hours ago

answered 18 hours ago

freakish

10.3k1526

share|cite|improve this answer

share|cite|improve this answer

edited 18 hours ago

edited 18 hours ago

edited 18 hours ago

answered 18 hours ago

freakish

10.3k1526

answered 18 hours ago

freakish

10.3k1526

answered 18 hours ago

freakish

10.3k1526

10.3k1526

add a comment | 

add a comment | 

 

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