Krull dimension, how to compute?
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0
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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
asked Nov 19 at 10:26
roi_saumon
33817
add a comment |
up vote
0
down vote
favorite
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
asked Nov 19 at 10:26
roi_saumon
33817
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
Nov 19 at 10:33
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
Nov 19 at 11:10
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
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
asked Nov 19 at 10:26
roi_saumon
33817
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
asked Nov 19 at 10:26
roi_saumon
33817
asked Nov 19 at 10:26
roi_saumon
33817
asked Nov 19 at 10:26
roi_saumon
33817
asked Nov 19 at 10:26
roi_saumon
33817
asked Nov 19 at 10:26
roi_saumon
33817
33817
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
Nov 19 at 10:33
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
Nov 19 at 11:10
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
Nov 19 at 10:33
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
Nov 19 at 11:10
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
Nov 19 at 10:33
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
Nov 19 at 10:33
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
Nov 19 at 11:10
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
Nov 19 at 11:10
add a comment |
1 Answer
1
active
oldest
votes
up vote
2
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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 $0$.
answered Nov 19 at 10:37
freakish
10.7k1527
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
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 $0$.
answered Nov 19 at 10:37
freakish
10.7k1527
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 $0$.
answered Nov 19 at 10:37
freakish
10.7k1527
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 $0$.
answered Nov 19 at 10:37
freakish
10.7k1527
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 $0$.
answered Nov 19 at 10:37
freakish
10.7k1527
edited Nov 19 at 10:42
answered Nov 19 at 10:37
freakish
10.7k1527
answered Nov 19 at 10:37
freakish
10.7k1527
answered Nov 19 at 10:37
freakish
10.7k1527
10.7k1527
add a comment |
add a comment |
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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
Nov 19 at 10:33
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
Nov 19 at 11:10