Krull dimension, how to compute?
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
36017
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
36017
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
36017
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
36017
asked Nov 19 at 10:26
roi_saumon
36017
asked Nov 19 at 10:26
roi_saumon
36017
asked Nov 19 at 10:26
roi_saumon
36017
asked Nov 19 at 10:26
roi_saumon
36017
36017
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
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
10.9k1527
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
10.9k1527
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
10.9k1527
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
10.9k1527
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
10.9k1527
edited Nov 19 at 10:42
answered Nov 19 at 10:37
freakish
10.9k1527
answered Nov 19 at 10:37
freakish
10.9k1527
answered Nov 19 at 10:37
freakish
10.9k1527
10.9k1527
add a comment |
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Some of your past answers have not been well-received, and you're in danger of being blocked from answering.
Please pay close attention to the following guidance:
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3004760%2fkrull-dimension-how-to-compute%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
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