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Is a local ring Artinian, if the maximal ideal is nilpotent?
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This answer suggests the idea, that a local ring $(R, mathfrakm)$ whose maximal ideal is nilpotent is in fact an Artinian ring.
Is this true? If so, how is it proven?
abstract-algebra commutative-algebra artinian
asked Jul 16 at 9:39
red_trumpet
216111
add a comment |Â
up vote
4
down vote
favorite
This answer suggests the idea, that a local ring $(R, mathfrakm)$ whose maximal ideal is nilpotent is in fact an Artinian ring.
Is this true? If so, how is it proven?
abstract-algebra commutative-algebra artinian
asked Jul 16 at 9:39
red_trumpet
216111
add a comment |Â
up vote
4
down vote
favorite
up vote
4
down vote
favorite
This answer suggests the idea, that a local ring $(R, mathfrakm)$ whose maximal ideal is nilpotent is in fact an Artinian ring.
Is this true? If so, how is it proven?
abstract-algebra commutative-algebra artinian
asked Jul 16 at 9:39
red_trumpet
216111
This answer suggests the idea, that a local ring $(R, mathfrakm)$ whose maximal ideal is nilpotent is in fact an Artinian ring.
Is this true? If so, how is it proven?
abstract-algebra commutative-algebra artinian
asked Jul 16 at 9:39
red_trumpet
216111
asked Jul 16 at 9:39
red_trumpet
216111
asked Jul 16 at 9:39
red_trumpet
216111
asked Jul 16 at 9:39
red_trumpet
216111
216111
add a comment |Â
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
4
down vote
accepted
You need that $R$ is noetherian, else there are counterexamples.
E.g., take $R = K[x_i]_i in mathbbN/(x_i | i in mathbbN)^2$.
If $R$ is Noetherian, this is a special case of Lemma 10.59.4 here which states that a Noetherian ring of dimension $0$ is Artinian.
answered Jul 16 at 9:49
Louis
2,28011228
I just added a proof that $R$ has dimension zero, so I will not become confused (again) in the future.
– red_trumpet
Aug 14 at 13:06
add a comment |Â
up vote
3
down vote
There is a version of this theorem even for noncommutative rings. It is called the Hopkins-Levitzki theorem.
It says, in a nutshell, that if $R/J(R)$ is semisimple and $J(R)$ is nilpotent, then $R$ is right Artinian iff right Noetherian. Yours is a special case where $R/J(R)$ is a field.
Here is the DaRT search which currently yields two examples of local, non-Artinian rings with nilpotent Jacobson radicals.
answered Jul 16 at 13:38
rschwieb
100k1193227
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
4
down vote
accepted
You need that $R$ is noetherian, else there are counterexamples.
E.g., take $R = K[x_i]_i in mathbbN/(x_i | i in mathbbN)^2$.
If $R$ is Noetherian, this is a special case of Lemma 10.59.4 here which states that a Noetherian ring of dimension $0$ is Artinian.
answered Jul 16 at 9:49
Louis
2,28011228
I just added a proof that $R$ has dimension zero, so I will not become confused (again) in the future.
– red_trumpet
Aug 14 at 13:06
add a comment |Â
up vote
4
down vote
accepted
You need that $R$ is noetherian, else there are counterexamples.
E.g., take $R = K[x_i]_i in mathbbN/(x_i | i in mathbbN)^2$.
If $R$ is Noetherian, this is a special case of Lemma 10.59.4 here which states that a Noetherian ring of dimension $0$ is Artinian.
answered Jul 16 at 9:49
Louis
2,28011228
I just added a proof that $R$ has dimension zero, so I will not become confused (again) in the future.
– red_trumpet
Aug 14 at 13:06
add a comment |Â
up vote
4
down vote
accepted
up vote
4
down vote
accepted
You need that $R$ is noetherian, else there are counterexamples.
E.g., take $R = K[x_i]_i in mathbbN/(x_i | i in mathbbN)^2$.
If $R$ is Noetherian, this is a special case of Lemma 10.59.4 here which states that a Noetherian ring of dimension $0$ is Artinian.
answered Jul 16 at 9:49
Louis
2,28011228
You need that $R$ is noetherian, else there are counterexamples.
E.g., take $R = K[x_i]_i in mathbbN/(x_i | i in mathbbN)^2$.
If $R$ is Noetherian, this is a special case of Lemma 10.59.4 here which states that a Noetherian ring of dimension $0$ is Artinian.
answered Jul 16 at 9:49
Louis
2,28011228
answered Jul 16 at 9:49
Louis
2,28011228
answered Jul 16 at 9:49
Louis
2,28011228
answered Jul 16 at 9:49
Louis
2,28011228
2,28011228
I just added a proof that $R$ has dimension zero, so I will not become confused (again) in the future.
– red_trumpet
Aug 14 at 13:06
add a comment |Â
I just added a proof that $R$ has dimension zero, so I will not become confused (again) in the future.
– red_trumpet
Aug 14 at 13:06
I just added a proof that $R$ has dimension zero, so I will not become confused (again) in the future.
– red_trumpet
Aug 14 at 13:06
I just added a proof that $R$ has dimension zero, so I will not become confused (again) in the future.
– red_trumpet
Aug 14 at 13:06
add a comment |Â
up vote
3
down vote
There is a version of this theorem even for noncommutative rings. It is called the Hopkins-Levitzki theorem.
It says, in a nutshell, that if $R/J(R)$ is semisimple and $J(R)$ is nilpotent, then $R$ is right Artinian iff right Noetherian. Yours is a special case where $R/J(R)$ is a field.
Here is the DaRT search which currently yields two examples of local, non-Artinian rings with nilpotent Jacobson radicals.
answered Jul 16 at 13:38
rschwieb
100k1193227
add a comment |Â
up vote
3
down vote
There is a version of this theorem even for noncommutative rings. It is called the Hopkins-Levitzki theorem.
It says, in a nutshell, that if $R/J(R)$ is semisimple and $J(R)$ is nilpotent, then $R$ is right Artinian iff right Noetherian. Yours is a special case where $R/J(R)$ is a field.
Here is the DaRT search which currently yields two examples of local, non-Artinian rings with nilpotent Jacobson radicals.
answered Jul 16 at 13:38
rschwieb
100k1193227
add a comment |Â
up vote
3
down vote
up vote
3
down vote
There is a version of this theorem even for noncommutative rings. It is called the Hopkins-Levitzki theorem.
It says, in a nutshell, that if $R/J(R)$ is semisimple and $J(R)$ is nilpotent, then $R$ is right Artinian iff right Noetherian. Yours is a special case where $R/J(R)$ is a field.
Here is the DaRT search which currently yields two examples of local, non-Artinian rings with nilpotent Jacobson radicals.
answered Jul 16 at 13:38
rschwieb
100k1193227
There is a version of this theorem even for noncommutative rings. It is called the Hopkins-Levitzki theorem.
It says, in a nutshell, that if $R/J(R)$ is semisimple and $J(R)$ is nilpotent, then $R$ is right Artinian iff right Noetherian. Yours is a special case where $R/J(R)$ is a field.
Here is the DaRT search which currently yields two examples of local, non-Artinian rings with nilpotent Jacobson radicals.
answered Jul 16 at 13:38
rschwieb
100k1193227
answered Jul 16 at 13:38
rschwieb
100k1193227
answered Jul 16 at 13:38
rschwieb
100k1193227
answered Jul 16 at 13:38
rschwieb
100k1193227
100k1193227
add a comment |Â
add a comment |Â
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