Properties of Noetherian local ring (Sharp, Exercise 8.33) [closed]
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Let $(R,M)$ be a Noetherian local ring. Show that
i) If there exists a non-maximal prime ideal of $R$, then $M^{n+1} subset M^{n}$ for every $n in mathbb{N}$
ii) If $I$ is a proper ideal of $R$ and $sqrt{I} neq M$ then $I +M^{n+1} subset I +M^{n}$ for every $n in mathbb{N}$
I think $M^{n+1} subset M^{n}$ is obvious without condition about the existence of non-maximal prime ideal, doesn't it? I have no idea for this problem. Can anyone help me?
commutative-algebra
edited Nov 19 at 21:59
user26857
39.2k123882
asked Nov 19 at 15:04
Desunkid
16510
closed as off-topic by user26857, Shailesh, Leucippus, Rebellos, Brahadeesh Nov 21 at 9:09
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – user26857, Shailesh, Leucippus, Rebellos, Brahadeesh
If this question can be reworded to fit the rules in the help center, please edit the question.
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up vote
-1
down vote
favorite
Let $(R,M)$ be a Noetherian local ring. Show that
i) If there exists a non-maximal prime ideal of $R$, then $M^{n+1} subset M^{n}$ for every $n in mathbb{N}$
ii) If $I$ is a proper ideal of $R$ and $sqrt{I} neq M$ then $I +M^{n+1} subset I +M^{n}$ for every $n in mathbb{N}$
I think $M^{n+1} subset M^{n}$ is obvious without condition about the existence of non-maximal prime ideal, doesn't it? I have no idea for this problem. Can anyone help me?
commutative-algebra
edited Nov 19 at 21:59
user26857
39.2k123882
asked Nov 19 at 15:04
Desunkid
16510
closed as off-topic by user26857, Shailesh, Leucippus, Rebellos, Brahadeesh Nov 21 at 9:09
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – user26857, Shailesh, Leucippus, Rebellos, Brahadeesh
If this question can be reworded to fit the rules in the help center, please edit the question.
$M^{n+1} subset RM^n=M^n$ is that right?
– Desunkid
Nov 19 at 15:10
that seems right... isn't part (ii) then obvious as well? we are probably missing something though
– mathworker21
Nov 19 at 15:11
1
I think here $subset$ is meant as " is strictly included" : some authors use this convention and add a second bar under $subset$ to mean "is included"
– Max
Nov 19 at 15:39
1
@Desunkid But $M supset M^2 supset cdots supset M^n supset M^{n+1} supset cdots$ is a descending chain.
– André 3000
Nov 19 at 17:41
1
@Desunkid For i) use Nakayama. For ii) use i) by replacing $R$ with $R/I$.
– user26857
Nov 19 at 21:54
|
show 1 more comment
up vote
-1
down vote
favorite
up vote
-1
down vote
favorite
Let $(R,M)$ be a Noetherian local ring. Show that
i) If there exists a non-maximal prime ideal of $R$, then $M^{n+1} subset M^{n}$ for every $n in mathbb{N}$
ii) If $I$ is a proper ideal of $R$ and $sqrt{I} neq M$ then $I +M^{n+1} subset I +M^{n}$ for every $n in mathbb{N}$
I think $M^{n+1} subset M^{n}$ is obvious without condition about the existence of non-maximal prime ideal, doesn't it? I have no idea for this problem. Can anyone help me?
commutative-algebra
edited Nov 19 at 21:59
user26857
39.2k123882
asked Nov 19 at 15:04
Desunkid
16510
Let $(R,M)$ be a Noetherian local ring. Show that
i) If there exists a non-maximal prime ideal of $R$, then $M^{n+1} subset M^{n}$ for every $n in mathbb{N}$
ii) If $I$ is a proper ideal of $R$ and $sqrt{I} neq M$ then $I +M^{n+1} subset I +M^{n}$ for every $n in mathbb{N}$
I think $M^{n+1} subset M^{n}$ is obvious without condition about the existence of non-maximal prime ideal, doesn't it? I have no idea for this problem. Can anyone help me?
commutative-algebra
commutative-algebra
edited Nov 19 at 21:59
user26857
39.2k123882
asked Nov 19 at 15:04
Desunkid
16510
edited Nov 19 at 21:59
user26857
39.2k123882
asked Nov 19 at 15:04
Desunkid
16510
edited Nov 19 at 21:59
user26857
39.2k123882
edited Nov 19 at 21:59
user26857
39.2k123882
edited Nov 19 at 21:59
user26857
39.2k123882
39.2k123882
asked Nov 19 at 15:04
Desunkid
16510
asked Nov 19 at 15:04
Desunkid
16510
asked Nov 19 at 15:04
Desunkid
16510
16510
closed as off-topic by user26857, Shailesh, Leucippus, Rebellos, Brahadeesh Nov 21 at 9:09
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – user26857, Shailesh, Leucippus, Rebellos, Brahadeesh
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by user26857, Shailesh, Leucippus, Rebellos, Brahadeesh Nov 21 at 9:09
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – user26857, Shailesh, Leucippus, Rebellos, Brahadeesh
If this question can be reworded to fit the rules in the help center, please edit the question.
$M^{n+1} subset RM^n=M^n$ is that right?
– Desunkid
Nov 19 at 15:10
that seems right... isn't part (ii) then obvious as well? we are probably missing something though
– mathworker21
Nov 19 at 15:11
1
I think here $subset$ is meant as " is strictly included" : some authors use this convention and add a second bar under $subset$ to mean "is included"
– Max
Nov 19 at 15:39
1
@Desunkid But $M supset M^2 supset cdots supset M^n supset M^{n+1} supset cdots$ is a descending chain.
– André 3000
Nov 19 at 17:41
1
@Desunkid For i) use Nakayama. For ii) use i) by replacing $R$ with $R/I$.
– user26857
Nov 19 at 21:54
|
show 1 more comment
$M^{n+1} subset RM^n=M^n$ is that right?
– Desunkid
Nov 19 at 15:10
that seems right... isn't part (ii) then obvious as well? we are probably missing something though
– mathworker21
Nov 19 at 15:11
1
I think here $subset$ is meant as " is strictly included" : some authors use this convention and add a second bar under $subset$ to mean "is included"
– Max
Nov 19 at 15:39
1
@Desunkid But $M supset M^2 supset cdots supset M^n supset M^{n+1} supset cdots$ is a descending chain.
– André 3000
Nov 19 at 17:41
1
@Desunkid For i) use Nakayama. For ii) use i) by replacing $R$ with $R/I$.
– user26857
Nov 19 at 21:54
$M^{n+1} subset RM^n=M^n$ is that right?
– Desunkid
Nov 19 at 15:10
$M^{n+1} subset RM^n=M^n$ is that right?
– Desunkid
Nov 19 at 15:10
that seems right... isn't part (ii) then obvious as well? we are probably missing something though
– mathworker21
Nov 19 at 15:11
that seems right... isn't part (ii) then obvious as well? we are probably missing something though
– mathworker21
Nov 19 at 15:11
1
1
I think here $subset$ is meant as " is strictly included" : some authors use this convention and add a second bar under $subset$ to mean "is included"
– Max
Nov 19 at 15:39
I think here $subset$ is meant as " is strictly included" : some authors use this convention and add a second bar under $subset$ to mean "is included"
– Max
Nov 19 at 15:39
1
1
@Desunkid But $M supset M^2 supset cdots supset M^n supset M^{n+1} supset cdots$ is a descending chain.
– André 3000
Nov 19 at 17:41
@Desunkid But $M supset M^2 supset cdots supset M^n supset M^{n+1} supset cdots$ is a descending chain.
– André 3000
Nov 19 at 17:41
1
1
@Desunkid For i) use Nakayama. For ii) use i) by replacing $R$ with $R/I$.
– user26857
Nov 19 at 21:54
@Desunkid For i) use Nakayama. For ii) use i) by replacing $R$ with $R/I$.
– user26857
Nov 19 at 21:54
|
show 1 more comment
1 Answer
1
active
oldest
votes
up vote
1
down vote
accepted
The $subset$ subset symbol is definitely meant to indicate strict inclusion here.
For the first part, since trivially $I^n subseteq I^m$ for any ideal $I$ and $n geq m$, the problem is just to prove that $M^{n+1} not= M^{n}$ given the assumption that the ring is not $0$-dimensional.
We will show contrapositively that $M^{n+1} = M^n$ implies that the ring is $0$-dimensional.
Recall that Nakayama's lemma, in one of its incarnations, states the following:
Nakayama's lemma: Let $I$ be an ideal contained in the Jacobson radical of $R$ and $M$ be a finitely generated ideal. If $IM = M$ then $M = 0$.
In a local ring, the Jacobson radical is the unique maximal ideal, which contains our non-maximal ideal $M$. Moreover $M$ is f.g. by the Noetherian assumption. So if we have $MM^n = M^n$ then Nakayama says that $M^n = 0$.
If the maximal ideal is nilpotent, then every nonunit is nilpotent, and this implies that the ring has a unique prime ideal (good elementary exercise).
Hence the ring is $0$-dimensional.
For the second part, note that $(R/I, M/I)$ is a local Noetherian ring, too. Again we show the contrapositive: if $I + M^{n+1} = I + M^n$ then $sqrt{I} = M$. By the first part, if $I + M^{n+1} = I + M^n$ then $R/I$ is $0$-dimensional. So $M$ is the only prime ideal of $R$ containing $I$. But the radical of an ideal is the intersection of the primes containing it, so $sqrt{I} = M$.
answered Nov 20 at 16:55
Badam Baplan
4,326722
It's definitely a bad idea to solve someone homework. He had already two hints in the comments. If he needed more details could have ask.
– user26857
Nov 20 at 18:00
1
My intuition from the abundant confusion surrounding this question and from looking at the user's asking history was that this is somebody hopping around in self-directed study rather than doing homework, and that details would be helpful. Granted I might be very wrong about that. I tend to agree that giving out homework solutions sans dialogue doesn't help anybody. Office hours are good.
– Badam Baplan
Nov 20 at 18:54
Thank you @Badam Baplan. It's not my homework, I'd study about Nakayama's lemma and Krull intersection theorem and do some exercises about these theorems.
– Desunkid
Nov 21 at 1:16
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
The $subset$ subset symbol is definitely meant to indicate strict inclusion here.
For the first part, since trivially $I^n subseteq I^m$ for any ideal $I$ and $n geq m$, the problem is just to prove that $M^{n+1} not= M^{n}$ given the assumption that the ring is not $0$-dimensional.
We will show contrapositively that $M^{n+1} = M^n$ implies that the ring is $0$-dimensional.
Recall that Nakayama's lemma, in one of its incarnations, states the following:
Nakayama's lemma: Let $I$ be an ideal contained in the Jacobson radical of $R$ and $M$ be a finitely generated ideal. If $IM = M$ then $M = 0$.
In a local ring, the Jacobson radical is the unique maximal ideal, which contains our non-maximal ideal $M$. Moreover $M$ is f.g. by the Noetherian assumption. So if we have $MM^n = M^n$ then Nakayama says that $M^n = 0$.
If the maximal ideal is nilpotent, then every nonunit is nilpotent, and this implies that the ring has a unique prime ideal (good elementary exercise).
Hence the ring is $0$-dimensional.
For the second part, note that $(R/I, M/I)$ is a local Noetherian ring, too. Again we show the contrapositive: if $I + M^{n+1} = I + M^n$ then $sqrt{I} = M$. By the first part, if $I + M^{n+1} = I + M^n$ then $R/I$ is $0$-dimensional. So $M$ is the only prime ideal of $R$ containing $I$. But the radical of an ideal is the intersection of the primes containing it, so $sqrt{I} = M$.
answered Nov 20 at 16:55
Badam Baplan
4,326722
It's definitely a bad idea to solve someone homework. He had already two hints in the comments. If he needed more details could have ask.
– user26857
Nov 20 at 18:00
1
My intuition from the abundant confusion surrounding this question and from looking at the user's asking history was that this is somebody hopping around in self-directed study rather than doing homework, and that details would be helpful. Granted I might be very wrong about that. I tend to agree that giving out homework solutions sans dialogue doesn't help anybody. Office hours are good.
– Badam Baplan
Nov 20 at 18:54
Thank you @Badam Baplan. It's not my homework, I'd study about Nakayama's lemma and Krull intersection theorem and do some exercises about these theorems.
– Desunkid
Nov 21 at 1:16
add a comment |
up vote
1
down vote
accepted
The $subset$ subset symbol is definitely meant to indicate strict inclusion here.
For the first part, since trivially $I^n subseteq I^m$ for any ideal $I$ and $n geq m$, the problem is just to prove that $M^{n+1} not= M^{n}$ given the assumption that the ring is not $0$-dimensional.
We will show contrapositively that $M^{n+1} = M^n$ implies that the ring is $0$-dimensional.
Recall that Nakayama's lemma, in one of its incarnations, states the following:
Nakayama's lemma: Let $I$ be an ideal contained in the Jacobson radical of $R$ and $M$ be a finitely generated ideal. If $IM = M$ then $M = 0$.
In a local ring, the Jacobson radical is the unique maximal ideal, which contains our non-maximal ideal $M$. Moreover $M$ is f.g. by the Noetherian assumption. So if we have $MM^n = M^n$ then Nakayama says that $M^n = 0$.
If the maximal ideal is nilpotent, then every nonunit is nilpotent, and this implies that the ring has a unique prime ideal (good elementary exercise).
Hence the ring is $0$-dimensional.
For the second part, note that $(R/I, M/I)$ is a local Noetherian ring, too. Again we show the contrapositive: if $I + M^{n+1} = I + M^n$ then $sqrt{I} = M$. By the first part, if $I + M^{n+1} = I + M^n$ then $R/I$ is $0$-dimensional. So $M$ is the only prime ideal of $R$ containing $I$. But the radical of an ideal is the intersection of the primes containing it, so $sqrt{I} = M$.
answered Nov 20 at 16:55
Badam Baplan
4,326722
It's definitely a bad idea to solve someone homework. He had already two hints in the comments. If he needed more details could have ask.
– user26857
Nov 20 at 18:00
1
My intuition from the abundant confusion surrounding this question and from looking at the user's asking history was that this is somebody hopping around in self-directed study rather than doing homework, and that details would be helpful. Granted I might be very wrong about that. I tend to agree that giving out homework solutions sans dialogue doesn't help anybody. Office hours are good.
– Badam Baplan
Nov 20 at 18:54
Thank you @Badam Baplan. It's not my homework, I'd study about Nakayama's lemma and Krull intersection theorem and do some exercises about these theorems.
– Desunkid
Nov 21 at 1:16
add a comment |
up vote
1
down vote
accepted
up vote
1
down vote
accepted
The $subset$ subset symbol is definitely meant to indicate strict inclusion here.
For the first part, since trivially $I^n subseteq I^m$ for any ideal $I$ and $n geq m$, the problem is just to prove that $M^{n+1} not= M^{n}$ given the assumption that the ring is not $0$-dimensional.
We will show contrapositively that $M^{n+1} = M^n$ implies that the ring is $0$-dimensional.
Recall that Nakayama's lemma, in one of its incarnations, states the following:
Nakayama's lemma: Let $I$ be an ideal contained in the Jacobson radical of $R$ and $M$ be a finitely generated ideal. If $IM = M$ then $M = 0$.
In a local ring, the Jacobson radical is the unique maximal ideal, which contains our non-maximal ideal $M$. Moreover $M$ is f.g. by the Noetherian assumption. So if we have $MM^n = M^n$ then Nakayama says that $M^n = 0$.
If the maximal ideal is nilpotent, then every nonunit is nilpotent, and this implies that the ring has a unique prime ideal (good elementary exercise).
Hence the ring is $0$-dimensional.
For the second part, note that $(R/I, M/I)$ is a local Noetherian ring, too. Again we show the contrapositive: if $I + M^{n+1} = I + M^n$ then $sqrt{I} = M$. By the first part, if $I + M^{n+1} = I + M^n$ then $R/I$ is $0$-dimensional. So $M$ is the only prime ideal of $R$ containing $I$. But the radical of an ideal is the intersection of the primes containing it, so $sqrt{I} = M$.
answered Nov 20 at 16:55
Badam Baplan
4,326722
The $subset$ subset symbol is definitely meant to indicate strict inclusion here.
For the first part, since trivially $I^n subseteq I^m$ for any ideal $I$ and $n geq m$, the problem is just to prove that $M^{n+1} not= M^{n}$ given the assumption that the ring is not $0$-dimensional.
We will show contrapositively that $M^{n+1} = M^n$ implies that the ring is $0$-dimensional.
Recall that Nakayama's lemma, in one of its incarnations, states the following:
Nakayama's lemma: Let $I$ be an ideal contained in the Jacobson radical of $R$ and $M$ be a finitely generated ideal. If $IM = M$ then $M = 0$.
In a local ring, the Jacobson radical is the unique maximal ideal, which contains our non-maximal ideal $M$. Moreover $M$ is f.g. by the Noetherian assumption. So if we have $MM^n = M^n$ then Nakayama says that $M^n = 0$.
If the maximal ideal is nilpotent, then every nonunit is nilpotent, and this implies that the ring has a unique prime ideal (good elementary exercise).
Hence the ring is $0$-dimensional.
For the second part, note that $(R/I, M/I)$ is a local Noetherian ring, too. Again we show the contrapositive: if $I + M^{n+1} = I + M^n$ then $sqrt{I} = M$. By the first part, if $I + M^{n+1} = I + M^n$ then $R/I$ is $0$-dimensional. So $M$ is the only prime ideal of $R$ containing $I$. But the radical of an ideal is the intersection of the primes containing it, so $sqrt{I} = M$.
answered Nov 20 at 16:55
Badam Baplan
4,326722
answered Nov 20 at 16:55
Badam Baplan
4,326722
answered Nov 20 at 16:55
Badam Baplan
4,326722
answered Nov 20 at 16:55
Badam Baplan
4,326722
4,326722
It's definitely a bad idea to solve someone homework. He had already two hints in the comments. If he needed more details could have ask.
– user26857
Nov 20 at 18:00
1
My intuition from the abundant confusion surrounding this question and from looking at the user's asking history was that this is somebody hopping around in self-directed study rather than doing homework, and that details would be helpful. Granted I might be very wrong about that. I tend to agree that giving out homework solutions sans dialogue doesn't help anybody. Office hours are good.
– Badam Baplan
Nov 20 at 18:54
Thank you @Badam Baplan. It's not my homework, I'd study about Nakayama's lemma and Krull intersection theorem and do some exercises about these theorems.
– Desunkid
Nov 21 at 1:16
add a comment |
It's definitely a bad idea to solve someone homework. He had already two hints in the comments. If he needed more details could have ask.
– user26857
Nov 20 at 18:00
1
My intuition from the abundant confusion surrounding this question and from looking at the user's asking history was that this is somebody hopping around in self-directed study rather than doing homework, and that details would be helpful. Granted I might be very wrong about that. I tend to agree that giving out homework solutions sans dialogue doesn't help anybody. Office hours are good.
– Badam Baplan
Nov 20 at 18:54
Thank you @Badam Baplan. It's not my homework, I'd study about Nakayama's lemma and Krull intersection theorem and do some exercises about these theorems.
– Desunkid
Nov 21 at 1:16
It's definitely a bad idea to solve someone homework. He had already two hints in the comments. If he needed more details could have ask.
– user26857
Nov 20 at 18:00
It's definitely a bad idea to solve someone homework. He had already two hints in the comments. If he needed more details could have ask.
– user26857
Nov 20 at 18:00
1
1
My intuition from the abundant confusion surrounding this question and from looking at the user's asking history was that this is somebody hopping around in self-directed study rather than doing homework, and that details would be helpful. Granted I might be very wrong about that. I tend to agree that giving out homework solutions sans dialogue doesn't help anybody. Office hours are good.
– Badam Baplan
Nov 20 at 18:54
My intuition from the abundant confusion surrounding this question and from looking at the user's asking history was that this is somebody hopping around in self-directed study rather than doing homework, and that details would be helpful. Granted I might be very wrong about that. I tend to agree that giving out homework solutions sans dialogue doesn't help anybody. Office hours are good.
– Badam Baplan
Nov 20 at 18:54
Thank you @Badam Baplan. It's not my homework, I'd study about Nakayama's lemma and Krull intersection theorem and do some exercises about these theorems.
– Desunkid
Nov 21 at 1:16
Thank you @Badam Baplan. It's not my homework, I'd study about Nakayama's lemma and Krull intersection theorem and do some exercises about these theorems.
– Desunkid
Nov 21 at 1:16
add a comment |
$M^{n+1} subset RM^n=M^n$ is that right?
– Desunkid
Nov 19 at 15:10
that seems right... isn't part (ii) then obvious as well? we are probably missing something though
– mathworker21
Nov 19 at 15:11
1
I think here $subset$ is meant as " is strictly included" : some authors use this convention and add a second bar under $subset$ to mean "is included"
– Max
Nov 19 at 15:39
1
@Desunkid But $M supset M^2 supset cdots supset M^n supset M^{n+1} supset cdots$ is a descending chain.
– André 3000
Nov 19 at 17:41
1
@Desunkid For i) use Nakayama. For ii) use i) by replacing $R$ with $R/I$.
– user26857
Nov 19 at 21:54