Mixing
Properties of Noetherian local ring (Sharp, Exercise 8.33)
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 yesterday
user26857
39.1k123882
asked yesterday
Desunkid
16510
|
show 1 more comment
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 yesterday
user26857
39.1k123882
asked yesterday
Desunkid
16510
$M^{n+1} subset RM^n=M^n$ is that right?
– Desunkid
yesterday
that seems right... isn't part (ii) then obvious as well? we are probably missing something though
– mathworker21
yesterday
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
yesterday
1
@Desunkid But $M supset M^2 supset cdots supset M^n supset M^{n+1} supset cdots$ is a descending chain.
– André 3000
yesterday
1
@Desunkid For i) use Nakayama. For ii) use i) by replacing $R$ with $R/I$.
– user26857
yesterday
|
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 yesterday
user26857
39.1k123882
asked yesterday
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 yesterday
user26857
39.1k123882
asked yesterday
Desunkid
16510
edited yesterday
user26857
39.1k123882
asked yesterday
Desunkid
16510
edited yesterday
user26857
39.1k123882
edited yesterday
user26857
39.1k123882
edited yesterday
user26857
39.1k123882
39.1k123882
asked yesterday
Desunkid
16510
asked yesterday
Desunkid
16510
asked yesterday
Desunkid
16510
16510
$M^{n+1} subset RM^n=M^n$ is that right?
– Desunkid
yesterday
that seems right... isn't part (ii) then obvious as well? we are probably missing something though
– mathworker21
yesterday
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
yesterday
1
@Desunkid But $M supset M^2 supset cdots supset M^n supset M^{n+1} supset cdots$ is a descending chain.
– André 3000
yesterday
1
@Desunkid For i) use Nakayama. For ii) use i) by replacing $R$ with $R/I$.
– user26857
yesterday
|
show 1 more comment
$M^{n+1} subset RM^n=M^n$ is that right?
– Desunkid
yesterday
that seems right... isn't part (ii) then obvious as well? we are probably missing something though
– mathworker21
yesterday
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
yesterday
1
@Desunkid But $M supset M^2 supset cdots supset M^n supset M^{n+1} supset cdots$ is a descending chain.
– André 3000
yesterday
1
@Desunkid For i) use Nakayama. For ii) use i) by replacing $R$ with $R/I$.
– user26857
yesterday
$M^{n+1} subset RM^n=M^n$ is that right?
– Desunkid
yesterday
$M^{n+1} subset RM^n=M^n$ is that right?
– Desunkid
yesterday
that seems right... isn't part (ii) then obvious as well? we are probably missing something though
– mathworker21
yesterday
that seems right... isn't part (ii) then obvious as well? we are probably missing something though
– mathworker21
yesterday
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
yesterday
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
yesterday
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
yesterday
@Desunkid But $M supset M^2 supset cdots supset M^n supset M^{n+1} supset cdots$ is a descending chain.
– André 3000
yesterday
1
1
@Desunkid For i) use Nakayama. For ii) use i) by replacing $R$ with $R/I$.
– user26857
yesterday
@Desunkid For i) use Nakayama. For ii) use i) by replacing $R$ with $R/I$.
– user26857
yesterday
|
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 9 hours ago
Badam Baplan
3,931722
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
8 hours ago
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
7 hours ago
Ok. Then let's hope he'll appreciate your effort and learn something useful from the answer.
– user26857
6 hours ago
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
1 hour ago
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 9 hours ago
Badam Baplan
3,931722
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
8 hours ago
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
7 hours ago
Ok. Then let's hope he'll appreciate your effort and learn something useful from the answer.
– user26857
6 hours ago
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
1 hour ago
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 9 hours ago
Badam Baplan
3,931722
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
8 hours ago
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
7 hours ago
Ok. Then let's hope he'll appreciate your effort and learn something useful from the answer.
– user26857
6 hours ago
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
1 hour ago
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 9 hours ago
Badam Baplan
3,931722
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 9 hours ago
Badam Baplan
3,931722
answered 9 hours ago
Badam Baplan
3,931722
answered 9 hours ago
Badam Baplan
3,931722
answered 9 hours ago
Badam Baplan
3,931722
3,931722
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
8 hours ago
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
7 hours ago
Ok. Then let's hope he'll appreciate your effort and learn something useful from the answer.
– user26857
6 hours ago
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
1 hour ago
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
8 hours ago
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
7 hours ago
Ok. Then let's hope he'll appreciate your effort and learn something useful from the answer.
– user26857
6 hours ago
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
1 hour ago
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
8 hours ago
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
8 hours ago
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
7 hours ago
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
7 hours ago
Ok. Then let's hope he'll appreciate your effort and learn something useful from the answer.
– user26857
6 hours ago
Ok. Then let's hope he'll appreciate your effort and learn something useful from the answer.
– user26857
6 hours ago
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
1 hour ago
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
1 hour ago
add a comment |
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%2f3005038%2fproperties-of-noetherian-local-ring-sharp-exercise-8-33%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
$M^{n+1} subset RM^n=M^n$ is that right?
– Desunkid
yesterday
that seems right... isn't part (ii) then obvious as well? we are probably missing something though
– mathworker21
yesterday
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
yesterday
1
@Desunkid But $M supset M^2 supset cdots supset M^n supset M^{n+1} supset cdots$ is a descending chain.
– André 3000
yesterday
1
@Desunkid For i) use Nakayama. For ii) use i) by replacing $R$ with $R/I$.
– user26857
yesterday