Is a quotient of ring of polynomials Cohen-Macaulay? [closed]
$begingroup$
Let $R = mathbb{R}[x_1 , ldots , x_n]$ be polynomial ring and $I subset R$ be a principal ideal with $I = langle f rangle$. I know that $R$ is a CM ring. So my question is that:
- Is the quotient ring $R/I$ also CM for any $I subset R$ ?
- If $R/I$ is CM, then the quotient $R/I^2$ also CM where $I^2 = langle f^2 rangle$ ?
polynomials commutative-algebra cohen-macaulay
edited Dec 28 '18 at 10:10
Jyrki Lahtonen
109k13169375
asked Dec 27 '18 at 6:09
Philip JohnsonPhilip Johnson
193
$endgroup$
closed as off-topic by Saad, Leucippus, José Carlos Santos, KReiser, Paul Frost Dec 28 '18 at 12:18
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 provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Saad, Leucippus, José Carlos Santos, KReiser, Paul Frost
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
$begingroup$
Let $R = mathbb{R}[x_1 , ldots , x_n]$ be polynomial ring and $I subset R$ be a principal ideal with $I = langle f rangle$. I know that $R$ is a CM ring. So my question is that:
- Is the quotient ring $R/I$ also CM for any $I subset R$ ?
- If $R/I$ is CM, then the quotient $R/I^2$ also CM where $I^2 = langle f^2 rangle$ ?
polynomials commutative-algebra cohen-macaulay
edited Dec 28 '18 at 10:10
Jyrki Lahtonen
109k13169375
asked Dec 27 '18 at 6:09
Philip JohnsonPhilip Johnson
193
$endgroup$
closed as off-topic by Saad, Leucippus, José Carlos Santos, KReiser, Paul Frost Dec 28 '18 at 12:18
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 provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Saad, Leucippus, José Carlos Santos, KReiser, Paul Frost
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
$begingroup$
Let $R = mathbb{R}[x_1 , ldots , x_n]$ be polynomial ring and $I subset R$ be a principal ideal with $I = langle f rangle$. I know that $R$ is a CM ring. So my question is that:
- Is the quotient ring $R/I$ also CM for any $I subset R$ ?
- If $R/I$ is CM, then the quotient $R/I^2$ also CM where $I^2 = langle f^2 rangle$ ?
polynomials commutative-algebra cohen-macaulay
edited Dec 28 '18 at 10:10
Jyrki Lahtonen
109k13169375
asked Dec 27 '18 at 6:09
Philip JohnsonPhilip Johnson
193
$endgroup$
Let $R = mathbb{R}[x_1 , ldots , x_n]$ be polynomial ring and $I subset R$ be a principal ideal with $I = langle f rangle$. I know that $R$ is a CM ring. So my question is that:
- Is the quotient ring $R/I$ also CM for any $I subset R$ ?
- If $R/I$ is CM, then the quotient $R/I^2$ also CM where $I^2 = langle f^2 rangle$ ?
polynomials commutative-algebra cohen-macaulay
polynomials commutative-algebra cohen-macaulay
edited Dec 28 '18 at 10:10
Jyrki Lahtonen
109k13169375
asked Dec 27 '18 at 6:09
Philip JohnsonPhilip Johnson
193
edited Dec 28 '18 at 10:10
Jyrki Lahtonen
109k13169375
asked Dec 27 '18 at 6:09
Philip JohnsonPhilip Johnson
193
edited Dec 28 '18 at 10:10
Jyrki Lahtonen
109k13169375
edited Dec 28 '18 at 10:10
Jyrki Lahtonen
109k13169375
edited Dec 28 '18 at 10:10
Jyrki Lahtonen
109k13169375
109k13169375
asked Dec 27 '18 at 6:09
Philip JohnsonPhilip Johnson
193
asked Dec 27 '18 at 6:09
Philip JohnsonPhilip Johnson
193
asked Dec 27 '18 at 6:09
Philip JohnsonPhilip Johnson
193
193
closed as off-topic by Saad, Leucippus, José Carlos Santos, KReiser, Paul Frost Dec 28 '18 at 12:18
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 provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Saad, Leucippus, José Carlos Santos, KReiser, Paul Frost
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by Saad, Leucippus, José Carlos Santos, KReiser, Paul Frost Dec 28 '18 at 12:18
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 provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Saad, Leucippus, José Carlos Santos, KReiser, Paul Frost
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
The answer to both (1) and (2) is yes. More generally we have the following:
Proposition. If $R$ is any CM domain with principal ideal $I$, then $R/I$ is CM.
Proof. First recall the definition for (non-local) CM rings, from Stacks Project:
Definition 10.103.6. A Noetherian ring $R$ is called Cohen-Macaulay if all its local rings are Cohen-Macaulay.
and a fact from the Wikipedia page Cohen-Macaulay ring:
($bullet$) For a non-zero-divisor $u$ in the maximal ideal of a Noetherian local ring $R$, $R$ is Cohen–Macaulay if and only if $R/(u)$ is Cohen–Macaulay.
For $mathfrak p in operatorname{Spec}R$ we know $R_{mathfrak p}$ is CM and $I_mathfrak p$ is principal. By ($bullet$), we find that for all $mathfrak p supset I$ the quotient $R_{mathfrak p}/I_mathfrak p$ is CM. Such $mathfrak p$ correspond precisely to the prime ideals $bar{mathfrak p}$ of $R/I$. Finally since $(R/I)_{bar{mathfrak p}} = R_{mathfrak p}/I_{mathfrak p}$, the proposition follows. $square$
edited Dec 29 '18 at 15:59
user26857
39.3k124183
answered Dec 28 '18 at 11:19
BenBen
4,063617
$endgroup$
$begingroup$
As far as I can see $R$ in the question is not local.
$endgroup$
– user26857
Dec 28 '18 at 23:07
$begingroup$
@user26857 my understanding is that a noetherian ring is CM if all its local rings are CM local rings, so the non-local case will follow
$endgroup$
– Ben
Dec 29 '18 at 8:10
$begingroup$
@user26857 I edited to address your comment more clearly.
$endgroup$
– Ben
Dec 29 '18 at 9:15
$begingroup$
My understanding is the same. But the answer, in its previous form, didn't point this out, and I think it should. (+1)
$endgroup$
– user26857
Dec 29 '18 at 15:57
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
The answer to both (1) and (2) is yes. More generally we have the following:
Proposition. If $R$ is any CM domain with principal ideal $I$, then $R/I$ is CM.
Proof. First recall the definition for (non-local) CM rings, from Stacks Project:
Definition 10.103.6. A Noetherian ring $R$ is called Cohen-Macaulay if all its local rings are Cohen-Macaulay.
and a fact from the Wikipedia page Cohen-Macaulay ring:
($bullet$) For a non-zero-divisor $u$ in the maximal ideal of a Noetherian local ring $R$, $R$ is Cohen–Macaulay if and only if $R/(u)$ is Cohen–Macaulay.
For $mathfrak p in operatorname{Spec}R$ we know $R_{mathfrak p}$ is CM and $I_mathfrak p$ is principal. By ($bullet$), we find that for all $mathfrak p supset I$ the quotient $R_{mathfrak p}/I_mathfrak p$ is CM. Such $mathfrak p$ correspond precisely to the prime ideals $bar{mathfrak p}$ of $R/I$. Finally since $(R/I)_{bar{mathfrak p}} = R_{mathfrak p}/I_{mathfrak p}$, the proposition follows. $square$
edited Dec 29 '18 at 15:59
user26857
39.3k124183
answered Dec 28 '18 at 11:19
BenBen
4,063617
$endgroup$
$begingroup$
As far as I can see $R$ in the question is not local.
$endgroup$
– user26857
Dec 28 '18 at 23:07
$begingroup$
@user26857 my understanding is that a noetherian ring is CM if all its local rings are CM local rings, so the non-local case will follow
$endgroup$
– Ben
Dec 29 '18 at 8:10
$begingroup$
@user26857 I edited to address your comment more clearly.
$endgroup$
– Ben
Dec 29 '18 at 9:15
$begingroup$
My understanding is the same. But the answer, in its previous form, didn't point this out, and I think it should. (+1)
$endgroup$
– user26857
Dec 29 '18 at 15:57
add a comment |
$begingroup$
The answer to both (1) and (2) is yes. More generally we have the following:
Proposition. If $R$ is any CM domain with principal ideal $I$, then $R/I$ is CM.
Proof. First recall the definition for (non-local) CM rings, from Stacks Project:
Definition 10.103.6. A Noetherian ring $R$ is called Cohen-Macaulay if all its local rings are Cohen-Macaulay.
and a fact from the Wikipedia page Cohen-Macaulay ring:
($bullet$) For a non-zero-divisor $u$ in the maximal ideal of a Noetherian local ring $R$, $R$ is Cohen–Macaulay if and only if $R/(u)$ is Cohen–Macaulay.
For $mathfrak p in operatorname{Spec}R$ we know $R_{mathfrak p}$ is CM and $I_mathfrak p$ is principal. By ($bullet$), we find that for all $mathfrak p supset I$ the quotient $R_{mathfrak p}/I_mathfrak p$ is CM. Such $mathfrak p$ correspond precisely to the prime ideals $bar{mathfrak p}$ of $R/I$. Finally since $(R/I)_{bar{mathfrak p}} = R_{mathfrak p}/I_{mathfrak p}$, the proposition follows. $square$
edited Dec 29 '18 at 15:59
user26857
39.3k124183
answered Dec 28 '18 at 11:19
BenBen
4,063617
$endgroup$
$begingroup$
As far as I can see $R$ in the question is not local.
$endgroup$
– user26857
Dec 28 '18 at 23:07
$begingroup$
@user26857 my understanding is that a noetherian ring is CM if all its local rings are CM local rings, so the non-local case will follow
$endgroup$
– Ben
Dec 29 '18 at 8:10
$begingroup$
@user26857 I edited to address your comment more clearly.
$endgroup$
– Ben
Dec 29 '18 at 9:15
$begingroup$
My understanding is the same. But the answer, in its previous form, didn't point this out, and I think it should. (+1)
$endgroup$
– user26857
Dec 29 '18 at 15:57
add a comment |
$begingroup$
The answer to both (1) and (2) is yes. More generally we have the following:
Proposition. If $R$ is any CM domain with principal ideal $I$, then $R/I$ is CM.
Proof. First recall the definition for (non-local) CM rings, from Stacks Project:
Definition 10.103.6. A Noetherian ring $R$ is called Cohen-Macaulay if all its local rings are Cohen-Macaulay.
and a fact from the Wikipedia page Cohen-Macaulay ring:
($bullet$) For a non-zero-divisor $u$ in the maximal ideal of a Noetherian local ring $R$, $R$ is Cohen–Macaulay if and only if $R/(u)$ is Cohen–Macaulay.
For $mathfrak p in operatorname{Spec}R$ we know $R_{mathfrak p}$ is CM and $I_mathfrak p$ is principal. By ($bullet$), we find that for all $mathfrak p supset I$ the quotient $R_{mathfrak p}/I_mathfrak p$ is CM. Such $mathfrak p$ correspond precisely to the prime ideals $bar{mathfrak p}$ of $R/I$. Finally since $(R/I)_{bar{mathfrak p}} = R_{mathfrak p}/I_{mathfrak p}$, the proposition follows. $square$
edited Dec 29 '18 at 15:59
user26857
39.3k124183
answered Dec 28 '18 at 11:19
BenBen
4,063617
$endgroup$
The answer to both (1) and (2) is yes. More generally we have the following:
Proposition. If $R$ is any CM domain with principal ideal $I$, then $R/I$ is CM.
Proof. First recall the definition for (non-local) CM rings, from Stacks Project:
Definition 10.103.6. A Noetherian ring $R$ is called Cohen-Macaulay if all its local rings are Cohen-Macaulay.
and a fact from the Wikipedia page Cohen-Macaulay ring:
($bullet$) For a non-zero-divisor $u$ in the maximal ideal of a Noetherian local ring $R$, $R$ is Cohen–Macaulay if and only if $R/(u)$ is Cohen–Macaulay.
For $mathfrak p in operatorname{Spec}R$ we know $R_{mathfrak p}$ is CM and $I_mathfrak p$ is principal. By ($bullet$), we find that for all $mathfrak p supset I$ the quotient $R_{mathfrak p}/I_mathfrak p$ is CM. Such $mathfrak p$ correspond precisely to the prime ideals $bar{mathfrak p}$ of $R/I$. Finally since $(R/I)_{bar{mathfrak p}} = R_{mathfrak p}/I_{mathfrak p}$, the proposition follows. $square$
edited Dec 29 '18 at 15:59
user26857
39.3k124183
answered Dec 28 '18 at 11:19
BenBen
4,063617
edited Dec 29 '18 at 15:59
user26857
39.3k124183
edited Dec 29 '18 at 15:59
user26857
39.3k124183
edited Dec 29 '18 at 15:59
user26857
39.3k124183
39.3k124183
answered Dec 28 '18 at 11:19
BenBen
4,063617
answered Dec 28 '18 at 11:19
BenBen
4,063617
answered Dec 28 '18 at 11:19
BenBen
4,063617
4,063617
$begingroup$
As far as I can see $R$ in the question is not local.
$endgroup$
– user26857
Dec 28 '18 at 23:07
$begingroup$
@user26857 my understanding is that a noetherian ring is CM if all its local rings are CM local rings, so the non-local case will follow
$endgroup$
– Ben
Dec 29 '18 at 8:10
$begingroup$
@user26857 I edited to address your comment more clearly.
$endgroup$
– Ben
Dec 29 '18 at 9:15
$begingroup$
My understanding is the same. But the answer, in its previous form, didn't point this out, and I think it should. (+1)
$endgroup$
– user26857
Dec 29 '18 at 15:57
add a comment |
$begingroup$
As far as I can see $R$ in the question is not local.
$endgroup$
– user26857
Dec 28 '18 at 23:07
$begingroup$
@user26857 my understanding is that a noetherian ring is CM if all its local rings are CM local rings, so the non-local case will follow
$endgroup$
– Ben
Dec 29 '18 at 8:10
$begingroup$
@user26857 I edited to address your comment more clearly.
$endgroup$
– Ben
Dec 29 '18 at 9:15
$begingroup$
My understanding is the same. But the answer, in its previous form, didn't point this out, and I think it should. (+1)
$endgroup$
– user26857
Dec 29 '18 at 15:57
$begingroup$
As far as I can see $R$ in the question is not local.
$endgroup$
– user26857
Dec 28 '18 at 23:07
$begingroup$
As far as I can see $R$ in the question is not local.
$endgroup$
– user26857
Dec 28 '18 at 23:07
$begingroup$
@user26857 my understanding is that a noetherian ring is CM if all its local rings are CM local rings, so the non-local case will follow
$endgroup$
– Ben
Dec 29 '18 at 8:10
$begingroup$
@user26857 my understanding is that a noetherian ring is CM if all its local rings are CM local rings, so the non-local case will follow
$endgroup$
– Ben
Dec 29 '18 at 8:10
$begingroup$
@user26857 I edited to address your comment more clearly.
$endgroup$
– Ben
Dec 29 '18 at 9:15
$begingroup$
@user26857 I edited to address your comment more clearly.
$endgroup$
– Ben
Dec 29 '18 at 9:15
$begingroup$
My understanding is the same. But the answer, in its previous form, didn't point this out, and I think it should. (+1)
$endgroup$
– user26857
Dec 29 '18 at 15:57
$begingroup$
My understanding is the same. But the answer, in its previous form, didn't point this out, and I think it should. (+1)
$endgroup$
– user26857
Dec 29 '18 at 15:57
add a comment |
