Associated prime ideal of a primary ideal vs. associated primes as annihilators [closed]
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Let A be a commutative ring and $Q$ be a primary ideal, then $sqrt{Q}$ is called the associated prime ideal of $Q$. Is $sqrt{Q}$ naturally an associated prime of some $A$-module $M$, i.e. $sqrt{Q}=ann(x)$ for some $xin M$?
commutative-algebra
asked Nov 19 at 3:42
No One
1,9771519
closed as off-topic by user26857, amWhy, Leucippus, KReiser, Cesareo Nov 22 at 1:45
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, amWhy, Leucippus, KReiser, Cesareo
If this question can be reworded to fit the rules in the help center, please edit the question.
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up vote
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Let A be a commutative ring and $Q$ be a primary ideal, then $sqrt{Q}$ is called the associated prime ideal of $Q$. Is $sqrt{Q}$ naturally an associated prime of some $A$-module $M$, i.e. $sqrt{Q}=ann(x)$ for some $xin M$?
commutative-algebra
asked Nov 19 at 3:42
No One
1,9771519
closed as off-topic by user26857, amWhy, Leucippus, KReiser, Cesareo Nov 22 at 1:45
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, amWhy, Leucippus, KReiser, Cesareo
If this question can be reworded to fit the rules in the help center, please edit the question.
Of course! Choose $M=A/Q$.
– user26857
Nov 19 at 8:27
@user26857 I guess in this case we should take $x=1+Q$. I know $sqrt{Q}supset ann(1+Q)$ but why $sqrt{Q}subset ann(1+Q)$? i.e. if $rin sqrt Q$ or $r^nin Q$, why does $r$ annihilate $1+Q$?
– No One
Nov 20 at 19:40
ann(1+Q)=Q, so the choice is wrong. Choose x=a+Q, where a belongs to the quotient ideal (Q:P), where P is the radical of Q.
– user26857
Nov 21 at 9:42
@user26857 I guess $a$ is an arbitrary element s.t. $aPsubset Q$? Is it still unclear to me how to show $ann(a+Q)=sqrt{Q}$. Could you add more details?
– No One
Nov 22 at 4:55
add a comment |
up vote
-1
down vote
favorite
up vote
-1
down vote
favorite
Let A be a commutative ring and $Q$ be a primary ideal, then $sqrt{Q}$ is called the associated prime ideal of $Q$. Is $sqrt{Q}$ naturally an associated prime of some $A$-module $M$, i.e. $sqrt{Q}=ann(x)$ for some $xin M$?
commutative-algebra
asked Nov 19 at 3:42
No One
1,9771519
Let A be a commutative ring and $Q$ be a primary ideal, then $sqrt{Q}$ is called the associated prime ideal of $Q$. Is $sqrt{Q}$ naturally an associated prime of some $A$-module $M$, i.e. $sqrt{Q}=ann(x)$ for some $xin M$?
commutative-algebra
commutative-algebra
asked Nov 19 at 3:42
No One
1,9771519
asked Nov 19 at 3:42
No One
1,9771519
asked Nov 19 at 3:42
No One
1,9771519
asked Nov 19 at 3:42
No One
1,9771519
asked Nov 19 at 3:42
No One
1,9771519
1,9771519
closed as off-topic by user26857, amWhy, Leucippus, KReiser, Cesareo Nov 22 at 1:45
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, amWhy, Leucippus, KReiser, Cesareo
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by user26857, amWhy, Leucippus, KReiser, Cesareo Nov 22 at 1:45
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, amWhy, Leucippus, KReiser, Cesareo
If this question can be reworded to fit the rules in the help center, please edit the question.
Of course! Choose $M=A/Q$.
– user26857
Nov 19 at 8:27
@user26857 I guess in this case we should take $x=1+Q$. I know $sqrt{Q}supset ann(1+Q)$ but why $sqrt{Q}subset ann(1+Q)$? i.e. if $rin sqrt Q$ or $r^nin Q$, why does $r$ annihilate $1+Q$?
– No One
Nov 20 at 19:40
ann(1+Q)=Q, so the choice is wrong. Choose x=a+Q, where a belongs to the quotient ideal (Q:P), where P is the radical of Q.
– user26857
Nov 21 at 9:42
@user26857 I guess $a$ is an arbitrary element s.t. $aPsubset Q$? Is it still unclear to me how to show $ann(a+Q)=sqrt{Q}$. Could you add more details?
– No One
Nov 22 at 4:55
add a comment |
Of course! Choose $M=A/Q$.
– user26857
Nov 19 at 8:27
@user26857 I guess in this case we should take $x=1+Q$. I know $sqrt{Q}supset ann(1+Q)$ but why $sqrt{Q}subset ann(1+Q)$? i.e. if $rin sqrt Q$ or $r^nin Q$, why does $r$ annihilate $1+Q$?
– No One
Nov 20 at 19:40
ann(1+Q)=Q, so the choice is wrong. Choose x=a+Q, where a belongs to the quotient ideal (Q:P), where P is the radical of Q.
– user26857
Nov 21 at 9:42
@user26857 I guess $a$ is an arbitrary element s.t. $aPsubset Q$? Is it still unclear to me how to show $ann(a+Q)=sqrt{Q}$. Could you add more details?
– No One
Nov 22 at 4:55
Of course! Choose $M=A/Q$.
– user26857
Nov 19 at 8:27
Of course! Choose $M=A/Q$.
– user26857
Nov 19 at 8:27
@user26857 I guess in this case we should take $x=1+Q$. I know $sqrt{Q}supset ann(1+Q)$ but why $sqrt{Q}subset ann(1+Q)$? i.e. if $rin sqrt Q$ or $r^nin Q$, why does $r$ annihilate $1+Q$?
– No One
Nov 20 at 19:40
@user26857 I guess in this case we should take $x=1+Q$. I know $sqrt{Q}supset ann(1+Q)$ but why $sqrt{Q}subset ann(1+Q)$? i.e. if $rin sqrt Q$ or $r^nin Q$, why does $r$ annihilate $1+Q$?
– No One
Nov 20 at 19:40
ann(1+Q)=Q, so the choice is wrong. Choose x=a+Q, where a belongs to the quotient ideal (Q:P), where P is the radical of Q.
– user26857
Nov 21 at 9:42
ann(1+Q)=Q, so the choice is wrong. Choose x=a+Q, where a belongs to the quotient ideal (Q:P), where P is the radical of Q.
– user26857
Nov 21 at 9:42
@user26857 I guess $a$ is an arbitrary element s.t. $aPsubset Q$? Is it still unclear to me how to show $ann(a+Q)=sqrt{Q}$. Could you add more details?
– No One
Nov 22 at 4:55
@user26857 I guess $a$ is an arbitrary element s.t. $aPsubset Q$? Is it still unclear to me how to show $ann(a+Q)=sqrt{Q}$. Could you add more details?
– No One
Nov 22 at 4:55
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Of course! Choose $M=A/Q$.
– user26857
Nov 19 at 8:27
@user26857 I guess in this case we should take $x=1+Q$. I know $sqrt{Q}supset ann(1+Q)$ but why $sqrt{Q}subset ann(1+Q)$? i.e. if $rin sqrt Q$ or $r^nin Q$, why does $r$ annihilate $1+Q$?
– No One
Nov 20 at 19:40
ann(1+Q)=Q, so the choice is wrong. Choose x=a+Q, where a belongs to the quotient ideal (Q:P), where P is the radical of Q.
– user26857
Nov 21 at 9:42
@user26857 I guess $a$ is an arbitrary element s.t. $aPsubset Q$? Is it still unclear to me how to show $ann(a+Q)=sqrt{Q}$. Could you add more details?
– No One
Nov 22 at 4:55