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Equivalent definitions of primary ideals [duplicate]

Primary ideals in Noetherian ringsHow to prove Ass$(R/Q)={P}$ if and only if $Q$ is $P$-primary when $R$ is Noetherian?Primary ideals of Noetherian rings which are not irreducibleMinimal and minimal associated prime idealsset of associated primes to a primary decomposition of a moduleInfinitely many primary decompositions of an idealIntersection of Primary IdealsHow to prove Ass$(R/Q)={P}$ if and only if $Q$ is $P$-primary when $R$ is Noetherian?Irreducible primary idealsExample where $I,J$ are $p$-primary ideals, but $I+J$ is not $p$-primary.Equivalence of two definitions of primary ideals$A$ has only finitely many minimal prime ideals $implies (0)$ is decomposable?

1

$begingroup$

This question already has an answer here:

• 
  Primary ideals in Noetherian rings
  
  1 answer
  
  


• 
  How to prove Ass$(R/Q)={P}$ if and only if $Q$ is $P$-primary when $R$ is Noetherian? [duplicate]
  
  1 answer
  
  

Here are two definitions of a primary ideal.

1. An ideal $Isubset A$ is primary if $Ine A$ and $xyin Iimplies$ either $xin I$ or $y^nin I$ for some $n> 0$.




2. An ideal $Isubset A$ is primary if $Ass(A/I)$ consists of a single element.

Are they equivalent?

$2.implies1.$ Suppose $Ass(A/I)$ consists of a single element $P$. If $I=A$, then I believe $Ass(A/I)=Ass({0})$. The zero ring has no prime ideals, so there are no associated primes of the element $0$ in the zero ring. This contradiction shows that $Ine A$. But I don't know how to deduce the other condition.

I don't have ideas for the other implication either.

abstract-algebra commutative-algebra ideals

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edited Mar 12 at 22:29

J. W. Tanner

3,4751320

asked Mar 12 at 21:42

user437309user437309

766314

$endgroup$

marked as duplicate by jgon, Shailesh, Leucippus, Alex Provost, Cesareo Mar 13 at 8:39

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

add a comment | 

1

$begingroup$

This question already has an answer here:

• 
  Primary ideals in Noetherian rings
  
  1 answer
  
  


• 
  How to prove Ass$(R/Q)={P}$ if and only if $Q$ is $P$-primary when $R$ is Noetherian? [duplicate]
  
  1 answer
  
  

Here are two definitions of a primary ideal.

1. An ideal $Isubset A$ is primary if $Ine A$ and $xyin Iimplies$ either $xin I$ or $y^nin I$ for some $n> 0$.




2. An ideal $Isubset A$ is primary if $Ass(A/I)$ consists of a single element.

Are they equivalent?

$2.implies1.$ Suppose $Ass(A/I)$ consists of a single element $P$. If $I=A$, then I believe $Ass(A/I)=Ass({0})$. The zero ring has no prime ideals, so there are no associated primes of the element $0$ in the zero ring. This contradiction shows that $Ine A$. But I don't know how to deduce the other condition.

I don't have ideas for the other implication either.

abstract-algebra commutative-algebra ideals

share|cite|improve this question

edited Mar 12 at 22:29

J. W. Tanner

3,4751320

asked Mar 12 at 21:42

user437309user437309

766314

$endgroup$

marked as duplicate by jgon, Shailesh, Leucippus, Alex Provost, Cesareo Mar 13 at 8:39

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

add a comment | 

$begingroup$

This question already has an answer here:

• 
  Primary ideals in Noetherian rings
  
  1 answer
  
  


• 
  How to prove Ass$(R/Q)={P}$ if and only if $Q$ is $P$-primary when $R$ is Noetherian? [duplicate]
  
  1 answer
  
  

Here are two definitions of a primary ideal.

1. An ideal $Isubset A$ is primary if $Ine A$ and $xyin Iimplies$ either $xin I$ or $y^nin I$ for some $n> 0$.




2. An ideal $Isubset A$ is primary if $Ass(A/I)$ consists of a single element.

Are they equivalent?

$2.implies1.$ Suppose $Ass(A/I)$ consists of a single element $P$. If $I=A$, then I believe $Ass(A/I)=Ass({0})$. The zero ring has no prime ideals, so there are no associated primes of the element $0$ in the zero ring. This contradiction shows that $Ine A$. But I don't know how to deduce the other condition.

I don't have ideas for the other implication either.

abstract-algebra commutative-algebra ideals

share|cite|improve this question

edited Mar 12 at 22:29

J. W. Tanner

3,4751320

asked Mar 12 at 21:42

user437309user437309

766314

$endgroup$

share|cite|improve this question

edited Mar 12 at 22:29

J. W. Tanner

3,4751320

asked Mar 12 at 21:42

user437309user437309

766314

share|cite|improve this question

edited Mar 12 at 22:29

J. W. Tanner

3,4751320

asked Mar 12 at 21:42

user437309user437309

766314

edited Mar 12 at 22:29

J. W. Tanner

3,4751320

edited Mar 12 at 22:29

J. W. Tanner

3,4751320

asked Mar 12 at 21:42

user437309user437309

766314

asked Mar 12 at 21:42

user437309user437309

766314

1 Answer
1

active

oldest

votes

1

$begingroup$

I assume $A$ is Noetherian.

$1implies 2$. Suppose $I$ is a proper ideal such that if $xyin I$, then either $xin I$ or $y^nin I$ for some $n>0$.

Then if $P$ is associated to $A/I$, there exists $xnot in I$ in $A$ such that $P$ is the annihilator of $x$ in $A/I$. I.e., $Pxsubseteq I$, but this implies that $$Isubseteq Psubseteq sqrt{I},$$ so $sqrt{I}=P$. Thus the only possible associated prime of $A/I$ is $sqrt{I}$, and since $A/I$ is nonzero, it has an associated prime. Thus $operatorname{Ass} A/I$ consists of a single element, $sqrt{I}$.

$2implies 1$. Suppose $operatorname{Ass} A/I$ consists of a single element. Then if $xyin I$, with $xnotin I$, then $y$ annihilates the image of $x$ in $A/I$, so $y$ lies in some associated prime, necessarily the unique associated prime of $A/I$, $P$. Since every ideal minimal over $I$ is associated to $A/I$, $P$ is the unique minimal prime over $I$, and therefore $P=sqrt{I}$. Thus since $yin P$, $y^nin I$ for some $n$.

share|cite|improve this answer

answered Mar 12 at 22:03

jgonjgon

15.7k32143

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  And I realized that this question is a duplicate, but oh well.
  $endgroup$
  – jgon
  Mar 12 at 22:06
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  How did you get the chain $Isubseteq Psubseteq sqrt{I}$? Shouldn't it be $Isubseteq sqrt{I}subseteq Psubseteq I$?
  $endgroup$
  – user437309
  Mar 12 at 23:00
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  $Isubseteq P$, since $P$ is the annihilator of an element of $A/I$. Then for all $yin P$, $yxin I$, and $xnot in I$, so $y^nin I$. Thus $yinsqrt{I}$. Thus $Psubseteq sqrt{I}$.
  $endgroup$
  – jgon
  Mar 12 at 23:10
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thanks for the clarification. I also don't understand why every ideal minimal over $I$ is associated to $A/I$ (I only know that any prime ideal minimal over $Ann(A/I)$ is associated to $A/I$), and how it follows that $P$ is the unique minimal prime over $I$...
  $endgroup$
  – user437309
  Mar 12 at 23:21
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @user437309 The annihilator of $A/I$ is just $I$.
  $endgroup$
  – jgon
  Mar 12 at 23:25
  

  
  
  
  

 | 
show 1 more comment

1

$begingroup$

I assume $A$ is Noetherian.

$1implies 2$. Suppose $I$ is a proper ideal such that if $xyin I$, then either $xin I$ or $y^nin I$ for some $n>0$.

Then if $P$ is associated to $A/I$, there exists $xnot in I$ in $A$ such that $P$ is the annihilator of $x$ in $A/I$. I.e., $Pxsubseteq I$, but this implies that $$Isubseteq Psubseteq sqrt{I},$$ so $sqrt{I}=P$. Thus the only possible associated prime of $A/I$ is $sqrt{I}$, and since $A/I$ is nonzero, it has an associated prime. Thus $operatorname{Ass} A/I$ consists of a single element, $sqrt{I}$.

$2implies 1$. Suppose $operatorname{Ass} A/I$ consists of a single element. Then if $xyin I$, with $xnotin I$, then $y$ annihilates the image of $x$ in $A/I$, so $y$ lies in some associated prime, necessarily the unique associated prime of $A/I$, $P$. Since every ideal minimal over $I$ is associated to $A/I$, $P$ is the unique minimal prime over $I$, and therefore $P=sqrt{I}$. Thus since $yin P$, $y^nin I$ for some $n$.

share|cite|improve this answer

answered Mar 12 at 22:03

jgonjgon

15.7k32143

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  And I realized that this question is a duplicate, but oh well.
  $endgroup$
  – jgon
  Mar 12 at 22:06
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  How did you get the chain $Isubseteq Psubseteq sqrt{I}$? Shouldn't it be $Isubseteq sqrt{I}subseteq Psubseteq I$?
  $endgroup$
  – user437309
  Mar 12 at 23:00
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  $Isubseteq P$, since $P$ is the annihilator of an element of $A/I$. Then for all $yin P$, $yxin I$, and $xnot in I$, so $y^nin I$. Thus $yinsqrt{I}$. Thus $Psubseteq sqrt{I}$.
  $endgroup$
  – jgon
  Mar 12 at 23:10
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thanks for the clarification. I also don't understand why every ideal minimal over $I$ is associated to $A/I$ (I only know that any prime ideal minimal over $Ann(A/I)$ is associated to $A/I$), and how it follows that $P$ is the unique minimal prime over $I$...
  $endgroup$
  – user437309
  Mar 12 at 23:21
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @user437309 The annihilator of $A/I$ is just $I$.
  $endgroup$
  – jgon
  Mar 12 at 23:25
  

  
  
  
  

 | 
show 1 more comment

1

$begingroup$

I assume $A$ is Noetherian.

$1implies 2$. Suppose $I$ is a proper ideal such that if $xyin I$, then either $xin I$ or $y^nin I$ for some $n>0$.

Then if $P$ is associated to $A/I$, there exists $xnot in I$ in $A$ such that $P$ is the annihilator of $x$ in $A/I$. I.e., $Pxsubseteq I$, but this implies that $$Isubseteq Psubseteq sqrt{I},$$ so $sqrt{I}=P$. Thus the only possible associated prime of $A/I$ is $sqrt{I}$, and since $A/I$ is nonzero, it has an associated prime. Thus $operatorname{Ass} A/I$ consists of a single element, $sqrt{I}$.

$2implies 1$. Suppose $operatorname{Ass} A/I$ consists of a single element. Then if $xyin I$, with $xnotin I$, then $y$ annihilates the image of $x$ in $A/I$, so $y$ lies in some associated prime, necessarily the unique associated prime of $A/I$, $P$. Since every ideal minimal over $I$ is associated to $A/I$, $P$ is the unique minimal prime over $I$, and therefore $P=sqrt{I}$. Thus since $yin P$, $y^nin I$ for some $n$.

share|cite|improve this answer

answered Mar 12 at 22:03

jgonjgon

15.7k32143

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  And I realized that this question is a duplicate, but oh well.
  $endgroup$
  – jgon
  Mar 12 at 22:06
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  How did you get the chain $Isubseteq Psubseteq sqrt{I}$? Shouldn't it be $Isubseteq sqrt{I}subseteq Psubseteq I$?
  $endgroup$
  – user437309
  Mar 12 at 23:00
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  $Isubseteq P$, since $P$ is the annihilator of an element of $A/I$. Then for all $yin P$, $yxin I$, and $xnot in I$, so $y^nin I$. Thus $yinsqrt{I}$. Thus $Psubseteq sqrt{I}$.
  $endgroup$
  – jgon
  Mar 12 at 23:10
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thanks for the clarification. I also don't understand why every ideal minimal over $I$ is associated to $A/I$ (I only know that any prime ideal minimal over $Ann(A/I)$ is associated to $A/I$), and how it follows that $P$ is the unique minimal prime over $I$...
  $endgroup$
  – user437309
  Mar 12 at 23:21
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @user437309 The annihilator of $A/I$ is just $I$.
  $endgroup$
  – jgon
  Mar 12 at 23:25
  

  
  
  
  

 | 
show 1 more comment

$begingroup$

I assume $A$ is Noetherian.

$1implies 2$. Suppose $I$ is a proper ideal such that if $xyin I$, then either $xin I$ or $y^nin I$ for some $n>0$.

Then if $P$ is associated to $A/I$, there exists $xnot in I$ in $A$ such that $P$ is the annihilator of $x$ in $A/I$. I.e., $Pxsubseteq I$, but this implies that $$Isubseteq Psubseteq sqrt{I},$$ so $sqrt{I}=P$. Thus the only possible associated prime of $A/I$ is $sqrt{I}$, and since $A/I$ is nonzero, it has an associated prime. Thus $operatorname{Ass} A/I$ consists of a single element, $sqrt{I}$.

$2implies 1$. Suppose $operatorname{Ass} A/I$ consists of a single element. Then if $xyin I$, with $xnotin I$, then $y$ annihilates the image of $x$ in $A/I$, so $y$ lies in some associated prime, necessarily the unique associated prime of $A/I$, $P$. Since every ideal minimal over $I$ is associated to $A/I$, $P$ is the unique minimal prime over $I$, and therefore $P=sqrt{I}$. Thus since $yin P$, $y^nin I$ for some $n$.

share|cite|improve this answer

answered Mar 12 at 22:03

jgonjgon

15.7k32143

$endgroup$

share|cite|improve this answer

answered Mar 12 at 22:03

jgonjgon

15.7k32143

answered Mar 12 at 22:03

jgonjgon

15.7k32143

answered Mar 12 at 22:03

jgonjgon

15.7k32143