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?
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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?
$begingroup$
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.
An ideal $Isubset A$ is primary if $Ine A$ and $xyin Iimplies$ either $xin I$ or $y^nin I$ for some $n> 0$.
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
$endgroup$
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.
An ideal $Isubset A$ is primary if $Ine A$ and $xyin Iimplies$ either $xin I$ or $y^nin I$ for some $n> 0$.
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
$endgroup$
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.
An ideal $Isubset A$ is primary if $Ine A$ and $xyin Iimplies$ either $xin I$ or $y^nin I$ for some $n> 0$.
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
$endgroup$
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.
An ideal $Isubset A$ is primary if $Ine A$ and $xyin Iimplies$ either $xin I$ or $y^nin I$ for some $n> 0$.
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.
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
abstract-algebra commutative-algebra ideals
abstract-algebra commutative-algebra ideals
edited Mar 12 at 22:29
J. W. Tanner
3,3451320
3,3451320
asked Mar 12 at 21:42
user437309user437309
763314
763314
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.
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 |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
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 $$Isubseteq Psubseteq sqrtI,$$ so $sqrtI=P$. Thus the only possible associated prime of $A/I$ is $sqrtI$, and since $A/I$ is nonzero, it has an associated prime. Thus $operatornameAss A/I$ consists of a single element, $sqrtI$.
$2implies 1$. Suppose $operatornameAss A/I$ 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 $P=sqrtI$. Thus since $yin P$, $y^nin I$ for some $n$.
$endgroup$
$begingroup$
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 $Isubseteq Psubseteq sqrtI$? Shouldn't it be $Isubseteq sqrtIsubseteq Psubseteq I$?
$endgroup$
– 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 $yinsqrtI$. Thus $Psubseteq sqrtI$.
$endgroup$
– 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 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
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 $$Isubseteq Psubseteq sqrtI,$$ so $sqrtI=P$. Thus the only possible associated prime of $A/I$ is $sqrtI$, and since $A/I$ is nonzero, it has an associated prime. Thus $operatornameAss A/I$ consists of a single element, $sqrtI$.
$2implies 1$. Suppose $operatornameAss A/I$ 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 $P=sqrtI$. Thus since $yin P$, $y^nin I$ for some $n$.
$endgroup$
$begingroup$
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 $Isubseteq Psubseteq sqrtI$? Shouldn't it be $Isubseteq sqrtIsubseteq Psubseteq I$?
$endgroup$
– 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 $yinsqrtI$. Thus $Psubseteq sqrtI$.
$endgroup$
– 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 $$Isubseteq Psubseteq sqrtI,$$ so $sqrtI=P$. Thus the only possible associated prime of $A/I$ is $sqrtI$, and since $A/I$ is nonzero, it has an associated prime. Thus $operatornameAss A/I$ consists of a single element, $sqrtI$.
$2implies 1$. Suppose $operatornameAss A/I$ 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 $P=sqrtI$. Thus since $yin P$, $y^nin I$ for some $n$.
$endgroup$
$begingroup$
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 $Isubseteq Psubseteq sqrtI$? Shouldn't it be $Isubseteq sqrtIsubseteq Psubseteq I$?
$endgroup$
– 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 $yinsqrtI$. Thus $Psubseteq sqrtI$.
$endgroup$
– 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 $$Isubseteq Psubseteq sqrtI,$$ so $sqrtI=P$. Thus the only possible associated prime of $A/I$ is $sqrtI$, and since $A/I$ is nonzero, it has an associated prime. Thus $operatornameAss A/I$ consists of a single element, $sqrtI$.
$2implies 1$. Suppose $operatornameAss A/I$ 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 $P=sqrtI$. Thus since $yin P$, $y^nin I$ for some $n$.
$endgroup$
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 $$Isubseteq Psubseteq sqrtI,$$ so $sqrtI=P$. Thus the only possible associated prime of $A/I$ is $sqrtI$, and since $A/I$ is nonzero, it has an associated prime. Thus $operatornameAss A/I$ consists of a single element, $sqrtI$.
$2implies 1$. Suppose $operatornameAss A/I$ 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 $P=sqrtI$. Thus since $yin P$, $y^nin I$ for some $n$.
answered Mar 12 at 22:03
jgonjgon
15.5k32143
15.5k32143
$begingroup$
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 $Isubseteq Psubseteq sqrtI$? Shouldn't it be $Isubseteq sqrtIsubseteq Psubseteq I$?
$endgroup$
– 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 $yinsqrtI$. Thus $Psubseteq sqrtI$.
$endgroup$
– 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
$begingroup$
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 $Isubseteq Psubseteq sqrtI$? Shouldn't it be $Isubseteq sqrtIsubseteq Psubseteq I$?
$endgroup$
– 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 $yinsqrtI$. Thus $Psubseteq sqrtI$.
$endgroup$
– 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
$begingroup$
And I realized that this question is a duplicate, but oh well.
$endgroup$
– jgon
Mar 12 at 22:06
$begingroup$
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 $Isubseteq Psubseteq sqrtI$? Shouldn't it be $Isubseteq sqrtIsubseteq Psubseteq I$?
$endgroup$
– user437309
Mar 12 at 23:00
$begingroup$
How did you get the chain $Isubseteq Psubseteq sqrtI$? Shouldn't it be $Isubseteq sqrtIsubseteq Psubseteq I$?
$endgroup$
– 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 $yinsqrtI$. Thus $Psubseteq sqrtI$.
$endgroup$
– jgon
Mar 12 at 23:10
$begingroup$
$Isubseteq P$, 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 $yinsqrtI$. Thus $Psubseteq sqrtI$.
$endgroup$
– 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
$begingroup$
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
$begingroup$
@user437309 The annihilator of $A/I$ is just $I$.
$endgroup$
– jgon
Mar 12 at 23:25
|
show 1 more comment