The intersection of the associated primes of a reduced ring












1












$begingroup$


Let $R $ be a commutative ring with identity. Recall that a prime ideal is called associated prime ideal whenever it is the annihilator of a nonzero element. Also a ring is called reduced whenever has no nonzero nilpotent.




Why is zero the intersection of all associated prime ideals of a reduced ring?







abstract-algebra algebraic-geometry ring-theory commutative-algebra







share|cite|improve this question











$endgroup$












  • $begingroup$
    a nonzero element.... of $R$? Associated primes are usually defined with respect to a given module $M$. Are we supposed to assume $M=R$ or something?
    $endgroup$
    – rschwieb
    Jan 11 at 17:04












  • $begingroup$
    Yes. A prime ideal is a associated prime ideal of an $R $-module, if there exists a non zero element of that module such that the prime ideal is its annihilator.
    $endgroup$
    – Deroty
    Jan 11 at 17:35










  • $begingroup$
    I already knew everything after "yes", and it is not relevant to my question. Does your "yes" mean you are confirming that you are talking about a nonzero element of $R$ or do you mean something else (all associated primes for all $R$ modules?)
    $endgroup$
    – rschwieb
    Jan 11 at 17:44












  • $begingroup$
    About non zero elements of $R $.
    $endgroup$
    – Deroty
    Jan 11 at 18:31
















1












$begingroup$


Let $R $ be a commutative ring with identity. Recall that a prime ideal is called associated prime ideal whenever it is the annihilator of a nonzero element. Also a ring is called reduced whenever has no nonzero nilpotent.




Why is zero the intersection of all associated prime ideals of a reduced ring?







abstract-algebra algebraic-geometry ring-theory commutative-algebra







share|cite|improve this question











$endgroup$












  • $begingroup$
    a nonzero element.... of $R$? Associated primes are usually defined with respect to a given module $M$. Are we supposed to assume $M=R$ or something?
    $endgroup$
    – rschwieb
    Jan 11 at 17:04












  • $begingroup$
    Yes. A prime ideal is a associated prime ideal of an $R $-module, if there exists a non zero element of that module such that the prime ideal is its annihilator.
    $endgroup$
    – Deroty
    Jan 11 at 17:35










  • $begingroup$
    I already knew everything after "yes", and it is not relevant to my question. Does your "yes" mean you are confirming that you are talking about a nonzero element of $R$ or do you mean something else (all associated primes for all $R$ modules?)
    $endgroup$
    – rschwieb
    Jan 11 at 17:44












  • $begingroup$
    About non zero elements of $R $.
    $endgroup$
    – Deroty
    Jan 11 at 18:31














1












1








1


1



$begingroup$


Let $R $ be a commutative ring with identity. Recall that a prime ideal is called associated prime ideal whenever it is the annihilator of a nonzero element. Also a ring is called reduced whenever has no nonzero nilpotent.




Why is zero the intersection of all associated prime ideals of a reduced ring?







abstract-algebra algebraic-geometry ring-theory commutative-algebra







share|cite|improve this question











$endgroup$




Let $R $ be a commutative ring with identity. Recall that a prime ideal is called associated prime ideal whenever it is the annihilator of a nonzero element. Also a ring is called reduced whenever has no nonzero nilpotent.




Why is zero the intersection of all associated prime ideals of a reduced ring?







abstract-algebra algebraic-geometry ring-theory commutative-algebra




abstract-algebra algebraic-geometry ring-theory commutative-algebra






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 13 at 8:26









user26857

39.5k124284




39.5k124284










asked Jan 11 at 14:15









DerotyDeroty

31919




31919












  • $begingroup$
    a nonzero element.... of $R$? Associated primes are usually defined with respect to a given module $M$. Are we supposed to assume $M=R$ or something?
    $endgroup$
    – rschwieb
    Jan 11 at 17:04












  • $begingroup$
    Yes. A prime ideal is a associated prime ideal of an $R $-module, if there exists a non zero element of that module such that the prime ideal is its annihilator.
    $endgroup$
    – Deroty
    Jan 11 at 17:35










  • $begingroup$
    I already knew everything after "yes", and it is not relevant to my question. Does your "yes" mean you are confirming that you are talking about a nonzero element of $R$ or do you mean something else (all associated primes for all $R$ modules?)
    $endgroup$
    – rschwieb
    Jan 11 at 17:44












  • $begingroup$
    About non zero elements of $R $.
    $endgroup$
    – Deroty
    Jan 11 at 18:31


















  • $begingroup$
    a nonzero element.... of $R$? Associated primes are usually defined with respect to a given module $M$. Are we supposed to assume $M=R$ or something?
    $endgroup$
    – rschwieb
    Jan 11 at 17:04












  • $begingroup$
    Yes. A prime ideal is a associated prime ideal of an $R $-module, if there exists a non zero element of that module such that the prime ideal is its annihilator.
    $endgroup$
    – Deroty
    Jan 11 at 17:35










  • $begingroup$
    I already knew everything after "yes", and it is not relevant to my question. Does your "yes" mean you are confirming that you are talking about a nonzero element of $R$ or do you mean something else (all associated primes for all $R$ modules?)
    $endgroup$
    – rschwieb
    Jan 11 at 17:44












  • $begingroup$
    About non zero elements of $R $.
    $endgroup$
    – Deroty
    Jan 11 at 18:31
















$begingroup$
a nonzero element.... of $R$? Associated primes are usually defined with respect to a given module $M$. Are we supposed to assume $M=R$ or something?
$endgroup$
– rschwieb
Jan 11 at 17:04






$begingroup$
a nonzero element.... of $R$? Associated primes are usually defined with respect to a given module $M$. Are we supposed to assume $M=R$ or something?
$endgroup$
– rschwieb
Jan 11 at 17:04














$begingroup$
Yes. A prime ideal is a associated prime ideal of an $R $-module, if there exists a non zero element of that module such that the prime ideal is its annihilator.
$endgroup$
– Deroty
Jan 11 at 17:35




$begingroup$
Yes. A prime ideal is a associated prime ideal of an $R $-module, if there exists a non zero element of that module such that the prime ideal is its annihilator.
$endgroup$
– Deroty
Jan 11 at 17:35












$begingroup$
I already knew everything after "yes", and it is not relevant to my question. Does your "yes" mean you are confirming that you are talking about a nonzero element of $R$ or do you mean something else (all associated primes for all $R$ modules?)
$endgroup$
– rschwieb
Jan 11 at 17:44






$begingroup$
I already knew everything after "yes", and it is not relevant to my question. Does your "yes" mean you are confirming that you are talking about a nonzero element of $R$ or do you mean something else (all associated primes for all $R$ modules?)
$endgroup$
– rschwieb
Jan 11 at 17:44














$begingroup$
About non zero elements of $R $.
$endgroup$
– Deroty
Jan 11 at 18:31




$begingroup$
About non zero elements of $R $.
$endgroup$
– Deroty
Jan 11 at 18:31










2 Answers
2






active

oldest

votes


















2












$begingroup$

Let $R$ be a commutative ring. We denote by ${rm Ass}(R)$ the set of associated primes of $R$ and by ${rm Ass}^f(R)$ the set of weakly associated primes of $R$.



(1) By Bourbaki, AC.IV.1 Exercice 17 b), we have $bigcap{rm Ass}^f(R)={rm Nil}(R)$. Thus, if $R$ is reduced, then $bigcap{rm Ass}^f(R)=0$.



(2) If $R$ is noetherian, then by Bourbaki, AC.IV.1 Exercice 17 g) we have ${rm Ass}^f(R)={rm Ass}(R)$, and therefore $bigcap{rm Ass}(R)={rm Nil}(R)$ by (1). Thus, if $R$ is reduced and noetherian, then $bigcap{rm Ass}(R)=0$.



(3) There exists a non-noetherian reduced ring $R$ such that $bigcap{rm Ass}(R)=0$. For this it suffices to exhibit a non-noetherian reduced ring $R$ such that ${rm Ass}^f(R)={rm Ass}(R)$. Such an example (with $R$ even a domain) is given in P.-J. Cahen, Ascending chain conditions and associated primes, Commutative ring theory (Fès, 1992), 41-46, Lecture Notes in Pure and Appl. Math. 153, Dekker, New York, 1994.



(4) There exists a reduced ring $R$ such that $bigcap{rm Ass}(R)neq 0$. For this, any nonzero reduced ring $R$ with ${rm Ass}(R)=emptyset$ will do.



(Thanks to user25867 for pointing out how to improve statement (1).)






share|cite|improve this answer











$endgroup$









  • 1




    $begingroup$
    I think the nilradical equals the intersection of all weakly associated primes for any commutative ring, no need to assume that R is reduced. And this is the way one should read Bernard's answer.
    $endgroup$
    – user26857
    Jan 18 at 11:21








  • 1




    $begingroup$
    @user26857: Thanks for pointing this out!
    $endgroup$
    – Fred Rohrer
    Jan 18 at 12:05



















1












$begingroup$

The intersection of all associated prime ideals is the nilradical of $R$, i.e. the set of nilpotent elements. If the ring is reduced, this set is ${0}$.






share|cite|improve this answer











$endgroup$









  • 2




    $begingroup$
    I've seen that the nilradical is this intersection for Noetherian rings but does it hold in general?
    $endgroup$
    – rschwieb
    Jan 11 at 14:28








  • 2




    $begingroup$
    I believe this is true with correct notion of associated primes (weakly associated prime ideals – which are the same if the ring is noetherian).
    $endgroup$
    – Bernard
    Jan 11 at 14:33










  • $begingroup$
    The intersection of all minimal prime ideals is the nilradical of $R $. I want to know why it is true for associated primes in reduced rings, with the mentioned definition?
    $endgroup$
    – Deroty
    Jan 11 at 14:33








  • 2




    $begingroup$
    That's because one shows $operatorname{Spec} A$ and $operatorname{Ass} A$ have the same minimal elements.
    $endgroup$
    – Bernard
    Jan 12 at 11:13








  • 1




    $begingroup$
    As far as I remember, you can replace associated with weakly associated.
    $endgroup$
    – Bernard
    Jan 12 at 12:50












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2 Answers
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active

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2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes









2












$begingroup$

Let $R$ be a commutative ring. We denote by ${rm Ass}(R)$ the set of associated primes of $R$ and by ${rm Ass}^f(R)$ the set of weakly associated primes of $R$.



(1) By Bourbaki, AC.IV.1 Exercice 17 b), we have $bigcap{rm Ass}^f(R)={rm Nil}(R)$. Thus, if $R$ is reduced, then $bigcap{rm Ass}^f(R)=0$.



(2) If $R$ is noetherian, then by Bourbaki, AC.IV.1 Exercice 17 g) we have ${rm Ass}^f(R)={rm Ass}(R)$, and therefore $bigcap{rm Ass}(R)={rm Nil}(R)$ by (1). Thus, if $R$ is reduced and noetherian, then $bigcap{rm Ass}(R)=0$.



(3) There exists a non-noetherian reduced ring $R$ such that $bigcap{rm Ass}(R)=0$. For this it suffices to exhibit a non-noetherian reduced ring $R$ such that ${rm Ass}^f(R)={rm Ass}(R)$. Such an example (with $R$ even a domain) is given in P.-J. Cahen, Ascending chain conditions and associated primes, Commutative ring theory (Fès, 1992), 41-46, Lecture Notes in Pure and Appl. Math. 153, Dekker, New York, 1994.



(4) There exists a reduced ring $R$ such that $bigcap{rm Ass}(R)neq 0$. For this, any nonzero reduced ring $R$ with ${rm Ass}(R)=emptyset$ will do.



(Thanks to user25867 for pointing out how to improve statement (1).)






share|cite|improve this answer











$endgroup$









  • 1




    $begingroup$
    I think the nilradical equals the intersection of all weakly associated primes for any commutative ring, no need to assume that R is reduced. And this is the way one should read Bernard's answer.
    $endgroup$
    – user26857
    Jan 18 at 11:21








  • 1




    $begingroup$
    @user26857: Thanks for pointing this out!
    $endgroup$
    – Fred Rohrer
    Jan 18 at 12:05
















2












$begingroup$

Let $R$ be a commutative ring. We denote by ${rm Ass}(R)$ the set of associated primes of $R$ and by ${rm Ass}^f(R)$ the set of weakly associated primes of $R$.



(1) By Bourbaki, AC.IV.1 Exercice 17 b), we have $bigcap{rm Ass}^f(R)={rm Nil}(R)$. Thus, if $R$ is reduced, then $bigcap{rm Ass}^f(R)=0$.



(2) If $R$ is noetherian, then by Bourbaki, AC.IV.1 Exercice 17 g) we have ${rm Ass}^f(R)={rm Ass}(R)$, and therefore $bigcap{rm Ass}(R)={rm Nil}(R)$ by (1). Thus, if $R$ is reduced and noetherian, then $bigcap{rm Ass}(R)=0$.



(3) There exists a non-noetherian reduced ring $R$ such that $bigcap{rm Ass}(R)=0$. For this it suffices to exhibit a non-noetherian reduced ring $R$ such that ${rm Ass}^f(R)={rm Ass}(R)$. Such an example (with $R$ even a domain) is given in P.-J. Cahen, Ascending chain conditions and associated primes, Commutative ring theory (Fès, 1992), 41-46, Lecture Notes in Pure and Appl. Math. 153, Dekker, New York, 1994.



(4) There exists a reduced ring $R$ such that $bigcap{rm Ass}(R)neq 0$. For this, any nonzero reduced ring $R$ with ${rm Ass}(R)=emptyset$ will do.



(Thanks to user25867 for pointing out how to improve statement (1).)






share|cite|improve this answer











$endgroup$









  • 1




    $begingroup$
    I think the nilradical equals the intersection of all weakly associated primes for any commutative ring, no need to assume that R is reduced. And this is the way one should read Bernard's answer.
    $endgroup$
    – user26857
    Jan 18 at 11:21








  • 1




    $begingroup$
    @user26857: Thanks for pointing this out!
    $endgroup$
    – Fred Rohrer
    Jan 18 at 12:05














2












2








2





$begingroup$

Let $R$ be a commutative ring. We denote by ${rm Ass}(R)$ the set of associated primes of $R$ and by ${rm Ass}^f(R)$ the set of weakly associated primes of $R$.



(1) By Bourbaki, AC.IV.1 Exercice 17 b), we have $bigcap{rm Ass}^f(R)={rm Nil}(R)$. Thus, if $R$ is reduced, then $bigcap{rm Ass}^f(R)=0$.



(2) If $R$ is noetherian, then by Bourbaki, AC.IV.1 Exercice 17 g) we have ${rm Ass}^f(R)={rm Ass}(R)$, and therefore $bigcap{rm Ass}(R)={rm Nil}(R)$ by (1). Thus, if $R$ is reduced and noetherian, then $bigcap{rm Ass}(R)=0$.



(3) There exists a non-noetherian reduced ring $R$ such that $bigcap{rm Ass}(R)=0$. For this it suffices to exhibit a non-noetherian reduced ring $R$ such that ${rm Ass}^f(R)={rm Ass}(R)$. Such an example (with $R$ even a domain) is given in P.-J. Cahen, Ascending chain conditions and associated primes, Commutative ring theory (Fès, 1992), 41-46, Lecture Notes in Pure and Appl. Math. 153, Dekker, New York, 1994.



(4) There exists a reduced ring $R$ such that $bigcap{rm Ass}(R)neq 0$. For this, any nonzero reduced ring $R$ with ${rm Ass}(R)=emptyset$ will do.



(Thanks to user25867 for pointing out how to improve statement (1).)






share|cite|improve this answer











$endgroup$



Let $R$ be a commutative ring. We denote by ${rm Ass}(R)$ the set of associated primes of $R$ and by ${rm Ass}^f(R)$ the set of weakly associated primes of $R$.



(1) By Bourbaki, AC.IV.1 Exercice 17 b), we have $bigcap{rm Ass}^f(R)={rm Nil}(R)$. Thus, if $R$ is reduced, then $bigcap{rm Ass}^f(R)=0$.



(2) If $R$ is noetherian, then by Bourbaki, AC.IV.1 Exercice 17 g) we have ${rm Ass}^f(R)={rm Ass}(R)$, and therefore $bigcap{rm Ass}(R)={rm Nil}(R)$ by (1). Thus, if $R$ is reduced and noetherian, then $bigcap{rm Ass}(R)=0$.



(3) There exists a non-noetherian reduced ring $R$ such that $bigcap{rm Ass}(R)=0$. For this it suffices to exhibit a non-noetherian reduced ring $R$ such that ${rm Ass}^f(R)={rm Ass}(R)$. Such an example (with $R$ even a domain) is given in P.-J. Cahen, Ascending chain conditions and associated primes, Commutative ring theory (Fès, 1992), 41-46, Lecture Notes in Pure and Appl. Math. 153, Dekker, New York, 1994.



(4) There exists a reduced ring $R$ such that $bigcap{rm Ass}(R)neq 0$. For this, any nonzero reduced ring $R$ with ${rm Ass}(R)=emptyset$ will do.



(Thanks to user25867 for pointing out how to improve statement (1).)







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Jan 18 at 12:05

























answered Jan 16 at 13:52









Fred RohrerFred Rohrer

386110




386110








  • 1




    $begingroup$
    I think the nilradical equals the intersection of all weakly associated primes for any commutative ring, no need to assume that R is reduced. And this is the way one should read Bernard's answer.
    $endgroup$
    – user26857
    Jan 18 at 11:21








  • 1




    $begingroup$
    @user26857: Thanks for pointing this out!
    $endgroup$
    – Fred Rohrer
    Jan 18 at 12:05














  • 1




    $begingroup$
    I think the nilradical equals the intersection of all weakly associated primes for any commutative ring, no need to assume that R is reduced. And this is the way one should read Bernard's answer.
    $endgroup$
    – user26857
    Jan 18 at 11:21








  • 1




    $begingroup$
    @user26857: Thanks for pointing this out!
    $endgroup$
    – Fred Rohrer
    Jan 18 at 12:05








1




1




$begingroup$
I think the nilradical equals the intersection of all weakly associated primes for any commutative ring, no need to assume that R is reduced. And this is the way one should read Bernard's answer.
$endgroup$
– user26857
Jan 18 at 11:21






$begingroup$
I think the nilradical equals the intersection of all weakly associated primes for any commutative ring, no need to assume that R is reduced. And this is the way one should read Bernard's answer.
$endgroup$
– user26857
Jan 18 at 11:21






1




1




$begingroup$
@user26857: Thanks for pointing this out!
$endgroup$
– Fred Rohrer
Jan 18 at 12:05




$begingroup$
@user26857: Thanks for pointing this out!
$endgroup$
– Fred Rohrer
Jan 18 at 12:05











1












$begingroup$

The intersection of all associated prime ideals is the nilradical of $R$, i.e. the set of nilpotent elements. If the ring is reduced, this set is ${0}$.






share|cite|improve this answer











$endgroup$









  • 2




    $begingroup$
    I've seen that the nilradical is this intersection for Noetherian rings but does it hold in general?
    $endgroup$
    – rschwieb
    Jan 11 at 14:28








  • 2




    $begingroup$
    I believe this is true with correct notion of associated primes (weakly associated prime ideals – which are the same if the ring is noetherian).
    $endgroup$
    – Bernard
    Jan 11 at 14:33










  • $begingroup$
    The intersection of all minimal prime ideals is the nilradical of $R $. I want to know why it is true for associated primes in reduced rings, with the mentioned definition?
    $endgroup$
    – Deroty
    Jan 11 at 14:33








  • 2




    $begingroup$
    That's because one shows $operatorname{Spec} A$ and $operatorname{Ass} A$ have the same minimal elements.
    $endgroup$
    – Bernard
    Jan 12 at 11:13








  • 1




    $begingroup$
    As far as I remember, you can replace associated with weakly associated.
    $endgroup$
    – Bernard
    Jan 12 at 12:50
















1












$begingroup$

The intersection of all associated prime ideals is the nilradical of $R$, i.e. the set of nilpotent elements. If the ring is reduced, this set is ${0}$.






share|cite|improve this answer











$endgroup$









  • 2




    $begingroup$
    I've seen that the nilradical is this intersection for Noetherian rings but does it hold in general?
    $endgroup$
    – rschwieb
    Jan 11 at 14:28








  • 2




    $begingroup$
    I believe this is true with correct notion of associated primes (weakly associated prime ideals – which are the same if the ring is noetherian).
    $endgroup$
    – Bernard
    Jan 11 at 14:33










  • $begingroup$
    The intersection of all minimal prime ideals is the nilradical of $R $. I want to know why it is true for associated primes in reduced rings, with the mentioned definition?
    $endgroup$
    – Deroty
    Jan 11 at 14:33








  • 2




    $begingroup$
    That's because one shows $operatorname{Spec} A$ and $operatorname{Ass} A$ have the same minimal elements.
    $endgroup$
    – Bernard
    Jan 12 at 11:13








  • 1




    $begingroup$
    As far as I remember, you can replace associated with weakly associated.
    $endgroup$
    – Bernard
    Jan 12 at 12:50














1












1








1





$begingroup$

The intersection of all associated prime ideals is the nilradical of $R$, i.e. the set of nilpotent elements. If the ring is reduced, this set is ${0}$.






share|cite|improve this answer











$endgroup$



The intersection of all associated prime ideals is the nilradical of $R$, i.e. the set of nilpotent elements. If the ring is reduced, this set is ${0}$.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Jan 11 at 21:51









user26857

39.5k124284




39.5k124284










answered Jan 11 at 14:26









BernardBernard

124k741117




124k741117








  • 2




    $begingroup$
    I've seen that the nilradical is this intersection for Noetherian rings but does it hold in general?
    $endgroup$
    – rschwieb
    Jan 11 at 14:28








  • 2




    $begingroup$
    I believe this is true with correct notion of associated primes (weakly associated prime ideals – which are the same if the ring is noetherian).
    $endgroup$
    – Bernard
    Jan 11 at 14:33










  • $begingroup$
    The intersection of all minimal prime ideals is the nilradical of $R $. I want to know why it is true for associated primes in reduced rings, with the mentioned definition?
    $endgroup$
    – Deroty
    Jan 11 at 14:33








  • 2




    $begingroup$
    That's because one shows $operatorname{Spec} A$ and $operatorname{Ass} A$ have the same minimal elements.
    $endgroup$
    – Bernard
    Jan 12 at 11:13








  • 1




    $begingroup$
    As far as I remember, you can replace associated with weakly associated.
    $endgroup$
    – Bernard
    Jan 12 at 12:50














  • 2




    $begingroup$
    I've seen that the nilradical is this intersection for Noetherian rings but does it hold in general?
    $endgroup$
    – rschwieb
    Jan 11 at 14:28








  • 2




    $begingroup$
    I believe this is true with correct notion of associated primes (weakly associated prime ideals – which are the same if the ring is noetherian).
    $endgroup$
    – Bernard
    Jan 11 at 14:33










  • $begingroup$
    The intersection of all minimal prime ideals is the nilradical of $R $. I want to know why it is true for associated primes in reduced rings, with the mentioned definition?
    $endgroup$
    – Deroty
    Jan 11 at 14:33








  • 2




    $begingroup$
    That's because one shows $operatorname{Spec} A$ and $operatorname{Ass} A$ have the same minimal elements.
    $endgroup$
    – Bernard
    Jan 12 at 11:13








  • 1




    $begingroup$
    As far as I remember, you can replace associated with weakly associated.
    $endgroup$
    – Bernard
    Jan 12 at 12:50








2




2




$begingroup$
I've seen that the nilradical is this intersection for Noetherian rings but does it hold in general?
$endgroup$
– rschwieb
Jan 11 at 14:28






$begingroup$
I've seen that the nilradical is this intersection for Noetherian rings but does it hold in general?
$endgroup$
– rschwieb
Jan 11 at 14:28






2




2




$begingroup$
I believe this is true with correct notion of associated primes (weakly associated prime ideals – which are the same if the ring is noetherian).
$endgroup$
– Bernard
Jan 11 at 14:33




$begingroup$
I believe this is true with correct notion of associated primes (weakly associated prime ideals – which are the same if the ring is noetherian).
$endgroup$
– Bernard
Jan 11 at 14:33












$begingroup$
The intersection of all minimal prime ideals is the nilradical of $R $. I want to know why it is true for associated primes in reduced rings, with the mentioned definition?
$endgroup$
– Deroty
Jan 11 at 14:33






$begingroup$
The intersection of all minimal prime ideals is the nilradical of $R $. I want to know why it is true for associated primes in reduced rings, with the mentioned definition?
$endgroup$
– Deroty
Jan 11 at 14:33






2




2




$begingroup$
That's because one shows $operatorname{Spec} A$ and $operatorname{Ass} A$ have the same minimal elements.
$endgroup$
– Bernard
Jan 12 at 11:13






$begingroup$
That's because one shows $operatorname{Spec} A$ and $operatorname{Ass} A$ have the same minimal elements.
$endgroup$
– Bernard
Jan 12 at 11:13






1




1




$begingroup$
As far as I remember, you can replace associated with weakly associated.
$endgroup$
– Bernard
Jan 12 at 12:50




$begingroup$
As far as I remember, you can replace associated with weakly associated.
$endgroup$
– Bernard
Jan 12 at 12:50


















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