Mixing
Let S be the set of non finitely generated ideals in R. Suppose A is a maximal element in S. Then A is a...
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This question already has an answer here:
An ideal that is maximal among non-finitely generated ideals is prime.
2 answers
This questions comes under the section about Zorn's Lemma. But I'm really lost to how maximality here implies prime.
Edit: R is commutative.
abstract-algebra
asked Jan 25 at 6:56
davidhdavidh
3138
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marked as duplicate by rschwieb abstract-algebra
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Jan 25 at 11:51
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.
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This question already has an answer here:
An ideal that is maximal among non-finitely generated ideals is prime.
2 answers
This questions comes under the section about Zorn's Lemma. But I'm really lost to how maximality here implies prime.
Edit: R is commutative.
abstract-algebra
asked Jan 25 at 6:56
davidhdavidh
3138
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marked as duplicate by rschwieb abstract-algebra
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Jan 25 at 11:51
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.
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This is a lemma used in proving a theorem of Cohen: a ring with all prime ideals f.g. is Noetherian. You may be able to find hints in texts on commutative algebra.
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– Lord Shark the Unknown
Jan 25 at 7:10
1
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@LordSharktheUnknown Just a crude idea: Doesn't it suffice to show that If $xy in A$ but $x,y notin A$, then $langle A, x rangle$ is a non-finitely generated ideal in $R$ that is larger than $A$? Is it really a difficult exercise? Because it looks kind of innocent.
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– stressed out
Jan 25 at 7:56
2
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How do you prove that is not f.g.? @stressedout
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– Lord Shark the Unknown
Jan 25 at 8:02
1
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@stressedout: if it's easy to show that $langle A, a rangle$ is not finitely generated; I don't know . . . is it?
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– Robert Lewis
Jan 25 at 8:03
1
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@stressedout: many hard problems look innocent! 😉
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– Robert Lewis
Jan 25 at 8:05
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show 5 more comments
$begingroup$
This question already has an answer here:
An ideal that is maximal among non-finitely generated ideals is prime.
2 answers
This questions comes under the section about Zorn's Lemma. But I'm really lost to how maximality here implies prime.
Edit: R is commutative.
abstract-algebra
asked Jan 25 at 6:56
davidhdavidh
3138
$endgroup$
This question already has an answer here:
An ideal that is maximal among non-finitely generated ideals is prime.
2 answers
This questions comes under the section about Zorn's Lemma. But I'm really lost to how maximality here implies prime.
Edit: R is commutative.
This question already has an answer here:
An ideal that is maximal among non-finitely generated ideals is prime.
2 answers
abstract-algebra
abstract-algebra
asked Jan 25 at 6:56
davidhdavidh
3138
asked Jan 25 at 6:56
davidhdavidh
3138
asked Jan 25 at 6:56
davidhdavidh
3138
asked Jan 25 at 6:56
davidhdavidh
3138
asked Jan 25 at 6:56
davidhdavidh
3138
3138
marked as duplicate by rschwieb abstract-algebra
Users with the abstract-algebra badge can single-handedly close abstract-algebra questions as duplicates and reopen them as needed.
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Jan 25 at 11:51
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marked as duplicate by rschwieb abstract-algebra
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Jan 25 at 11:51
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.
1
$begingroup$
This is a lemma used in proving a theorem of Cohen: a ring with all prime ideals f.g. is Noetherian. You may be able to find hints in texts on commutative algebra.
$endgroup$
– Lord Shark the Unknown
Jan 25 at 7:10
1
$begingroup$
@LordSharktheUnknown Just a crude idea: Doesn't it suffice to show that If $xy in A$ but $x,y notin A$, then $langle A, x rangle$ is a non-finitely generated ideal in $R$ that is larger than $A$? Is it really a difficult exercise? Because it looks kind of innocent.
$endgroup$
– stressed out
Jan 25 at 7:56
2
$begingroup$
How do you prove that is not f.g.? @stressedout
$endgroup$
– Lord Shark the Unknown
Jan 25 at 8:02
1
$begingroup$
@stressedout: if it's easy to show that $langle A, a rangle$ is not finitely generated; I don't know . . . is it?
$endgroup$
– Robert Lewis
Jan 25 at 8:03
1
$begingroup$
@stressedout: many hard problems look innocent! 😉
$endgroup$
– Robert Lewis
Jan 25 at 8:05
|
show 5 more comments
1
$begingroup$
This is a lemma used in proving a theorem of Cohen: a ring with all prime ideals f.g. is Noetherian. You may be able to find hints in texts on commutative algebra.
$endgroup$
– Lord Shark the Unknown
Jan 25 at 7:10
1
$begingroup$
@LordSharktheUnknown Just a crude idea: Doesn't it suffice to show that If $xy in A$ but $x,y notin A$, then $langle A, x rangle$ is a non-finitely generated ideal in $R$ that is larger than $A$? Is it really a difficult exercise? Because it looks kind of innocent.
$endgroup$
– stressed out
Jan 25 at 7:56
2
$begingroup$
How do you prove that is not f.g.? @stressedout
$endgroup$
– Lord Shark the Unknown
Jan 25 at 8:02
1
$begingroup$
@stressedout: if it's easy to show that $langle A, a rangle$ is not finitely generated; I don't know . . . is it?
$endgroup$
– Robert Lewis
Jan 25 at 8:03
1
$begingroup$
@stressedout: many hard problems look innocent! 😉
$endgroup$
– Robert Lewis
Jan 25 at 8:05
1
1
$begingroup$
This is a lemma used in proving a theorem of Cohen: a ring with all prime ideals f.g. is Noetherian. You may be able to find hints in texts on commutative algebra.
$endgroup$
– Lord Shark the Unknown
Jan 25 at 7:10
$begingroup$
This is a lemma used in proving a theorem of Cohen: a ring with all prime ideals f.g. is Noetherian. You may be able to find hints in texts on commutative algebra.
$endgroup$
– Lord Shark the Unknown
Jan 25 at 7:10
1
1
$begingroup$
@LordSharktheUnknown Just a crude idea: Doesn't it suffice to show that If $xy in A$ but $x,y notin A$, then $langle A, x rangle$ is a non-finitely generated ideal in $R$ that is larger than $A$? Is it really a difficult exercise? Because it looks kind of innocent.
$endgroup$
– stressed out
Jan 25 at 7:56
$begingroup$
@LordSharktheUnknown Just a crude idea: Doesn't it suffice to show that If $xy in A$ but $x,y notin A$, then $langle A, x rangle$ is a non-finitely generated ideal in $R$ that is larger than $A$? Is it really a difficult exercise? Because it looks kind of innocent.
$endgroup$
– stressed out
Jan 25 at 7:56
2
2
$begingroup$
How do you prove that is not f.g.? @stressedout
$endgroup$
– Lord Shark the Unknown
Jan 25 at 8:02
$begingroup$
How do you prove that is not f.g.? @stressedout
$endgroup$
– Lord Shark the Unknown
Jan 25 at 8:02
1
1
$begingroup$
@stressedout: if it's easy to show that $langle A, a rangle$ is not finitely generated; I don't know . . . is it?
$endgroup$
– Robert Lewis
Jan 25 at 8:03
$begingroup$
@stressedout: if it's easy to show that $langle A, a rangle$ is not finitely generated; I don't know . . . is it?
$endgroup$
– Robert Lewis
Jan 25 at 8:03
1
1
$begingroup$
@stressedout: many hard problems look innocent! 😉
$endgroup$
– Robert Lewis
Jan 25 at 8:05
$begingroup$
@stressedout: many hard problems look innocent! 😉
$endgroup$
– Robert Lewis
Jan 25 at 8:05
|
show 5 more comments
1 Answer
1
active
oldest
votes
$begingroup$
Let $mathcal{S}$ be the set of nonfinitely generated ideals, ordered by inclusion. If you have a chain $mathcal{C}$ in $mathcal{S}$, its union $K$ is an ideal. Let's prove that $K$ is not finitely generated. Otherwise, we'd find $Iinmathcal{C}$ such that $Ksubseteq I$ and therefore $K=I$, contrary to the assumption that $I$ is not finitely generated.
Let $M$ be a maximal element in $mathcal{S}$ (which exists by Zorn's lemma, but it is not necessarily a maximal ideal). Suppose $xyin M$, but $x,ynotin M$.
Then both $(M,x)$ and $(M,y)$ are ideals properly containing $M$, so they are finitely generated, by maximality of $M$ in $mathcal{S}$.
Can you finish? Further hint: the product of two finitely generated ideals is finitely generated.
answered Jan 25 at 8:39
egregegreg
184k1486205
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$begingroup$
I'm still not seeing why (M,x) would be finitely generated...
$endgroup$
– davidh
Jan 25 at 8:57
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@davidh Because $(M,x)$ is strictly larger than $M$. So, if it were not finitely generated, you'd get a contradiction with the maximality of $M$.
$endgroup$
– stressed out
Jan 25 at 8:58
1
$begingroup$
@egreg So the product of these two ideals would be M, but M is non f.g. Thus contradiction.
$endgroup$
– davidh
Jan 25 at 10:01
1
$begingroup$
@davidh Exactly! Just fill in the details.
$endgroup$
– egreg
Jan 25 at 10:02
1
$begingroup$
@RobertLewis Take a finite set of generators of $K$; each one belongs to an ideal in the chain; on the other hand this is a chain, so you find an ideal in the chain that contains all of them (here finiteness is crucial). Then such ideal contains $K$, but is also contained in it by definition of $K$.
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– egreg
Jan 30 at 10:13
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show 7 more comments
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Let $mathcal{S}$ be the set of nonfinitely generated ideals, ordered by inclusion. If you have a chain $mathcal{C}$ in $mathcal{S}$, its union $K$ is an ideal. Let's prove that $K$ is not finitely generated. Otherwise, we'd find $Iinmathcal{C}$ such that $Ksubseteq I$ and therefore $K=I$, contrary to the assumption that $I$ is not finitely generated.
Let $M$ be a maximal element in $mathcal{S}$ (which exists by Zorn's lemma, but it is not necessarily a maximal ideal). Suppose $xyin M$, but $x,ynotin M$.
Then both $(M,x)$ and $(M,y)$ are ideals properly containing $M$, so they are finitely generated, by maximality of $M$ in $mathcal{S}$.
Can you finish? Further hint: the product of two finitely generated ideals is finitely generated.
answered Jan 25 at 8:39
egregegreg
184k1486205
$endgroup$
$begingroup$
I'm still not seeing why (M,x) would be finitely generated...
$endgroup$
– davidh
Jan 25 at 8:57
$begingroup$
@davidh Because $(M,x)$ is strictly larger than $M$. So, if it were not finitely generated, you'd get a contradiction with the maximality of $M$.
$endgroup$
– stressed out
Jan 25 at 8:58
1
$begingroup$
@egreg So the product of these two ideals would be M, but M is non f.g. Thus contradiction.
$endgroup$
– davidh
Jan 25 at 10:01
1
$begingroup$
@davidh Exactly! Just fill in the details.
$endgroup$
– egreg
Jan 25 at 10:02
1
$begingroup$
@RobertLewis Take a finite set of generators of $K$; each one belongs to an ideal in the chain; on the other hand this is a chain, so you find an ideal in the chain that contains all of them (here finiteness is crucial). Then such ideal contains $K$, but is also contained in it by definition of $K$.
$endgroup$
– egreg
Jan 30 at 10:13
|
show 7 more comments
$begingroup$
Let $mathcal{S}$ be the set of nonfinitely generated ideals, ordered by inclusion. If you have a chain $mathcal{C}$ in $mathcal{S}$, its union $K$ is an ideal. Let's prove that $K$ is not finitely generated. Otherwise, we'd find $Iinmathcal{C}$ such that $Ksubseteq I$ and therefore $K=I$, contrary to the assumption that $I$ is not finitely generated.
Let $M$ be a maximal element in $mathcal{S}$ (which exists by Zorn's lemma, but it is not necessarily a maximal ideal). Suppose $xyin M$, but $x,ynotin M$.
Then both $(M,x)$ and $(M,y)$ are ideals properly containing $M$, so they are finitely generated, by maximality of $M$ in $mathcal{S}$.
Can you finish? Further hint: the product of two finitely generated ideals is finitely generated.
answered Jan 25 at 8:39
egregegreg
184k1486205
$endgroup$
$begingroup$
I'm still not seeing why (M,x) would be finitely generated...
$endgroup$
– davidh
Jan 25 at 8:57
$begingroup$
@davidh Because $(M,x)$ is strictly larger than $M$. So, if it were not finitely generated, you'd get a contradiction with the maximality of $M$.
$endgroup$
– stressed out
Jan 25 at 8:58
1
$begingroup$
@egreg So the product of these two ideals would be M, but M is non f.g. Thus contradiction.
$endgroup$
– davidh
Jan 25 at 10:01
1
$begingroup$
@davidh Exactly! Just fill in the details.
$endgroup$
– egreg
Jan 25 at 10:02
1
$begingroup$
@RobertLewis Take a finite set of generators of $K$; each one belongs to an ideal in the chain; on the other hand this is a chain, so you find an ideal in the chain that contains all of them (here finiteness is crucial). Then such ideal contains $K$, but is also contained in it by definition of $K$.
$endgroup$
– egreg
Jan 30 at 10:13
|
show 7 more comments
$begingroup$
Let $mathcal{S}$ be the set of nonfinitely generated ideals, ordered by inclusion. If you have a chain $mathcal{C}$ in $mathcal{S}$, its union $K$ is an ideal. Let's prove that $K$ is not finitely generated. Otherwise, we'd find $Iinmathcal{C}$ such that $Ksubseteq I$ and therefore $K=I$, contrary to the assumption that $I$ is not finitely generated.
Let $M$ be a maximal element in $mathcal{S}$ (which exists by Zorn's lemma, but it is not necessarily a maximal ideal). Suppose $xyin M$, but $x,ynotin M$.
Then both $(M,x)$ and $(M,y)$ are ideals properly containing $M$, so they are finitely generated, by maximality of $M$ in $mathcal{S}$.
Can you finish? Further hint: the product of two finitely generated ideals is finitely generated.
answered Jan 25 at 8:39
egregegreg
184k1486205
$endgroup$
Let $mathcal{S}$ be the set of nonfinitely generated ideals, ordered by inclusion. If you have a chain $mathcal{C}$ in $mathcal{S}$, its union $K$ is an ideal. Let's prove that $K$ is not finitely generated. Otherwise, we'd find $Iinmathcal{C}$ such that $Ksubseteq I$ and therefore $K=I$, contrary to the assumption that $I$ is not finitely generated.
Let $M$ be a maximal element in $mathcal{S}$ (which exists by Zorn's lemma, but it is not necessarily a maximal ideal). Suppose $xyin M$, but $x,ynotin M$.
Then both $(M,x)$ and $(M,y)$ are ideals properly containing $M$, so they are finitely generated, by maximality of $M$ in $mathcal{S}$.
Can you finish? Further hint: the product of two finitely generated ideals is finitely generated.
answered Jan 25 at 8:39
egregegreg
184k1486205
edited Jan 25 at 9:40
answered Jan 25 at 8:39
egregegreg
184k1486205
answered Jan 25 at 8:39
egregegreg
184k1486205
answered Jan 25 at 8:39
egregegreg
184k1486205
184k1486205
$begingroup$
I'm still not seeing why (M,x) would be finitely generated...
$endgroup$
– davidh
Jan 25 at 8:57
$begingroup$
@davidh Because $(M,x)$ is strictly larger than $M$. So, if it were not finitely generated, you'd get a contradiction with the maximality of $M$.
$endgroup$
– stressed out
Jan 25 at 8:58
1
$begingroup$
@egreg So the product of these two ideals would be M, but M is non f.g. Thus contradiction.
$endgroup$
– davidh
Jan 25 at 10:01
1
$begingroup$
@davidh Exactly! Just fill in the details.
$endgroup$
– egreg
Jan 25 at 10:02
1
$begingroup$
@RobertLewis Take a finite set of generators of $K$; each one belongs to an ideal in the chain; on the other hand this is a chain, so you find an ideal in the chain that contains all of them (here finiteness is crucial). Then such ideal contains $K$, but is also contained in it by definition of $K$.
$endgroup$
– egreg
Jan 30 at 10:13
|
show 7 more comments
$begingroup$
I'm still not seeing why (M,x) would be finitely generated...
$endgroup$
– davidh
Jan 25 at 8:57
$begingroup$
@davidh Because $(M,x)$ is strictly larger than $M$. So, if it were not finitely generated, you'd get a contradiction with the maximality of $M$.
$endgroup$
– stressed out
Jan 25 at 8:58
1
$begingroup$
@egreg So the product of these two ideals would be M, but M is non f.g. Thus contradiction.
$endgroup$
– davidh
Jan 25 at 10:01
1
$begingroup$
@davidh Exactly! Just fill in the details.
$endgroup$
– egreg
Jan 25 at 10:02
1
$begingroup$
@RobertLewis Take a finite set of generators of $K$; each one belongs to an ideal in the chain; on the other hand this is a chain, so you find an ideal in the chain that contains all of them (here finiteness is crucial). Then such ideal contains $K$, but is also contained in it by definition of $K$.
$endgroup$
– egreg
Jan 30 at 10:13
$begingroup$
I'm still not seeing why (M,x) would be finitely generated...
$endgroup$
– davidh
Jan 25 at 8:57
$begingroup$
I'm still not seeing why (M,x) would be finitely generated...
$endgroup$
– davidh
Jan 25 at 8:57
$begingroup$
@davidh Because $(M,x)$ is strictly larger than $M$. So, if it were not finitely generated, you'd get a contradiction with the maximality of $M$.
$endgroup$
– stressed out
Jan 25 at 8:58
$begingroup$
@davidh Because $(M,x)$ is strictly larger than $M$. So, if it were not finitely generated, you'd get a contradiction with the maximality of $M$.
$endgroup$
– stressed out
Jan 25 at 8:58
1
1
$begingroup$
@egreg So the product of these two ideals would be M, but M is non f.g. Thus contradiction.
$endgroup$
– davidh
Jan 25 at 10:01
$begingroup$
@egreg So the product of these two ideals would be M, but M is non f.g. Thus contradiction.
$endgroup$
– davidh
Jan 25 at 10:01
1
1
$begingroup$
@davidh Exactly! Just fill in the details.
$endgroup$
– egreg
Jan 25 at 10:02
$begingroup$
@davidh Exactly! Just fill in the details.
$endgroup$
– egreg
Jan 25 at 10:02
1
1
$begingroup$
@RobertLewis Take a finite set of generators of $K$; each one belongs to an ideal in the chain; on the other hand this is a chain, so you find an ideal in the chain that contains all of them (here finiteness is crucial). Then such ideal contains $K$, but is also contained in it by definition of $K$.
$endgroup$
– egreg
Jan 30 at 10:13
$begingroup$
@RobertLewis Take a finite set of generators of $K$; each one belongs to an ideal in the chain; on the other hand this is a chain, so you find an ideal in the chain that contains all of them (here finiteness is crucial). Then such ideal contains $K$, but is also contained in it by definition of $K$.
$endgroup$
– egreg
Jan 30 at 10:13
|
show 7 more comments
1
$begingroup$
This is a lemma used in proving a theorem of Cohen: a ring with all prime ideals f.g. is Noetherian. You may be able to find hints in texts on commutative algebra.
$endgroup$
– Lord Shark the Unknown
Jan 25 at 7:10
1
$begingroup$
@LordSharktheUnknown Just a crude idea: Doesn't it suffice to show that If $xy in A$ but $x,y notin A$, then $langle A, x rangle$ is a non-finitely generated ideal in $R$ that is larger than $A$? Is it really a difficult exercise? Because it looks kind of innocent.
$endgroup$
– stressed out
Jan 25 at 7:56
2
$begingroup$
How do you prove that is not f.g.? @stressedout
$endgroup$
– Lord Shark the Unknown
Jan 25 at 8:02
1
$begingroup$
@stressedout: if it's easy to show that $langle A, a rangle$ is not finitely generated; I don't know . . . is it?
$endgroup$
– Robert Lewis
Jan 25 at 8:03
1
$begingroup$
@stressedout: many hard problems look innocent! 😉
$endgroup$
– Robert Lewis
Jan 25 at 8:05