Proof of theorem: Every proper ideal a is contained in some maximal ideal.Every proper ideal $I$ in a nonzero commutative unitary ring $R$ is contained in a maximal ideal.every ideal is contained in a maximal idealZorn's lemma converse? (Context: Maximal proper subgroups)Every modular right ideal is contained in a modular maximal idealNilradical equals intersection of all prime idealsNon maximal prime ideal in $C[0,1]$$R$ integral domain where every prime ideal is principal then set of non-principal ideals has maximal elementWhy noetherian ring satisfies the maximal condition?Are non-coprime ideals always contained in some prime ideal?Every prime ideal of a comm. ring $A$ is maximal iff $A/textNilrad(A)$ is absolutely flat
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Proof of theorem: Every proper ideal a is contained in some maximal ideal.
Every proper ideal $I$ in a nonzero commutative unitary ring $R$ is contained in a maximal ideal.every ideal is contained in a maximal idealZorn's lemma converse? (Context: Maximal proper subgroups)Every modular right ideal is contained in a modular maximal idealNilradical equals intersection of all prime idealsNon maximal prime ideal in $C[0,1]$$R$ integral domain where every prime ideal is principal then set of non-principal ideals has maximal elementWhy noetherian ring satisfies the maximal condition?Are non-coprime ideals always contained in some prime ideal?Every prime ideal of a comm. ring $A$ is maximal iff $A/textNilrad(A)$ is absolutely flat
$begingroup$
I found a proof of this theorem in modern algebra 2.30
Proof: Set $S := text ideals mathfrak b mid mathfrak b ⊃ mathfrak a text and mathfrak b ∌ 1 $. Then $mathfrak a ∈ S$, and $S$ is partially ordered by inclusion. Given a totally ordered subset $mathfrak b_λ$ of $S$, set $mathfrak b := bigcup_lambda mathfrak b_λ$. Then $mathfrak b$ is clearly an ideal, and $1 notin mathfrak b$; so $mathfrak b$ is an upper bound of $mathfrak b_λ$ in $S$. Hence by Zorn’s Lemma, $S$ has a maximal element, and it is the desired maximal ideal.
I'm not satisfied with the proof, with follow questions:
- Why $S$ is a set, $S$ may not be a set as Russell's paradox.
- How to apply Zorn’s Lemma, can I apply Zorn’s Lemma to the range of real number $(0,1)$, and say that range $(0,1)$ has a maximal element ?
I'm guessing that missing key to the proof is that to prove all partially ordered chain is finite.
abstract-algebra
edited 17 hours ago
Bernard
122k741116
asked 17 hours ago
Zang MingJieZang MingJie
1406
$endgroup$
add a comment |
$begingroup$
I found a proof of this theorem in modern algebra 2.30
Proof: Set $S := text ideals mathfrak b mid mathfrak b ⊃ mathfrak a text and mathfrak b ∌ 1 $. Then $mathfrak a ∈ S$, and $S$ is partially ordered by inclusion. Given a totally ordered subset $mathfrak b_λ$ of $S$, set $mathfrak b := bigcup_lambda mathfrak b_λ$. Then $mathfrak b$ is clearly an ideal, and $1 notin mathfrak b$; so $mathfrak b$ is an upper bound of $mathfrak b_λ$ in $S$. Hence by Zorn’s Lemma, $S$ has a maximal element, and it is the desired maximal ideal.
I'm not satisfied with the proof, with follow questions:
- Why $S$ is a set, $S$ may not be a set as Russell's paradox.
- How to apply Zorn’s Lemma, can I apply Zorn’s Lemma to the range of real number $(0,1)$, and say that range $(0,1)$ has a maximal element ?
I'm guessing that missing key to the proof is that to prove all partially ordered chain is finite.
abstract-algebra
edited 17 hours ago
Bernard
122k741116
asked 17 hours ago
Zang MingJieZang MingJie
1406
$endgroup$
$begingroup$
$S$ is a subset of the powerset of the ring $R$. $(0,1)$ doesn't satisfy the hypotheses of Zorn's lemma
$endgroup$
– Max
17 hours ago
$begingroup$
You apparently misunderstand the significance of Russel's paradox. The significance is that you must have a consistent system of axioms for set theory, or bad things will happen. Any modern system of set theory, such as Zermelo-Fraenkel or Von Neumann–Bernays–Gödel, is axiomatized such that Russell's paradox does not happen, and you will probably use one or the other of those systems for every problem you work on.
$endgroup$
– rschwieb
16 hours ago
add a comment |
$begingroup$
I found a proof of this theorem in modern algebra 2.30
Proof: Set $S := text ideals mathfrak b mid mathfrak b ⊃ mathfrak a text and mathfrak b ∌ 1 $. Then $mathfrak a ∈ S$, and $S$ is partially ordered by inclusion. Given a totally ordered subset $mathfrak b_λ$ of $S$, set $mathfrak b := bigcup_lambda mathfrak b_λ$. Then $mathfrak b$ is clearly an ideal, and $1 notin mathfrak b$; so $mathfrak b$ is an upper bound of $mathfrak b_λ$ in $S$. Hence by Zorn’s Lemma, $S$ has a maximal element, and it is the desired maximal ideal.
I'm not satisfied with the proof, with follow questions:
- Why $S$ is a set, $S$ may not be a set as Russell's paradox.
- How to apply Zorn’s Lemma, can I apply Zorn’s Lemma to the range of real number $(0,1)$, and say that range $(0,1)$ has a maximal element ?
I'm guessing that missing key to the proof is that to prove all partially ordered chain is finite.
abstract-algebra
edited 17 hours ago
Bernard
122k741116
asked 17 hours ago
Zang MingJieZang MingJie
1406
$endgroup$
I found a proof of this theorem in modern algebra 2.30
Proof: Set $S := text ideals mathfrak b mid mathfrak b ⊃ mathfrak a text and mathfrak b ∌ 1 $. Then $mathfrak a ∈ S$, and $S$ is partially ordered by inclusion. Given a totally ordered subset $mathfrak b_λ$ of $S$, set $mathfrak b := bigcup_lambda mathfrak b_λ$. Then $mathfrak b$ is clearly an ideal, and $1 notin mathfrak b$; so $mathfrak b$ is an upper bound of $mathfrak b_λ$ in $S$. Hence by Zorn’s Lemma, $S$ has a maximal element, and it is the desired maximal ideal.
I'm not satisfied with the proof, with follow questions:
- Why $S$ is a set, $S$ may not be a set as Russell's paradox.
- How to apply Zorn’s Lemma, can I apply Zorn’s Lemma to the range of real number $(0,1)$, and say that range $(0,1)$ has a maximal element ?
I'm guessing that missing key to the proof is that to prove all partially ordered chain is finite.
abstract-algebra
abstract-algebra
edited 17 hours ago
Bernard
122k741116
asked 17 hours ago
Zang MingJieZang MingJie
1406
edited 17 hours ago
Bernard
122k741116
asked 17 hours ago
Zang MingJieZang MingJie
1406
edited 17 hours ago
Bernard
122k741116
edited 17 hours ago
Bernard
122k741116
edited 17 hours ago
Bernard
122k741116
122k741116
asked 17 hours ago
Zang MingJieZang MingJie
1406
asked 17 hours ago
Zang MingJieZang MingJie
1406
asked 17 hours ago
Zang MingJieZang MingJie
1406
1406
$begingroup$
$S$ is a subset of the powerset of the ring $R$. $(0,1)$ doesn't satisfy the hypotheses of Zorn's lemma
$endgroup$
– Max
17 hours ago
$begingroup$
You apparently misunderstand the significance of Russel's paradox. The significance is that you must have a consistent system of axioms for set theory, or bad things will happen. Any modern system of set theory, such as Zermelo-Fraenkel or Von Neumann–Bernays–Gödel, is axiomatized such that Russell's paradox does not happen, and you will probably use one or the other of those systems for every problem you work on.
$endgroup$
– rschwieb
16 hours ago
add a comment |
$begingroup$
$S$ is a subset of the powerset of the ring $R$. $(0,1)$ doesn't satisfy the hypotheses of Zorn's lemma
$endgroup$
– Max
17 hours ago
$begingroup$
You apparently misunderstand the significance of Russel's paradox. The significance is that you must have a consistent system of axioms for set theory, or bad things will happen. Any modern system of set theory, such as Zermelo-Fraenkel or Von Neumann–Bernays–Gödel, is axiomatized such that Russell's paradox does not happen, and you will probably use one or the other of those systems for every problem you work on.
$endgroup$
– rschwieb
16 hours ago
$begingroup$
$S$ is a subset of the powerset of the ring $R$. $(0,1)$ doesn't satisfy the hypotheses of Zorn's lemma
$endgroup$
– Max
17 hours ago
$begingroup$
$S$ is a subset of the powerset of the ring $R$. $(0,1)$ doesn't satisfy the hypotheses of Zorn's lemma
$endgroup$
– Max
17 hours ago
$begingroup$
You apparently misunderstand the significance of Russel's paradox. The significance is that you must have a consistent system of axioms for set theory, or bad things will happen. Any modern system of set theory, such as Zermelo-Fraenkel or Von Neumann–Bernays–Gödel, is axiomatized such that Russell's paradox does not happen, and you will probably use one or the other of those systems for every problem you work on.
$endgroup$
– rschwieb
16 hours ago
$begingroup$
You apparently misunderstand the significance of Russel's paradox. The significance is that you must have a consistent system of axioms for set theory, or bad things will happen. Any modern system of set theory, such as Zermelo-Fraenkel or Von Neumann–Bernays–Gödel, is axiomatized such that Russell's paradox does not happen, and you will probably use one or the other of those systems for every problem you work on.
$endgroup$
– rschwieb
16 hours ago
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
$S$ is a subset of the powerset of the elements in the ring. As such it can be easily constructed directly using the axioms of ZF set theory (the power set axiom and one of the axioms of comprehension / specification, specifically), starting with the set of all the elements of the ring (which by definition of a ring must be a set).
We cannot apply Zorn's lemma to $(0,1)$ with the standard ordering, because the hypothesis of Zorn's lemma isn't satisfied. Zorn's lemma says
If $(P, geq)$ is a partial order such that every totally ordered subset of $P$ has an upper bound in $P$, then $P$ has a maximal element.
The interval $(0,1)$ with the standard ordering has itself as a totally ordered subset, and that subset doesn't have an upper bound in $(0,1)$. So Zorn's lemma cannot be applied.
For sets of ideals ordered by inclusion, on the other hand, a totally ordered subset is just a chain of ideals contained in one another. In that case, the union of those ideals is still an ideal, and the union of a totally ordered chain of ideals in $S$ is still an ideal in $S$. So any totally ordered subset of $S$ has an upper bound in $S$, and therefore $S$ must have a maximal element.
answered 17 hours ago
ArthurArthur
117k7116200
$endgroup$
add a comment |
$begingroup$
$S$ is a set by the Axiom schema of specification.- You cannot apply Zorn's lemma to the interval $(0,1)$ for the usual order because not every chain in $(0,1)$ has an upperbound
in (0,1).
answered 17 hours ago
BernardBernard
122k741116
$endgroup$
add a comment |
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2 Answers
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2 Answers
2
active
oldest
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active
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active
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votes
$begingroup$
$S$ is a subset of the powerset of the elements in the ring. As such it can be easily constructed directly using the axioms of ZF set theory (the power set axiom and one of the axioms of comprehension / specification, specifically), starting with the set of all the elements of the ring (which by definition of a ring must be a set).
We cannot apply Zorn's lemma to $(0,1)$ with the standard ordering, because the hypothesis of Zorn's lemma isn't satisfied. Zorn's lemma says
If $(P, geq)$ is a partial order such that every totally ordered subset of $P$ has an upper bound in $P$, then $P$ has a maximal element.
The interval $(0,1)$ with the standard ordering has itself as a totally ordered subset, and that subset doesn't have an upper bound in $(0,1)$. So Zorn's lemma cannot be applied.
For sets of ideals ordered by inclusion, on the other hand, a totally ordered subset is just a chain of ideals contained in one another. In that case, the union of those ideals is still an ideal, and the union of a totally ordered chain of ideals in $S$ is still an ideal in $S$. So any totally ordered subset of $S$ has an upper bound in $S$, and therefore $S$ must have a maximal element.
answered 17 hours ago
ArthurArthur
117k7116200
$endgroup$
add a comment |
$begingroup$
$S$ is a subset of the powerset of the elements in the ring. As such it can be easily constructed directly using the axioms of ZF set theory (the power set axiom and one of the axioms of comprehension / specification, specifically), starting with the set of all the elements of the ring (which by definition of a ring must be a set).
We cannot apply Zorn's lemma to $(0,1)$ with the standard ordering, because the hypothesis of Zorn's lemma isn't satisfied. Zorn's lemma says
If $(P, geq)$ is a partial order such that every totally ordered subset of $P$ has an upper bound in $P$, then $P$ has a maximal element.
The interval $(0,1)$ with the standard ordering has itself as a totally ordered subset, and that subset doesn't have an upper bound in $(0,1)$. So Zorn's lemma cannot be applied.
For sets of ideals ordered by inclusion, on the other hand, a totally ordered subset is just a chain of ideals contained in one another. In that case, the union of those ideals is still an ideal, and the union of a totally ordered chain of ideals in $S$ is still an ideal in $S$. So any totally ordered subset of $S$ has an upper bound in $S$, and therefore $S$ must have a maximal element.
answered 17 hours ago
ArthurArthur
117k7116200
$endgroup$
add a comment |
$begingroup$
$S$ is a subset of the powerset of the elements in the ring. As such it can be easily constructed directly using the axioms of ZF set theory (the power set axiom and one of the axioms of comprehension / specification, specifically), starting with the set of all the elements of the ring (which by definition of a ring must be a set).
We cannot apply Zorn's lemma to $(0,1)$ with the standard ordering, because the hypothesis of Zorn's lemma isn't satisfied. Zorn's lemma says
If $(P, geq)$ is a partial order such that every totally ordered subset of $P$ has an upper bound in $P$, then $P$ has a maximal element.
The interval $(0,1)$ with the standard ordering has itself as a totally ordered subset, and that subset doesn't have an upper bound in $(0,1)$. So Zorn's lemma cannot be applied.
For sets of ideals ordered by inclusion, on the other hand, a totally ordered subset is just a chain of ideals contained in one another. In that case, the union of those ideals is still an ideal, and the union of a totally ordered chain of ideals in $S$ is still an ideal in $S$. So any totally ordered subset of $S$ has an upper bound in $S$, and therefore $S$ must have a maximal element.
answered 17 hours ago
ArthurArthur
117k7116200
$endgroup$
$S$ is a subset of the powerset of the elements in the ring. As such it can be easily constructed directly using the axioms of ZF set theory (the power set axiom and one of the axioms of comprehension / specification, specifically), starting with the set of all the elements of the ring (which by definition of a ring must be a set).
We cannot apply Zorn's lemma to $(0,1)$ with the standard ordering, because the hypothesis of Zorn's lemma isn't satisfied. Zorn's lemma says
If $(P, geq)$ is a partial order such that every totally ordered subset of $P$ has an upper bound in $P$, then $P$ has a maximal element.
The interval $(0,1)$ with the standard ordering has itself as a totally ordered subset, and that subset doesn't have an upper bound in $(0,1)$. So Zorn's lemma cannot be applied.
For sets of ideals ordered by inclusion, on the other hand, a totally ordered subset is just a chain of ideals contained in one another. In that case, the union of those ideals is still an ideal, and the union of a totally ordered chain of ideals in $S$ is still an ideal in $S$. So any totally ordered subset of $S$ has an upper bound in $S$, and therefore $S$ must have a maximal element.
answered 17 hours ago
ArthurArthur
117k7116200
answered 17 hours ago
ArthurArthur
117k7116200
answered 17 hours ago
ArthurArthur
117k7116200
answered 17 hours ago
ArthurArthur
117k7116200
117k7116200
add a comment |
add a comment |
$begingroup$
$S$ is a set by the Axiom schema of specification.- You cannot apply Zorn's lemma to the interval $(0,1)$ for the usual order because not every chain in $(0,1)$ has an upperbound
in (0,1).
answered 17 hours ago
BernardBernard
122k741116
$endgroup$
add a comment |
$begingroup$
$S$ is a set by the Axiom schema of specification.- You cannot apply Zorn's lemma to the interval $(0,1)$ for the usual order because not every chain in $(0,1)$ has an upperbound
in (0,1).
answered 17 hours ago
BernardBernard
122k741116
$endgroup$
add a comment |
$begingroup$
$S$ is a set by the Axiom schema of specification.- You cannot apply Zorn's lemma to the interval $(0,1)$ for the usual order because not every chain in $(0,1)$ has an upperbound
in (0,1).
answered 17 hours ago
BernardBernard
122k741116
$endgroup$
$S$ is a set by the Axiom schema of specification.- You cannot apply Zorn's lemma to the interval $(0,1)$ for the usual order because not every chain in $(0,1)$ has an upperbound
in (0,1).
answered 17 hours ago
BernardBernard
122k741116
answered 17 hours ago
BernardBernard
122k741116
answered 17 hours ago
BernardBernard
122k741116
answered 17 hours ago
BernardBernard
122k741116
122k741116
add a comment |
add a comment |
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$begingroup$
$S$ is a subset of the powerset of the ring $R$. $(0,1)$ doesn't satisfy the hypotheses of Zorn's lemma
$endgroup$
– Max
17 hours ago
$begingroup$
You apparently misunderstand the significance of Russel's paradox. The significance is that you must have a consistent system of axioms for set theory, or bad things will happen. Any modern system of set theory, such as Zermelo-Fraenkel or Von Neumann–Bernays–Gödel, is axiomatized such that Russell's paradox does not happen, and you will probably use one or the other of those systems for every problem you work on.
$endgroup$
– rschwieb
16 hours ago