Maximal ideal with respect to property that it does not contain an ideal A
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let A=($a1, a2, .... ,an$) be a non-zero finitely generated ideal of R.
Prove that there is an ideal B which is maximal with respect to property that it does not contain A.
I don't know how to proceed. Maybe by Zorn's lemma.
If I define order by containment of ideals. Then in a chain what will be the maximal element.
If ∪J { J ∈ Chain } how to check that A is not contained inside ∪J.
ring-theory maximal-and-prime-ideals
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up vote
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let A=($a1, a2, .... ,an$) be a non-zero finitely generated ideal of R.
Prove that there is an ideal B which is maximal with respect to property that it does not contain A.
I don't know how to proceed. Maybe by Zorn's lemma.
If I define order by containment of ideals. Then in a chain what will be the maximal element.
If ∪J { J ∈ Chain } how to check that A is not contained inside ∪J.
ring-theory maximal-and-prime-ideals
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
let A=($a1, a2, .... ,an$) be a non-zero finitely generated ideal of R.
Prove that there is an ideal B which is maximal with respect to property that it does not contain A.
I don't know how to proceed. Maybe by Zorn's lemma.
If I define order by containment of ideals. Then in a chain what will be the maximal element.
If ∪J { J ∈ Chain } how to check that A is not contained inside ∪J.
ring-theory maximal-and-prime-ideals
let A=($a1, a2, .... ,an$) be a non-zero finitely generated ideal of R.
Prove that there is an ideal B which is maximal with respect to property that it does not contain A.
I don't know how to proceed. Maybe by Zorn's lemma.
If I define order by containment of ideals. Then in a chain what will be the maximal element.
If ∪J { J ∈ Chain } how to check that A is not contained inside ∪J.
ring-theory maximal-and-prime-ideals
ring-theory maximal-and-prime-ideals
edited Nov 16 at 5:38
asked Nov 16 at 4:39
infintedimensional
428
428
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You are right, the idea, is the Zorn, consider the set $J$ of ideals which does not contain $A$, let $(B_i)$ be a family of well ordered (by the inclusion) element of $J$, $B=cup_{iin I}B_i$ is the sup. Suppose that $B$ contains $A$, $a_jin B_{i_j}$ let $k=sup(i_1,...,i_n)$, since $B_{i_j}subset B_k$, $a_1,...,a_nin B_k$ contradiction. Thus $J$ has a maximum element.
can you please explain last line saying k = sup(i1, ....in )
– infintedimensional
Nov 16 at 4:49
since $J$ is totally ordered, $k$ exists.
– Tsemo Aristide
Nov 16 at 4:50
got it. Thank you !
– infintedimensional
Nov 16 at 4:50
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
You are right, the idea, is the Zorn, consider the set $J$ of ideals which does not contain $A$, let $(B_i)$ be a family of well ordered (by the inclusion) element of $J$, $B=cup_{iin I}B_i$ is the sup. Suppose that $B$ contains $A$, $a_jin B_{i_j}$ let $k=sup(i_1,...,i_n)$, since $B_{i_j}subset B_k$, $a_1,...,a_nin B_k$ contradiction. Thus $J$ has a maximum element.
can you please explain last line saying k = sup(i1, ....in )
– infintedimensional
Nov 16 at 4:49
since $J$ is totally ordered, $k$ exists.
– Tsemo Aristide
Nov 16 at 4:50
got it. Thank you !
– infintedimensional
Nov 16 at 4:50
add a comment |
up vote
1
down vote
You are right, the idea, is the Zorn, consider the set $J$ of ideals which does not contain $A$, let $(B_i)$ be a family of well ordered (by the inclusion) element of $J$, $B=cup_{iin I}B_i$ is the sup. Suppose that $B$ contains $A$, $a_jin B_{i_j}$ let $k=sup(i_1,...,i_n)$, since $B_{i_j}subset B_k$, $a_1,...,a_nin B_k$ contradiction. Thus $J$ has a maximum element.
can you please explain last line saying k = sup(i1, ....in )
– infintedimensional
Nov 16 at 4:49
since $J$ is totally ordered, $k$ exists.
– Tsemo Aristide
Nov 16 at 4:50
got it. Thank you !
– infintedimensional
Nov 16 at 4:50
add a comment |
up vote
1
down vote
up vote
1
down vote
You are right, the idea, is the Zorn, consider the set $J$ of ideals which does not contain $A$, let $(B_i)$ be a family of well ordered (by the inclusion) element of $J$, $B=cup_{iin I}B_i$ is the sup. Suppose that $B$ contains $A$, $a_jin B_{i_j}$ let $k=sup(i_1,...,i_n)$, since $B_{i_j}subset B_k$, $a_1,...,a_nin B_k$ contradiction. Thus $J$ has a maximum element.
You are right, the idea, is the Zorn, consider the set $J$ of ideals which does not contain $A$, let $(B_i)$ be a family of well ordered (by the inclusion) element of $J$, $B=cup_{iin I}B_i$ is the sup. Suppose that $B$ contains $A$, $a_jin B_{i_j}$ let $k=sup(i_1,...,i_n)$, since $B_{i_j}subset B_k$, $a_1,...,a_nin B_k$ contradiction. Thus $J$ has a maximum element.
answered Nov 16 at 4:45
Tsemo Aristide
54.3k11344
54.3k11344
can you please explain last line saying k = sup(i1, ....in )
– infintedimensional
Nov 16 at 4:49
since $J$ is totally ordered, $k$ exists.
– Tsemo Aristide
Nov 16 at 4:50
got it. Thank you !
– infintedimensional
Nov 16 at 4:50
add a comment |
can you please explain last line saying k = sup(i1, ....in )
– infintedimensional
Nov 16 at 4:49
since $J$ is totally ordered, $k$ exists.
– Tsemo Aristide
Nov 16 at 4:50
got it. Thank you !
– infintedimensional
Nov 16 at 4:50
can you please explain last line saying k = sup(i1, ....in )
– infintedimensional
Nov 16 at 4:49
can you please explain last line saying k = sup(i1, ....in )
– infintedimensional
Nov 16 at 4:49
since $J$ is totally ordered, $k$ exists.
– Tsemo Aristide
Nov 16 at 4:50
since $J$ is totally ordered, $k$ exists.
– Tsemo Aristide
Nov 16 at 4:50
got it. Thank you !
– infintedimensional
Nov 16 at 4:50
got it. Thank you !
– infintedimensional
Nov 16 at 4:50
add a comment |
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