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.










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    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.










    share|cite|improve this question


























      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.










      share|cite|improve this question















      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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      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.






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          • 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











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          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.






          share|cite|improve this answer





















          • 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















          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.






          share|cite|improve this answer





















          • 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













          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.






          share|cite|improve this answer












          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.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          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


















          • 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


















           

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