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Prime Ideal & Noetherian Integral Domain

  1. Nov 16, 2014 #1
    I am reading a graduate-level Abstract Algebra lemma on noetherian integral domain, I am bring it up here hoping for help. The original passage is in one big-fat paragraph but I broke it down here for your easy reading. Let me know if I forget to include any underlying lemmas, and especially, let me know if I should have posted this in Abstract Algebra forum instead. Thank you for your time and help.

    ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

    LEMMA:
    Let M be an R-module. Let T be maximal among the ideals of R such that M possesses a submodule L for which L/LT is not noetherian. Then T is a prime ideal of R.

    PROOF:
    (1) We are assuming that M possesses a submodule L for which L/LT is not noetherian. Thus, as L/LR = L/L is noetherian, T≠R.

    (2) Let us assume, by way of contradiction, that T is not prime. Then R possesses ideals U and V such that T ⊂ U,T ⊂ V , and UV ⊆ T.

    (3) The (maximal) choice of T forces L/LU and LU/LUV to be noetherian. [QUESTION: I understand that while T is maximal but U and V are strictly larger than T, and the only way to resolve this paradox is to take U and V as structures different from T. But I am lost on how all these "force L/LU and LU/LUV to be noetherian."]

    (4) Thus, by Lemma below, L/LUV is noetherian. [QUESTION: Does it mean that since L/LU and LU/LUV are noetherian from above, therefore L, LU and LUV are noetherian, and therefore L/LUV is noetherian?]

    (5) On the other hand, as UV ⊆ T, LUV ⊆ LT. Thus, L/LT is a factor module of L/LUV. [QUESTION: Here, I am begging explanation on how L/LT is a factor module of L/LUV, step-by-step if possible.]

    (6) Thus, as L/LUV is noetherian, L/LT is noetherian; cf. Lemma below. This contradiction finishes the proof.

    ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

    This is the lemma quoted above: Let M be an R-module, and let L be a submodule of M. Then M is noetherian if and only if L and M/L are noetherian.
     
  2. jcsd
  3. Nov 21, 2014 #2
    Thanks for the post! This is an automated courtesy bump. Sorry you aren't generating responses at the moment. Do you have any further information, come to any new conclusions or is it possible to reword the post?
     
  4. Nov 21, 2014 #3

    mathwonk

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    i have answered this post in detail in the algebra forum. it is a double post.
     
  5. Nov 23, 2014 #4
    Yes, Mathwonk is right. Since I did not get any response from this homework/textbook section, I moved the posting to Abstract Algebra section, thinking that the subject might be too advanced for homework/textbook section. Since then, Mathwonk has given me very generous responses. Thank you to both of you.

    One question for either Greg or Mathwonk: In Math Stack Exchange forum, you can alert another member of your new posting by typing in "@username" at the beginning of your posting. Can you do the same here? Thanks again.
     
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