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A.Magnus
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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.
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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.
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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.
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
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.
