I am reading a lemma on noetherian integral domains but I am stuck, 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. Thanks beforehand for all your time and help, let me know if I forget to include any underlying lemmas.


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, $T \neq R$. [QUESTION: Here, I understand $LR=L$, but I am totally lost on how $L/LR=L/L$ is suddenly notherian.]

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

(3) The (maximal) choice of $T$ forces $L/LU$ and $LU/LUV$ to be noetherian. [QUESTION: I am lost on how the maximal choice of $T$ forces $L/LU$ 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, therefore $L$, $LU$ and $LUV$ are noetherian, and therefore $L/LUV$ is noetherian?]

(5) On the other hand, as $UV \subseteq T, LUV \subseteq LT$. 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 $L/LUV$.]

(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.

  • $\begingroup$ (1) The zero module is Noetherian (3) $U$ and $V$ are strictly larger than $T$, which is maximal with respect to some property (4,5) It might be good to review the isomorphism theorems. $\endgroup$ – Hoot Nov 10 '14 at 4:52
  • $\begingroup$ @Hoot Thanks for quick response. (1) Does $L/L$ = {0} and therefore noetherian? (2) I understand that $V$ and $U$ are larger than $T$ but $T$ is the maximal, therefore $V$ and $U$ must be different from $T$. But how does this translates into $L/LU$ and $LU/LUV$ being noetherian? (4) and (5) I am still totally lost here. Thanks again. $\endgroup$ – Amanda.M Nov 10 '14 at 14:01

Since posting the lemma above, I have managed to come up with my own and here is the break down of the lemma into dumb line-by-line, step-by-step. Thanks to all who have spent time reading and responding.


PROOF: (1) Here we are given M possesses a submodule $L$ for which $L/LT$ is not noetherian. Since $L$ is submodule of $M$ and $M$ is $R$-module, $L = LR$. Thus $L/LR = L/L$ = {0}. Because singleton zero is noetherian, $L/LR$ is noetherian.

(2) Since $L/LT$ is not noetherian while $L/LR$ is noetherian, we conclude that $T\neq R$.

(3) Let’s assume, by way of contradiction, that $T$ is not prime.

(4) Since $T$ is not a prime, then $R$ must possess ideals $U$ and $V$ such that $T \subset U, T \subset V$, and $UV \subseteq T$.

(5) We are given that $T$ is maximal, but at the same time from #(4) above we have concluded that there are ideal $U$ and ideal $V$ which are strictly larger than $T$.

(6) The only way to reconcile the above paradox is to conclude that $U$ and $V$ must be of different structure than $T$. Thus as $L/LT$ is not noetherian, $L/LU$ must be noetherian.

(7) Since $L/LU$ is noetherian, $L$ and $LU$ must be noetherian; cf. Lemma in the original posting.

(8) Notice here that $L$ is submodule of $M$, $LU \subset L$, and that $L/LU$ is noetherian. At the same time, notice that $LU$ is submodule of $M$, and that $LUV \subset LU$.

(9) The above two facts then lead us to conclude that $LU/LUV$ is noetherian too, as in the case of $L/LU$.

(10) Since $LU/LUV$ is noetherian, thus by Lemma above both $LU$ and $LUV$ are noetherian too.

(11) From #(7) we know that $L$ is noetherian, and from #(10) we know that $LUV$ is noetherian. Thus $L/LUV$ is noetherian by Lemma above.

(12) On the other hand, #(4) gives us that $UV \subset T$. Since $L$ is submodule to both $UV$ and $T$, therefore $LUV \subset LT$.

(13) In $L/LT$ from #(2), we notice that $LT \subset L$, where $L$ is a submodule of $M$ and $T$ is an ideal of $R$. In $L/LUV$ from #(11), we also notice that $LUV \subset L$, where $L$ is a submodule of $M$ and $UV$ is an ideal of $R$.

(14) From #(13) above, therefore we conclude that $L/LT$ is isomorphic to $L/LUV$.

(15) The above isomorphism leads us further to conclude that since $L/LUV$ is noetherian, $L/LT$ must be noetherian too.

(16) But this conclusion contradicts the given fact from #(1) that $L/LT$ is not noetherian.

(17) Because of this contradiction, we conclude that the assumption we made on #(1), that $T$ is not prime, is false. Therefore $T$ is prime. $\Box$


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