1
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ (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
0
$\begingroup$

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. $\Box$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.