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I have a problem with this Noether's theorem from Faith's book Rings and Things,page 33...

If $R$ is a ring satisfying ACC on ideals then every ideal contains a product of prime ideals. An ideal $I$ of a ring $R$ is prime iff $I \neq R$ and $A,B \supset I$ then $AB \subset I \iff A \subset I \text{ or } B\subset I$. Proof is as follows :Let $I$ be a maximal counterexample. Then $I$ is not a prime ideal hence $I$ is properly contained in ideals $A,B$ such that $AB \subset I$. By the maximality of $I$, $A$ and $B$ are product of primes, hence so is $AB$. This proves the theorem. My question is if we can find I without ACC using Zorn's lemma.

The problem is as follows. Why do we need in the assumptions of the Theorem 2.18B the ACC condition?? If there is no counterexample, then the theorem is true. If there is a counterexample, then either it is maximal or not. If it is maximal then we are done by the second part of the proof. If it is not maximal, then there is a proper superset ideal J above it. If ACC holds, then every such construction stops and Zorn Lemma does not come into the game at all.

However, without ACC we can construct a well ordered chain of proper extensions of counterexamples $$ I_1 \subset I_2 \subset I_3 \cdots=\bigcup_i I_i=I $$ and by Zorn Lemma there is a maximal element,the union, which is our maximal counterexample, and hence we do not need ACC here, in the assumptions of the theorem! Note, I don't want to avoid Zorn's lemma but on the contrary use it instead of ACC. (Ascending chain condition on ideals) Thank you,JG

Viktor Vaughn
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user122424
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  • Wouldn't it be miles easier to just write down the contents of Theorem 2.18B in your post, instead of having whoever wants to help you go through an elaborate measure of clicking unfamiliar links? – Asaf Karagila Jan 19 '14 at 18:37
  • Also, if you are asking about the necessity of Zorn's lemma for proving the equivalence between different characterizations of Noetherian rings/modules/whatnot, it has been covered before on the site. (See http://math.stackexchange.com/questions/59003/can-one-avoid-ac-in-the-proof-that-in-noetherian-rings-there-is-a-maximal-elemen, for example) – Asaf Karagila Jan 19 '14 at 18:37
  • I don't want to avoid Zorn's lemma but on the contrary use it instead of ACC. (Ascending chain condition on ideals) – user122424 Jan 19 '14 at 18:49
  • The theorem 2.18B states this.If R is a ring satisfying ACC on ideals then every ideal contains a product of prime ideals.Ideal of a ring R is prime iff I =/= R and A,B superset I then AB \subset I <=> A \subset I or B\subset I. Proof from the pdf is as follows :Let I be a maximal counterexample. Then I is not a prime ideal hence I is properly contained in ideals A,B such that AB \subset I.By the maximality on I A and B are product of primes ,hence so is AB. This proves the theorem. My question is if we can find I without ACC using Zorn's lemma. – user122424 Jan 19 '14 at 18:59
  • Don't post it in the comments. [Edit] it into the question. – Asaf Karagila Jan 19 '14 at 18:59

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An equivalent characterization of the Noetherian condition (ACC on ideals) is that every nonempty set of ideals has a maximal element. Contrast this with Zorn's Lemma, which only gives that every nonempty set of ideals, which satisfies a certain property (namely chains have upper bounds), has a maximal element. It is true that for various properties (e.g. existence of maximal ideals), the hypotheses of Zorn's Lemma hold, so the result holds for any ring. However, not every desirable property will satisfy the hypotheses (e.g. existence of associated primes), and in these cases we must appeal to Noetherianness.

In the case you mention, it is not so clear that the union of the ascending chain will still be a counterexample (why can't the union contain a product of primes?) which is presumably why the authors chose to assume Noetherianness here.

zcn
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