Are Dummit and Foote making a mistake in proving Cohen's theorem?

Question

Exercise 11 on page 669 (this is Chapter 15) wants to prove Cohen's theorem that if every prime ideal of a ring is f.g. then every ideal is f.g. that is the ring is noetherian. The highbrow (perhaps?) solution is that the set of finitely generated ideals of a ring is an Oka family, and maximal elements in the complement of an Oka family of ideals are prime. Now, suppose $I$ is an ideal maximal wrt to being non finitely generated. Such an ideal exists if the family of ideals that is nonfinitely generated is nonempty by Zorn's lemma. Now suppose $x,y\notin I$. I will show that $xy\notin I$. Now D&F suggests the following line of attack.

$(a)$ Prove that if the collection of ideals of R that are not f.g. is nonempty, then it contains a maximal element $I$, and $R/I$ is noetherian.

Check.

$(b)$ Prove that there are finitely generated ideals $J_1,J_2$ containing $I$ with $J_1J_2\subseteq I$ and that $J_1J_2$ is f.g. Observe that $I$ is not a prime ideal.

Here I am stumped. This is false! [Yes, the point is that we're assuming prime ideals are f.g. We will thus reach a contradiction.]

$(c)$ Prove that $I/J_1J_2$ is a finitely generated $R/I$ submodule of $J_1/J_1J_2$

$(d)$ Show this implies $I$ is finitely generated over $R$, which is a contradiction. Deduce that $R$ is noetherian.

I am also not seeing where they are using the hypothesis that prime ideals are finitely generated. I proved that non f.g. ideals implies non f.g. primes, but here they seem to be reaching the conclusion that $R$ is noetherian, somehow. [They are proving that if all primes are f.g. then $R$ is noetherian, not that if $R$ is not noetherian then there is a non f.g. prime]

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Now, let me use D&F idea to show $I$ is actually prime. Take $x,y\notin I$. Then $J_1=I+(x)$ and $J_2=I+(y)$ lie strictly over $I$ so are f.g., hence $J_1J_2$ is. Now, if $xy\in I$, then $J_1J_2\subseteq I$. Since $I$ is maximal wrt to being non f.g. clearly $R/I$ is noetherian. Note that if $r-r'\in I$ and $p+ax\in J_1$ then $(r-r')(p+ax)=(r-r')p+(r-r)'ax\subset I^2+ I(x)\subseteq J_1J_2$ so we have a well defined action of $R/I$ on $J_1/J_1J_2$ and $I/J_1J_2$. Since $R/I$ noetherian and $J_1/J_1J_2$ is f.g., it is noetherian and hence $I/J_1J_2$ is f.g. as a $R/I$ module, i.e. $I+J_1J_2=I$ is a f.g. $R$ module, which is impossible. Hence $xy\notin I$; so $I$ is prime.

So, am I completely missing the point somewhere or is this a honest mistake in D&Fs exposition?

Answer 1

$I$ not prime (because it is maximal among the non-f.g. ideals, and all prime ideals are f.g. by hypothesis) implies that there are two ideals $I_1,I_2$ such that $I_1I_2\subseteq I$, but $I_1\nsubseteq I$, and $I_2\nsubseteq I$. Now take $J_i=I_i+I$ and observe that $J_1J_2\subseteq I$, and $I\subsetneq J_i$, so $J_i$ are f.g.

Answer 2

The ideal $I$ cannot be prime, because by (a) $I$ is not finitely generated (the ideal $I$ is maximal in the collection of non-finitely generated ideals, but it is not necessarily a maximal ideal!). If $I$ were prime, $I$ would be finitely generated by assumption on $R$, a contradiction. Therefore $I$ is not prime ; this means there exists two principal ideals $J_1 = (x_1)$, $J_2 = (x_2)$ such that $x_1 x_2 \equiv 0 \pmod{I}$, i.e. $J_1 J_2 \subseteq I$. This is what D&F meant by "Observe that $I$ is not a prime ideal", and it is a correct observation.

P.S. : You are proving a statement by contradiction, so of course you might encounter statements which are false in general, but that is precisely the point.

Hope that helps,

Answer 3

We're assuming $R$ is a ring in which every prime ideal is f.g., so $I$ being non f.g. should be non prime (this is the "Observe $I$ is not prime" line). This guarantees the existence of such $J_1,J_2$ as in the proof in the post. The point is that, as proved, any such ideal is prime. Inevitably we reach a contradiction, namely that $I$ is both f.g. and non f.g., which means that the set of ideals which are not finitely generated must be empty all along, so $R$ is noetherian. Alternatively, we're proving that in any non noetherian ring not only there are non f.g. ideals, but there are always non f.g. prime ideals. I hope the confusion I caused is cleared out.