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This questions comes under the section about Zorn's Lemma. But I'm really lost to how maximality here implies prime.

Edit: R is commutative.

gws
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    This is a lemma used in proving a theorem of Cohen: a ring with all prime ideals f.g. is Noetherian. You may be able to find hints in texts on commutative algebra. – Angina Seng Jan 25 '19 at 07:10
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    @LordSharktheUnknown Just a crude idea: Doesn't it suffice to show that If $xy \in A$ but $x,y \not\in A$, then $\langle A, x \rangle$ is a non-finitely generated ideal in $R$ that is larger than $A$? Is it really a difficult exercise? Because it looks kind of innocent. – stressed out Jan 25 '19 at 07:56
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    How do you prove that is not f.g.? @stressedout – Angina Seng Jan 25 '19 at 08:02
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    @stressedout: if it's easy to show that $\langle A, a \rangle$ is not finitely generated; I don't know . . . is it? – Robert Lewis Jan 25 '19 at 08:03
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    @stressedout: many hard problems look innocent! – Robert Lewis Jan 25 '19 at 08:05
  • Haha. Thanks guys (@LordSharktheUnknown & @RobertLewis). On a second thought, I realized that adding one element to a non-finitely generated ideal does not necessarily keep it not f.g. :P – stressed out Jan 25 '19 at 08:07
  • @stressedout What would be an example of adding one element to a non f.g. ideal would make it f.g.? – gws Jan 25 '19 at 08:11
  • @davidh adding $1$, for example. – stressed out Jan 25 '19 at 08:12
  • @davidh The main point is that adding an element to an ideal which is maximal among the nonfinitely generated ideals certainly yields a finitely generated ideal. – egreg Jan 25 '19 at 08:40
  • Please search for your question before asking. Also please put the question in the body of the post, even if it means repeating some of your title. – rschwieb Jan 25 '19 at 11:52

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Let $\mathcal{S}$ be the set of nonfinitely generated ideals, ordered by inclusion. If you have a chain $\mathcal{C}$ in $\mathcal{S}$, its union $K$ is an ideal. Let's prove that $K$ is not finitely generated. Otherwise, we'd find $I\in\mathcal{C}$ such that $K\subseteq I$ and therefore $K=I$, contrary to the assumption that $I$ is not finitely generated.

Let $M$ be a maximal element in $\mathcal{S}$ (which exists by Zorn's lemma, but it is not necessarily a maximal ideal). Suppose $xy\in M$, but $x,y\notin M$.

Then both $(M,x)$ and $(M,y)$ are ideals properly containing $M$, so they are finitely generated, by maximality of $M$ in $\mathcal{S}$.

Can you finish? Further hint: the product of two finitely generated ideals is finitely generated.

egreg
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  • I'm still not seeing why (M,x) would be finitely generated... – gws Jan 25 '19 at 08:57
  • @davidh Because $(M,x)$ is strictly larger than $M$. So, if it were not finitely generated, you'd get a contradiction with the maximality of $M$. – stressed out Jan 25 '19 at 08:58
  • Oh ok, so they are like "pushed out of" S. – gws Jan 25 '19 at 08:59
  • So is the contradiction that these two ideals would the whole ring? – gws Jan 25 '19 at 09:01
  • @davidh No, that's not right. I extended the hint a bit. – egreg Jan 25 '19 at 09:36
  • @egreg I'm asking this question because I already had a similar solution for this question on my mind (which I'm not going to post because you already posted a perfect one which is similar to mine step by step), but just to make sure that I'm right, is it correct that $(M,x)(M,y)=(M,xy)=M$? – stressed out Jan 25 '19 at 09:46
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    @egreg So the product of these two ideals would be M, but M is non f.g. Thus contradiction. – gws Jan 25 '19 at 10:01
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    @davidh Exactly! Just fill in the details. – egreg Jan 25 '19 at 10:02
  • @egreg Great! Thanks so much – gws Jan 25 '19 at 10:04
  • @egreg: how do we know that if $K$ were finitely generated we would have $K \subset I$? I think I' m not getting something . . . – Robert Lewis Jan 30 '19 at 04:07
  • @egreg: OK, I think I see this; but would still like to read whatever you might say. Thanks! – Robert Lewis Jan 30 '19 at 04:08
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    @RobertLewis Take a finite set of generators of $K$; each one belongs to an ideal in the chain; on the other hand this is a chain, so you find an ideal in the chain that contains all of them (here finiteness is crucial). Then such ideal contains $K$, but is also contained in it by definition of $K$. – egreg Jan 30 '19 at 10:13