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I was working through exercise 3.15 in Aluffi's Chapter 0, which has a step-by-step plan of a proof of the fact that a commutative ring is Noetherian if all the prime ideals are finitely generated. The plan is as follows:

3.15. Recall that a (commutative) ring $R$ is Noetherian if every ideal of $R$ is finitely generated. Assume the seemingly weaker condition that every prime ideal of $R$ is finitely generated. Let $\mathcal F$ be the family of ideals that are not finitely generated in $R$. You will prove $\mathcal F=\emptyset$.

  • If $\mathcal F\ne\emptyset$, prove that it has a maximal element $I$.
  • Prove that $R/I$ is Noetherian.
  • Prove that there are ideals $J_1$, $J_2$ containing $I$, such that $J_1J_2\subseteq I$.
  • Give a structure of $R/I$ module to $I/J_1J_2$ and $J_1/J_1J_2$.
  • Prove that $$I/J_1J_2$$ is a finitely generated $R/I$-module.
  • Prove that $I$ is finitely generated, thereby reaching a contradiction.

However, it seems that half of the steps are actually redundant. This makes me unsure whether my proof is correct, so if anybody could help find the error (if there is one), that would be great.

Proof

The union of every chain of $\mathcal F$ is not finitely generated since otherwise some element in the chain would be finitely generated. So every chain has an upper bound and we have a maximal element $I$.

$R/I$ is Noetherian since the preimage of any non-finitely generated ideal would be a non-finitely generated ideal in $R$ larger than $I$.

$I$ is not prime since otherwise it would be finitely generated by the statement of the theorem. Hence we have $a,b\notin I$ s.t. $ab\in I$.Then $I$ is a proper subset of ideals $I+(a)$ and $I+(b)$. This is where my proof differs from the plan: but these 2 larger ideals must be finitely generated, or that would contradict the maximality of $I$. Now note that $(I+(a))(I+(b))=I+(a)I+(b)I+(ab)=I+(ab)=I$, and $I$ is finitely generated as a product of finitely generated ideals. We arrived at a contradiction, so the assumption $\mathcal F\neq \emptyset$ is false.

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  • $\begingroup$ @ N.S. assume $J=\bigcup_k I_k$, where the ideals $I_k$ form a chain and are not finitely generated, is finitely generated. Then the generators must all belong to some $I_m$ in the chain, so either $I_m$ is finitely generated, or $J$ is strictly larger than $I_m$, which is larger than the ideals generated by said generators. $\endgroup$ – Serg Sep 20 '19 at 18:57
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You have an error in the equation $$(I+(a))(I+(b))=I+(a)I+(b)I+(ab).$$ The first term on the right side should be $I^2$, not $I$. As a result you cannot be sure that $(I+(a))(I+(b))$ is all of $I$.

Writing $J_1=I+(a)$ and $J_2=I+(b)$, you can fix the proof by showing that although $J_1J_2$ may not be all of $I$, $I/J_1J_2$ is at least finitely generated, and so you can still conclude $I$ is finitely generated. But this is just what the plan of Exercise 3.15 suggests.

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Your proof is not different than the one given, $J_1=I+(a), J_2=I+(b)$ gives $I \subseteq J_1, J_2$ and $J_1 J_2 \subset I$.

Also note that your conclusion that $$(I+(a))(I+(b))=I$$ contains a mistake. Indeed

$$(I+(a))(I+(b))=I^2+(a)I+(b)I+(ab)\subseteq I+(ab)=I$$

For example, look at $R= \mathbb Z$, $I=36 \mathbb Z, a=12, b=18$. Then $$(I+(a))(I+(b))=12 \mathbb Z \times 18 \mathbb Z \neq 36 \mathbb Z $$

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  • $\begingroup$ It is different from the given proof; the whole point is that they're trying to shortcut the last three steps by proving that $J_1J_2$ is actually equal to $I$. $\endgroup$ – Eric Wofsey Sep 20 '19 at 19:03
  • $\begingroup$ @EricWofsey The hint about $J_1, J_2$ is really hinting to those two ideals, and I already pointed out that the stronger claim cannot hold.... $\endgroup$ – N. S. Sep 20 '19 at 19:06

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