I am working on Exercise 3.15 from Aluffi's book Algebra: Chapter 0.

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 $\mathcal F\ne0$, 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.

Thus, a ring is Noetherian if and only if its prime ideals are finitely generated.

I want to follow the hint, but I got stuck in the 4th step. Actually, I don't know how to come up with these steps as well as the motivation, could someone help me? thanks.

  • $\begingroup$ @Krish very comprehensive andw concise! thanks,I will try the other steps $\endgroup$
    – Mugenen
    Commented Dec 7, 2017 at 13:22
  • $\begingroup$ @Krish sorry,now I find I couldnt understood the existence of the surjection,I think the surjection you mention maps (r+I) to (r+J1J2),but how to prove it is well defined? $\endgroup$
    – Mugenen
    Commented Dec 7, 2017 at 13:42
  • 1
    $\begingroup$ Dear @likemath, I made a terrible mistake. What I wrote is wrong. I am really very sorry for that. Here is a counter-example: $4\mathbb Z \subset 2\mathbb Z$, but there can't be any surjection from $\mathbb Z/2\mathbb Z$ to $\mathbb Z/4\mathbb Z$. (Thanks for pointing out the mistake, I'll delete my previous comment.) $\endgroup$
    – Krish
    Commented Dec 7, 2017 at 15:18
  • $\begingroup$ For some basic information about writing math at this site see e.g. basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to. In this case I have edited the post. Please edit it further if needed (in particular, if you want to stress some parts). And if you are satisfied with the result, you can probably remove the picture. $\endgroup$ Commented Dec 7, 2017 at 18:38
  • $\begingroup$ @MartinSleziak thanks! $\endgroup$
    – Mugenen
    Commented Dec 8, 2017 at 0:07

1 Answer 1


Some general theory: if $M$ is an $R$-module and $A$ is an ideal of $R$ such that $AM=\{0\}$ (that is, $A$ is contained in the annihilator of $M$), then $M$ can be given a structure of module over $R/I$ by $$ (r+A)x=rx $$ for $r\in R$ and $x\in M$. The proof this is a well-defined action of $R/A$ on $M$ that makes $M$ into a module over $R/A$ is straightforward.

Note. A module $M$ with $AM=0$ is finitely generated as $R$-module if and only if it is finitely generated as a module over $R/A$.

Let's prove that the annihilator of the $R$-module $I/J_1J_2$ contains $I$. Indeed, if $r,x\in I$, then $rx\in I^2\subseteq J_1J_2$, so that $r(x+J_1J_2)=0+J_1J_2$ and we have the thesis.

By general theory, we can define an $R/I$-module structure on $I/J_1J_2$.

Similarly, if $r\in I$ and $x\in J_1$, then $rx\in IJ_1\subseteq J_2J_1$, so $r(x+J_1J_2)=0+J_1J_2$.

A brief series of hints about the various steps.

  1. The union of a chain of non finitely generated ideals is not finitely generated; Zorn's lemma provides a maximal element $I$ in $\mathcal{F}$.

  2. Every ideal propertly containing $I$ is finitely generated, by maximality. Hence every nonzero ideal of $R/I$ is finitely generated and therefore $R/I$ is Noetherian.

  3. Since $I$ is not finitely generated it is not prime. Hence there are ideal $K_1$ and $K_2$ not contained in $I$ such that $K_1K_2\subseteq I$. Setting $J_1=K_1+I$ and $J_2=K_2+I$, we have that $J_1$ and $J_2$ properly contain $I$ and $J_1J_2\subseteq I$.

  4. See above.

  5. Since $J_1$ properly contains $I$, it is finitely generated, so also $J_1/J_1J_2$ is finitely generated as an $R$-module, hence as a module over $R/I$ as well. Since $R/I$ is Noetherian and $I/J_1J_2$ is a submodule over $J_1/J_1J_2$, also $I/J_1J_2$ is finitely generated as $R/I$ module, hence as $R$-module.

  6. Consider $J_1J_2\subseteq I$ and observe that, since $J_1J_2$ is finitely generated and $I/J_1J_2$ is finitely generated, also $I$ is finitely generated.

  • $\begingroup$ thanks,I will try the other steps $\endgroup$
    – Mugenen
    Commented Dec 7, 2017 at 13:21
  • $\begingroup$ Oh..I failed..I do not know how to continue the other two steps,could you give me some hints?And I also dont know why we need give the two $R/I-module$ structure.. $\endgroup$
    – Mugenen
    Commented Dec 8, 2017 at 3:03
  • $\begingroup$ @likemath I added a sketch. The $R/I$-module structure is important because $R/I$ is Noetherian. $\endgroup$
    – egreg
    Commented Dec 8, 2017 at 10:26
  • $\begingroup$ thank you very much,I got it. $\endgroup$
    – Mugenen
    Commented Dec 8, 2017 at 12:15

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .