A ring is Noetherian if its prime ideals are finitely generated

Question

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.

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 $R/A$.

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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 $r(x+J_1J_2)=0+J_1J_2$.

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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.