Let $R$ be a reduced ring with all non-prime ideals finitely generated. Then is $R$ Noetherian ? If not, then is it true at least in the local case ?


Without reduced assumption, it is not true even in local case, although it is true when the ring is an integral domain; see [this previous question](https://mathoverflow.net/questions/303973/rings-with-all-non-prime-ideals-finitely-generated). 


NOTE : All our rings are commutative with unity. 



Reduced ring with all non-prime ideals finitely generated
**Question: Let $R$ be a reduced ring with all
non-prime ideals finitely generated. Then is $R$ Noetherian?**


The answer is Yes.


To lessen my typing, let me use the abbreviation
**NFG** to mean not-finitely-generated.
The result proved here is 


**Theorem.**
If $R$ is a commutative unital ring whose
NFG ideals are prime, then $R$ has at most
one NFG ideal. If one exists, it
is a nonmaximal, nonzero prime which squares to zero.
[In particular, a reduced, commutative, unital
ring whose NFG ideals are prime is Noetherian.]


Throughout the rest of this post $R$ always
denotes a commutative, unital
ring whose NFG ideals are prime.
I will also assume that $R$ has at least one
NFG ideal. I now choose one
and label it $\mathfrak p$.
I choose $\mathfrak p$ arbitrarily, except
that if there exists some NFG ideal
that does not square to zero, I choose
$\mathfrak p$ so that it does not square to zero.
That way, by proving that, indeed, $\mathfrak p^2=(0)$,
I will have established that all NFG ideals square to $(0)$.


**Lemma 1.**
$\mathfrak p$ is contained in the Jacobson radical, $J(R)$, of $R$


**Proof of Lemma.**
Let $\mathfrak m$ be a maximal ideal of $R$.
If $\mathfrak p\not\subseteq \mathfrak m$,
then $\mathfrak p\cdot \mathfrak m\subseteq
\mathfrak p\cap \mathfrak m$, and $\mathfrak p\not\subseteq \mathfrak p\cap\mathfrak m$, and $\mathfrak m\not\subseteq \mathfrak p\cap\mathfrak m$,
so $\mathfrak p\cap\mathfrak m$ is not prime.
The ideal $\mathfrak p\cap \mathfrak m$ is therefore finitely generated.


As $R$-modules, we have
$\mathfrak p/(\mathfrak p\cap \mathfrak m)\cong
(\mathfrak p+\mathfrak m)/\mathfrak m = R/\mathfrak m$.
The latter is simple, hence $1$-generated.
Altogether we have that, as $R$-modules,
$\mathfrak p$ is $1$-generated over
$\mathfrak p\cap \mathfrak m$, and that
$\mathfrak p\cap \mathfrak m$ is finitely generated.
This is enough to prove that $\mathfrak p$ is
finitely generated as an $R$-module, hence as an ideal.
This contradicts the assumption that $\mathfrak p$ is NFG.


We have shown that $\mathfrak p$ is contained in an
arbitrarily chosen maximal ideal, hence it is
contained in $J(R)$. (End proof of Lemma 1.)


**Lemma 2.** 
If $a\in \mathfrak p$ and $a\cdot \mathfrak p\neq (0)$, then the annihilator
$\mathfrak q:=(0:a)$ is NFG and $\mathfrak q$ is a proper
subideal of $\mathfrak p$.
Moreover, $R/\mathfrak q$ is Noetherian.


**Proof of Lemma.**
We cannot have $a\in a\cdot \mathfrak p$ for the following reason.
It leads to $a = ab$ for some
$b\in \mathfrak p$, hence to $(1-b)a=0$ for some 
$b\in\mathfrak p\subseteq J(R)$.
This yields $a=0$, since $1-J(R)$ consists of units.
But $a=0$ contradicts $a\cdot \mathfrak p\neq (0)$.


Therefore, we have $a\notin a\cdot \mathfrak p = (a)\mathfrak p$.
In otherwords, the product $(a)\mathfrak p$ contains neither
of the factors $(a)$ or $\mathfrak p$. (The factors each contain
$a$ but the product does not.) Hence $(a)\mathfrak p=a\mathfrak p$
is not prime, and therefore $a\mathfrak p$ is finitely generated.


The map $x\mapsto ax$ is an $R$-module homomorphism of the
nonfinitely generated $R$-module $\mathfrak p$
onto the finitely generated $R$-module $a\mathfrak p$.
Necessarily this map has NFG kernel, which is
$\mathfrak q:=(0:a)\cap \mathfrak p$.


The element $a$ does not annihilate $\mathfrak p$, i.e.
$\mathfrak p\not\subseteq (0:a)$, so we derive that $\mathfrak p$
properly contains $(0:a)\cap \mathfrak p = \mathfrak q$. 
Now the fact
that
$(0:a)\mathfrak p\subseteq (0:a)\cap \mathfrak p\subseteq \mathfrak q$,
implies that $(0:a)\subseteq \mathfrak q\subsetneq \mathfrak p$.
Thus $\mathfrak q=(0:a)\cap \mathfrak p = (0:a)$.


To summarize the progress so far:
if $a\in \mathfrak p$ and $a\mathfrak p\neq (0)$,
then $\mathfrak q:= (0:a)$ is a proper NFG
subideal of $\mathfrak p$.


To finish, we need to prove the last sentence of the lemma.
Note that the class of rings
whose NFG ideals are prime is closed under
the formation of quotients. Moreover, it follows
from the earlier part of this lemma that no
non-Noetherian domain
belongs to this class. (Starting with an
NFG ideal $\mathfrak p$ we produced nontrivial
zero divisors by establishing that for any $a\in\mathfrak p$
either $a\mathfrak p=(0)$ or $(0:a)$ is NFG.
Either case produces nontrivial zero divisors.)
Hence, the domain $R/\mathfrak q$ must be Noetherian.
(End proof of Lemma 2.)


**Lemma 3.** 
$\mathfrak p^2=(0)$ and $\mathfrak p$ is the unique
NFG ideal of $R$.


**Proof of Lemma.**
To obtain a contradiction to $\mathfrak p^2=(0)$,
suppose that $a, b\in \mathfrak p$ satisfy
$ab\neq 0$. Then $a\mathfrak p\neq (0)$,
so by Lemma 2 $\mathfrak q:=(0:a)$ is a proper NFG
subideal of $\mathfrak p$, which does not contain $b$.
Moreover, $R/\mathfrak q$ is Noetherian.


In particular, $\mathfrak p/\mathfrak q$ is a finitely generated
ideal of $R/\mathfrak q$,
so $\mathfrak p$ is finitely generated over 
the subideal $\mathfrak q$ in $R$.
It follows that $\mathfrak p$ is also finitely generated
over the larger subideal $(b^2)+\mathfrak q$. If this latter ideal 
$(b^2)+\mathfrak q$ was f.g. in $R$,
then $\mathfrak p$ would also be f.g.
in $R$, which is false.
Hence $(b^2)+\mathfrak q$ is NFG, and therefore prime, in $R$. 
But $b^2\in (b^2)+\mathfrak q$,
so by primality $b\in (b^2)+\mathfrak q$.
It must be possible to express $b$ as $b=b^2c+q$
where $c\in R$ and $q\in\mathfrak q$. Rewritten,
$b(1-bc)=q\in \mathfrak q$.


But $b\in\mathfrak p\subseteq J(R)$, so $(1-bc)$ is a unit,
and we derive that $b=(1-bc)^{-1}q\in\mathfrak q$,
a contradiction to $b\notin (0:a)=\mathfrak q$. What we have contradicted
was the initial assumption that $\mathfrak p^2\neq (0)$.


Now suppose that $R$ has NFG ideals $\mathfrak p$
and $\mathfrak r$. By the first part of the lemma,
$\mathfrak p^2=(0)=\mathfrak r^2$.
Since $\mathfrak p^2=(0)\subseteq \mathfrak r$,
and $\mathfrak r$ is prime, we get $\mathfrak p\subseteq \mathfrak r$.
Similarly, $\mathfrak r\subseteq \mathfrak p$, so
$\mathfrak p=\mathfrak r$. 
(End proof of Lemma 3.)


The theorem is essentially proved now, but let me write it out.


**Theorem.**
If $R$ is a commutative unital ring whose
NFG ideals are prime, then $R$ has at most
one NFG ideal. If one exists, it
is a nonmaximal, nonzero prime which squares to zero.
[In particular, a reduced, commutative, unital
ring whose NFG ideals are prime is Noetherian.]


**Proof of Theorem.**
The lemmas show that 
if $R$ is a commutative unital ring whose
NFG ideals are prime, and $R$ has at least one
NFG ideal $\mathfrak p$, then $\mathfrak p$ is the unique NFG
ideal of $R$ and $\mathfrak p^2=(0)$. The ideal $\mathfrak p$ 
cannot be $(0)$, since $\mathfrak p$ is NFG and $(0)$ is not.
Thus $R/\mathfrak p$ is a domain that
acts on the module $\mathfrak p$
is such a way that this module is NFG, but all proper
submodules are f.g.
It follows that $\mathfrak p$ is not a maximal ideal of $R$,
since then $R/\mathfrak p$ would be a field,
and fields have no NFG modules
whose proper submodules are all f.g.
(This last observation also appears
in the answer to
[this question](https://mathoverflow.net/questions/302468/local-ring-all-whose-non-maximal-ideals-are-finitely-generated).)


The square-bracketed claim at the end of the theorem statement is clear.
(End proof of Theorem.)


**Comment.** There exists a ring $R$ whose NFG ideals are prime,
which actually contains an NFG ideal (i.e., 
there exists a non-Noetherian ring whose NFG ideals are prime). One is constructed in the answer to [this question](https://mathoverflow.net/questions/303973/rings-with-all-non-prime-ideals-finitely-generated).


