# Reduced ring with all non-prime ideals finitely generated

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.

NOTE : All our rings are commutative with unity.

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.)

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.

This is not another answer, but an addendum to the answer I already gave. I don't want to edit these new remarks into the original answer, and in fact do not want to change that answer in any way, since that answer is referenced in a meta thread.

I will continue to write NFG for not-finitely-generated.

Call a commutative ring $R$ almost Noetherian if its NFG ideals are prime. There have been MO questions about the implication almost Noetherian implies Noetherian'' for rings $R$ satisfying additional assumptions. The questions can be found on this page, and here, and here. The additional assumptions include combinations of: $R$ is local, $R$ is a domain, $R$ is reduced, or that the NFG ideals of $R$ are not just prime but are maximal.

It occurred to me that my answer on this page almost fully describes the structure of almost Noetherian rings. I will complete the description with this addendum.

Theorem. A ring $R$ is almost Noetherian (recall: this means that NFG ideals are prime) if it is either (i) Noetherian or (ii) has exactly one NFG ideal $\mathfrak p$ and

• $R/\mathfrak p$ is a Noetherian integral domain
• $\mathfrak p^2=(0)$
• when $\mathfrak p$ is viewed as an $R/\mathfrak p$-module it is a faithful module. Moreover, as a module, $\mathfrak p$ is NFG, but all proper submodules are finitely generated. \\\

My original answer on this page proves that a ring $R$ that is almost Noetherian and not Noetherian satisfies all of the bulleted properties, EXCEPT that I did not prove that $R/\mathfrak p$ acts faithfully on $\mathfrak p$. I will prove that here, and then prove that, conversely, any ring satisfying all conditions is almost Noetherian.

I will assume all properties proved in my original answer.

Lemma. If $R$ is almost Noetherian and $\mathfrak p$ is an NFG ideal of $R$, then $R/\mathfrak p$ acts faithfully on $\mathfrak p$. (This strengthens the earlier conclusion $\mathfrak p^2=(0)$ to $(0:\mathfrak p)=\mathfrak p$.)

Proof. Lemma 3 of the original solution asserts that $\mathfrak p^2=(0)$ and $\mathfrak p$ is the unique NFG ideal of $R$. The first part of this statement guarantees that $(0:\mathfrak p)\supseteq \mathfrak p$. To obtain a contradiction, assume that containment is strict. The second part of the statement of Lemma 3 guarantees that $(0:\mathfrak p)$ is finitely generated as an $R/\mathfrak p$-module. But $R/\mathfrak p$ is a Noetherian integral domain, so submodules of f.g. modules are again f.g. This contradicts the possibility that $(0:\mathfrak p)$ is f.g. and contains the NFG submodule $\mathfrak p$. The conclusion is that $\mathfrak p=(0:\mathfrak p)$. (End proof of Lemma.)

Theorem. Assume that $R$ is a ring that has an ideal $\mathfrak p$ such that

• $R/\mathfrak p$ is a Noetherian integral domain
• $\mathfrak p=(0:\mathfrak p)$, and
• if $\mathfrak p$ is viewed as an $R/\mathfrak p$-module, then $\mathfrak p$ is NFG, but all proper submodules are finitely generated.

Then $R$ is almost Noetherian.

Proof. The idea of the proof is simple. The ideal $\mathfrak p$ is NFG by the last item. I will show that any ideal that properly contains $\mathfrak p$ or is properly contained in $\mathfrak p$ is finitely generated. Then I will show that there are no other ideals. (That is, every ideal of $R$ is comparable with $\mathfrak p$). This will establish that $R$ has exactly one NFG ideal, and that it is prime.

The fact that ideals properly contained in $\mathfrak p$ are finitely generated follows from the third bulleted item of the theorem. The fact that $\mathfrak p$ is NFG follows from the same item. The fact that $\mathfrak p$ is prime follows from the first bulleted item (the quotient mod $\mathfrak p$ is a domain). The first bulleted item also guarantees that every ideal containing $\mathfrak p$ is finitely generated over $\mathfrak p$.

It is enough, now, to show that for any ideal $I\lhd R$, if $I\not\subseteq \mathfrak p$, then $I\supsetneq \mathfrak p$. For from this one can argue: if $I\not\subseteq \mathfrak p$, then $I$ properly contains $\mathfrak p$, so $I$ is f.g. over $\mathfrak p$, so $I=\mathfrak p+F$ for some finitely generated ideal $F\not\subseteq \mathfrak p$. By the same reasoning, $F\not\subseteq \mathfrak p$, so $F\supsetneq \mathfrak p$, so $F = \mathfrak p+F = I$, and $I$ is f.g. Consequently $\mathfrak p$ is the only NFG ideal, and it is prime, so $R$ is almost Noetherian.

To establish the necessary condition, identified in the last paragraph, choose any $I\not\subseteq \mathfrak p$. Choose $r\in I-\mathfrak p$. I wish to show, by contradiction, that $r\cdot \mathfrak p=\mathfrak p$. Assume instead that $r\cdot \mathfrak p\neq \mathfrak p$. The image of the module endomorphism $\varepsilon_r:\mathfrak p\to r\mathfrak p: x\mapsto rx$ is finitely generated, and $\mathfrak p$ is NFG, so $\ker(\varepsilon_r)$ is NFG. The only NFG submodule of $\mathfrak p$ is $\mathfrak p$ itself, so $\ker(\varepsilon_r)=\mathfrak p$, so $r\cdot\mathfrak p=(0)$, implying that $r\in(0:\mathfrak p)$. But $r\notin\mathfrak p=(0:\mathfrak p)$, a contradiction. The contradiction shows that if $r\in I-\mathfrak p$, then $r\cdot \mathfrak p = \mathfrak p$. This shows that $I\supseteq I\mathfrak p\supseteq r\cdot\mathfrak p=\mathfrak p$, as desired. (End of proof.)

This makes it easy to construct examples of almost Noetherian rings that are not Noetherian. Start with any Noetherian domain $S$, and any faithful NFG module $M$ whose proper submodules are finitely generated. Then the idealization $R:=S\oplus M$ is almost Noetherian. This describes all almost Noetherian rings $R$ that are split extensions of $R/\mathfrak p$ by $\mathfrak p$. (I was able to construct an almost Noetherian ring of characteristic $p^2$, so they are not all split.)