# On rings with a unique maximal ideal

I would be grateful if you guide me through the following question:

Suppose a commutative ring with identity, $R$, has a unique maximal ideal, say $M$. If $M$ is principal, can we show that every ideal is finitely generated?

Well, I am considering the following facts:

$1)$ Every ideal $I$ of $R$ lies in a maximal ideal, and since we have a unique maximal ideal, then $I$ lies in $M=(m)$.

$2)$ We can also consider Nakayama's lemma: $IM$ is an ideal of $M$, and if IM=M then $M=0$. In particular we can set I=M.

$3)$ Another fact I can think of, is that it follows easily that all nonunits form an ideal in such a ring, which is $M$ itself.

• In (2), $M = 0$ is not a contradiction, but it would mean that $R$ is a field, and thus answer your question in the affirmative. Out of curiosity, is this a homework problem?
– zcn
Commented Jan 22, 2014 at 4:26
• In fact, I would like to know: are you being asked to prove this exact statement? Or did you come up with the statement yourself?
– zcn
Commented Jan 22, 2014 at 4:32
• I see, this is completely equivalent to the statement you posted. I appreciate your honesty - this homework problem is not so easy! As a start, it is enough to prove that every (nonmaximal) prime ideal is finitely generated
– zcn
Commented Jan 22, 2014 at 4:40
• I'm not sure but I don't think the claim's true without requiring the ring is Noetherian, and I'm almost sure I've a coutner example somewhere in books/papers/stuff. Anyway, check the Noetherian thing. Commented Jan 22, 2014 at 5:23
• @Sean, I'm not sure I understood: the question was "is it true that...", and what I wrote above is that it is not true (if I remember correctly). So if the question asked to prove a local ring with principal maximal ideal is a PIR then, I think, that can't be done. Commented Jan 22, 2014 at 5:34

This question came up on Math Overflow several years ago. I answered it -- negatively -- here.

I am pretty sure that the question has been asked several times on this site as well. I will try to search for it. (But I am not good at searching on this site. Help would be appreciated...)

• Thanks for the reference - I'll keep that example in mind. I'd like to note though, that the reasoning I gave shows that it is true if the ring is complete (or just $M$-adically separated)
– zcn
Commented Jan 22, 2014 at 6:01
• "The reasoning I gave..." I agree. Commented Jan 22, 2014 at 6:05
• I just knew a choice of valuation group would do the trick, but I did not know what would work. Glad to see the method applied. My advisor showed me how to pull something similar for an example in my dissertation... Commented Jan 22, 2014 at 11:19

As in the problem statement, write the maximal ideal as $M = (m)$. Notice that since $M$ is finitely generated, by Nakayama's Lemma, $M \neq M^2$, and in general $M^i \neq M^j$ for any $i \neq j$. Assume that the following holds:

Lemma: $\cap_{n=1}^\infty M^n = 0$.

Assuming the lemma, let $I \neq 0$ be an proper ideal of $R$. Since $I \subseteq M$ but $I \not \subseteq \cap_{n=1}^\infty M^n$, there exists an $n$ such that $I \subseteq M^n$, but $I \not \subseteq M^{n+1}$. Pick $x \in I \setminus M^{n+1}$. As $x \in M^n$, write $x = m^ny$ for some $y \in R$. If $y \in M$, then $x \in M^{n+1}$, a contradiction, so $y$ is a unit, i.e. $(x) = (m^n) = M^n$. But then $(x) \subseteq I \subseteq M^n = (x)$, so $I = (x)$ is in fact principal.

EDIT: (placed later so as to not disrupt continuity) Together with the Krull Intersection Theorem, the reasoning above in fact proves that a local ring $R$ with principal maximal ideal is Noetherian iff it is $M$-adically separated.

• Your Lemma holds if $R$ is Noetherian by the Krull Intersection Theorem (see e.g. $\S$ 8.12.2 of math.uga.edu/~pete/integral.pdf). It does not hold in general: see my answer. Commented Jan 22, 2014 at 6:01
• Of course, the lemma is exactly (a weak version of) Krull Intersection. Your example explains why it was so impossible to prove without Noetherian hypotheses
– zcn
Commented Jan 22, 2014 at 6:03