# 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 Jan 22 '14 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 Jan 22 '14 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 Jan 22 '14 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. – DonAntonio Jan 22 '14 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. – DonAntonio Jan 22 '14 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 Jan 22 '14 at 6:01
• "The reasoning I gave..." I agree. – Pete L. Clark Jan 22 '14 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... – rschwieb Jan 22 '14 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. – Pete L. Clark Jan 22 '14 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 Jan 22 '14 at 6:03