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.
 A: 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...)
A: 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.
