$A$ local Noetherian ring with principal maximal ideal implies PIR? 
Suppose that $A$ is a local Noetherian ring with principal maximal ideal. Can we prove that every ideal of $A$ is principal?

I tried to exploit the Noetherian property on the set of non-principal ideals obtaining that (if the statement is false) there must exist  a prime non-principal ideal but I can't conclude nothing.
 A: Here is an elementary proof (which is of course much easier than the proof of the deeper theorem of Kaplansky), which does not invoke any theorems in commutative algebra besides Nakayama:
Let $I \subset (m)$ be a non-zero ideal. Define $n := \min \{ s \geq 1 | \exists x \in I \text{ such that } x \in (m)^s \setminus (m)^{s+1} \}$.
First, we have to show that $n < \infty$, i.e. we have to show that $I$ is not contained in $\bigcap\limits_{s \geq 1} (m)^s$, i.e. we have to show $\bigcap\limits_{s \geq 1} (m)^s=0$. This is clear with Krull's intersection theorem, but I don't want to invoke any theorem, so I will give an argument in this case:
Let $a$ be contained in that intersection, in particular $a=mb$ for some $b \in A$. I claim that $b$ is also contained in that intersection: If not, $b$ is non-zero in some $(m)^s/(m)^{s+1}$. Since this is a one-dimensional vector-space, we obtain that $b$ generates $(m)^s/(m)^{s+1}$. By Nakayama, $b$ generates $(m)^s$, i.e. $a$ generates $(m)^{s+1}$. In particular $a \notin (m)^{s+2}$, contradiction!
Thus, we have shown $(m) \bigcap\limits_{s \geq 1} (m)^s = \bigcap\limits_{s \geq 1} (m)^s$, i.e. $\bigcap\limits_{s \geq 1} (m)^s=0$ by Nakayama. This is where we need the noetherian hypothesis, because we need to guarantee that $\bigcap\limits_{s \geq 1} (m)^s$ is a priori finitely generated to invoke Nakayama.
Now it is very easy to show that $I=(m)^n=(m^n)$ holds: By the minimality of $n$, we have $I \subset (m)^n$ and we have some $x \in I$ with $x \notin (m)^{n+1}$. Again invoking Nakayama, we get $(x)=(m)^n$, i.e. $I \subset (m)^n=(x) \subset I$. The proof ends here.
