# Are local, Noetherian rings with principal maximal ideal PIR?

A question asked by a friend. I believe it's false, but lack a decisive counterexample.

This question shows that it is true for valuation rings, but I know too little about them.

In the wider context, a solution to this problem would provide another proof that Artinian local rings whose maximal ideal is principal are principal ideal rings by shifting from Artinianness to Noetherianness instead of exploiting the nilpotence of the maximal ideal.

I'm tagging this commutative-rings because those are the only ones I really care about, but a noncommutative example would be just as decisive.

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I think projecteuclid.org/… gives an answer. – darij grinberg Sep 24 '10 at 15:13
(Actually it's Proposition 4 and the proof is ridiculously simple - it mainly uses Krull's intersection theorem.) – darij grinberg Sep 24 '10 at 15:14
In the commutative case, yes. I don't actually know what a non-commutative local ring is (that's just my ignorance though). Darij is right, you use the Krull intersection theorem. Let $\pi$ be a generator for the maximal ideal. Since $\bigcap_{n\geq 1}(\pi^n)=0$, for any non-zero, proper ideal $I$, there is a smallest positive integer $k$ with $I\subseteq (\pi^k)$ and $I$ not contained in $(\pi^{k+1})$. Then $I=(\pi^k)$, for if $a\in I$ and $a\notin(\pi^{k+1})$, then $a=\pi^ku$ with $u\notin(\pi)$, i.e. $u$ a unit. Then $(a)=(\pi^k)\subseteq I$. – Keenan Kidwell Sep 24 '10 at 15:19
There are generalizations of this result to the non-commutative case. I recommend the exercises in section 7 of Lam's "First Course in Noncommutative Rings". [Exercise 12 gives a ring which is local, right noetherian and right ideals are principal, but left ideals are not, and it is not left noetherian.] – Pace Nielsen Sep 24 '10 at 16:16
Oops. Well, I'm still a commutative algebra noob. Thanks for the references! – Andrew Homan Sep 24 '10 at 16:43