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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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    $\begingroup$ I think projecteuclid.org/… gives an answer. $\endgroup$
    – darij grinberg
    Commented Sep 24, 2010 at 15:13
  • $\begingroup$ (Actually it's Proposition 4 and the proof is ridiculously simple - it mainly uses Krull's intersection theorem.) $\endgroup$
    – darij grinberg
    Commented Sep 24, 2010 at 15:14
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    $\begingroup$ 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 $(a)=(\pi^k)\subseteq I$. $\endgroup$
    – Keenan Kidwell
    Commented Sep 24, 2010 at 15:19
  • $\begingroup$ Oops. Well, I'm still a commutative algebra noob. Thanks for the references! $\endgroup$
    – Andrew Homan
    Commented Sep 24, 2010 at 16:43
  • $\begingroup$ @PaceNielsen: I don't think all right ideals are right principal in Lam's Exercise 12 (you mean Section 19, Chapter 7, I believe). In particular, $\bigcap_{i \geq 0} (rad R)^i \neq 0$, so by my answer this intersection can't be right principal. The exercise asks to prove that all right ideals are two-sided ideals, but says nothing about principal. $\endgroup$
    – Joshua Grochow
    Commented Dec 21, 2018 at 17:52

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Even in the noncommutative case, we can show that a local ring whose maximal ideal $\mathfrak{m}$ is left principal is a left PIR if and only if $\bigcap_{i \geq 0} \mathfrak{m}^i = 0$ (the conclusion of Krull's intersection theorem).

Recall that a noncommutative ring $R$ is local if the non-units form a left ideal $\mathfrak{m}$ (and in this case $\mathfrak{m}$ is in fact a two-sided ideal). Suppose this maximal ideal is left-principal, say $\mathfrak{m} = Rt$.

The "if" direction follows the same proof as in the commutative case.

For the "only if" direction, suppose that $R$ is a left PIR. Let $I = \bigcap_{i \geq 0} \mathfrak{m}^i$. Since $R$ is a PIR, we have $I = Rx$ for some $x$. Similar to the proof of Krull's intersection theorem in the commutative case, show that $tI =I$ (if $z \in I$, then $z = tz'$ for some $z'$; if $z' \notin I$, then $z' \notin \mathfrak{m}^j$ for some $j$, but then $z \notin \mathfrak{m}^{j+1}$, contradicting $z \in I$). Then there is some $y \in I$ such that $x = ty$. Since $y \in I=Rx$, we have $y=rx$ for some $r$, and thus $x = trx$, or $(1-tr)x=0$. Since $tr \in \mathfrak{m}$, $1-tr$ is a unit, and thus $x=0$ and $I = Rx = 0$.

(I don't even think you need Noetherian to prove this; in case $R$ ends up being a left PIR, Noetherian seems to follow.)

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