Intersection of powers of the maximal ideal

Question

There certainly exist commutative rings $R$ with unity containing a maximal ideal $m \subsetneq R$ such that $$\bigcap_{k=1}^\infty m^k \neq \{0\}.$$ (Thanks to Dietrich Burde, rschwieb and Lord Shark the Unknown.)

Question: What happens if we further assume that

(i) $R$ is an integral domain,

$\rightarrow$ (EDIT3:) No, see here: Can an ideal in a commutative integral domain be its own square?.

or that

(ii) the quotient $R/m$ is finite?

$\rightarrow$ (EDIT2:) No, take $R = \mathbb{C} \times \mathbb{F}_2$ and $m = \mathbb{C} \times \{ 0 \}$.

Or more general, is there a way to describe those rings $R$ such that $\bigcap_{k=1}^\infty m^k = \{0\}$ holds?

$F_2\times F_2$ where $F_2$ is the field of two elements is an example. I think I also gave two more relevant examples of local rings with idempotent maximal ideals here — rschwieb, May 17 '17 at 19:41

score 3 · Answer 1 · answered May 17 '17 at 19:29

3

How about the ring of continuous functions from $\Bbb R$ to $\Bbb R$ and let $m$ be the functions vanishing at $0$? Then $f(x)=|x|$ is a $k$-th power of a function in $m$ for any $k$.

answered May 17 '17 at 19:29

Angina Seng

158,341

Intersection of powers of the maximal ideal

1 Answers1