$\bigcap_{n\in\mathbb{N}} I^n = (0)$ if and only if no zero divisor of $R$ is of the form $1-z$ with $z\in I$.

Question

Full problem, suppose that $R$ is a commutative Noetherian ring and $I$ is an ideal of $R$. We wish to prove that $$\bigcap_{n=1}^{\infty} I^n=(0)$$ if and only if no zerodivisor of $R$ is of the form $1-z$ with $z\in I$.

First I'll suppose that the intersection is $(0)$. Let $z\in I$ and let $0\neq r\in R$ such that $r(1-z)=0$. Then $r=rz$ and so $r\in I$. Is this useful here? I'm not sure how to use the Noetherian condition of $R$ since the chain of $I^n$ is descending, not ascending.

Any help would be much appreciated! I'm studying for a qual and need all the help I can get.

Answer 1

Yes, the observation that $r=rz$ $\implies$ $r\in I$ is very useful, since we can use the equation again to get that $r\in I^2$, and thus $r\in I^3$, and so on. Hence $r\in \bigcap_{n=1}^\infty I^n$, so $r=0$, contradiction. No need to use Noetherianness here. (It may be necessary for the converse, but I haven't thought that far, since it's not clear from your question if you're also asking about that.)

Edit

Worked out my thoughts on the converse. I was being dumb. It's Nakayama's lemma (the general, not local ring version).

Let $$I^\infty = \bigcap_{n=1}^\infty I^n.$$ Observe that clearly $I(I^\infty) = I^\infty$. Then, since $R$ is Noetherian, $I^\infty$ is finitely generated, so Nakayama's lemma (Statement 1) applies.

Thus there exists $r\in R$ with $r-1\in I$ such that $rI^\infty =0$. But then $r-1=i$ for some $i\in I$, and $r=1+i$. Then $r$ is not a zero divisor by assumption, hence the fact that $rI^\infty = 0$ implies that $I^\infty=0$ as desired.

Answer 2

If there exists some y $\in$ I such that (1-y) is a zero divisor, then there exists an x $\not=$ 0 such that x(1-y) = 0. Then x = xy $\in$ I. Thus x $\in$ I $\cap$ I$^2$. It follows that x $\in$ $\displaystyle\bigcap_n$ I$^n$.