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

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.

• Maybe take a look at Krulls intersection theorem? – red_trumpet Dec 7 '18 at 18:38
• That assumes that R is local, though @red_trumpet – Gengar Dec 7 '18 at 18:47
• Right, I missed that :( – red_trumpet Dec 7 '18 at 19:37

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.

• Do you have any ideas on the other direction? Thank you so much! – Gengar Dec 7 '18 at 18:41
• @Gengar, for the converse red_trumpet's suggestion to use Krull's intersection theorem looks viable. I'll flesh out the details in my answer. – jgon Dec 7 '18 at 18:47
• I appreciate it so much @jgon – Gengar Dec 7 '18 at 18:47
• @Gengar actually I have to go, and I can't complete my thoughts rn, I'll post what I have. – jgon Dec 7 '18 at 19:09
• @Gengar, got it, I was being silly. – jgon Dec 8 '18 at 0:30

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$$^n$$.

• Yeah, jgon already said that – Gengar Dec 7 '18 at 19:53