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Let $I$ be an ideal in a Noetherian ring. Show that either $I$ contains an $R$-regular element or else $aI=0$ for some $0\neq a\in R$.

How would I prove this? Also what does $aI=0$ mean?

Username Unknown
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  • $aI$ is an ideal, and $0$ is an ideal. So $aI =0$ is just equality as ideals/sets. In other words, $ai = 0$ for every $i \in I$. You should define what an $R$-regular element is, though. – Sarvesh Ravichandran Iyer Oct 31 '18 at 04:28
  • @астон вілла олоф мэллбэрг: "regular element" means "non-zero-divisor".$;$https://en.wikipedia.org/wiki/Regular_element – quasi Oct 31 '18 at 07:36
  • @Username Unknown: I don't think the claim is true. For a counterexample, see exercises $6$ and $7$ on pages $62$-$63$ of the text:$;$Kaplansky -- Commutative Rings, 2nd Ed (1974). – quasi Oct 31 '18 at 07:40
  • @quasi I see. I asked in case this could be proven by definition or so. – Sarvesh Ravichandran Iyer Oct 31 '18 at 09:31
  • @quasi Kaplansky wanted to provide a counter-example for non-Noetherian rings. For Noetherian rings this holds for the set of zerodivisors is the union of associated primes. – user26857 Oct 31 '18 at 19:18
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    @user26857: Yes, I see, thanks. In the construction for Kaplansky's exercise 7, while the ideal in question is finitely generated, the contaning ring is non-Noetherian. – quasi Oct 31 '18 at 23:39

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Yes this is an important property of Noetherian rings. It is Theorem 82 in Kaplansky's Commutative Rings, which he prefaces as "a result that is among the most useful in the theory of commutative rings."

In the literature you often encounter this property as Property (A)

A ring is said to have Property (A) if every finitely generated dense ideal contains a regular element. (Or equivalently, if every f.g. ideal consisting entirely of zero divisors has a nonzero annihilator.)

Kaplansky shows that Noetherian rings have Property (A).

It also comes up in the study of the integral closure of $R[x]$ and the total ring of fractions of a reduced ring.

user26857
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Badam Baplan
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  • So to prove this statement I just need to say that it comes from thm 80 and 81 in Kaplansky? – Username Unknown Oct 31 '18 at 13:33
  • If you say it comes from thm 80 and 81 in Kaplansky, I don't think you've proved the statement so much as acknowledged that it has been proved by somebody else. I wonder, in what context did you encounter the problem, and who is the proof for? – Badam Baplan Oct 31 '18 at 13:40