I'm reading Neukirch's Algebraic Number Theory and trying to do the exercises. I think I may have found another error, but am not sure...

Exercise 7. In a noetherian ring $R$ in which every prime ideal is maximal, each descending chain of ideals $a_1 \supseteq a_2 \supseteq \ldots$ becomes stationary

Hint: Show as in (3.4) that (0) is a product $p_1 \ldots p_r$ of prime ideals and that the chain $R \supseteq p_1 \supseteq p_1p_2 \supseteq \ldots \supseteq p_1\ldots p_r = (0)$ can be refined into a composition series.

So, my main issue is that the ring $\mathbb{Z}$ is noetherian and every prime ideal is maximal, yet the chain $$(2) \supsetneq (4) \supsetneq (8) \supsetneq \ldots \supsetneq (2^n) \supsetneq \ldots$$ Never becomes stationary.

Additionally, I don't know what (correct) statement he could be asking me to prove. (I also realize there may well be a translational error.) So am I reading it wrong or what?


The (0) ideal in $\mathbb{Z}$ is prime, but not maximal. So $\mathbb{Z}$ does not satisfy the hypotheses.

  • $\begingroup$ That makes so much more sense. I'm just so used to seeing "Every nonzero prime ideal..." that I completely missed the subtlety. Thanks! $\endgroup$ Jun 13 '13 at 6:55
  • $\begingroup$ I'll accept in a few minutes when it lets me. $\endgroup$ Jun 13 '13 at 6:56
  • $\begingroup$ By the way, his hint is still not quite right anyways: in a typical Artin ring, the zero ideal is not a product of primes - that's just the nilradical! (The hint isn't useless, but it is false as stated.) $\endgroup$ Jul 26 '14 at 14:14
  • $\begingroup$ The hint is right. In a Noetherian ring, the nilradical is nilpotent. So if the nilradical is a product of primes, so is the zero ideal. $\endgroup$ Feb 10 '15 at 11:40

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