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Analogous to the ascending chain condition we can define a descending chain condition: if $I_1 \supseteq I_2 \supseteq I_3 \supseteq \cdots$ is a descending chain of ideals then there exists a positive integer $N$ where $I_N = I_{N+1} = I_{N+2} = \cdots$. Show that it is not necessarily true that a Noetherian ring has the descending chain condition.

Proof:

Let $Z$ be a ring under multiplication and addition. We know that $Z$ is commutative under multiplication since we know standard multiplication to be commutative. So we have a commutative ring then clearly $Z$ is Noetherian. Let ideal $I_1=aZ$ contain ideal $I_2=bZ$, where $a|b$. Also, $I_2$ to contain third ideal, $I_3=cZ$, where $b|c$. And, $I_3$ contain fourth ideal, $I_4=dZ$, where $c|d$ and so on...

If we continue this pattern, each successive ideal must have a generator that is a multiple of the generator of the ideal immediately previous in the chain. Since the integers are infinite, we can keep constructing multiples forever and our chain will never end. Therefore, the Noetherian ring fails the descending chain condition. $\square$ (checking if my proof works)

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  • $\begingroup$ Did you mean for “$Z$” (or “$\Z$”) to denote the ring of integers $ℤ$? (If so you shouldn’t write “Let $Z$ be a ring …”, but “Let $ℤ$ be the ring of integers.”) Maybe you should get more specific and really use integers like $2, 4, …$ for your proof. $\endgroup$
    – k.stm
    Dec 12, 2014 at 7:47
  • $\begingroup$ I did mean Z to be the ring of integers $\endgroup$
    – user188222
    Dec 12, 2014 at 8:14

1 Answer 1

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Consider $\mathbb Z$ which is a PID, hence Noetherian. Notice that $$(2) \supsetneq (4) \supsetneq (8) \supsetneq (16) \supsetneq \cdots$$ is not stationary.

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