# Thread: Decending chain condition and Integral Domain

1. ## Decending chain condition and Integral Domain

Does every Integral Domain satisfy Decending chain condition? Or there should be a stronger restriction?

2. Originally Posted by ynj
Does every Integral Domain satisfy Decending chain condition? Or there should be a stronger restriction?
you mean descending chain conditions on its ideals?

3. yes

4. Originally Posted by ynj
yes
the answer is no and an example is $\displaystyle \mathbb{Z}: \ <2> \supset <4> \supset <8> \supset \cdots$

5. well..then something wrong with my text..
Can any one give a sufficiency condition about DCC on ideals??

6. actually an integral domain satisfies the descending chain condition if and only if it's a field. the proof is very easy! one side is clear. now suppose an integral domain D satisfies the descending

chain condition and let $\displaystyle x \neq 0$ be in D. consider the chain $\displaystyle <x> \supseteq <x^2> \supseteq <x^3> \supseteq \cdots .$ so there exists an n > 0 such that $\displaystyle <x^n>=<x^{n+1}>$ and hence $\displaystyle x^n=dx^{n+1},$ for some d in D. but then

$\displaystyle x^n(dx -1)=0$ and thus $\displaystyle dx=1$ because $\displaystyle x \neq 0$ and D is a domain. so every non-zero element of D is invertible, i.e. D is a field.

7. Originally Posted by NonCommAlg
actually an integral domain satisfies the descending chain condition if and only if it's a field. the proof is very easy! one side is clear. now suppose an integral domain D satisfies the descending

chain condition and let $\displaystyle x \neq 0$ be in D. consider the chain $\displaystyle <x> \supseteq <x^2> \supseteq <x^3> \supseteq \cdots .$ so there exists an n > 0 such that $\displaystyle <x^n>=<x^{n+1}>$ and hence $\displaystyle x^n=dx^{n+1},$ for some d in D. but then

$\displaystyle x^n(dx -1)=0$ and thus $\displaystyle dx=1$ because $\displaystyle x \neq 0$ and D is a domain. so every non-zero element of D is invertible, i.e. D is a field.
emm..I think so...but every field have no nontrivial ideal...so I think this problem may not be discussed in integral domain and then become complex......
Thank you for your solution

8. Originally Posted by ynj
emm..I think so...but every field have no nontrivial ideal...so I think this problem may not be discussed in integral domain and then become complex......
Thank you for your solution
descending chain condition is for all ideals, proper or not proper. so, you can say this: an integral domain satisfies the descending chain condition if and only if it's a field.

by the way rings satisfying the descending chain condition (on their ideals) have a name. they're called Artinian.