Decending chain condition and Integral Domain

• Aug 18th 2009, 12:15 AM
ynj
Decending chain condition and Integral Domain
Does every Integral Domain satisfy Decending chain condition? Or there should be a stronger restriction?
• Aug 18th 2009, 12:19 AM
NonCommAlg
Quote:

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?
• Aug 18th 2009, 12:25 AM
ynj
yes
• Aug 18th 2009, 12:26 AM
NonCommAlg
Quote:

Originally Posted by ynj
yes

the answer is no and an example is $\mathbb{Z}: \ <2> \supset <4> \supset <8> \supset \cdots$
• Aug 18th 2009, 12:29 AM
ynj
well..then something wrong with my text..
Can any one give a sufficiency condition about DCC on ideals??
• Aug 18th 2009, 12:36 AM
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 $x \neq 0$ be in D. consider the chain $ \supseteq \supseteq \supseteq \cdots .$ so there exists an n > 0 such that $=$ and hence $x^n=dx^{n+1},$ for some d in D. but then

$x^n(dx -1)=0$ and thus $dx=1$ because $x \neq 0$ and D is a domain. so every non-zero element of D is invertible, i.e. D is a field.
• Aug 18th 2009, 12:41 AM
ynj
Quote:

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

$x^n(dx -1)=0$ and thus $dx=1$ because $x \neq 0$ 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......