# 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 \$\displaystyle \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 \$\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.
• 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 \$\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......