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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.