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Show that an integral domain with the property that every strictly decreasing chain of ideals $I_1 supset I_2supset cdots $ must be finite in length is a field.

Attempt: We need to show that every element in the integral domain $D$ is invertible. Let us suppose $exists~ a in D$ such that $a$ is not invertible.

Clearly, $ langle a rangle supset langle a^2 rangle supsetcdots supsetlangle a^n rangle $ where $langle a^{n } rangle $ is the last ideal in the chain.

Now, $langle a^{n+1} rangle $ should be equal to one of the $langle a^i rangle ;~ 1 leq i leq n$

But $langle a^{n+1} rangle $ also satisfies $langle a^n rangle supseteq langle a^{n+1} rangle $ and $langle a^i rangle supsetlangle a^n rangle $

$implies langle a^j rangle =langle a^n rangle ~~forall~~j geq n$

Now, since $a$ is not a unit, $langle a rangle $ does not contain the unity and hence, $langle a rangle neq D$

How do I move forward and have I interpreted the problem correctly?

Thank you for your help..

If $a$ is not a unit, then $langle a^{n+1} rangle ne langle a^n rangle$ because if $a^n in langle a^{n+1} rangle$, i.e., $a^n = r a^{n+1}$ for some $r in D$, then $a^n(1 - ra) = 0$, and that implies $1 = ra$ (because $D$ is a domain), contradicting the assumption that $a$ is not a unit.

Hint: $langle a^nrangle=langle a^{n+1}rangleimplies a^{n+1}inlangle a^nrangleimplies ??$