Show that an integral domain with the property that every strictly decreasing chain of ideals $I_1 \supset I_2\supset \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 \supset\cdots \supset\langle 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 \supset\langle 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.