in generell I want to show that if a Ring $R$ satisfies the descending chaincondition for cyclic ideals, so every chain of cyclic ideals $(r_1)\supset (r_2)\supset \dots$ in $R$ becomes stationary, then every prime ideal is maximal.
If we take a prime ideal $\mathcal{p} \subset R$, then $R/\mathcal{p}$ is an integral domain. For $\mathcal{p}$ to be maximal, $R/\mathcal{p}$ has to be a field.
Now my question: Can we show that $R/\mathcal{p}$ satisfies the descending chaincondition for cyclic ideals, if $R$ does?
In this case it would be rather easy to show that every element in $R/\mathcal{p}$ has an inverse and therefore $R/\mathcal{p}$ would be a field.
If that's not the case, then what approach could I try instead?
Thanks in advance!
Rather than worrying about the chain condition in the quotient, one could just note that for all $a\in R$, there exist positive integers $m,n$ with $m>n$ and an element $x\in R$ such that $a^n=a^{m}x$. This is a simple consequence of applying the descending chain condition to the chain $aR\supseteq a^2R\supseteq a^3R\supseteq\ldots$. (*)
Because you can do this in $R$, you can do it in any quotient of $R$.
Of course, in the quotients that are domains, $a$ is zero or cancels down to $1=a^{m-n}x$, proving $a$ is a unit.
(*) Rings satisfying this condition for all elements are called strongly $\pi$-regular. Rings satisfying the descending chain condition on principal right ideals are called left perfect rings. (Yes, I do mean left. It is a very curious crossover characterization.)