$R$ be an infinite commutative ring with unity such that for every non-zero ideal $I$ , $R/I$ is finite ; then is $R$ a PID or at least Noetherian? Let $R$ be an infinite commutative ring with unity such that for every non-zero ideal $I$ of $R$ , $R/I$ is finite; then is $R$ a PID or at least Noetherian ? 
I can only prove that $R$ must be an integral domain.
 A: It is noetherian. Let $(\,\{0\}\subsetneq\,)\, I_1\subsetneq I_2\subsetneq I_3\subsetneq\cdots$ an ascending chain of ideals.
Then, by the correspondence between ideals containing $I_1$ and ideals in $\dfrac{R}{I_1}$, $$\dfrac{I_2}{I_1}\subsetneq \dfrac{I_3}{I_1}\subsetneq\cdots$$ must be an infinite ascending chain in $\dfrac{R}{I_1}$. Which is absurd, because it is a finite ring.
A: Other answers already show that $R$ is noetherian. However, $R$ does not have to be a PID. If $R$ is a ring of algebraic integers, then $R/I$ is finite for all nonzero ideals $I$. However, $R$ is a PID if and only if its class group is trivial. Thus, any ring of integers with non-trivial class group, like $\mathbb Z[\sqrt{-5}]$, is a counterexample. 
A: Noetherian:
Note that if $I_i\subset I_{i+1}$ then $R/I_i \supset R/I_{i+1}$, where all relations are strict.
Since $R/I$ is always finite and $R/I_{i+1}$ must always take something away from $R/I_i$, eventually for any increasing chain of ideals $R/I_i$ must become the zero ring. This means the chain terminates.
