Commutative ring and factor by prime ideals Let $R$ be a commutative ring with unity such that for every prime ideal $P$ of $R$ , $R/P$ is a finite ring , so a finite field ; also suppose $R$ is artinian ring ; then can we say that $R$ is finite ?  
 A: Being Artinian, you can factor it into a product of local Artinian rings, and they all have to have finite residue fields. Clearly this allows us to reduce the problem to local Artinian rings, since the product will be finite iff the summands are finite.
Let $R$ be local, Artinian, with maximal ideal $M$, such that $|R/M|=n<\infty$. Actually $R/M$ is the only isotype of simple $R$ module that exists! Since $R$ is Artinian, it has a finite composition series
$$
\{0\}\subseteq I_1\subseteq\ldots\subseteq I_k\subseteq R
$$
Each factor of which is isomorphic to $R/M$. 
Basic counting says that $|R|=[R:I_k][I_k:I_{k-1}]\cdots[I_1:\{0\}]$. There are $k+1$ factors here, each one with finite size $|R/M|$.
So we see then that $|R|=|R/M|^{k+1}<\infty$
For Noetherian rings, the answer is still true, because such a ring is actually still Artinian: if all primes are maximal, the nilradical equals the Jacobson radical and it is nilpotent. Furthermore $R/J(R)$ is semisimple Artinian, and at this point Hopkins-Levitzki says $R$ was Artinian all along.
If you drop Noetherianity, then it is obviously false. You can just take $R=\prod _{i=1}^\infty F_2$.
