Commutative, infinite, Noetherian ring with no non-trivial nilpotent and no non-trivial idempotent Let $R$ be a commutative, infinite reduced (no non-zero nilpotent) Noetherian ring with no non-trivial idempotent (so that the prime spectrum is connected under Zariski topology). Then is it true that $R/P$ is infinite for every associated prime ideal $P$ of $R$ ? 
I can show that if $I$ is a prime ideal of a ring $R$ as above such that $R/I$ is finite, then $I$ is not a minimal prime ideal , hence there are prime ideals properly contained in $I$ and for any such prime ideal $J \subsetneq I$, we have $R/J$ is a non-field domain hence $R/J$ is infinite. I don't know whether this is relevant here or not. 
NOTE : By associated prime ideal of $R$, I mean a prime ideal of the form $Ann (r) =\{ x \in R: rx=0\}$ for some $r \in R$ (as in the Primary decomposition theory )
Please help 
 A: Yes, $R/P$ is infinite for all prime ideals of the form $P=Ann(r)$ for some $r\in R$.
Let $P = Ann(r)$ be a prime ideal for a non-zero element $r$.  Assume that $R/P$ is finite.  Then it is a field, since all finite integral domains are fields.  Moreover, its cardinal is a certain power $q$ of a prime number $p$, since all finite fields have this property.
Now, $R$ is reduced, so $r$ does not belong to $P$.  Therefore its image $\bar r$ in the quotient $R/P$ is non-zero, so that $\bar r^{q-1} = \bar 1$. In other words, $r^{q-1} -1 \in P$.  By definition of $P=Ann(r)$, this implies that $(r^{q-1} -1)r =0$, so $r^{q} = r$.
We can deduce from this that $r^{q-1}$ is an idempotent:
$$
 (r^{q-1})^2 = r^{2q-2} = r^q \cdot r^{q-2} = r\cdot r^{q-2} = r^{q-1}.
$$
Since $R$ contains no non-trivial idempotents, $r^{q-1}$ is either $1$ or $0$.  Since $R$ is reduced and $r$ is non-zero, no power of $r$ can be zero, so $r^{q-1} = 1$.  But then $r$ is invertible; thus $P = Ann(r) = \{0\}$, so that $R/P$ and $R$ are isomorphic.  Since the former is finite and the latter is infinite, we have a contradiction.
