Is this quotient ring finite? Let $R$ be a commutative ring with unity satisfying $r^{10}=r^2$  $\forall$$r$$\in$$R$, and $P$ a prime ideal of $R$. The original question is to find the possible orders of the quotient ring $R/P$.
My thoughts: If $R/P$ is finite it is a finite domain which is a field and since $r^8=1$ iff $r\neq0$, isomorphic to the Galois field $GF(9), GF(5), GF(3),$ or $GF(2)$. It remains to show that $R/P$ is finite. Can I use the fact that $P$ is a prime ideal? Or should I somehow make use of the given equation? Thank you in advance.
 A: Let $D=R/P$. Then $r^{10}=r^2$ for all $r \in R$ implies $z^{10}=z^2$ for all $z \in D$ because of the canonical projection $R \to D$.
In a field, a polynomial equation of degree $n$ has at most $n$ roots. Thus, there are at most $10$ solutions for  $z^{10}=z^2$ in the quotient field of $D$. Therefore, $D$ has at most $10$ elements.
A: $r^{10}=r^2$ for all $r \in R$ so the same is true in the quotient $R/P$.
 Since $P$ is prime the quotient $R/P$ is an integral domain and thus has a field of fractions $K$ which inherits the property $x^{10} = x^2$ for all $x \in K$ which implies $x^8=1$ for all $0 \neq x \in K$ (there are no zero divisors).
Since a polynomial equation over a field has at most as many zeros as the degree of the polynomial the field $K$ consequently has at most $9$ elements and $|K^\times| = |K|-1$ divides $8$ since $K^\times $ is a cyclic group.
By simply checking we end up with the list $|K| \in \{2,3,5,9\}$. Finally, $R/P$ is a subring of $K$ whose field of fractions is $K$ so $R/P$ already coincides with $K$.
