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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.

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    $\begingroup$ The characteristic of $R/P$ is necessarily either $2$, $3$ or $5$, since if it's neither $2$ nor $3$, then $2^8-1 = 255 = 0$ and $3^8-1 = 6560 = 0$, and $\gcd(255, 6560) = 5$, which means that $5 = 0$. Don't know whether that helps prove finiteness, though. $\endgroup$
    – Arthur
    Nov 10, 2017 at 8:59
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    $\begingroup$ Hint: Look at the field of fractions of the quotient. It inherits the property that $r^8=1$ for all nonzero $r$ and thus... $\endgroup$ Nov 10, 2017 at 9:13
  • $\begingroup$ @SebastianSchoennenbeck So the field of fractions is finite! Thank you! Can you add that as an answer? $\endgroup$
    – Hyobin Lee
    Nov 10, 2017 at 9:36
  • $\begingroup$ @HyobinLee Sure thing. $\endgroup$ Nov 10, 2017 at 10:12

2 Answers 2

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$r^{10}=r^2$ 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 $K$.

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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.

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    $\begingroup$ In general, any integral domain, in which every element suffices a fixed polynomial equation, is a finite field. The possible fields can be obtained by looking at the factors of the polynomial. $\endgroup$
    – MooS
    Nov 10, 2017 at 10:11
  • $\begingroup$ Thank you for suggesting another point of view. $\endgroup$
    – Hyobin Lee
    Nov 10, 2017 at 10:19

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