11
$\begingroup$

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.

$\endgroup$
4
  • 3
    $\begingroup$ 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 '17 at 8:59
  • 4
    $\begingroup$ 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 '17 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 '17 at 9:36
  • $\begingroup$ @HyobinLee Sure thing. $\endgroup$ Nov 10 '17 at 10:12
11
$\begingroup$

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

$\endgroup$
2
$\begingroup$

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.

$\endgroup$
2
  • 1
    $\begingroup$ 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 '17 at 10:11
  • $\begingroup$ Thank you for suggesting another point of view. $\endgroup$
    – Hyobin Lee
    Nov 10 '17 at 10:19

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.