Since $ \scriptstyle Z_p $ is a field, and $ \scriptstyle f(x) $ is irreducible over $ \scriptstyle Z_p $, by Corollary 1 of Theorem 17.5, $ \scriptstyle Z_p[x]/\langle f(x)\rangle $ is a field. In particular,
$ \scriptstyle Z_p[x]/\langle f(x)\rangle\ =\ \textstyle\{\scriptstyle a_{n-1}x^{n-1}+a_{n-2}x^{n-2}+\ldots+a_0+\langle f(x)\rangle\ \textstyle\mid\scriptstyle\ a_{n-1},\ldots,a_0\in Z_p\textstyle\} $.
$ \scriptstyle\mid Z_p\mid\ =\ p $ and $ \scriptstyle\mid\{a_{n-1},a_{n-2},\ldots,a_0\}\mid\ =\ n $, so there are $ \scriptstyle p^n $ permutations of $ \scriptstyle a_{n-1},\ldots,a_0 $, and therefore, $ \scriptstyle p^n $ elements in $ \scriptstyle Z_p[x]/\langle f(x)\rangle $.
- --Nick Rupley 04:04, 8 April 2009 (UTC)
