Congruence in rings Let $R$ be a commutative (and probably unitary, if you like) ring and $p$ a prime number. If $x_1,\ldots,x_n\in R$ are elements of $R$, then we have $(x_1+\cdots+x_n)^p\equiv x_1^p+\cdots+x_n^p$ mod $pR$. Why is this true? I tried to show that in $R/pR$ their congruence classes are equal, but without sucess.

Just compute ;-) ... we have - as $R$ is commutative - by the multinomial theorem 
$$ (x_1 + \cdots + x_n)^p = \sum_{\nu_1 + \cdots + \nu_n = p} \frac{p!}{\nu_1! \cdots \nu_n!} x_1^{\nu_1} \cdots x_n^{\nu_n} $$
If all $\nu_i <p $, the denominator contains no factor $p$ (as $p$ is prime), hence $\frac{p!}{\nu_1! \cdots \nu_n!} \equiv 0 \pmod p$, that is the only terms which survice reduction mod $pR$ are those where one $\nu_i = p$, hence the others are $0$, so
$$ (x_1 + \cdots + x_n)^p = \sum_{\nu_1 + \cdots + \nu_n = p} \frac{p!}{\nu_1! \cdots \nu_n!} x_1^{\nu_1} \cdots x_n^{\nu_n} \equiv x_1^p + \cdots + x_n^p \pmod{pR}. $$
