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.
 A: 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
$$ (x_1 + \cdots + x_n)^p = \sum_{\nu_1 + \cdots + \nu_n = p} \frac{p!}{\nu_1! \cdots \nu_n!} x_1^{\nu_1} \cdots x_n^{\nu_n} \equiv x_1^p + \cdots + x_n^p \pmod{pR}. $$
A: HINT:
When $n=2$ the question reduces immediately to checking that the binomial coefficients $\binom pk$ are multiples of $p$ for $k\in\{1,...,p-1\}$.
Then use induction on $n$.
A: This follows by induction if we know it for the case $n=2$.
But the binomial theorem is true in any commutative ring, so:
$$(x_1+x_2)^p = \sum_{k=0}^p \binom{p}{k}x_1^kx_2^{p-k} = x_1^p + x_2^p + \sum_{k=1}^{p-1}\binom p k x_1^kx_2^{p-k}$$
Now all you have to prove that for $1\leq k \leq p-1$, $\binom{p}{k}$ is divisible by $p$. This is fairly easy.
