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.

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.
