Freshman's dream: $(x+y)^p = x^p+y^p$ in a commutative ring of characteristic $p$ a prime.

Question

It is easy to prove that $p\mid \binom pk$ for $k=1,2,\ldots,p-1$, so that for $x,y\in R$ a commutative ring of characteristic $p$ a prime, $$(x+y)^p = \sum_{k=0}^p \binom pk x^ky^{p-k} = x^p + y^p. $$ But is there reason a priori that the $\binom pk$ should be multiples of $p$?

Answer 1

I am just rewriting/rephasing TonyK's comments. By definition, for $k>1$we have that $$\binom pk = \dfrac{p!}{k!(p-k)!} = \frac{1.2.3...p}{(1.2.3...k)(1.2.3...(p-k))}.$$ Any factor at the bottom is clearly less than $p$ and since $p$ is a prime none of the factors of the bottom will divide $p$. Therefore, $\binom pk$ must be a multiple of $p$. I think the whole story is simply the consequence of the primeness of $p$. I convince my self by the following identity: $$ \binom pk. k!= ^pP_k .$$ The right side is clearly a multiple of $p$, and because of the primeness of $p$, we cannot cancel this $p$ guy. I know there is nothing important here. Just rephasing the same thing. So, this should not be a good answer to your question. Finally, I don't see any interesting reason that forces $\binom nk$ to be a multiple of $p$ besides the primeness of $p$.