Why is $(a+b)^p = a^p+b^p$, where $a,b \in R$, a commutative ring with prime characteristic $p$?

Question

Here is the answer from lecture notes. $(a+b)^p = \sum {{p}\choose{k}}a^kb^{p-k}$ and all the terms divide $p$ except $a^p$ and $b^p$ terms. So reducing (mod p) all terms are zero except the ones above.

So I don't understand how by reducing (mod p) makes those terms $0$? By definition of prime characteristic, $p.1=0$ where $1$ and $0$ are the multiplicative and additive identities. So I am not able to make the connection between the definition and the statement above. Thanks and appreciate a hint.

Answer 1

$$ {{p}\choose{k}} = \frac{p\cdot(p-1)\cdot\ldots\cdot{(p-k+1)}}{k!} $$ As $p$ is prime and $0 < k < p$, $p$ doesn't divide $k!$, so this expression is a multiple of $p$ and thus it's zero in characteristic $p$.

Therefore, the sum $$ \sum_{k=0}^p {{p}\choose{k}}a^kb^{p-k} = a^p + b^p + \sum_{k=1}^{p-1} {{p}\choose{k}}a^kb^{p-k} = a^p + b^p + \sum_{k=1}^{p-1}0\cdot a^kb^{p-k} = a^p+b^p $$

Answer 2

Because for all $k$ with $0<k<p$ the binomial coefficient $\binom{p}{k}=\frac{p!}{k!(p-k)!}$ is a multiple of $p$, because the numerator divisible by $p$, but the denominator isn't.