Question about proof that $(a + b)^p = a^p+b^p$ in a field of characteristic $p$. Definition charactersitic $p$: a field has characterstic $p$ is its prime field is isomorphic to $\mathbb{Z}/p\mathbb{Z}$.
Proposition: let $K$ be a field of characteristic $p$. Then for any $a, b \in K$ we have $(a + b)^p = a^p + b^p$.
Proof: (from my text) because $p \choose i$ is divisible by $p$ for $0 < 1 < p$ it is zero $\mod p$, so $p \choose i$  $a^{p - i}b^i$ is zero (this is the part I don't understand), so it follows from Newton's binomial law.
I understand that the prime field of $K$ is isomorphic to $\mathbb{Z}/p\mathbb{Z}$, so if $a, b$ are in the prime field, then $p \choose i$  $a^{p - i}b^i$ is zero, but that is not necessarily the case is it?
 A: By your definition, the prime field of $K$ is $\mathbb Z/p\mathbb Z$. Let $1$ be the identity element in $K$, $1_p$ the identity element in $\mathbb Z/p\mathbb Z$ and let
$$\varphi: \mathbb Z/p\mathbb Z\to K$$
be a field immersion.
We know that $\varphi$ is injective, meaning that the counterimage of 0 is only $0_p$. This means that $\varphi(1_p)$ is invertible in $K$ and
$$
\varphi(1_p) = \varphi(1_p \cdot 1_p) = \varphi(1_p)\varphi(1_p) \implies \varphi(1_p) = 1.
$$
Given any $a\in K$, the sum of $p$ copies of $a$ is
$$ a + \dots + a = (1 + \dots + 1)a = (\varphi(1_p) + \dots + \varphi(1_p)) a = \varphi(1_p + \dots + 1_p)a = \varphi(0) a = 0.$$
This means that the sum of $p$ copies of $a$ yields zero. 
Since $\binom pi$ is a multiple of $p$, then if $x\in K$, you have
$$\binom pix = p(..) x = 0 $$ 
A: You're missing the definition of "characteristic $p$". This mean essentially that $p=0$ in $\mathbb{K}$. Well, if $p=0$, then $pk=0$ for any $k\in \mathbb{K}$. Since $p|$${p}\choose{i}$, then ${p}\choose{i}$ $=pm=0$ for some $m\in \mathbb{K}$.
