Proving the Frobenius map is an endomorphism 
I have prime $p$, and $K$ a field such that $p \cdot 1 = 1+1+\cdots+1 = 0$. I am asked to prove that $F: K \rightarrow K$,  $a \mapsto a^p$ is a ring homomorphism. 

I can prove this for everything except addition, $F(a+b)=F(a)+F(b)$. For addition, it seems to me that $F(a+b)=(a+b)^p=(a+b)^0=1$ but $F(a)+F(b)=a^p+b^p=a^0+b^0=1+1=2$. Can someone point out where I'm going wrong?
 A: If we multiply by $p$ we get zero (this is known as "the ring $K$ has characteristic $p$"). But if we raise to the $p$-th power we may not necessarily get $1$. Then you have to convince yourself that $F(a+b)=(a+b)^p=a^p+b^p$. If you write out $(a+b)^p$ for specific primes $p$ you will notice that all terms but $a^p$ and $b^p$ are multiples of $p$ - thus zero.
For instance for $p=2$ we have
$$
(a+b)^2=a^2+2ab+b^2
$$
where $2ab=0$ in characteristic $2$. Similarly for $p=3$ we get
$$
(a+b)^3=a^3+3a^2b+3ab^2+b^3
$$
where $3a^2b+3ab^2$ vanishes since they are multiples of $3$ hence equal to zero in characteristic $3$.
A: What you're trying to prove is that if the characteristic of the field is $p$, then $(a+b)^p = a^b + b^p$. The binomial coefficients give an expansion of the left hand side readily as
$$
(a+b)^p = \sum_{k = 0}^p \binom{p}{k} a^kb^{p-k} = \sum_{k = 0}^p \frac{p!}{k!(p-k)!} a^kb^{p-k}
$$
Notice that the first term in that sum is $(p!/0!p!) b^p = b^p$ and the last one is $(p!/p!0!) a^p = a^p$, so what you're really trying to prove is that if $1 \leq k \leq p-1$, then $\frac{p!}{k!(p-k)!} = 0$ (in your field of characteristic $p$). This is equivalent to saying that $p$ divides $\frac{p!}{k!(p-k)!}$, which is the same as saying that $\frac{p!}{k!(p-k)!}$ has at least one power of $p$ in its prime factorization. Clearly it has one: the one that arises from the $p$ in the top term. Can that cancel out with anything in the bottom?
