[proofplan]
We verify the ring-homomorphism identities directly. Multiplication and preservation of the unit follow from associativity and the unit law in the field. Additivity follows by expanding $(a+b)^p$ with the [binomial theorem](/theorems/750) and using that each intermediate [binomial coefficient](/page/Binomial%20Coefficient) is divisible by $p$, hence acts as zero in a field of characteristic $p$. Finally, injectivity follows because a nonzero element of a field has a multiplicative inverse, so its $p$-th power cannot be zero.
[/proofplan]
[step:Define the Frobenius map and verify the multiplicative identities]
Define
\begin{align*}
\operatorname{Frob}: k \to k,\qquad a \mapsto a^p.
\end{align*}
For $a,b\in k$, associativity and commutativity of multiplication in the field $k$ give
\begin{align*}
\operatorname{Frob}(ab)=(ab)^p=a^p b^p=\operatorname{Frob}(a)\operatorname{Frob}(b).
\end{align*}
Also,
\begin{align*}
\operatorname{Frob}(1_k)=1_k^p=1_k.
\end{align*}
Thus $\operatorname{Frob}$ preserves multiplication and the multiplicative identity.
[/step]
[step:Use the binomial theorem and characteristic $p$ to prove additivity]
Let $a,b\in k$. For each integer $j$ with $1\leq j\leq p-1$, the prime number $p$ divides the binomial coefficient $\binom{p}{j}$. Indeed,
\begin{align*}
\binom{p}{j}=\frac{p!}{j!(p-j)!}.
\end{align*}
Since $1\leq j\leq p-1$, neither $j!$ nor $(p-j)!$ is divisible by $p$, so the single factor $p$ in $p!$ remains in the integer $\binom{p}{j}$.
For an integer $m\geq 0$ and an element $x\in k$, write $m\cdot x$ for the sum of $m$ copies of $x$, with $0\cdot x=0_k$. Since $\operatorname{char}(k)=p$, we have $p\cdot 1_k=0_k$. Hence if $p$ divides $m$, then $m\cdot 1_k=0_k$, and therefore $m\cdot x=(m\cdot 1_k)x=0_k$ for every $x\in k$.
Applying the binomial theorem for commutative rings in the commutative ring $k$, we obtain
\begin{align*}
(a+b)^p=\sum_{j=0}^{p}\binom{p}{j}a^j b^{p-j}.
\end{align*}
The terms with $1\leq j\leq p-1$ vanish because $\binom{p}{j}$ is divisible by $p$. Therefore
\begin{align*}
(a+b)^p=a^p+b^p.
\end{align*}
Thus
\begin{align*}
\operatorname{Frob}(a+b)=\operatorname{Frob}(a)+\operatorname{Frob}(b).
\end{align*}
[guided]
We need to prove that $\operatorname{Frob}$ preserves addition, so we must compare $\operatorname{Frob}(a+b)=(a+b)^p$ with $\operatorname{Frob}(a)+\operatorname{Frob}(b)=a^p+b^p$. The tool that expands $(a+b)^p$ is the binomial theorem, and the characteristic hypothesis is exactly what kills the middle terms.
Let $a,b\in k$. For each integer $j$ with $1\leq j\leq p-1$, the binomial coefficient $\binom{p}{j}$ is divisible by $p$. To see this, use the formula
\begin{align*}
\binom{p}{j}=\frac{p!}{j!(p-j)!}.
\end{align*}
Because $p$ is prime and $1\leq j\leq p-1$, none of the factors in $j!$ or $(p-j)!$ is divisible by $p$. Thus the factor $p$ appearing in $p!$ is not cancelled in the integer binomial coefficient, so $p\mid \binom{p}{j}$.
Now define the notation $m\cdot x$ for an integer $m\geq 0$ and an element $x\in k$ to mean the sum of $m$ copies of $x$, with $0\cdot x=0_k$. The condition $\operatorname{char}(k)=p$ means precisely that $p\cdot 1_k=0_k$. Therefore, whenever $m$ is divisible by $p$, we have $m\cdot 1_k=0_k$. Multiplying by any $x\in k$ gives
\begin{align*}
m\cdot x=(m\cdot 1_k)x=0_kx=0_k.
\end{align*}
This applies to $m=\binom{p}{j}$ for every $1\leq j\leq p-1$.
Since $k$ is a field, its multiplication is commutative, so the binomial theorem for commutative rings applies in $k$:
\begin{align*}
(a+b)^p=\sum_{j=0}^{p}\binom{p}{j}a^j b^{p-j}.
\end{align*}
The two endpoint terms are $b^p$ when $j=0$ and $a^p$ when $j=p$. Every intermediate term has coefficient divisible by $p$, hence is zero in $k$. Therefore
\begin{align*}
(a+b)^p=a^p+b^p.
\end{align*}
By the definition of $\operatorname{Frob}$, this says
\begin{align*}
\operatorname{Frob}(a+b)=\operatorname{Frob}(a)+\operatorname{Frob}(b).
\end{align*}
Thus $\operatorname{Frob}$ preserves addition.
[/guided]
[/step]
[step:Prove injectivity from the existence of inverses in a field]
Let $a\in k$ and suppose $\operatorname{Frob}(a)=0_k$. Then
\begin{align*}
a^p=0_k.
\end{align*}
If $a\neq 0_k$, then $a$ has an inverse $a^{-1}\in k$. Multiplying $a^p=0_k$ by $(a^{-1})^p$ gives
\begin{align*}
(a^{-1})^p a^p=0_k.
\end{align*}
By associativity and commutativity,
\begin{align*}
(a^{-1})^p a^p=(a^{-1}a)^p=1_k^p=1_k.
\end{align*}
Hence $1_k=0_k$, contradicting that $k$ is a field. Therefore every $a\in k$ satisfying $\operatorname{Frob}(a)=0_k$ must equal $0_k$. If $a,b\in k$ satisfy $\operatorname{Frob}(a)=\operatorname{Frob}(b)$, then additivity gives $\operatorname{Frob}(a-b)=0_k$, so $a-b=0_k$ and hence $a=b$. Thus $\operatorname{Frob}$ is injective.
[/step]
[step:Combine the identities to obtain the claimed homomorphism]
The preceding steps show that $\operatorname{Frob}$ preserves addition, multiplication, and the multiplicative identity $1_k$. Hence $\operatorname{Frob}:k\to k$ is a unital ring homomorphism. The preceding injectivity argument shows that it is injective. This proves the theorem.
[/step]
