Let a field $F$ with $\operatorname{char}(F)=p>0$. Let a map $f:F\to F$ defined be $x\mapsto x^p$. Show that $f$ is a ring homomorphism.
Obviousely $0\mapsto 0$ and $(xy)^p=x^py^p$. How can I show $(x+y)^p=x^p+y^p$? I need to show here for any $0<i<p$ $0=x^iy^{p-i}$.
The key here is that the binomial theorem holds in any commutative ring with unit, though some of the terms may drop out in the event the ring is of finite, prime characteristic. To wit:
$(x + y)^2 = (x + y)(x + y)$ $= x(x + y) + y(x + y) = x^2 + xy + yx + y^2 = x^2 + xy + xy + y^2 = x^2 + 2xy + y^2; \tag 1$
$(x + y)^3 = (x + y)(x + y)^2 = (x + y)(x^2 + 2xy + y^2) = x(x^2 + 2xy + y^2) + y(x^2 + 2xy + y^2)$ $= x^3 + 2x^2y + xy^2 + yx^2 + 2xy^2 + y^3 = x^3 + 3x^2y + 3xy^2 + y^3; \tag 2$
$(x + y)^4 = (x + y)(x + y)^3 = (x + y)(x^3 + 3x^2y + 3xy^2 + y^3)$ $= x(x^3 + 3x^2y + 3xy^2 + y^3) + y(x^3 + 3x^2y + 3xy^2 + y^3)$ $= x^4 + 3x^3y + 3x^2y^2 + xy^3 + yx^3 + 3x^2y^2 + 3xy^3 + y^4$ $= x^4 + 4x^3y + 6x^2y^2 + 4xy^3 + y^4; \tag 3$
$(x + y)^5 = (x + y)(x + y)^4 = (x + y)(x^4 + 4x^3y + 6x^2y^2 + 4xy^3 + y^4)$ $= x( x^4 + 4x^3y + 6x^2y^2 + 4xy^3 + y^4) + y(x^4 + 4x^3y + 6x^2y^2 + 4xy^3 + y^4)$ $= x^5 + 4x^4y + 6x^3y^2 + 4x^2y^3 + xy^4 + yx^4 + 4x^3y^2 + 6x^2y^3 + 4xy^4 + y^5$ $= x^5 + 5x^4y + 10x^3y^2 + 10x^2y^3 + 5xy^4 + y^5; \tag 4$
careful scrutiny of (1)-(4) reveals that for every case of prime exponent, viz. $n = 2, 3, 5$, $n$ divides every coefficient occurring in the binomial expansion save those of $x^n$ and $y^n$, which are themselves equal to $1$; this suggests we examine the binomial coefficents
$\begin{pmatrix} p \\ k \end{pmatrix} = \dfrac{p!}{k!(p - k)!}, \; 0 \le k \le p, \; \text{a prime}, \tag 5$
since, as is well-known for any $n \in \Bbb N$ we have
$(x + y)^n = \displaystyle \sum_0^n\begin{pmatrix} n \\ k \end{pmatrix} x^{n - k}y^k = \displaystyle \sum_0^n \dfrac{n!}{k!(n - k)!}x^{n - k}y^k; \tag 6$
if we write
$\dfrac{p!}{k!(p - k)!} = \dfrac{p(p - 1)(p - 2) \ldots (p - k + 1)}{k!}, \; p \in \Bbb P, \tag 7$
we observe that for $1 \le k \le p - 1$, $p$ and $k!$ have no factor greater than $1$ in common, since $p$ is prime and with $k < p$,
$k! = k(k - 1)(k - 2) \ldots (1) \tag 8$
has no factor divisible by $p$; it follows then that $p \in \Bbb P$ implies
$p \mid \dfrac{p!}{k!(p - k)!}, \; 1 \le k \le p - 1; \tag 9$
returning then to (6) with $n = p$ we see that, in a ring of characteristic $p \in \Bbb P$,
$(x + y)^p = x^p + y^p; \tag{10}$
now having
$(xy)^p = x^py^p \tag{11}$
already at our disposal, (10) allows us to conclude that the map
$x \to x^p \tag{12}$
is in fact a ring homomorphism in characteristic $p$; thus a field homomorphism in characteristic $p \in \Bbb P$ as well. $OE\Delta$.
