Frobenius Mapping as a Ring Homomorphism

Question

I'm studying Lang's Algebra, where it is observed that, for the finite field $\mathbf{F}_q$ with $q=p^n$ elements ($p$ prime), the Frobenius mapping $$ z\mapsto z^p$$ is a ring homomorphism with kernel 0. This doesn't seem obvious to me. Why is the Frobenius mapping a ring morphism in the case of $\mathbf{F}_q$?

Answer 1

The key is that $(a+b)^p=a^p+b^p$ where $p$ is the characteristic of the field.
Clearly, $\varphi=\, z\mapsto z^p$ also preserves multiplication, hence it is a ring homomorphism.

Now, $q=p^n$ for some $n$, thus $z\mapsto z^q\, =\varphi^n$ is also a homomorphism.

Note also that, as ${\bf F}_q\setminus \{0 \}$ is a group under multiplication, all nonzero $z\in{\bf F}_q$ satisfies $z^{q-1}=1$, hence $z^q=z$ for all $z$, and so $\varphi^n$ is the identity.

Answer 2

Well, the multiplicative aspect is easy to see:

$(xy)^q = x^qy^q; \tag 1$

as for the additive aspect, if we exploit the binomial theorem, which is valid over any unital commutative ring, we have

$(x + y)^q = \displaystyle \sum_0^q \dfrac{q!}{i!(q - i)!} x^{q - i}y^i = x^q + y^q, \tag 2$

since

$q \mid \dfrac{q!}{i!(q - i)!}, \; 1 \le i \le q - 1, \tag 3$

which of course implies

$\dfrac{q!}{i!(q - i)!} \equiv 0 \mod q. \tag 4$

Thus we see $z \to z^q$ is a ring homomorphism; furthermore, since $\Bbb F_q$ is a field,

$z = 0 \Longleftrightarrow z^q = 0, \tag 5$

and thus we have

$\ker(z \to z^q) = 0. \tag 6$

Thus the Frobenius is a field automorphism of $\Bbb F_q$.