How do I prove this field homomorphism is an isomorphism? The question is as follows. Let $F$ be a finite field with unit $1$ not equal to zero. Let the function $f: F \to F$ be given by $f(x) = x^3$, where the $\operatorname{char}(F) = 3$. Prove it is a ring isomorphism.


I can prove it is a ring homomorphism by showing $f(xy) = (xy)^3 = x^3y^3 = f(x)f(y)$ by commutativity. Similarly, since the characteristic is three, 

$$(x+y)^3 = x^3 + y^3$$

$$f(a+b) = f(a) + f(b)$$ 

$$f(1) = 1,\ \ \textrm{and}\ \  f(0) = 0$$

Now, how do I show $f$ is a bijection?
That homomorphism is necessarily injective, but not necessarily surjective. For example, the field $F=\mathbb{F}_3(t)$ has characteristic $3$, and the homomorphism $\varphi:F\to F$ defined by $\varphi(a)=a^3$ for all $a\in F$ is not surjective because there is no $a\in F$ with $\varphi(a)=t$.

For a more in-depth analysis, take a look at my answer on the thread How to prove that the Frobenius homomorphism is surjective?
To get started, $\ker(f)$ is an ideal in the ring $F$.