Frobenius map as a ring homomorphism

Question

I have to prove that the frobenius map given by $Fr: GF27 \to GF27$ where $Fr([y]) = [y]^3$ for $[y] \in GF27$. Here $GF27$ denotes a field made as $\mathbb{Z}_3/f(x)$ where $f$ is an irreducible polynomial of degree 3. I have to show $Fr([y_1] + [y_2]) = Fr([y_1])+Fr([y_2])$ and $Fr([y_1]\cdot[y_2]) = Fr([y_1])\cdot Fr([y_2])$ for $[y_1], [y_2] \in GF27$. I do not have access to Freshmans dream, is there an easy way to prove this without that access, or do i have to show it the hard way doing all the calculations and showing it that way?

Answer 1

Any function $x\mapsto x^k$ with arbitrary $k$ preserves multiplication in a commutative ring.

And, use the binomial theorem to prove $x\mapsto x^p$ preserves addition in a ring with characteristic $p$, where $p$ is a prime (specifically $3$ in your example).

$$(x+y)^p=\sum_{i=0}^p{p\choose i}x^iy^{p-i}$$ and $p$ divides ${p\choose i}$ whenever $1\le i< p$, so the intermediate terms all vanish.