I'm working on the field $F = Z_{2}[x]/P(X) $ where $P(X) = X^3 + X + 1$ is an irreducible polynomial.
Function $g$ is defined as follows and I'm asked to show it is one-to-one:
$g: F \rightarrow F $, $g(x) = x^2$
My solution
1) To show it's 1-1, I've to show $g(x) = g(y) \implies x = y$
$g(x + y) = g(x - y)= g(x) - g(y)$
So, if $g(x) = g(y)$ then $g(x-y) = 0$
2) Now I've to show $g(x) = 0 \implies x =0$
Clearly, $(F-\{0\}, .)$ is a multiplicative group of order 7. So, every element except $1$ must be a generator. Hence, for all elements $x$ from $F-\{0\}$, $g(x) \neq 0$. Thus, $g(x) = 0$ iff $x = 0$
3) From 1, we know that if $g(x) = g(y)$ then $g(x-y) = 0$ and from 2, we know that only $g(0) = 0$. Hence, $x - y = 0$ and $x = y$
Does this look correct ?
Point(1) seems a little under-explained though, apparently, you assume as clear that $\;(a\pm b)^{2^n}=a^{2^n}\pm b^{2^n}\;$ whenever char$\,\Bbb F=2\;$ , but it looks fine.
I think a shorter, perhaps even neater proof can be given as follows:
$$f(x)=f(y)\implies x^2=y^2\implies (x-y)^2=x^2-y^2=0\implies x-y=0\implies x=y$$
since over any field, $\;k^n=0\iff k=0\;$...and we're done.