I did this problem, but now I'm left with more questions! Suppose $f(x)$ is a monic irreducible polynomial of degree $3$ over $GF(2)$. Prove that if $a$ is a root of $f$ in an extension of $GF(2)$, then $f(x)=(x-a)(x-a^2)(x-a^4)$
So I let $F = \dfrac{GF(2)}{<f>}$ so $|F| =2^3$
I understand that automorphisms maps roots to roots, so I defined this automorphism
$$\phi:F\rightarrow F$$ $$\phi(x)=x^2 \ \ \ \ \ \ \forall x \in F$$
Hence $\phi(a) =a^2 ,\phi^2(a)=a^4$
But I feel that in general $GF(p^n)$ , $\phi(x) =x^p$ is maybe not an ...isomorphism? Is it, or not?
Proof? :
Clearly $\phi$ respects addition and mulptiplication. Injection: If $\phi(x) = \phi(y)$ $\rightarrow x^p-y^p=(x-y)^p = 0 \rightarrow x=y$ Domain and codomain have same cardinality, hence bijection.
Is there a flaw in this proof?
Thanks in advanced :D
That proof looks good!
Having another way to see the result may help your confidence in it: the multiplicative group of nonzero elements of $\mathbf{F}_{p^n}$ is a cyclic group of order $p^n - 1$. In particular, its order relatively prime to $p$, and thus every nonzero element has a unique $p$-th root. (and $0$ has a unique $p$-th root too, obviously)