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Given a finite field $|K|=q$ and an irreducible $f \in K[x]$ with $\deg(f)=n$ with $\alpha$ as a root. My candidates for the roots are $\alpha, \dots , \alpha^{q^{n-1}}$.

Assuming $\alpha^{q^i} = \alpha^{q^j}$ I want to conclude that $\alpha = \alpha^{q^{j-i}}$ or more precisly I want to conclude that $f \mid x^{q^{j-i}}-x$.

Its a basic algebra course and we had not had Galois theory yet. So assume we have to lead $\alpha^{q^i} = \alpha^{q^j}$ to a contradiction by hand.

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    @Ragib: That answer uses some Galois Theory, though, which is probably why the caveat. I do wish he had asked in that question instead of posting a new one. I've provided an eleementary proof here.2012-07-04

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Judging from your comment to Dilip's answer to your previous question you have not fully absorbed one aspect of finite field arithmetic. Namely the fact that if $\mathrm{char} K=p$ (or equivalently $q=p^m$ for some positive integer $m$) implies the rules $ (a+b)^p=a^p+b^p,\qquad (a-b)^p=a^p-b^p. $ I am willing to bet that this really is somewhere in your lecture notes, in which case this is just a refresher. Going from particular to general, let us first study the case $q=2^m$. Then $K$ has $\mathbb{F}_2=\mathbb{Z}/2\mathbb{Z}$ as a subfield, so in the field $K$ we have $2=1+1=0$, because that's the way it goes in that subfield. This implies that $ (a+b)^2=a^2+2ab+b^2=a^2+0ab+b^2=a^2+b^2 $ as claimed. Similarly, if $|K|=3^m$ it has $\mathbb{F}_3=\mathbb{Z}/3\mathbb{Z}$ as a subfield, and consequently $3=1+1+1=0$. This then implies $ (a+b)^3=a^3+3a^2b+3ab^2+b^3=a^3+0a^2b+0ab^2+b^3=a^3+b^3. $ The general case $|K|=p^m, p>3$ follows from a similar calculation with the binomial formula, after we first observe that the binomial coefficient $ {p\choose k}=\frac{p!}{(p-k)!k!} $ is divisible by $p$ (the numerator is manifestly divisible by $p$, but the denominator is not, because $p$ is a prime and all the factors in the factorials in the denominator are smaller than $p$). The claim for the differences then follows from this rule by replacing $b$ with $-b$.

As a consequence of this fact we also get that in any extension field of $K$, e.g. in $K(\alpha)$ we have $ (a-b)^q=(a-b)^{p^m}=\left((a-b)^p\right)^{p^{m-1}}=(a^p-b^p)^{p^{m-1}}=\cdots a^q-b^q $ proving the formula you had trouble with.

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Let $\alpha$ be a root in an algebraic extension $L$ of $K$.

Since $(L^*,\cdot)$ is a group, you get by Lagrange Theorem than

$\alpha^{q^i}=\alpha$

Thus, $\alpha$ is a root of $x^{q^i}-x$. Now since $f$ is the minimal polynomial of $\alpha$ over $K$, you get that $f | x^{q^i}-x$.

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Consider the field $L=K[x]/(f)\cong K(\alpha)$ where $\alpha$ is the image of $x$. Then the field $L$ has $q^n$ elements and the multiplicative group $L^*$ is cyclic with $q^n-1$ elements. So any element of $L$ satisfies $x^{q^n}-x$.

So $\alpha$ satisfies $x^{q^n}-x$, but $\alpha$ does not have to be the generator of the multiplicative group.

If you know that $L$ is the splitting field of $f(x)$ then you could conclude that $f\ |\ x^{q^n}-x$

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Note that every element of $K(\alpha)$ satisfies $r^{q^n} = r$, as noted.

Assume $\alpha^{q^i} = \alpha^{q^j}$ with $i\leq j$. Then: $\alpha = \alpha^{q^{n}}=(\alpha^{q^i})^{q^{n-i}} = (\alpha^{q^j})^{q^{n-i}} = \alpha^{q^{n+j-i}} = \alpha^{q^nq^{j-i}} = (\alpha^{q^n})^{q^{j-i}} = \alpha^{q^{j-i}},$ which is the equality you want.