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This problem comes from Question 14 of Chapter 5, Section 6 in I.N. Herstein's Abstract Algebra, second ed.

Given a field $F$ with characteristic $p\neq 0$, show that all roots of $x^m-x$ within this field are distinct (no root has multiplicity greater than 1), where $m = p^n$ for some $n$.

I know that if $F$ is a finite field of order $p$ that $$x^p-x=\prod_{a\in F}(x-a)$$ and this result should lead to having $$x^p-x=x(x-1)(x-2)(x-3)\cdots(x-(p-1))$$ as I believe I can form a finite subfield of $F$ out of these elements $\{1,2,3,\ldots,p-1,0\}$, but I'm not sure of this claim. If I can do that then I do know that $$(x^{p^k}-x)(1+x^{p^k-1}+x^{2p^k-2}+\ldots+x^{p^k\cdot p-p})=x^{p^{k+1}}-x$$ so I only need to show that $1+x^{p^k-1}+x^{2p^k-2}+\ldots+x^{p^{k+1}-p}$ is irreducible or at least is not divisible by $x-a$ for any $a\in F$.

I realized that if I let $p(x) = 1+x^{p^k-1}+x^{2p^k-2}+\ldots+x^{p^{k+1}-p}$ and assumed that it was divisible by some $x-a$ we would have $$p(x)=q(x)(x-a)=xq(x)-aq(x)$$ and therefore if $$q(x) = \sum_{i=0}^{\deg q(x)}(q_ix^i)$$ we would have $p_i=q_{i-1}-aq_i$ for all $i$ (where $p_i$ is the $i^\text{th}$ coefficient in $p(x)$). From this I got that $q_0a=1$ so $q_0 = a^{-1}$ and then $q_1=a^{-2}$ and I could work my way up but I had issues figuring out how to extrapolate when I git $q_{p-1}$ and $q_{2p-2}$.

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The usual way to show this is to define the formal derivative $f'$ of a polynomial $f$. For your polynomial $f$, the formal derivative is exactly what you think it would be by applying the power-rule on $f$. Then there's a theorem, the actual proof of which is not too bad but should be believable without proof from your experience in calculus, that goes

For a field $K$ let $F$ be the splitting field of $f \in K[x]$. Then $f$ has distinct roots in $F$ if $f$ and $f'$ have no common roots.

For your particular polynomial $x^{p^n} -x$, the formal derivative will be $p^n x^{p^n-1}-1$ which equals $-1$ since we're in a field of characteristic $p$, so the formal derivative doesn't even have roots, and we're good.

I think what you're attempting in your solution just unfolds the details of the proof of the above theorem.

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    $\begingroup$ Is there a name for this theorem about the formal derivative? I would like to either prove this statement or at least cite the theorem in my solution as it does not appear in my text book. $\endgroup$ – Benji Altman Jun 29 '18 at 22:53
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    $\begingroup$ @BenjiAltman Not that I'm aware of. I've added a link to a proof in my answer. Theorem 2.1 $\endgroup$ – Mike Pierce Jun 29 '18 at 22:55
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Well, $\dfrac{\mathrm d}{\mathrm dx} (x^m-x) = mx^{m-1}-1 = -1 \ne 0$, so the polynomial is separable, so the roots are distinct.

Bonus: the roots of $x^m - x$ actually form a field.

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You only have to show that $f(x)=x^{p^{n}}-x$ and $f'(x)$ have no common root, which results from the fact that in a field of characteristic $p$, $f'(x)=-1$.

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