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Problem: Factoring $x^p-x= (x)(x-1)\cdots(x-(p-1)) \in \Bbb Z_p[x]$.

Clearly $0$ is a root, and so consider $(x^{p-1}-1) \mod p$, by Fermat's little theorem, for each $a \neq$ 0 , $(x-a)$ is a root, and each root must have multiplicity $1$ since $(x^{p-1}-1)$ has at most $p-1$ roots ($Z_p[x]$ is an integral domain).

At each step above, by the factor theorem for commutative rings: $(x^{p-1}-1)=q(x)(x-k)$ for $1 \leq k \leq p-1$ and $q(x)\in\Bbb Z_{p}[x]$, but I am not sure how to conclude that overall $(x^{p-1}-1)= (x-1)\cdots(x-(p-1))$

Bernard
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IntegrateThis
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  • So, by Fermat you can definitely conclude that overall your equation is correct with maybe an extra $f(x)$ on the right-hand side. It only rests to prove that $f(x)=1$. Do you have an idea? – Stan Tendijck Mar 02 '19 at 21:04
  • @StanTendijck, the only idea I have is expanding it out. – IntegrateThis Mar 02 '19 at 21:06
  • Ok, so that is not something you want to be doing. It might be worthwhile to compare the leading coefficient, i.e., the term for the highest order in $x$ and determining what the highest order, i.e. the order of the polynomial is. – Stan Tendijck Mar 02 '19 at 22:39

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Hint:

Show that the $(x-a),\; a\in\mathbf Z_p$, are coprime, so that $x^p-x$ will be divisible by their product. Finally, compare the degrees and the leading coefficients.

Bernard
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  • Not sure what $(x-a), (x-b)$ for $ 1 \leq a,b \leq p-1$ being co-prime means? Does this mean that no prime number divides those? I know what co-prime means generally, but in this case I am confused. – IntegrateThis Mar 02 '19 at 21:15
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    They're coprime as polynomials (actually, they're distinct irreducible polynomials). Don't forget that when $k$ is field, $k[x]$ is a P.I.D. – Bernard Mar 02 '19 at 21:18