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I'm trying to understand this:

Working in $\mathbb{Z}_p[x]$, $(x-c)^p \equiv x-c \pmod{x^p-x}$.

From Fermat Little theorem we know that $(x-c)^p \equiv (x-c) \pmod{p}$ but I don't see a connection. Actually, I'm not sure if I'm mixing things up because $\mathbb{Z}_p[x]$ is a polynomial ring.

Also I tried to show it from: $x^p \equiv x \pmod{x^p-x}$ holds because $x^p-p \mid x^p-x$. But I don't know how substitute $x-c$ instead of $x$.

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Have you tried just using the binomial theorem? It is actually true that (x - c)^p - (x - c) = x^p - x in F_p[x]. See if you can prove this stronger statement. –  Qiaochu Yuan Dec 15 '10 at 21:14
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2 Answers

up vote 9 down vote accepted

Working in $\mathbb{Z}_p[x]$, you have the "Freshman's Dream" binomial theorem: $$(a+b)^p = a^p + b^p.$$ (In fact, this holds in any commutative ring of characteristic $p$, simply by the Binomial Theorem, as mentioned by Qiaochu: all the mixed coefficients are multiples of $p$).

So $(x-c)^p = x^p - c^p$. By Fermat's Little Theorem, you also know that $c^p \equiv c \pmod{p}$, so $(x-c)^p = x^p - c$ in $\mathbb{Z}_p[x]$. That's pretty much all you need.


Addendum

Let me add that you were very perceptive to notice that even though the function $x\mapsto (x-c)^p$ is the same as the function $x\mapsto x-c$ in $\mathbb{Z}_p$, this does not mean that the polynomials $(x-c)^p$ and $x-c$ are equal. We are used to (in the integers, rationals, reals, etc) the fact that two polynomials give the same function if and only if they are identical, but this does not occur over finite rings; for example, Fermat's Little Theorem tells you that the polynomial $x^p$ and the polynomial $x$ both define the same function.

However, what you can prove is that if $p(x)$ and $q(x)$ are two polynomials with coefficients in $\mathbb{Z}_p$, and they define the same function (that is, $p(a) = q(a)$ for all $a\in\mathbb{Z}_p$), then they differ by a multiple of the polynomial $x^p-x$; i.e., $p(x)\equiv q(x)\pmod{x^p-x}$ in $\mathbb{Z}_p[x]$. What you have here is a special case. See if you can figure out how to prove the more general statement.

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For reference, Freshman's Dream: en.wikipedia.org/wiki/Freshman's_dream –  user2468 Dec 15 '10 at 21:47
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Alternatively, by Fermat's little Theorem every $\rm\ a\in \mathbb F_p\ $ is a root of $\rm\ f(x)^p - f(x)\:,\ $ therefore $\rm\ x\ (x-1)\ (x-2)\cdots (x-p+1)\ |\ f(x)^p-f(x)\:.\ $ Making the specialization $\rm\ f(x) = x\ $ this yields $\rm\ x \ (x-1)\ (x-2)\cdots (x-p+1)\ |\ x^p-x\:.\ $ This divisibility must be an equality since both are monic polynomials of equal degree. Therefore we deduce that $\rm\: x^p - x\ |\ f(x)^p - f(x),\ \forall\: f\in \mathbb F_p\:[x]\:.\ $ The sought result is the special case $\rm\ f(x) = x-c\:. $

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