How to show that $h(x^p) \equiv h(x)^p \pmod{p}$? [duplicate]

Possible Duplicate:
Why $g(x^{p})=(g(x))^{p}$ in the reduction mod $p$?

Let $h(x) \in \mathbb{Z}[x]$ and $p$ be a prime. We know that for any integer $\alpha$ we have that $\alpha^p \equiv \alpha \pmod{p}$. How can we use this to show that $h(x^p) \equiv h(x)^p \pmod{p}$? It seems to me that we have to reduce the indeterminate $x$ modulo $p,$ which does not make sense.

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marked as duplicate by Jyrki Lahtonen♦, Leonid Kovalev, tomasz, Chris Eagle, J. M. is back.Aug 17 '12 at 18:34

For $(h(x))^p$, imagine expanding using the Multinomial Theorem. All "mixed" multinomial coefficients are divisible by $p$. Same as one of standard proofs of Fermat's Theorem. – André Nicolas Jul 27 '12 at 16:32
@Andre Nicolas: I actually made a mistake: f and g need not have a factor in common. So i will retain only the first part of the question, which is already answered. If you could write it as an answer, i will accept it. – Manos Jul 27 '12 at 16:44
@Belgi I recalled seeing this question on M.SE before, but I wasn't able to search for it. I just figured out that search function is terrible with math. – user2468 Jul 27 '12 at 18:35
@J.D. I think so to, I had it easy since it was my question :) – Belgi Jul 27 '12 at 18:37

Imagine expanding $$\left(a_n x^n +a_{n-1}x^{n-1} +\cdots +a_0\right)^p$$ by using the Multinomial Theorem. All of the "mixed" multinomial coefficients are divisible by $p$, so modulo $p$ we end up with $$a_n^p (x^n)^p +a_{n-1}^p(x^{n-1})^p +\cdots + a_0^p,$$ which, modulo $p$, is the same polymomial as $$a_n(x^p)^n +a_{n-1}(x^p)^{n-1}+\cdots +a_0.$$

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By the binomial theorem, \begin{align} (ax+by)^p &=\sum_{k=0}^p\binom{p}{k}(ax)^{p-k}(by)^k\\ &\equiv ax^p+by^p\pmod{p}\tag{1} \end{align} because $\left.p\;\middle|\;\binom{p}{k}\right.$ except when $k=0$ or $k=p$, and because $a^p=a\pmod{p}$ for all $a\in\mathbb{Z}$.

Inductively using $(1)$, we get that $$\left(\sum_{k=0}^na_kx^k\right)^p\equiv\sum_{k=0}^na_kx^{kp}\pmod{p}\tag{2}$$

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Hint  Freshman's Dream $\rm\Rightarrow\, z\to z^{\,p}$ is a ring hom. Hom's always commute with polynomials, being compositions of basic ring operations, which, by definition, hom's commute with.

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