## f(x)ᵖ = f(xᵖ) over ZZ/(p)

Prove that $f(x)^p = f(x^p)$ for all $f(x) \in \mathbb{F}_p[x]$.

Let $f(x) = \sum c_ix^i$. Remember that the elements of $\mathbb{F}_p$ are precisely the roots of $x^p-x$; in particular, $\alpha^p = \alpha$ for all $\alpha \in \mathbb{F}_p$.

Then $f(x)^p = (\sum c_ix^i)^p$ $= \sum (c_ix^i)^p$ $= \sum c_i^p(x^i)^p$ $= \sum c_i (x^p)^i$ $= f(x^p)$ as desired.

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### Comments

• Hachem  On August 21, 2012 at 12:22 am

can we prove de same for p^k, precisely i mean :
(f(x))^(p^k)=f(x^(p^k)) for all f(x) in F_(p^k)[X] ???

Thank you for the nice site.

• nbloomf  On August 21, 2012 at 10:17 am

I think so. If $q = p^n$, then all the middle coefficients of $(a+b)^p$ in $\mathbb{F}_q$ are 0 (being divisible by $p$. So $(a+b)^q = a^q + b^q$, and the proof generalizes. We could probably also show this using the fact that the elements of $\mathbb{F}_q$ are precisely the roots of $x^q-x$.