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.

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