If F is a field of characteristic p, with prime subfield K = GF(p) and u in F is a root of f(x) (over K), then [itex]u^p[/itex] is a root of f(x).(adsbygoogle = window.adsbygoogle || []).push({});

Now, I know that [itex] x^p \equiv x (\text{mod } p)[/itex], so isn't it immediately true that [itex] f(x^p)=f(x) [/itex] (over K)? So, [itex] 0=f(u)=f(u^p) [/itex].

I only ask because this type of problem (as I understand it) is much more elementary than the material that the text is covering. In the section from which this problem comes, the main results are (1) that if F = GF(q), then the irreducible factors of [itex]x^{q^{m}}-x[/itex] over F are precisely the irreducible polynomials over F of degree dividing m, (2) the Mobius inversion formula, and (3) a formula for finding the number of monic irreducible polynomials of degree m over GF(q).

Please tell me if there is some critical misunderstanding that I am having.

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# Roots of polynomials over GF(p)

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