Roots of polynomials over GF(p)

  1. 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).

    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.
  2. jcsd
  3. In general, we dont have f(u)=f(up) in your field F, but only for u in K, and you also need to use that (a+b)p=ap+bp for any a,b in F. Then it should be easy and fun to write out 0=f(u)p and see where that leads.
  4. The result that [itex](a+b)^p=a^p+b^p[/itex] (which is all that is needed) was demonstrated (via exercise) much earlier in the chapter, though. Something is strange about the placement of this problem.
  5. Hurkyl

    Hurkyl 15,987
    Staff Emeritus
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    It is true that, for every element [itex]a[/itex] of GF(p), we have [itex]a^p = a[/itex]. As a consequence, it's also true that, for every integer [itex]n[/itex], we have [itex]n^p \equiv n \pmod p[/itex]. Also as a consequence, for every polynomial with coefficients from GF(p) and element [itex]a[/itex] of GF(p), we have [itex]f(a) = f(a^p)[/itex]. Similarly, for every polynomial with integer coefficients and any integer [itex]n[/itex], we have [itex]f(n^p) \equiv f(n) \pmod p[/itex].

    (Yes, I mean = where I've written =, and [itex]\equiv[/itex] where I've written [itex]\equiv[/itex])

    If [itex]x[/itex] is an indeterminate over GF(p), then it's not true that [itex]x^p = x[/itex]. Consequently, if [itex]t[/itex] is an indeterminate over Z, it's not true that [itex]t^p \equiv t \pmod p[/itex].

    (While [itex]x[/itex] and [itex]x^p[/itex] both determine the same function on GF(p), they are different polynomials. And, e.g., they give different functions on GF([itex]p^2[/itex]))

    If [itex]u \notin \text{GF}(p)[/itex], then we have no reason to expect [itex]u^p = u[/itex]. If [itex]u[/itex] is not an integer, we have no reason to expect [itex]u^p \equiv p \pmod p[/itex] (if that equation even makes sense).

    For a finite field [itex]\text{GF}(p^e)[/itex], the operation [itex]a \mapsto a^p[/itex] is actually generalization of the idea of conjugation. For the special case of [itex]e = 2[/itex], you even have that [itex](a^p)^p = a[/itex]. (for larger [itex]e[/itex], you have to iterate [itex]e[/itex] times rather than just twice)
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