
#1
Mar1612, 10:27 AM

P: 71

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
Mar1612, 01:03 PM

P: 144

In general, we dont have f(u)=f(u^{p}) in your field F, but only for u in K, and you also need to use that (a+b)^{p}=a^{p}+b^{p} 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.




#3
Mar1612, 09:29 PM

P: 71

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.




#4
Mar1712, 12:34 PM

Emeritus
Sci Advisor
PF Gold
P: 16,101

Roots of polynomials over GF(p)
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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