Is \( u^p \) a root of \( f(x) \) over \( GF(p) \) if \( u \) is a root?

[joeblow]

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 u^p is a root of f(x).

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

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 x^{q^{m}}-x 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.

 

[Norwegian]

In general, we don't 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.

 

[joeblow]

The result that (a+b)^p=a^p+b^p (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.

 

[Hurkyl]

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

(Yes, I mean = where I've written =, and \equiv where I've written \equiv)If x is an indeterminate over GF(p), then it's not true that x^p = x. Consequently, if t is an indeterminate over Z, it's not true that t^p \equiv t \pmod p.

(While x and x^p both determine the same function on GF(p), they are different polynomials. And, e.g., they give different functions on GF(p^2))If u \notin \text{GF}(p), then we have no reason to expect u^p = u. If u is not an integer, we have no reason to expect u^p \equiv p \pmod p (if that equation even makes sense).For a finite field \text{GF}(p^e), the operation a \mapsto a^p is actually generalization of the idea of conjugation. For the special case of e = 2, you even have that (a^p)^p = a. (for larger e, you have to iterate e times rather than just twice)