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)

 

[blue_raver22]

I can understand your confusion and I am happy to clarify this concept for you. The statement "if u is a root of f(x) over K, then u^p is a root of f(x)" is a fundamental property of finite fields and polynomials over them. This property is known as the Frobenius automorphism and it is a key tool in understanding the structure of finite fields and their roots.

To understand why this property holds, we need to first understand the structure of finite fields. A finite field of characteristic p, denoted GF(p), is a field that contains p elements. The elements of GF(p) can be represented as integers from 0 to p-1, and arithmetic operations are performed modulo p. For example, in GF(5), the elements are 0, 1, 2, 3, and 4, and the addition and multiplication operations are performed modulo 5.

Now, let's consider a polynomial f(x) over GF(p). This polynomial can be written as f(x) = a_0 + a_1x + a_2x^2 + ... + a_nx^n, where a_i are elements of GF(p). When we raise this polynomial to the pth power, we get f(x)^p = (a_0 + a_1x + a_2x^2 + ... + a_nx^n)^p. By the binomial theorem, we can expand this polynomial as f(x)^p = a_0^p + a_1^px^p + a_2^px^2p + ... + a_n^px^{np}. Since we are working in GF(p), the pth power of any element is equal to itself (as you mentioned, x^p ≡ x (mod p)). Therefore, we can rewrite f(x)^p as f(x)^p = a_0 + a_1x^p + a_2x^{2p} + ... + a_nx^{np}. This is equivalent to f(x) (since we are working in GF(p), all the exponents are reduced modulo p). Hence, f(x)^p = f(x) over GF(p).

Now, let's go back to the statement "if u is a root of f(x) over K, then u^p is a root of f(x)". Since f(x)^p = f(x) over