Let $F$ be a finite field with $\text{char}(F) = p$. Now, if $u$ is a primitive element, show that $u^p$ is also primitive.

Question

I need help in understanding how to prove this. I know that if $u$ is a primitive element of a Finite field, $F$, then $u$ generates $F^*$.

Answer 1

We need to show that if $u$ is a generator of $F^{*}$, then $u^{p}$ is also a generator of $F^{*}$. Since $F$ is a finite field, $|F| = p^{n}$ for some $n\geq 1$, so that $|F^{*}| = p^{n}-1$. Now, this is coprime to $p$, so...

Answer 2

The order of $u^p$ is given by $\mid u^p\mid=\dfrac{\mid u\mid}{\operatorname{gcd}(\mid u\mid,p)}=\dfrac{p^n-1}{\operatorname {gcd}(p^n-1,p)}=p^n-1$.

Answer 3

Consider the function

$\theta: F \to F \tag 1$

given by

$\theta(u) = u^p, \; \forall u \in F; \tag 2$

clearly

$\theta(uv) = (uv)^p = u^pv^p = \theta(u)\theta(v); \tag 3$

also,

$\theta(u + v) = (u + v)^p = \displaystyle \sum_0^p \dfrac{p!}{j!(p - j)!} u^{p - j} v^j, \tag 4$

since the ordinary, vanilla-flavored binomial theorem applies in any field $F$; furthermore, it is a well-known fact from elementary number theory that

$p \mid \dfrac{p!}{j!(p - j)!}, \; 1 \le j \le p - 1, \tag 5$

which yields

$\displaystyle \sum_0^p \dfrac{p!}{j!(p - j)!} u^{p - j} v^j = u^p + v^p; \tag 7$

combining this with (4) we obtain

$\theta(u + v) = u^p + v^p = \theta(u) + \theta(v); \tag 8$

by virtue of (3) and (8), we see that $\theta$ is in fact a homomorphism from $F$ to itself; $\theta$ is in fact injective, since $F$ is a field, whence

$\theta(u) = u^p = 0 \Longleftrightarrow u = 0; \tag 9$

now $\theta$ being an injective function from the finite field $F$ to itself, $\theta$ is also surjective, that is,

$\forall w \in F \exists v \in F, \; w = \theta(v) = v^p; \tag{10}$

being an injective, surjective homomorphism, $\theta$ is in fact a field automorphism of $F$; as such, $\theta$ restricts to a group automorphism

$\theta: F^\times \to F^\times; \tag{11}$

it follows that, since $u$ generates the group $F^\times$, $\theta(u) = u^p$ generates the group $\theta(F^\times) = F^\times$; $\theta(u) = u^p$ is also a primitive for the field $F$.

Answer 4

I am not completely sure what you mean by "primitive element" of a field. Usually we speak of a primitive element of a field extension, so I am going to assume that you mean that $u$ generates $F$ over its prime field $F_0$.

Now since the Frobenius map $\operatorname{Fr}$ defined by $x \mapsto x^p$ is an automorphism of $F$ leaving $F_0$ fixed, it permutes the roots of an irreducible polynomial. In particular, some iterate of $\operatorname{Fr}$ applied to $u$ yields $u$ again. So we have $(u^p)^{p^i} = u$ for some integer $i$. Hence the field $F_0(u^p)$ contains $u$, hence we have $F_0(u^p)=F_0(u)=F$, i.e. $u^p$ is a primitive element of the field extension $F/F_0$.

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EDIT: If by any chance by "primitive element" you mean "generator of the multiplicative group", you can use (part of) the same argument: the equation $(u^p)^{p^i} = u$ for some positive integer $i$, shows that $u^p$ generates the same subgroup of $F^\ast$ as does $u$.