Show the norm map is surjective

Let $K/F$ be an extension of finite field. Show that the norm map $N_{K/F}$ is surjective.

Here is what I have so far:

Since $F$ is a finite field and $K/F$ is a finite extension of degree $n$, so $\operatorname{Gal}(K/F)=\langle\sigma\rangle$, where $\sigma(a)= a^{q}$ with $q=p^{m}=|F|$. In addition, by primitive element theorem, $K=F(\alpha)$ for some $\alpha \in K$.

We want to show $N_{K/F}(\alpha)$ generates $F$. By the definition of norm, we have $$N_{K/F}(\alpha)=\alpha^{1+q+\cdots+q^{n-1}}$$ and since $(1+q+\cdots+q^{n-1})(q-1)=q^n-1$, we have the order of $N_{K/F}(\alpha)$ is divided by $q-1$.

But I want to show $o(N_{K/F}(\alpha))=q-1$ in order to conclude surjectivity. So any hint for how to proceed? Any other methods are also perferred.

• How could the order be any larger? Kudos for showing your work. May 11, 2012 at 0:11
• @DylanMoreland Yeah. I'm stucking here. I believe it couldn't be larger, but I can't see it. May 11, 2012 at 0:12
• Actually, don't we actually have the opposite? If $a$ is an element of a group and $a^n = 1$, then I know that the order of $a$ divides $n$. But I don't get the reverse. [By the way, it's best to use \cdots when writing things like $x_1 + \cdots + x_n$.] May 11, 2012 at 0:16
• Another way to see this is to consider the kernel of the norm map. What can you say about an element in the kernel? How big can the kernel be? May 11, 2012 at 0:17

I think you have a lot of the right stuff written down. Let's take $\alpha$ to be a generator for the cyclic [see Lemma 1.6 here for a proof] group $K^*$, so $\alpha$ has order $q^n - 1$. Then, as you say, its norm is $N_{K/F}(\alpha) = \alpha^{1 + q + \cdots + q^{n - 1}} \in F^*.$ This norm generates $F^*$ because its order is precisely $q - 1$. This is just group theory: if an element $x$ in a group has order $mk$ then $x^m$ has order $k$.
• Elements of the kernel are roots of $x^{1+q+\dots+q^{n-1}}-1$.
Hint: $N_{K/F}(\alpha) \in K^{\rm{Gal}(K/F)}= F$.