10
$\begingroup$

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.

$\endgroup$
4
  • $\begingroup$ How could the order be any larger? Kudos for showing your work. $\endgroup$ May 11, 2012 at 0:11
  • $\begingroup$ @DylanMoreland Yeah. I'm stucking here. I believe it couldn't be larger, but I can't see it. $\endgroup$
    – CC_Azusa
    May 11, 2012 at 0:12
  • 1
    $\begingroup$ 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 $x_1 + \cdots + x_n$.] $\endgroup$ May 11, 2012 at 0:16
  • 2
    $\begingroup$ 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? $\endgroup$
    – Mike B
    May 11, 2012 at 0:17

2 Answers 2

18
$\begingroup$

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 $k$.

You could also focus on the size of the kernel, as Mike B suggests. As Brandon points out, these are the roots of a certain polynomial, which gives you an upper bound. And there is an obvious lower bound.

A very useful generalization of this is Hilbert's Theorem 90.

$\endgroup$
2
  • 3
    $\begingroup$ Elements of the kernel are roots of $x^{1+q+\dots+q^{n-1}}-1$. $\endgroup$
    – bzc
    May 11, 2012 at 1:03
  • $\begingroup$ @BrandonCarter Ah, I think it follows from facts about cyclic groups as well, but I like your way even better. $\endgroup$ May 11, 2012 at 1:04
1
$\begingroup$

Hint: $N_{K/F}(\alpha) \in K^{\rm{Gal}(K/F)}= F$.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .