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Let $\mathbb{F}_p =:K\leqslant F:=\mathbb{F}_{p^n}$. Define the norm of $f\in F$ to be $N_{F/K}(f) = f^{p^{n-1}+p^{n-2}+\ldots +p+1}$

Show that if $f\in\mbox{ker}N$, then $f = g^{p-1}$, for some $g\in F^\times$

One can verify that the norm is always an element of $K$.
We also know that $N(f^p) = N(f)$ and the book [R. Lidl, H. Niederreiter, Finite Fields] says the implication is obvious [analogous to necessity in Thm 2.25 p.56].

Is it obvious? We have $N(f^p)=N(f)=1$ i.e $$(f^p)^{(p^n-1)/(p-1)} = f^{(p^n-1)/(p-1)}=1 $$ how do I explicitly get $g$ here, it's obvious, after all.

AlvinL
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1 Answers1

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Consider $\gamma \colon F^\times \to F^\times$, $f \mapsto f^{p-1}$. Then $\def\im{\mathop{\rm im}}\im \gamma \subseteq \ker N$. Now $$f\in \ker \gamma \iff f^{p-1} = 1 \iff f^p = f \iff f \in K^\times $$ that is $\ker \gamma = K^\times$. Hence, $\im \gamma$ has $\frac{p^n -1}{p-1}$ elements. As $N$ is onto $F^\times$, $\ker N$ also has $\frac{p^n-1}{p-1}$ elements. That is $\im \gamma = \ker N$.

Now let $f \in \ker N = \im\gamma$. There is $g \in F^\times$ with $\gamma(g) = f$, that is $g^{p-1} = f$.

martini
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