Extension of finite fields under norm map If $L/K$ is a finite Galois extension with group $G$, we can define the norm of an element $a\in L$ as
\begin{equation}
N_{L/K}(a)=\prod_{\sigma\in G}\sigma(a).
\end{equation}
And obviously, this product lies in $K$.
It is a natural question to ask whether the image of $L$ under this map is the whole $K$ or not.
I computed a few examples for the special case of finite extensions of finite fields, for example:
If $L=\mathbb{F}_{3}(\sqrt{2})=\mathbb{F}_{9}$, then we find
\begin{align}
N(0)&=0\\
N(1)&=1\\
N(2)&=1\\
N(1+\sqrt{2})&=2\\
N(2+\sqrt{2})&=2\\
N(3+\sqrt{2})&=1\\
N(1+\sqrt{2})&=2\\
N(2+\sqrt{2})&=2\\
N(3+\sqrt{2})&=1\\
\end{align}
It is always the case for finite fields?
If $L/K$ is an finite extension of finite fields, then $N_{L/K}(L)=K$?
 A: If $K = \Bbb F_q \subset \Bbb F_{q^n} = L$, then $N(x) = x(x^q)(x^{q^2})\ldots(x^{q^{n-1}}) = x^{1+q+\ldots+q^{n-1}} = x^{(q^n-1)/(q-1)}$.
So for a given $a \in K$ there can be at most $(q^n-1)/(q-1)$ solutions to $N(x)=a$. Moreover, $N$ is a group morphism $L^* \to K^*$, so its kernel is at most $(q^n-1)/(q-1)$, and $|\ker N| \times |\operatorname{im} N| = |L^*|$.
From $|\ker N| \le (q^n-1)/(q-1)$ , $ |\operatorname{im} N| \le |K^*| = (q-1)$ and $|L^*| = q^n-1$, all the inequalities must be equalities, and so $N$ has to be surjective.
A: Here is a purely Galois theoretic solution of the surjectivity of the norm for finite fields. Consider the extension F$_{q^n}$/F$_q$ and let N be the norm map between the multiplicative groups. It is well known that Gal(F$_{q^n}$/F$_q$) is cyclic, generated by the Frobenius automorphism Fr defined by Fr($x$) = $x^q$. According to Hilbert's thm. 90 for cyclic extensions, the kernel Ker N consists of all the elements of the form  f($y$) := Fr($y$)/$y$, with $y$ running through F$_{q^n}$*. Since Ker f = F$_q$ *, it follows that Im f = Ker N  has order ${q^n}$ - 1/$q$ - 1, so Im N has order $q$ - 1, and the surjectivity of N is proved.
Note that exactly the same argument applied to the additive structure (the additive version of Hilbert 90 is true) shows the surjectivity of the trace map for finite fields.
NB.  In general the norm map is not surjective. Obvious counter-example : in the extension C/R, Im N consist of sums of two squares, so Im N is of index 2 in R *. If we replace these archimedean local fields by p-adic local fields, local CFT tells us that for an abelian extension L/K of such fields, Gal(L/K) is isomorphic to K */ N L * .
A: Yes, this is always the case, because of Chevalley-Warning theorem ( https://en.wikipedia.org/wiki/Chevalley%E2%80%93Warning_theorem ).
The norm $N_{L/K}$ is a homogeneous polynomial of degree $n$ in $n$ variables where $n=[L:K]$. So for any $\lambda\in K$ the polynomial function $f: L\times K\to K$ defined by $f(x,a) = N_{L/K}(x)-\lambda a^n$ is homogeneous of degree $n$ with $n+1$ variables : it has a non-trivial zero $(x,a)$, and then $N_{L/K}(x/a)=\lambda$ (you can't have $a=0$ because then $N_{L/K}(x)=0$ so $x=0$).
