Is the norm of a degree 2 extension surjective?

Question

Consider the finite field $K=\mathbb{F}_q$. Let $L=\mathbb{F}_{q^2}$. Then $[L:K]=2$. Let $Gal(L|K)=\{1,\sigma\}$.

The norm defined as

$$\text{N}_{L|K}(x)=x\sigma(x)$$ is a map from $L$ to $K$. Is it true that $\text{N}_{L|K}$ is surjective ?

Answer 1

The map $\sigma$ is $x\mapsto x^q$. So, $N_{L/K}(x)=x^{q+1}$. So, $\ker N_{L/K}$ are the elements of $\mathbb{F}_{q^2}^\times$ satisfying $x^{q+1}=1$. Note that there are at most $q+1$ of these. Thus, the image of $N_{L/K}$ must have size at least $|\mathbb{F}_{q^2}^\times|$ divided by $q+1$. But, $|\mathbb{F}_{q^2}^\times|=(q^2-1)=(q-1)(q+1)$. So, the image has size at least $q-1$. Since $|\mathbb{F}_{q}|=q-1$ this implies it's surjective.

Answer 2

If $\mathbf F_{q^{n}}/ \mathbf F_q$ is an extension of degree $n$ of finite fields, it is known that :
1) The extension is simple, i.e. $\mathbf F_{q^{n}}$ is of the form $\mathbf F_q (x)$
2) Its Galois group is cyclic of order $n$, generated by the Frobenius automorphism $\sigma$ defined by $\sigma (x) = x^q$
3) The groups $\mathbf F_{q^{n}}^*$ is cyclic of order $q^n - 1$, generated by $x$

The norm map $N$ is defined by $N(x)=x. \sigma (x). ... \sigma^{n-1}(x)= x^{1+q+ ... +q^{n-1}} = x^{\frac {q^n -1}{q- 1}}$, hence its kernel, characterized by $x^{\frac {q^n -1}{q- 1}}=1$, is a cyclic group of order $\frac {q^n -1}{q- 1}$, and its image has order $q-1$ . This shows that $N$ is surjective. Note that the surjectivity of the trace map can be shown in the same way.

See also https://math.stackexchange.com/a/1728615/300700, Field Norm Surjective for Finite Extensions of $\mathbb{F}_{p^k}$ and Show the norm map is surjective