## When the norm map is surjective

$\mathbf{Problem:}$ Suppose $F$ is finite field with $q=p^m$ elements, $E$ an Galois extension of $F$ such that $[E:F]=n$. Prove that $N_{E/F}:E^\ast\to F^\ast$ is surjective.

$\mathbf{Proof:}$ From theorem 4.26 we know that $\text{Gal }E/F$ is cyclic over $F$ generated by $\eta:a\to a^q$.

$\displaystyle N_{E/F}(a)=aa^qa^{q^2}\cdots a^{q^{n-1}}=a^{\frac{q^n-1}{q-1}}$.

$N_{E/F}:E^\ast\to F^\ast$ is a homomorphism. It is well known that  any finite subgroup of the multiplicative group of field is cyclic, which means $E^\ast$ and $F\ast$ is cyclic.

Since $|E\ast|=q^n-1$ and $|F^\ast|=q-1$, and $N_{E/F}(a)=a^m$, where $m={\frac{q^n-1}{q-1}}$, then $|\text{Ker } N|=(q^n-1,m)=m$ and $|\text{Im }N|=\frac{q^n-1}{(q^n-1,m)}=q-1$. Thus $N$ is surjective.

$\text{Q.E.D}\hfill \square$

$\mathbf{Remark:}$ Jacobson p300.