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.

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