cyclic extensions, the kernel Ker N consists of all the elements of the form f(y) := Fr(y )/y , with y running through E*. Since Ker f = F q *, it follows that Im f = Ker NĀ ...
Norms in finite fields Here is a purely Galois theoretic solution of the surjectivity of the norm for finite fields. Consider the extension E/F q of degree m, and let N be the norm map between the multiplicative groups. It is well known that Gal(E/Fq) is cyclic, generated by the Frobenius automorphism Fr defined by Fr(x) = š„ š . 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 E*. Since Ker f = F q *, it follows that Im f = Ker N has order šš - 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 Class Field Theory tells us that for an abelian extension L/K of such fields, Gal(L/K) is isomorphic to K */ N L * .
