 Solution

For $K=\mathbb{Q}(\zeta )$, every Galois automorphism sends the primitive root $\zeta$ to another primitive $m$th root of unity, uniquely ${\zeta }^a$ for $a\in (\mathbb{Z}/m\mathbb{Z}{)}^{\times }$. The Cyclotomic character of a finite cyclotomic extension isIf ${\chi }_m(\sigma )=a$ and ${\chi }_m(\tau )=b$, then $(\sigma \tau )(\zeta )=\sigma ({\zeta }^b)={\zeta }^{ab}$, proving the group homomorphism property. Its kernel is trivial: an automorphism fixing $\zeta$ fixes the field it generates. Thus the character is injective. Here its domain is the finite cyclotomic Galois group; the corresponding map from an absolute Galois group need not be injective.