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I do not understand the following argument by Marcus:

Let $K|\mathbb Q$ an abelian extension where $K$ is a number field. We identify $\mathbb Z_m^*$ with the Galois group of $\mathbb Q(\omega)$ over $\mathbb Q$.[This is fine]

Then $G$ is a homomorphic image of $\mathbb Z_m^*$.[What does it mean? It means maybe that $G$ is isomorphic to a quotient of $\mathbb Z_m^*$? Because this would be fine for me.]

Hence characters of $G$ can then be regarded as characters $mod$ $m$.[Why? What does it mean?]

Thus we consider the group of characters of $G$ as a subgroup of the group of characters of $\mathbb Z_m^*$. [why? Because if $G$ is isomorphic to subgroup of $\mathbb Z_m^*$ this is fine]

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Let $S$ be a subgroup of a finite abelian group $T$. We can identify $X(T/S)$ as a subgroup of $X(T)$: namely, $X(T/S)$ consists of those characters $\chi \in X(T)$ for which $\chi(s) = 1$ for all $s \in S$. This follows from the fact that every character of $S$ extends to a character of $T$.

Now let $T = \textrm{Gal}(\mathbb{Q}(\omega)/\mathbb{Q})$, where $\omega$ is a primitive $m$th root of unity, and let $S = \textrm{Gal}(\mathbb{Q}(\omega)/K)$. Then restriction $\sigma \mapsto \sigma|K$ induces an isomorphism of $T/S$ with the Galois group $\textrm{Gal}(K/\mathbb{Q})$. Now we can identify the characters of $\textrm{Gal}(K/\mathbb{Q})$ with those characters of $T$ which are trivial on $S$.

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  • $\begingroup$ About your first part, one should prove then the map $f:X(T/S)\to X(T)$ such that $f(\chi)=\chi\times \pi$ is injective? where $\pi$ is the canonical projection. By $\times$ I mean composition of maps. Anyway, is this proof trivial because $\pi$ admits a right inverse? $\endgroup$ – Richard Jan 23 '17 at 18:20
  • $\begingroup$ Yes, it should be an injection. And then you have to be able to say what the image in $X(T)$ is. $\endgroup$ – D_S Jan 23 '17 at 18:22
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I guess that it is assumed that $K$ is a subfield of $\mathbb{Q}[\zeta_m]$. Since the Galois group of $\mathbb{Q}[\zeta_m]/\mathbb{Q}$ is $\mathbb{Z}_m^*$, the Galois group of the subextension $K/\mathbb{Q}$ is a homomorphic image of it, namely there is a surjection $Gal(\mathbb{Q}[\zeta_m]/\mathbb{Q})\to Gal(K/\mathbb{Q})$, which takes the Galois action on $\mathbb{Q}[\zeta_m]$ and restricts it to $K$. A character on $Gal(K/\mathbb{Q})$ can then be lifted to a character on $Gal(\mathbb{Q}[\zeta_m]/\mathbb{Q})$.

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  • $\begingroup$ sorry but how do you lift the character? Because you pick a character of the quotient of $G=Gal(Q(\omega)|Q)$ isomorphic to $Gal(K|Q)$, and then what? Why should it define a subgroup of $\mathbb Z_m^*$? Thank you very much $\endgroup$ – Richard Jan 23 '17 at 17:53
  • $\begingroup$ @Richard A character is just a homomorphism to $\mathbb{C}^*$. So if you have a homomorphism from the quotient group, then composing it with the projection produces a homomorphism from $Gal(\mathbb{Q}[\zeta_m]/\mathbb{Q})$. $\endgroup$ – Ofir Jan 23 '17 at 18:00
  • $\begingroup$ ok this is fine, but why doing this for all the characters of the quotient produce a subgroup? $\endgroup$ – Richard Jan 23 '17 at 18:02
  • $\begingroup$ @Richard You should check that if $\psi:G\to H$ is a surjective homomorphism, then $\psi^*:Hom(H,\mathbb{C}^*)\to Hom(G,\mathbb{C}^*)$ is injective, and moreover, it preserves the group product, namely if $\chi,\tau\in Hom(H,\mathbb{C}^*)$, then $psi^*(\chi\tau)=\psi^*(\chi) \psi^*(\tau)$. $\endgroup$ – Ofir Jan 23 '17 at 18:06
  • $\begingroup$ ok. But does it mean that the group of characters of $Gal(K|Q)$ is isomorphic to a subgroup of the group of characters of $\mathbb Z_m^*$ ?Because for sure we have correspondence by what you said $\endgroup$ – Richard Jan 23 '17 at 18:12

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