Why is the image of $k^+$ dense in the character group?

Solution 1:

If $C$ is the closure of the image of $\phi$, then $\widehat{k^+}/C$ is a locally compact Hausdorff abelian group under the quotient topology. To show $C$ is all of $\widehat{k^+}$, we just need to show that every character on $\widehat{k^+}/C$ is trivial. Let $\psi: \widehat{k^+}/C \to S^1$ be a character. Then we get a character on $\widehat{k^+}$ by composing:

$$\widehat{k^+} \stackrel{quot}{\to} \widehat{k^+}/C \stackrel{\psi}{\to} S^1.$$

By the Pontryagin duality isomorphism $k^+ \cong \widehat{\widehat{k^+}}$, this character is of the form $Y \mapsto Y(\zeta)$ for some $\zeta \in k^+$. Moreover, since this character factors through the quotient, it vanishes on $C$, which is equivalent to its vanishing on the image of $\phi$, which is precisely to say that $X(\eta\zeta) = 1$ for every $\eta \in k^+$. If we suppose Tate's implication holds, we infer that $\zeta = 0$, so that the character $\psi$ had to be trivial, as desired.