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Let $p$ be a prime and denote by $\mathbb{Z}_p^{\times}$ the group of $p$-adic units. Suppose that $\chi$ is a character $\chi: \mathbb{Z}_p^{\times} \rightarrow \mathbb{C}^{\times}$. Then it is well known that

$\mathbb{Z}_p^{\times} \simeq (\mathbb{Z}/ p \mathbb{Z})^{\times} \times (1+ p \mathbb{Z}_p) $.

Do we have that the kernel of $\chi$ contains $(1+ p \mathbb{Z}_p)$ ? Any help or reference would be appreciated.

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    $\begingroup$ No, Pontryagin duality ensures continuous homomorphisms to the unit circle separate points. And actually it's not needed: just consider for $n\ge 2$ the reduction map $Z_p^*\to (Z/p^nZ)^*$. $\endgroup$
    – YCor
    Commented Aug 11, 2021 at 7:02

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YCor's example in the comments shows that you can have continuous characters with image an arbitrarily large finite group. In fact, there are even injective homomorphisms $\mathbb{Z}_p^\times \to \mathbb{C}^\times$ (if you don't ask for continuity): Observe that $\mathbb{C}^\times \cong \mathbb{Q}/\mathbb{Z}\times \bigoplus \mathbb{Q}$ as abstract abelian group, where $\bigoplus \mathbb{Q}$ is a $\mathbb{Q}$-vector space of continuum dimension.

For odd $p$, we have $\mathbb{Z}_p^\times \cong \mathbb{F}_p^\times \times \mathbb{Z}_p$. You can map the cyclic group $\mathbb{F}_p^\times$ injectively into the $\mathbb{Q}/\mathbb{Z}$ factor, and $\mathbb{Z}_p$ injectively into the continuum-dimensional $\mathbb{Q}$-vector space, since $\mathbb{Z}_p$ is torsion free and $\mathbb{Z}_p\otimes \mathbb{Q}$ is also a continuum-dimensional $\mathbb{Q}$-vector space.

For $p=2$, we abstractly have $\mathbb{Z}_2^\times \cong \{\pm 1\} \times \mathbb{Z}_2$, so the same argument works.

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