Let $F$ be a p-adic field. A character on $F^\times$ is defined as a continuous group homomorphism $F^\times\longrightarrow\mathbb{C}^\times$. But is there any way to construct a non-continuous group homomorphism $F^\times\longrightarrow\mathbb{C}^\times$? Is there even any such homomorphism?
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1$\begingroup$ @EmilJeřábek you are constructing homomorphisms on the additive group $F$, but the question is about homomorphisms on the multiplicative group $F^\times$, which is not a vector space, so your basis argument does not adapt to $F^\times$. $\endgroup$– KConradCommented Jul 15, 2023 at 17:18
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2$\begingroup$ It is easy to show that there are $2^c$ homomorphisms (but only $c$ continuous ones). On the other hand there is no way to "construct" an explicit non-continuous one. $\endgroup$– YCorCommented Jul 15, 2023 at 17:28
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$\begingroup$ @KConrad Oh, you are right, I misread the question. $\endgroup$– Emil JeřábekCommented Jul 15, 2023 at 17:29
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1 Answer
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Since $\mathbf C^\times$ is a divisible group, Zorn’s lemma tells us for every abelian group $G$ and subgroup $H$ that each group homomorphism $H\to \mathbf C^\times$ extends (somehow) to a group homomorphism $G\to\mathbf C^\times$.
Use $G = F^\times$ and $H = \langle u\rangle$, where $u$ is in $\mathcal O_F^\times$ and not a root of unity, so $H$ is isomorphic to $\mathbf Z$. Start with the homomorphism $H\to \mathbf C^\times$ where $u^k \to 2^k$ and extend to a homomorphism with domain $F^\times$ by Zorn’s lemma. This homomorphism is not continuous since $\mathcal O_F^\times$ is compact but the homomorphism on this subgroup of $F^\times$ is discontinuous because its image in $\mathbf C^\times$ is unbounded and thus not compact.
