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So for a one dimensional Galois representation $\rho: G_{\Bbb Q} \to \mathbb C^{\times}$, I know that it must factor through the abelianization of $G_{\Bbb Q}$, which by the Kronecker-Weber theorem is the Galois group of the maximal cyclotomic extension of $\Bbb Q$. I want to conclude from this that $\rho$ factors through a Dirichlet character, that is, a representation of the Galois group of some finite cyclotomic extension. I've seen this question: Complex Galois Representations are Finite

This definitely gives me the answer, but my question is, can I do this without assuming $\rho$ is continuous? In general how important is the continuity assumption when talking about Galois representations? I'm wondering if this is a purely algebraic fact or if it only applies to continuous representations. I would absolutely accept a reference in lieu of a written answer, surely this is written somewhere but I haven't been able to find it.

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    $\begingroup$ Continuity merely means that the representation factors through a finite quotient, that is, comes from a finite subextension. $\endgroup$ Jan 9 '14 at 4:22
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    $\begingroup$ The topology on the Galois group is nothing but a convenient (a very convenient!) way of saying that sort of thing. $\endgroup$ Jan 9 '14 at 4:23
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    $\begingroup$ Are there examples where the representation doesn't factor through a finite quotient when the representation isn't continuous? $\endgroup$
    – Dylan Yott
    Jan 9 '14 at 4:37
  • $\begingroup$ By definition, continuity is equivalent to factoring through a finite quotient. $\endgroup$ Jan 9 '14 at 4:43
  • $\begingroup$ Okay, so then my question reduces to if there exists an abstract group homomorphism from $\hat{\Bbb Z}$ to $\Bbb C^\times$ which doesn't factor through a finite quotient. It's enough to find a homomorphism from $\Bbb Z_p$ to $\Bbb C^\times$ which doesn't factor through a finite quotient. From here I'm not sure how to proceed. I'm not used to forgetting about the topology on $\Bbb Z_p$. $\endgroup$
    – Dylan Yott
    Jan 9 '14 at 4:57
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As an abstract group, $\Bbb Z_p$ is $q$-divisible for any prime $q \neq p$ and uncountable, while no element is infinitely divisible by $p$, so it is a direct sum of uncountably many copies of $\Bbb Z_{(p)}$ (the localisation of $\Bbb Z$ at $p$).

There are uncountably many group morphisms $\Bbb Z_{(p)} \to \Bbb C^*$ (choose the value at $1$, then for each prime $q \neq p$, you have $q$ choices for the value at $q^{-1}$, again $q$ choices for the value at $q^{-2}$, and so on), so yes, there are many group morphisms $\Bbb Z_p \to \Bbb C^*$, and many group morphisms $\hat {\Bbb Z} \to \Bbb C^*$

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    $\begingroup$ Why does $q$-divisible for $q\ne p$ and uncountable imply $\Bbb Z_p$ is a direct sum of copies of $\Bbb Z_{(p)}$? (While I am familiar with the theory of finitely generated modules over PIDs, this group is infinitely generated and I don't know the machinery for that stuff.) $\endgroup$
    – anon
    Jan 30 '14 at 7:33
  • $\begingroup$ @anon : hmm I was thinking that for $G$ abelian + torsionfree + $q$-divisible for $q \neq p$ + no element being infinitely $p$-divisible ; we can pick an element that is not a $p$-multiple, then find an embedding of $\Bbb Z_{(p)}$ sending $1$ to that element, which gives us a quotient group $G/\Bbb Z_{(p)}$ that has all the same properties. However I'm stuck proving that the last properties transfers to the quotient =( . So you are right this is not a complete answer. $\endgroup$
    – mercio
    Feb 2 '14 at 11:45
  • $\begingroup$ It is not true that $\mathbb{Z}_p$ is a direct sum of copies of $\mathbb{Z}_{(p)}$. Indeed, $\mathbb{Z}_p/p\mathbb{Z}_p$ is just $\mathbb{Z}/p$, but if $\mathbb{Z}_p$ were a direct sum of copies of $\mathbb{Z}_{(p)}$ then this quotient would be an uncountable direct sum of copies of $\mathbb{Z}/p$. $\endgroup$ May 1 '16 at 21:46
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A variant on mercio's answer is that $\mathbb Z_p/\mathbb Z$ is a uniquely divisible abelian group, i.e. a $\mathbb Q$-vector space. It has cardinality, and hence dimension, equal to the continuum. Thus it is isomorphic to $\mathbb C$ as a $\mathbb Q$-vector space. Any such isomorphism has infinite order (obviously) and is also incredibly far from being continuous.

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    $\begingroup$ The relevant quotient is $\mathbb Z_p/\mathbb Z_{(p)},$ where $\mathbb Z_{(p)}$ is the localization of $\mathbb Z$ at the prime ideal $(p)$. We have that $\mathbb Z_{(p)} = \mathbb Z_p \cap \mathbb Q$, and so $$\mathbb Z_p/\mathbb Z_{(p)} = \mathbb Z_p/(\mathbb Z_p \cap \mathbb Q) = (\mathbb Z_p + \mathbb Q)/\mathbb Q = \mathbb Q_p/\mathbb Q,$$ and hence is a $\mathbb Q$-vector space. On the other hand $ \mathbb Z_p/\mathbb Z$ is not torsion-free; it contains $$\mathbb Z_{(p)}/\mathbb Z = (\mathbb Z_{(p)} + \mathbb Z[1/p])/\mathbb Z[1/p] = \mathbb Q/\mathbb Z[1/p]$$ as its torsion subgroup. $\endgroup$
    – tracing
    Dec 25 '14 at 1:08
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The comment by @Mariano Suárez-Alvarez "By definition, continuity is equivalent to factoring through a finite quotient." is completely wrong. In particular, there are uncountably many surjective homomorphisms from the Galois group of $\mathbf{Q}$ to a cyclic group of order $2$, but only countably many of them are continuous, namely the ones factoring through the Galois group of the coutably many quadratic fields. So even for characters of degree $2$, it is important to consider continuity.

Added: This (naturally) provides an answer to your question: Even for characters $G_{\mathbb{Q}} \rightarrow \mu_2 \subset \mathbb{C}^{\times}$ which have finite image, some continuity is required to deduce that one has a Dirichlet character (there are only countably many such characters).

BTW: The moderator who summarily deleted the previous version of this answer as "This does not provide an answer to the question" should refrain from making mathematical judgements on questions which they apparently have no mathematical background. It is honestly embarassing that an answer both correcting important misconceptions in the comments and also answering the actual question would be deleted.

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    $\begingroup$ This does not provide an answer to the question. Once you have sufficient reputation you will be able to comment on any post; instead, provide answers that don't require clarification from the asker. - From Review $\endgroup$ May 1 '16 at 16:07
  • $\begingroup$ This answer was never "summarily deleted" by any moderator. I really suggest you avoid telling people what to refrain from (even if you have your intel right) $\endgroup$ May 1 '16 at 20:17
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    $\begingroup$ @MarianoSuárez-Alvarez, the fact that this answer was deleted was appalling. Either it reflects that this site values process over mathematics, or it incidates the presence of a rogue user with moderator powers who should be reigned in. At the very least, if a user thinks that a (mathematically pertinent) answer should be a comment, it should be converted into a comment, not deleted. $\endgroup$
    – user335783
    May 1 '16 at 20:45
  • $\begingroup$ @MarianoSuárez-Alvarez, I really suggest you avoid telling people mathematically incorrect statements on subjects not in your field of expertise. Everyone makes mistakes, but you have the extra burden of being a moderator with a high "reputation," which users might mistakenly imagine has some correlation with expertise. $\endgroup$
    – user335783
    May 1 '16 at 20:48
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    $\begingroup$ @MarianoSuárez-Alvarez, I don't understand your comment. Are you claiming that a finite index normal subgroup of $G_{\mathbb{Q}}$ is open? $\endgroup$
    – user335783
    May 1 '16 at 22:02

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