Are there any textbooks which discuss/classify the injective group homomorphisms from $\mathbb Q$ (under addition) into $\mathbb C \setminus \{0\}$ (under multiplication)?

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    $\begingroup$ Isn't such a homomorphism $\phi$ determined by $\phi(1)$ since $\phi(a/b)=\phi(1)^{a/b}$ ? $\endgroup$
    – lhf
    Feb 2 '16 at 13:11
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    $\begingroup$ @lhf No, $\phi_{a,k} \colon x \mapsto \exp ((a+2\pi i k)x),\; k \in \mathbb{Z}$, are different homomorphisms with $\phi_{a,k}(1) = \exp(a)$ for all $k$. $\endgroup$ Feb 2 '16 at 13:17
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    $\begingroup$ Can you write a bit about why you're interested in $\mathbb{Q}$ and not $\mathbb{R}$? $\endgroup$ Feb 2 '16 at 13:21
  • $\begingroup$ I'd remark that, since $(\mathbb{C} \setminus \{0\},\times,1)$ is isomorphic to $(\mathbb{R}_{> 0} ,\times,1) \times (\mathbb{R}/2\pi\mathbb{Z},+,0),$ hence if you're not too fussed about injectivity, you can reduce the problem to finding all group homomorphisms $\mathbb{Q} \rightarrow (\mathbb{R}_{> 0} ,\times,1)$ and all group homomorphisms $\mathbb{Q} \rightarrow \mathbb{R}/2\pi\mathbb{Z}.$ I don't know if this is any easier though. Note also that $(\mathbb{R}_{> 0} ,\times,1) \cong (\mathbb{R},+,0).$ $\endgroup$ Feb 2 '16 at 13:23
  • $\begingroup$ @DanielFischer, right! Thanks. $\endgroup$
    – lhf
    Feb 2 '16 at 13:53

The multiplicative group $\mathbb C^*$ is the product of $\mathbb R^+_0$ with the unit circle $C$. Any homomorphism decomposes into a modulus and an argument that way, and its kernel is the intersection of the kernels of the two components. So the set you are looking for is the cartesian product of the homomorphisms to the unit circle with the homomorphisms to the positive reals, minus the pairs where neither member is injective.

Group homomorphisms $\mathbb Q\to\mathbb R^+_0$ are uniquely determined by how they map $1,$ since $n$-th roots are unique in $\mathbb R^+_0.$ Conversely, every positive real $r$ defines a homomorphism

$$\varphi_r:\mathbb Q\to\mathbb R^+_0:q\mapsto r^q.$$

Thus the second component of our cartesian product is indexed by the positive real numbers $r.$ The homomorphisms are injective for all $r\neq1.$

The interesting part of the question is to describe the injective homomorphisms from $\mathbb Q$ to the unit circle.

A homomorphism $\psi:\mathbb Q\to C$ is determined by its values on reciprocals of prime powers. There, of course, we do not have complete freedom of choice because we have to satisfy the constraints

$$\psi(p^{-n})=\psi(p^{-(n+1)})^p,\ n=0,1,\ldots$$

For a complete description of the unitary characters of the additive group of rational numbers I would like to refer to older answers in this forum such as Representation theory of the additive group of the rationals?

For a unitary character $\psi$ to be injective a necessary and sufficient condition is that $\pi^{-1}\arg\psi(1)\notin\mathbb Q.$

  • $\begingroup$ Does the situation change drastically if the injective condition is dropped? $\endgroup$
    – Jack M
    Feb 2 '16 at 15:09
  • $\begingroup$ No, it doesn't. The only non-injective ones map $1$ to a complex root of unity (and therefore they automatically have $r=1$). $\endgroup$ Feb 2 '16 at 18:54

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