Continuous homomorphisms from $\widehat{\mathbb Z}$ to $\overline{\mathbb F}_p^\times$

Question

I would like to compute the continuous homomorphisms from $\widehat{\mathbb Z}$ to $\overline{\mathbb F}_p^\times$, where $\widehat{\mathbb Z} = \varprojlim_N \mathbb Z/N\mathbb Z$ has the profinite topology and $\overline{\mathbb F}_p^\times$ the discrete topology. I believe such homomorphisms are built from maps $\mathbb Z/N\mathbb Z \to \mathbb F_{p^f}^\times \cong \mathbb Z/(p^f -1) \mathbb Z$, of which there are $\gcd(N, p^f-1)$, but I don't know how to organise these in the limit.

Answer 1

Let $A$ be any discrete torsion group. Then $\mathrm{Hom}_{\text{cont}}(\widehat{\Bbb Z},A) \cong A, f\mapsto f(1)$. Let's prove this. Note that clearly the map $\varphi: \mathrm{Hom}_{\text{cont}}(\widehat{\Bbb Z},A) \to A, f \mapsto f(1)$ is a well-defined homomorphism. Let's see that it is bijective.

Injectivity: $\Bbb Z \subset \widehat{\Bbb Z}$ is dense and $A$ is Hausdorff, thus any continuous map $\widehat{\Bbb Z} \to A$ is uniquely determined by its restriction to $\Bbb Z$. In other words, the restriction map $\mathrm{Hom}(\widehat{\Bbb Z},A) \to \mathrm{Hom}(\Bbb Z,A)$ is injective. Also the map $\mathrm{Hom}(\Bbb Z,A) \to A, f \mapsto f(1)$ is well-known to be an isomorphism. Thus the composition $\mathrm{Hom}(\widehat{\Bbb Z},A) \to \mathrm{Hom}(\Bbb Z,A) \to A$, which is just $\varphi$, is injective as well.

Surjectivity: Let $a \in A$ have order $N$. Then we have a homomorphism (continuous, as both sides are discrete) $\Bbb Z/N\Bbb Z \to A, \overline{1} \mapsto a$. We can compose this with the projection $\widehat{\Bbb Z} \to \Bbb Z/N\Bbb Z$ to obtain a continuous homomorphism $\widehat{\Bbb Z} \to A$ such that $1 \mapsto a$.

In summary, we have a bijection $\mathrm{Hom}_{\text{cont}}(\widehat{\Bbb Z},A) \cong A, f\mapsto f(1)$, the map in other direction is given by $a \mapsto (n \mapsto a^n)$. To make sense of $a^n$ for $n$ profinite, one can define $a^n=a^{n_m}$, where $m$ is the order of $a$ and $n_m = n \pmod{m}$.