Let $k$ be a number field, and let $\psi = \otimes_v \psi_v$ be the standard unitary character of $\mathbb A_k/k$ defined by Tate. Explicitly,
$$ \psi_v(x) = \begin{cases} e^{-2\pi i x} & \textrm{ if $k_v$ is real} \\ e^{-4\pi \operatorname{Re}(x)} & \textrm{ if $k_v$ is complex} \\ e^{2 \pi i [\operatorname{Tr}_{k_v/\mathbb Q_p}(x)]} & \textrm{ if $k_v$ contains $\mathbb Q_p$}\end{cases}$$
where $[x]$ denotes the composition $\mathbb Q_p \rightarrow \mathbb Q_p/\mathbb Z_p \xrightarrow{\cong} \mathbb Q/\mathbb Z \subset \mathbb R/\mathbb Z$.
Using $\psi_v$ to identify $k_v$ with its Pontryagin dual, one can show that $a \mapsto (x \mapsto \psi(ax))$ defines an isomorphism of $\mathbb A_k$ with its Pontryagin dual.
I would like to modify this isomorphism to show that this restricts to an isomorphism of $k$ with the Pontryagin dual of $\mathbb A_k/k$. The problem becomes to show that if $a \in \mathbb A_k$, and $x \mapsto \psi(ax)$ is a character of $\mathbb A_k$ which is trivial on $k$, then actually $a \in k$.
Let $$\chi : \Bbb{A_Q\to S^1 \subset C^*}, \qquad\chi(x) = \exp(-2i \pi x_\infty) \prod_p\exp(2i\pi( x_p \bmod \Bbb{Z}_p))$$
$\chi(\sum_p \frac{n_p}{p^{e_p}}) = \exp(-2i \pi \sum_p \frac{n_p}{p^{e_p}}) \prod_p \exp(2i \pi \frac{n_p}{p^{e_p}}) = 1$ thus its kernel contains $\Bbb{Q} + \hat{\Bbb{Z}}\times 0 + 0\times \Bbb{Z}$. This is the whole kernel because $\chi(x)$ is a finite product so any $p^e$-th root of unity appearing in the $p$ part must be cancelled by some piece of the $\infty$ part.
For a character $\phi : \Bbb{Q}_p \to S^1$, $\lim_{v_p(r)\to \infty} \phi(p^{-n})^r = 1$ implies $\phi(p^{-n}) = \exp(2i \pi \frac{A_n}{p^{d_n}})$, that $\phi(p^{-n})=\phi(p^{-n-1})^p$ means $a_p=\lim_{n \to \infty} \frac{A_n p^n}{p^{d_n}}$ converges in $\Bbb{Q}_p$ so that $\phi(x_p) = \exp(2i\pi (a_p x_p \bmod \Bbb{Z}_p))$.
Thus any character $\phi: \Bbb{A_Q}\to S^1$ is of the form $\phi(x)=\prod_{v \le \infty} \phi(x_v)=\exp(-2i\pi a_\infty x_\infty)\prod_p \exp(2i\pi (a_p x_p \bmod \Bbb{Z}_p))=\chi(ax)$ and the continuity implies $a_p \in \Bbb{Z}_p$ for almost every $p$ ie. $a\in \Bbb{A_Q}$.
As $\{ x \in \ker(\chi), \Bbb{Q}x\subset \ker(\chi)\} = \Bbb{Q}$ we find $x \mapsto \chi(ax)$ is a character $\Bbb{A_Q/Q}\to S^1$ iff $a\in \Bbb{Q}$.
$n = [k:\Bbb{Q}]$ $$\Bbb{A}_k = \Bbb{A_Q} \otimes_{\Bbb{Q}} k= \sum_{j=1}^n \beta_j \Bbb{A_Q}, \quad \beta_j\in k$$
Thus any character $\Bbb{A}_k \to S^1$ is of the form $$\psi(\sum_{j=1}^n \beta_j x_j) = \prod_{j=1}^n \chi(a_j x_j), \qquad a_1,\ldots,a_n \in \Bbb{A_Q}$$
It is a character $\psi : \Bbb{A}_k/k \to S^1 $ iff the $a_j \in \Bbb{Q}$.
On $\Bbb{A}_k$ the trace $\to \Bbb{A_Q}$ is defined by $Tr(\sum_j \beta_j x_j) = Tr_{k/\Bbb{Q}}(\beta_j) x_j$.
The $\Bbb{Q}$-linear map $\alpha \in k \mapsto (Tr_{k/\Bbb{Q}}(\alpha \beta_1),\ldots,Tr_{k/\Bbb{Q}}(\alpha \beta_n))$ is injective $k \to \Bbb{Q}^n$, thus it is surjective and for any $a_1,\ldots,a_n\in \Bbb{Q}^n$ there exists $\alpha \in k$ such that $Tr_{k/\Bbb{Q}}(\alpha \beta_j) = a_j $ and hence $$\psi(\mathrm{x}) = \chi(Tr(\alpha \mathrm{x}))$$
The map $\alpha \mapsto (\mathrm{x} \mapsto \chi(Tr(\alpha \mathrm{x}))$ is an isomorphism $k \cong Hom(\Bbb{A}_k/k , S^1)$.