Let $\Bbb A_K'$ be the dual to the group of adeles $\Bbb A_K$ of some field $K$. Then $\Bbb A_K'$ is an $\Bbb A_K$ module by the prescription $$a\cdot \Psi(x) \mapsto \Psi(ax)$$
But why is $\Bbb A_K'$ in fact a 1-dimensional $\Bbb A_K$ module?
Note: I said 1-dimensional because that's what I have in my notes. I don't know what is a 1-dimensional module, but I believe it means rank 1.
EDIT: Actually, I know in fact that $\Bbb A_K'\cong \Bbb A_K$. But there too many (not difficult, but numerous) things to check in this proof. For example, one would first show that $\Bbb A_K \cong \Bbb A_{\Bbb Q} \otimes K$ and then, one could use that $\Bbb A_{\Bbb Q}'\cong \Bbb A_{\Bbb Q}$ because it is a restricted product and $\Bbb Q_\lambda' \cong \Bbb Q_\lambda$ and $\Bbb Z_\lambda^\perp \cong \Bbb Z_\lambda$, where the $_\lambda$ indiicates completion at some place. But I was looking for a straightforward proof of just the fact that I asked here: that $\Bbb A_K'$ is a one-dimensional $\Bbb A_K$ module.