The group $\mathbb{R}/\mathbb{Z}$ rather famously is isomorphic to the algebraic group given by the variety $X^2 + Y^2 =1$ over $\mathbb{R}^2$ with group operation given by $(X_1,Y_1)\cdot (X_2,Y_2) = (X_1X_2 - Y_1Y_2, X_1Y_2+X_2Y_1)$, and further homeomorphic in the analytic topology. Is $\mathbb{Q}_p/\mathbb{Z}_p$ isomorphic to an algebraic group over $\mathbb{Q}_p^n$, for $n>0$ or some power of an extension?
The topology seems to be an obstruction for a perfect analogue as $\mathbb{Q}_p/\mathbb{Z}_p$ is discrete and a variety of dimension higher than 0 over $\mathbb{Q}_p$ is presumably not in the analytic topology.
My attempt: Suppose $G$ is a group variety over some power of an extension of $\mathbb{Q}_p$ isomorphic to $\mathbb{Q}_p/\mathbb{Z}_p$. Endowing it with the analytic topology from $\mathbb{Q}_p$ we note that this is Hausdorff and a group topology. But any Hausdorff group topology on $\mathbb{Q}_p/\mathbb{Z}_p$ is discrete, hence our group variety must be discrete and thus $0$-dimensional.
Are there any significant mistakes or omissions in the above argument?