Let $ \phi: A \rightarrow B$ be a separable isogeny between two abelian varieties over a field $k$. One knows that there is a dual isogeny $ \hat {\phi} : B \rightarrow A$ such that $ \hat{\phi} \circ \phi = $ multiplication by $ \mathrm{deg}(\phi)$.

When I studied elliptic curves and abelian varieties, most of the references deal with a base field which is perfect. In this case, the proof for the existence of the dual isogeny is as following:

One makes a base change to work on the algebraic closure $\overline{k}$ of $k$. One considers $\mathrm{ker} (\phi)$, the closed points of the fiber of $\phi$ at the origin of $B$. It's a finite group which acts on $A$ and we have $\mathrm{ ker } (\phi) \subset \mathrm{ker} ( \times \mathrm{deg} (\phi))$. Then using the results about the quotien of a scheme by a finite group, one get the dual isogeny $\hat{\phi}$. Finally, one uses the action of $\mathrm{Gal}(\overline{k}/k)$ to get the result on the original base filed $k$.

Now if $k$ is not perfect, I didn't figure out how to do this. I have a feeling that the reason we work on the algebraic closure $\overline{k}$ is to get a finite group action. Of course, one can think about the action of $\mathrm{ker} (\phi)(k)$, but I think it doesn't work, because it's too small (We need all its closed points).

I would like to know if the dual isogeny exists in a general base field. If yes, what's the idea to see it, (and some reference)