# Why is the kernel of the Verschiebung morphism either étale either connected?

Consider an elliptic curve $E$ over an algebraically closed field $k$ of characteristic $p>0$. We have the Fröbenius morphism $$F:E\rightarrow E^{(p)}$$

and the Verschiebung morphism $V:E^{(p)}\rightarrow E$ which actually is its dual, so that the composition $$E\xrightarrow FE^{(p)}\xrightarrow VE$$

is the multiplication by $p$ on $E$.

The book I am reading (Katz and Mazur's "Arithmetic moduli of elliptic curves") then claims

Therefore $\operatorname{Ker}(V)$ in $E^{(p)}$ is either étale, either connected, in which case it is equal to $\operatorname{Ker}(F^{(p)})$ in $E^{(p)}$, the unique connected subgroup-scheme of $E^{(p)}/k$ which has rank $p$.

I would like to find a justification for this last statement. How can one see that the kernel of $V$ is either étale, either connected? To what extent is it implied by the above composition? Also, why does $E^{(p)}/k$ only have one connected subgroup-scheme of rank $p$ up to isomorphism?

I thank you very much for your help.

About your first question, since the kernel has order $p$, it has no no-trivial proper subgroup schemes, so the connected component of the identity is either trivial, hence it is etalé, or total, thus connected.