Let $k$ be a perfect field of characteristic $p>0$. I'm trying to calculate the Dieudonné module $D(\alpha_{p^2,k})$ "explicitly". Let $M:=D(\alpha_{p^2})$.
Since $\alpha_{p^2}$ is of order $p^2$, $M$ has $W(k)$-length 2. In addition, $\alpha_{p^2}$ is killed by $p^2$ and not $p$, so $M\simeq W(k)/p^2W(k)$ as a $W(k)$-module.
The module $W(k)/p^2W(k)$ is $W(k)$-cyclic, so I think it's enough to specify the action of $F$ and $V$ on $\bar{1}:=1+p^2W(k)$. The only way $F$ can have kernel of length 1 is if $F(\bar{1}) = \bar{p}$, hence $F^2 = p^2 = 0$. On the other hand, $FV = VF = p$, so this seems to imply that $V = \operatorname{id}$. But $V$ should be nilpotent.
Question 1: What have I done wrong and how do I actually calculate V?
From a result in Pink's notes on finite group schemes https://people.math.ethz.ch/~pink/ftp/FGS/CompleteNotes.pdf (after Theorem 23.2), if I note that $\alpha_{p^2} = W_1^2 = \operatorname{ker}F^2|_{\mathbb G_a}$, then $$D(\alpha_{p^2}) \simeq E/(EF^2+EV)$$ where $E = W(k)[F,V]$ with the usual relations.
Question 2: Does this mean that $V = 0$ and $F^2 = 0$? If so, how do I reconcile this with $VF = FV= p$?
This is basically rephrasing Q1. Any help would be much appreciated!