An example of $p$-divisible group In page 76 of Demazure's book "Lectures on $p$-divisible groups". The formal group $G^{\lambda}$ defined by the exact sequence
$$ 0\to G^{\lambda} \to W(p) \xrightarrow{F^r-V^s} W(p)  $$
is claimed to be a $p$-divisible group over $\mathbb{F}_p$, where $W(p)=\varinjlim(\mathbf{Ker}p^n: W_{\mathbb{F}_p}\to W_{\mathbb{F}_p})$.
Question 1. Why this is a $p$-divisible group?
Question 2. It is claimed that its Dieudonné module is $\mathbb{Z}_p[F]/(F^{r-s}-V^s)$, how should I understand this module (the "$V$" is not well-defined inside  $\mathbb{Z}_p[F]$? ) and why this fact is true?
ps. As I understand, $W_{\mathbb{F}_p}$ can be written (as a group scheme): $$W_{\mathbb{F}_p}=\prod_{n\geq 0}\mathbb{A}^1_{\mathbb{F}_p}=\mathbf{Spec}  \mathbb{F}_p[X_0, X_1, \cdots].$$ (This question is part of the following question---not answered yet: An example of $p$-divisible group from the book of Demazure.)
 A: Notations:
$$D_k=W(k)[F, V]/(Fa=a^pF, aV=Va^p, FV=VF=p \text{ for all } a\in W(k) )$$
$$W_k=\mathsf{Spec}(k[t_0, t_1, \cdots])$$
$$K_n:=\mathbf{Ker}(p^n\cdot id_{W_k})\subset W_k$$
$$W(p)_{k}=\varinjlim_n K_n$$
$${}_{n}W_m=\mathsf{Spec}(k[t_0, \cdots, t_{n-1}]/\big( t_0^{p^m}, \cdots, t_{n-1}^{p^m}  \big))$$
Lemma:
Let $k$ be a perfect field and $f(x, y)\in W(k)[x,y]$ a polynomial with $deg_{y}f>0$. The formal group $G=\mathsf{Ker}(f(F, V):W(p)_k\to W(p)_k)$ is a $p$-divisible group with Dieudonné module
$$ M(G)=\varprojlim_n D_{k}/(F^n, f(F, V)).$$
Proof:
Put $G_n=\mathsf{Ker}(f(F, V):K_n\to K_n)$, then $G=\varinjlim_{n} 
 G_n$
(as $\{ K_n \}_{n\geq 0}$ form a filtered system, its injective limit preserves kernel).
We define $G_{n,m}$ by the following exact sequence
$$0\to G_{n,m} \to {}_{n}W_m \xrightarrow{f(F, V)} {}_{n}W_m. $$
Taking the inverse limit (remark that inverse limit commutes with kernel) and we have $G_{n}=\varprojlim G_{n,m}$.
We know that $M({}_{n}W_m)\simeq D_{k}(F^n, V^m)$, and hence we have
$$ M({}_{n}W_m)\xrightarrow{f(F, V)} M({}_{n}W_m) \to M(G_{n, m}) \to 0, $$
and hence $M(G_{n,m})= D_k/(F^n, V^m, f(F, V))$.
Then we have
$$M(G_n)=\varinjlim_m M(G_{n,m})=\varinjlim_m D_k/(F^n, V^m, f(F, V))$$
$$= D_k/(F^n, f(F, V)).$$
Hence
$$ M(G)=\varprojlim_n D_{k}/(F^n, f(F, V)).$$
This gives the Dieudonné module of $G$.
Conversely, by the equivalence of category given by Demazure (in page 71), we can construct a $p$-divisible group $G^{'}:=\varinjlim_n G_n^{'}$ where $G_n^{'}$ is defined to be the group scheme whose Dieudonné module is $M(G)/p^n$. Then we check that $G=G^{'}$.
(Here needs some details: We can see that $M(G_{n(r+s)}^{'})\twoheadrightarrow M(G_{n(r+s)}) \twoheadrightarrow M(G_n^{'})$ and hence $M(G)=M(G^{'})$, but then how to conclude? Shouldn't we have even $G^{'}_n=G_n$, as both should be $G[p^n]$?)
Example:
When $k=\mathbf{F}_p$, and $f(F, V)=F^r-V^s$, we have
$$ M(G)=\mathbf{Z}_p[F, V]/(FV-p, VF-p, F^r-V^s). $$
A: I think you have to first check the claim about Dieudenne module. for that you have to understand for each n $M(G[p^n])$ which you can easily compute from the exact sequence and the fact that you now the Dieudenne module of $ker(p^n:W_{F_p}\to W_{F_p})$.and then take the limit.$V$ is coming from here because it is defined on each $M(G[p^n])$ and hence on the limit.
now you can use the equivalence between p-divisible groups and (M, F, V) modules with $M$ free stated at the page 71 of the book.
