Lubin-Tate formal groups are $p$-divisible groups I am trying to understand how to see whether a given formal group is $p$-divisible.
Let $A$ be a complete noetherian local ring with maximal ideal $\mathfrak{m}$ and residue field $k$ of characteristic $p$ and let $F \in A[[X,Y]]$ be a formal group law. 

Is there a nice criterion to see whether the map $$[p]^* \colon A[[X]] \to A[[X]], f(X) \mapsto f([p](X))$$ is injective and makes $A[[X]]$ into a finite free module over $A[[X]]$? (This is the definition of $F$ being $p$-divisible.)

My main case of interest is when $A=\mathfrak{o}$ is the ring of integers in a finite extension of $\mathbb Q_p$. Lubin's answer to the question How is the $p$-adic Tate module of a formal group defined? seems to imply that 

$$F \,\,\text{    is } p\text{ -divisible } \iff [p] \text{  mod } \pi \neq 0 \text{ in } k[[X]]$$ 

where $\pi$ denotes a uniformizer of $\mathfrak o$. Is this true? I would very much be thankful for a proof of this!
 A: There are probably several proofs of this, and I would almost make a wager that the slickest of these would dispose of the matter with the right appeal to Nakayama’s Lemma. But let me be historical and tell you how I have looked at it.
I’m supposing that you’re aware that we’re always dealing with formal groups which are not isomorphic to the additive formal group $x+y$ over $k$. And I’m supposing that you know the result (due to Lazard, I suppose) that the first nonzero coefficient of $[p]_{\tilde F}$ (where $\tilde F$ is the formal group over $k$) must be in degree $p^h$, and that this $h$ is called the height of $F$.
This means that the first unit coefficient of $[p]_F$ appears in degree $p^h$.
Let $f=[p]$. Now, within $A[[x]]$ we have the subring $A[[f]]$, and I think you can easily see that $A[[f]]\cong A[[T]]$, power-series ring in one indeterminate over $A$. So, to avoid confusion, I’m going to name $f(x)=T$, so that we have $A[[T]]\subset A[[x]]$, and I am about to show that the big ring is free over $A[[T]]$. Consider $A[[T]][[X]]\big/\bigl(f(X)-T\bigr)$, which I think you see is isomorphic to $A[[x]]$ via $X\mapsto x$. But now I would call in Weierstrass Preparation, in this form: Let $\mathcal O$ be a complete local ring, and and let $\phi(X)\in\mathcal O[[X]]$ have first unit coefficient in degree $n$. Then there is, uniquely, a pair $(g,U)$ where $g\in\mathcal O[X]$, monic polynomial of degree $n$; and $U\in\mathcal O[[X]]^\times$, power series with constant term a unit of $\mathcal O$, such that $\phi=gU$.
And of course that does it for us, letting $\phi=f(X)-T$ above, and $\mathcal O=A[[T]]$.
If you have further questions, don’t hesitate to ask. As for W-Prep, I claim that every mathematician that sits down to prove it will give a different proof. It is definitely not as deep a fact as Hensel’s Lemma.
