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Let $K/\mathbb{Q}_p$ be a local field, $A$ the ring of integers of K, $\pi$ a uniformizer element for $A$, $F$ an n-dimensional formal group with coefficients in $A$ and $f$ an endomorphism of $F$. When we look at the reduction of $f$ $\pmod \pi$ it has the form $t(x^{p^h})$, where $X^{p^h}=(X_1^{p^h},\dots, X_N^{p^h})$ and the matrix, C(t), of linear coefficients of $t$ is not the zero matrix.

Do you know a proof-or a reference where I can look at- of the following theorem?

Theorem1:$f$ is an isogeny(i.e surjective and finite kernel) iff $c(t)$ is invertible. And the kernel is a porwer of $p$.

Thanks!!!

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    $\begingroup$ Take a look at M. Hazewinkel's Formal Groups and Applications. I think you'll find a proof or (at least) an idea of how you can prove the theorem. $\endgroup$
    – Octobris
    Commented Aug 14, 2013 at 7:50
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    $\begingroup$ Example (at $p=2$): Take the self-map of $\widehat{\Bbb G}_m \times \widehat{\Bbb G}_m$ given by multiplication by $2$ on the first factor and the identity on the second factor. Then $h=0$, and the matrix $C(t)$ is rank one, but the kernel is $\mu_2 \times \{1\}$. $\endgroup$
    – Tyler Lawson
    Commented May 26, 2014 at 21:55

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