Lubin-Tate modules and different uniformizers

Question

Suppose I have a local field $\mathcal{O}_K$ and two different prime elements $\pi$ and $\overline{\pi},$ i.e they differ by a unit $\overline{\pi} = u \pi$ for some $u \in \mathcal{O}_K^{\times}$ not equal to $1.$ Let us suppose I have a Lubin Tate module $F_\pi$ for the prime element $\pi$ and a Lubin-Tate module $F_{\overline{\pi}}$ for $\overline{\pi}.$ I have heard it been claimed that in this case, $F_{\overline{\pi}}$ and $F_{\pi}$ are never isomorphic. Is there an easy way to see why this is true?

Answer 1

Another way of seeing this is by way of the totally ramified extension $\mathcal L_\pi$ generated by the torsion points of $F_\pi$ over $K$. In $K$, $\pi$ is the only prime element that’s the norm from every finite extension of $K$ in $\mathcal L_\pi$. Isomorphic formal modules give the same totally ramified field, and so the same universal norm $\pi$.

Answer 2

If the formal groups were isomorphic then the (conjugation by the) corresponding change of variables would send $[\pi]_{F_\pi}$ into $[\pi]_{F_{\pi'}}$. Thus modulo the maximal ideal $(\pi)$ we have $[\pi]_{F_{\pi'}}(X)\equiv X^q\equiv [\pi']_{F_{\pi'}}(X)$. Hence the reduction of $[\pi-\pi']_{F_{\pi'}}$ would be zero, which is certainly wrong (easy).