 Solution

Define the Lubin–Tate torsion byThe scalar endomorphisms make this a module over ${\mathcal{O}}_K/{\pi }^n{\mathcal{O}}_K$.

The Newton polygon or Weierstrass preparation theorem applied to the Lubin–Tate congruences shows that $[{\pi }^n{]}_f$ has exactly $q^n$ distinct roots in $\overline{\mathfrak{m}}$. More precisely, the quotient of the distinguished factors for $[{\pi }^n{]}_f$ and $[{\pi }^{n-1}{]}_f$ has degree $q^{n-1}(q-1)$, and its roots are precisely the points killed by ${\pi }^n$ but not by ${\pi }^{n-1}$.

Choose such a point ${\omega }_n$. If $a\in {\mathcal{O}}_K$, write $a={\pi }^{r}u$ with $u$ a unit. Since $[u{]}_f$ is an automorphism, $[a{]}_f({\omega }_n)=0$ exactly when $r\ge n$. Thusis injective. Both sides have $q^n$ elements, so it is an isomorphism of modules. Therefore ${\mu }_{f,n}$ is a free module of rank one.