Past exam of the mathematics course of the University of Cambridge / 2022 / iii / Paper 136 / 5

Seek $F=L+F_2+F_3+\cdots$, with each $F_d$ homogeneous of degree $d$. Suppose the terms below degree $d$ have been chosen. Comparing degree $d$ ingives a linear equation for $F_d$ whose coefficient is $\pi -{\pi }^d=\pi (1-{\pi }^{d-1})$. The already known term is divisible by $\pi$ because $f(X),g(X)\equiv X^q(\mathrm{mod}\pi )$ and every residue ${\bar{a}}_i\in {\mathbb{F}}_q$ satisfies ${\bar{a}}_i^q={\bar{a}}_i$. Since $1-{\pi }^{d-1}$ is a unit, this determines a unique integral $F_d$. Induction constructs a unique $F\in {\mathcal{O}}_K[[X_1,\dots ,X_n]]$. This is the Lubin–Tate functional equation lemma.

Take $f=g$ and $L=X_1+\cdots +X_n$. Permuting the variables produces another solution with the same linear term, so uniqueness givesFor $a\in {\mathcal{O}}_K$, the one-variable case of part i gives a unique ${\theta }_a(X)\equiv aX(\mathrm{mod}{X}^2)$ commuting with $f$. Applying uniqueness once more to the two ways of composing ${\theta }_a$ with $F$ givesThus $F$ is the addition law and the ${\theta }_a$ are scalar endomorphisms of the Lubin–Tate formal group.

Since $f(X)=X(\pi +X^{q-1})$, its iterates satisfyThe polynomial ${\Phi }_m$ is Eisenstein: it is monic, every nonleading coefficient is divisible by $\pi$, and its constant term is $\pi$. It is also separable, since $q-1$ is prime to the residue characteristic and the iterates have nonzero derivative.

If $\alpha \ne 0$ and $m$ is least with $f_m(\alpha )=0$, then ${\Phi }_m(\alpha )=0$. Eisenstein irreducibility makes ${\Phi }_m$ its minimal polynomial, so $K(\alpha )/K$ is totally ramified and separable. For $\alpha =0$ the extension is trivial and has the same properties. This is the Eisenstein layers of Lubin–Tate torsion argument.