Let $L/K$ be a finite Galois extension of $p$-adic fields, and let $G = \operatorname{Gal}(L/K)$. Let $v$ be the valuation so that $v(\pi_K) = 1$. The valuation of $v$ extends to $L$ so that $v(\pi_L) = 1/e_{L/K}$. We can also write $\mathcal{O}_ L = \mathcal{O}_ K[\alpha]$.
There is then a decreasing filtration on $G$. First we have the inertia subgroup $$ I = \lbrace \sigma \in G : \sigma(x) \equiv x \pmod{\pi_L} \text{ for all } x \in \mathcal{O}_ L \rbrace. $$ Then we can extend it to a sequence $$ G_{(i)} = \lbrace \sigma \in G : v(\sigma(x) - x) \ge i \text{ for all } x \in \mathcal{O}_ L \rbrace. $$ Note that $i$ is a real number.
There is an issue with taking a tower of extensions. That is, if $\tilde{L} / L / K$ is a tower of Galois extensions, we have $$ \operatorname{Gal}(\tilde{L}/K) \twoheadrightarrow \operatorname{Gal}(L/K). $$ This is not compatible with the lower number in the sense that $G_{(i)}$ does not map to the $G_{(i)}$. So we define for $i \in \mathbb{R}_ {\ge 0}$ $$ \phi_{L/K}(i) = \sum_{\sigma \in G} \min(i, i(\sigma)), \quad i(\sigma) = v(\sigma(\alpha)-\alpha). $$ Here, note that if $\sigma = 1$ then the term is just $i$, so this is a piecewise linear, monotone strictly increasing, concave function.
Definition 1. We define $G^{(u)} = G_{(\phi_{L/K}^{-1}(i))}$.
Let us write $$ i_{L/K} = \max_{\sigma \neq 1} i(\sigma), \quad u_{L/K} = \phi_{L/K}(i_{L/K}). $$ Then $i_{L/K}$ is the largest $i$ such that $G_{(i)} \neq 1$ and $u_{L/K}$ is the largest $u$ such that $G^{(u)} \neq 1$. We can calculate that $$ u_{L/K} = \phi_{L/K}(i_{L/K}) = i_{L/K} + \sum_{\sigma \neq 1} i(\sigma) = i_{L/K} + v\Bigl(\prod_{\sigma \neq 1} (\sigma(\alpha)-\alpha) \Bigr) $$ That is, we have $$ v(D_{L/K}) = u_{L/K} - i_{L/K} \lt u_{L/K}. $$
It turns out that $u_{L/K} = 0$ if and only if it is unramified and $u_{L/K} = 1$ if and only if it is tame.
Example 2. For $L = \mathbb{Q}_ p(\zeta_{p^n}) / \mathbb{Q}_ p = K$, we have $G \cong (\mathbb{Z}/p^n \mathbb{Z})^\times$. We calculate that $$ i(\sigma) = v(\zeta_{p^n}^{m-1} - 1) = \frac{p^{v(m-1)}}{(p-1) p^{n-1}} $$ where $\sigma$ corresponds to $m$. Then we have $i_{L/K} = 1/(p-1)$ and $u_{L/K} = n$ and $v(\mathcal{D}_ {L/K}) = n - (p-1)^{-1}$.
Fontaine’s theorem
Theorem 3 (Fontaine). Let $K$ be a $p$-adic field with $e_K = e_{K/\mathbb{Q}_ p}$. Suppose $\Gamma$ is a finite commutative group scheme over $\mathcal{O}_ K$ that is killed by $p^n$. Let $L$ be the field obtained by adjoining the points of $\Gamma$ and taking its Galois closure, and write $G = \operatorname{Gal}(L/K)$. Then we have $$ u_{L/K} \le e_K \Bigl( n + \frac{1}{p-1} \Bigr). $$
So our previous example is tight up to a constant.
Proposition 4. Let $A$ be a finite flat $\mathcal{O}_ K$-algebra of the form $$ A = \mathcal{O}_ K[[x_1, \dotsc, x_m]] / (f_1, \dotsc, f_m), $$ and assume that there exists an $0 \neq a \in \mathcal{O}_ K$ annihilating $\Omega_{A/\mathcal{O}_ K}^1$ such that $\Omega_{A/\mathcal{O}_ K}^1$ is a flat $A/aA$-module.
- Let $S$ be a finite flat $\mathcal{O}_ K$-algebra and $I \subseteq S$ be a topologically nilpotnet pd-ideal. Then $$ \Hom_ {\mathcal{O}_ K}(A, S) = \im(\Hom_ {\mathcal{O}_ K}(A, S/aI) \to \Hom_ {\mathcal{O}_ K}(A, S/I)). $$
- Let $L$ be the Galois field extension of $K$. Then we have $$ u_ {L/K} \le u(a) + \frac{e_K}{p-1}. $$
This will be applied to $a = p^n$ and then the second part will give us Fontaine’s theorem. Here, we will have to actually embed $\Gamma$ into an abelian scheme and do something so that the assumptions are satisfied.
Definition 5. Let $S$ be a finite flat $\mathcal{O}_ K$-algebra, and let $I \subseteq S$ be an ideal. We say this is a pd-ideal if for all $x \in I$ we have $\gamma_n(x) = x^n / n! \in I$. In this case, we consider the ideal $$ I^{[m]} = (\gamma_{n_1}(x) \dotsm \gamma_{n_r}(x_n) : n_1 + \dotsb + c_r \ge m). $$ We say that $I$ is topologically nilpotent if $\bigcap_{m \ge 0} I^{[m]} = 0$.
Let us now prove this proposition. Let $\mathfrak{m}_ A$ be the maximal ideal of $A$ where $J = (f_1, \dotsc, f_m)$. Because $\Omega_{A/\mathcal{O}_ K}^1$ is a finite free $A/aA$-module, we have $$ \frac{\partial f_i}{\partial x_j} = a p_{ij} $$ for elements $p_{ij} \in A$, and moreover the matrix $(p_{ij})$ is invertible.
What we want to show is that every $\phi \colon A \to S/aI$ lifts uniquely to $A \to S$. This can be done inductively at each step to $A \to S/aI^{[n]}$. For the second part, we look at the following.
Lemma 6. Let $E/K$ be finite Galois and $t \in \mathbb{R}_ {\gt 0}$ and $\mathfrak{m}_ E^t = \lbrace x \in \mathcal{O}_ E : v(x) \ge t \rbrace$.
- If $t \gt u_{L/K}$ then any $\mathcal{O}_ K$-algebra homomorphism $\mathcal{O}_ L \to \mathcal{O}_ E / \mathfrak{m}_ E^t$ lifts to an $\mathcal{O}_ K$-algebra homomorphism $\mathcal{O}_ L \to \mathcal{O}_ E$.
- If every $\mathcal{O}_ K$-algebra homomorphism $\mathcal{O}_ L \to \mathcal{O}_ E / \mathfrak{m}_ E^t$ lifts to $\mathcal{O}_ L \to \mathcal{O}_ E$, then $t \gt u_{L/K} - e_{L/K}^{-1}$.