## 58.13 Ramification theory

In this section we continue the discussion of More on Algebra, Section 15.112 and we relate it to our discussion of the fundamental groups of schemes.

Let $(A, \mathfrak m, \kappa )$ be a normal local ring with fraction field $K$. Choose a separable algebraic closure $K^{sep}$. Let $A^{sep}$ be the integral closure of $A$ in $K^{sep}$. Choose maximal ideal $\mathfrak m^{sep} \subset A^{sep}$. Let $A \subset A^ h \subset A^{sh}$ be the henselization and strict henselization. Observe that $A^ h$ and $A^{sh}$ are normal rings as well (More on Algebra, Lemma 15.45.6). Denote $K^ h$ and $K^{sh}$ their fraction fields. Since $(A^{sep})_{\mathfrak m^{sep}}$ is strictly henselian by Lemma 58.12.2 we can choose an $A$-algebra map $A^{sh} \to (A^{sep})_{\mathfrak m^{sep}}$. Namely, first choose a $\kappa$-embedding1 $\kappa (\mathfrak m^{sh}) \to \kappa (\mathfrak m^{sep})$ and then extend (uniquely) to an $A$-algebra homomorphism by Algebra, Lemma 10.155.10. We get the following diagram

$\xymatrix{ K^{sep} & K^{sh} \ar[l] & K^ h \ar[l] & K \ar[l] \\ (A^{sep})_{\mathfrak m^{sep}} \ar[u] & A^{sh} \ar[u] \ar[l] & A^ h \ar[u] \ar[l] & A \ar[u] \ar[l] }$

We can take the fundamental groups of the spectra of these rings. Of course, since $K^{sep}$, $(A^{sep})_{\mathfrak m^{sep}}$, and $A^{sh}$ are strictly henselian, for them we obtain trivial groups. Thus the interesting part is the following

58.13.0.1
$$\label{pione-equation-inertia-diagram-pione} \vcenter { \xymatrix{ \pi _1(U^{sh}) \ar[r] \ar[rd]_1 & \pi _1(U^ h) \ar[d] \ar[r] & \pi _1(U) \ar[d] \\ & \pi _1(X^ h) \ar[r] & \pi _1(X) } }$$

Here $X^ h$ and $X$ are the spectra of $A^ h$ and $A$ and $U^{sh}$, $U^ h$, $U$ are the spectra of $K^{sh}$, $K^ h$, and $K$. The label $1$ means that the map is trivial; this follows as it factors through the trivial group $\pi _1(X^{sh})$. On the other hand, the profinite group $G = \text{Gal}(K^{sep}/K)$ acts on $A^{sep}$ and we can make the following definitions

$D = \{ \sigma \in G \mid \sigma (\mathfrak m^{sep}) = \mathfrak m^{sep}\} \supset I = \{ \sigma \in D \mid \sigma \bmod \mathfrak m^{sep} = \text{id}_{\kappa (\mathfrak m^{sep})}\}$

These groups are sometimes called the decomposition group and the inertia group especially when $A$ is a discrete valuation ring.

Lemma 58.13.1. In the situation described above, via the isomorphism $\pi _1(U) = \text{Gal}(K^{sep}/K)$ the diagram (58.13.0.1) translates into the diagram

$\xymatrix{ I \ar[r] \ar[rd]_1 & D \ar[d] \ar[r] & \text{Gal}(K^{sep}/K) \ar[d] \\ & \text{Gal}(\kappa (\mathfrak m^{sh})/\kappa ) \ar[r] & \text{Gal}(M/K) }$

where $K^{sep}/M/K$ is the maximal subextension unramified with respect to $A$. Moreover, the vertical arrows are surjective, the kernel of the left vertical arrow is $I$ and the kernel of the right vertical arrow is the smallest closed normal subgroup of $\text{Gal}(K^{sep}/K)$ containing $I$.

Proof. By construction the group $D$ acts on $(A^{sep})_{\mathfrak m^{sep}}$ over $A$. By the uniqueness of $A^{sh} \to (A^{sep})_{\mathfrak m^{sep}}$ given the map on residue fields (Algebra, Lemma 10.155.10) we see that the image of $A^{sh} \to (A^{sep})_{\mathfrak m^{sep}}$ is contained in $((A^{sep})_{\mathfrak m^{sep}})^ I$. On the other hand, Lemma 58.12.5 shows that $((A^{sep})_{\mathfrak m^{sep}})^ I$ is a filtered colimit of étale extensions of $A$. Since $A^{sh}$ is the maximal such extension, we conclude that $A^{sh} = ((A^{sep})_{\mathfrak m^{sep}})^ I$. Hence $K^{sh} = (K^{sep})^ I$.

Recall that $I$ is the kernel of a surjective map $D \to \text{Aut}(\kappa (\mathfrak m^{sep})/\kappa )$, see More on Algebra, Lemma 15.110.10. We have $\text{Aut}(\kappa (\mathfrak m^{sep})/\kappa ) = \text{Gal}(\kappa (\mathfrak m^{sh})/\kappa )$ as we have seen above that these fields are the algebraic and separable algebraic closures of $\kappa$. On the other hand, any automorphism of $A^{sh}$ over $A$ is an automorphism of $A^{sh}$ over $A^ h$ by the uniqueness in Algebra, Lemma 10.155.6. Furthermore, $A^{sh}$ is the colimit of finite étale extensions $A^ h \subset A'$ which correspond $1$-to-$1$ with finite separable extension $\kappa '/\kappa$, see Algebra, Remark 10.155.4. Thus

$\text{Aut}(A^{sh}/A) = \text{Aut}(A^{sh}/A^ h) = \text{Gal}(\kappa (\mathfrak m^{sh})/\kappa )$

Let $\kappa '/\kappa$ be a finite Galois extension with Galois group $G$. Let $A^ h \subset A'$ be the finite étale extension corresponding to $\kappa \subset \kappa '$ by Algebra, Lemma 10.153.7. Then it follows that $(A')^ G = A^ h$ by looking at fraction fields and degrees (small detail omitted). Taking the colimit we conclude that $(A^{sh})^{\text{Gal}(\kappa (\mathfrak m^{sh})/\kappa )} = A^ h$. Combining all of the above, we find $A^ h = ((A^{sep})_{\mathfrak m^{sep}})^ D$. Hence $K^ h = (K^{sep})^ D$.

Since $U$, $U^ h$, $U^{sh}$ are the spectra of the fields $K$, $K^ h$, $K^{sh}$ we see that the top lines of the diagrams correspond via Lemma 58.6.3. By Lemma 58.8.2 we have $\pi _1(X^ h) = \text{Gal}(\kappa (\mathfrak m^{sh})/\kappa )$. The exactness of the sequence $1 \to I \to D \to \text{Gal}(\kappa (\mathfrak m^{sh})/\kappa ) \to 1$ was pointed out above. By Proposition 58.11.3 we see that $\pi _1(X) = \text{Gal}(M/K)$. Finally, the statement on the kernel of $\text{Gal}(K^{sep}/K) \to \text{Gal}(M/K) = \pi _1(X)$ follows from Lemma 58.11.4. This finishes the proof. $\square$

Let $X$ be a normal integral scheme with function field $K$. Let $K^{sep}$ be a separable algebraic closure of $K$. Let $X^{sep} \to X$ be the normalization of $X$ in $K^{sep}$. Since $G = \text{Gal}(K^{sep}/K)$ acts on $K^{sep}$ we obtain a right action of $G$ on $X^{sep}$. For $y \in X^{sep}$ define

$D_ y = \{ \sigma \in G \mid \sigma (y) = y\} \supset I_ y = \{ \sigma \in D \mid \sigma \bmod \mathfrak m_ y = \text{id}_{\kappa (y)} \}$

similarly to the above. On the other hand, for $x \in X$ let $\mathcal{O}_{X, x}^{sh}$ be a strict henselization, let $K_ x^{sh}$ be the fraction field of $\mathcal{O}_{X, x}^{sh}$ and choose a $K$-embedding $K_ x^{sh} \to K^{sep}$.

Lemma 58.13.2. Let $X$ be a normal integral scheme with function field $K$. With notation as above, the following three subgroups of $\text{Gal}(K^{sep}/K) = \pi _1(\mathop{\mathrm{Spec}}(K))$ are equal

1. the kernel of the surjection $\text{Gal}(K^{sep}/K) \longrightarrow \pi _1(X)$,

2. the smallest normal closed subgroup containing $I_ y$ for all $y \in X^{sep}$, and

3. the smallest normal closed subgroup containing $\text{Gal}(K^{sep}/K_ x^{sh})$ for all $x \in X$.

Proof. The equivalence of (2) and (3) follows from Lemma 58.13.1 which tells us that $I_ y$ is conjugate to $\text{Gal}(K^{sep}/K_ x^{sh})$ if $y$ lies over $x$. By Lemma 58.11.4 we see that $\text{Gal}(K^{sep}/K_ x^{sh})$ maps trivially to $\pi _1(\mathop{\mathrm{Spec}}(\mathcal{O}_{X, x}))$ and therefore the subgroup $N \subset G = \text{Gal}(K^{sep}/K)$ of (2) and (3) is contained in the kernel of $G \longrightarrow \pi _1(X)$.

To prove the other inclusion, since $N$ is normal, it suffices to prove: given $N \subset U \subset G$ with $U$ open normal, the quotient map $G \to G/U$ factors through $\pi _1(X)$. In other words, if $L/K$ is the Galois extension corresponding to $U$, then we have to show that $X$ is unramified in $L$ (Section 58.11, especially Proposition 58.11.3). It suffices to do this when $X$ is affine (we do this so we can refer to algebra results in the rest of the proof). Let $Y \to X$ be the normalization of $X$ in $L$. The inclusion $L \subset K^{sep}$ induces a morphism $\pi : X^{sep} \to Y$. For $y \in X^{sep}$ the inertia group of $\pi (y)$ in $\text{Gal}(L/K)$ is the image of $I_ y$ in $\text{Gal}(L/K)$; this follows from More on Algebra, Lemma 15.110.11. Since $N \subset U$ all these inertia groups are trivial. We conclude that $Y \to X$ is étale by applying Lemma 58.12.4. (Alternative: you can use Lemma 58.11.4 to see that the pullback of $Y$ to $\mathop{\mathrm{Spec}}(\mathcal{O}_{X, x})$ is étale for all $x \in X$ and then conclude from there with a bit more work.) $\square$

Example 58.13.3. Let $X$ be a normal integral Noetherian scheme with function field $K$. Purity of branch locus (see below) tells us that if $X$ is regular, then it suffices in Lemma 58.13.2 to consider the inertia groups $I = \pi _1(\mathop{\mathrm{Spec}}(K_ x^{sh}))$ for points $x$ of codimension $1$ in $X$. In general this is not enough however. Namely, let $Y = \mathbf{A}_ k^ n = \mathop{\mathrm{Spec}}(k[t_1, \ldots , t_ n])$ where $k$ is a field not of characteristic $2$. Let $G = \{ \pm 1\}$ be the group of order $2$ acting on $Y$ by multiplication on the coordinates. Set

$X = \mathop{\mathrm{Spec}}(k[t_ it_ j, i, j \in \{ 1, \ldots , n\} ])$

The embedding $k[t_ it_ j] \subset k[t_1, \ldots , t_ n]$ defines a degree $2$ morphism $Y \to X$ which is unramified everywhere except over the maximal ideal $\mathfrak m = (t_ it_ j)$ which is a point of codimension $n$ in $X$.

Lemma 58.13.4. Let $X$ be an integral normal scheme with function field $K$. Let $L/K$ be a finite extension. Let $Y \to X$ be the normalization of $X$ in $L$. The following are equivalent

1. $X$ is unramified in $L$ as defined in Section 58.11,

2. $Y \to X$ is an unramified morphism of schemes,

3. $Y \to X$ is an étale morphism of schemes,

4. $Y \to X$ is a finite étale morphism of schemes,

5. for $x \in X$ the projection $Y \times _ X \mathop{\mathrm{Spec}}(\mathcal{O}_{X, x}) \to \mathop{\mathrm{Spec}}(\mathcal{O}_{X, x})$ is unramified,

6. same as in (5) but with $\mathcal{O}_{X, x}^ h$,

7. same as in (5) but with $\mathcal{O}_{X, x}^{sh}$,

8. for $x \in X$ the scheme theoretic fibre $Y_ x$ is étale over $x$ of degree $\geq [L : K]$.

If $L/K$ is Galois with Galois group $G$, then these are also equivalent to

1. for $y \in Y$ the group $I_ y = \{ g \in G \mid g(y) = y\text{ and } g \bmod \mathfrak m_ y = \text{id}_{\kappa (y)}\}$ is trivial.

Proof. The equivalence of (1) and (2) is the definition of (1). The equivalence of (2), (3), and (4) is Lemma 58.11.1. It is straightforward to prove that (4) $\Rightarrow$ (5), (5) $\Rightarrow$ (6), (6) $\Rightarrow$ (7).

Assume (7). Observe that $\mathcal{O}_{X, x}^{sh}$ is a normal local domain (More on Algebra, Lemma 15.45.6). Let $L^{sh} = L \otimes _ K K_ x^{sh}$ where $K_ x^{sh}$ is the fraction field of $\mathcal{O}_{X, x}^{sh}$. Then $L^{sh} = \prod _{i = 1, \ldots , n} L_ i$ with $L_ i/K_ x^{sh}$ finite separable. By Algebra, Lemma 10.147.4 (and a limit argument we omit) we see that $Y \times _ X \mathop{\mathrm{Spec}}(\mathcal{O}_{X, x}^{sh})$ is the integral closure of $\mathop{\mathrm{Spec}}(\mathcal{O}_{X, x}^{sh})$ in $L^{sh}$. Hence by Lemma 58.11.1 (applied to the factors $L_ i$ of $L^{sh}$) we see that $Y \times _ X \mathop{\mathrm{Spec}}(\mathcal{O}_{X, x}^{sh}) \to \mathop{\mathrm{Spec}}(\mathcal{O}_{X, x}^{sh})$ is finite étale. Looking at the generic point we see that the degree is equal to $[L : K]$ and hence we see that (8) is true.

Assume (8). Assume that $x \in X$ and that the scheme theoretic fibre $Y_ x$ is étale over $x$ of degree $\geq [L : K]$. Observe that this means that $Y$ has $\geq [L : K]$ geometric points lying over $x$. We will show that $Y \to X$ is finite étale over a neighbourhood of $x$. This will prove (1) holds. To prove this we may assume $X = \mathop{\mathrm{Spec}}(R)$, the point $x$ corresponds to the prime $\mathfrak p \subset R$, and $Y = \mathop{\mathrm{Spec}}(S)$. We apply More on Morphisms, Lemma 37.42.1 and we find an étale neighbourhood $(U, u) \to (X, x)$ such that $Y \times _ X U = V_1 \amalg \ldots \amalg V_ m$ such that $V_ i$ has a unique point $v_ i$ lying over $u$ with $\kappa (v_ i)/\kappa (u)$ purely inseparable. Shrinking $U$ if necessary we may assume $U$ is a normal integral scheme with generic point $\xi$ (use Descent, Lemmas 35.16.3 and 35.18.2 and Properties, Lemma 28.7.5). By our remark on geometric points we see that $m \geq [L : K]$. On the other hand, by More on Morphisms, Lemma 37.19.2 we see that $\coprod V_ i \to U$ is the normalization of $U$ in $\mathop{\mathrm{Spec}}(L) \times _ X U$. As $K \subset \kappa (\xi )$ is finite separable, we can write $\mathop{\mathrm{Spec}}(L) \times _ X U = \mathop{\mathrm{Spec}}(\prod _{i = 1, \ldots , n} L_ i)$ with $L_ i/\kappa (\xi )$ finite and $[L : K] = \sum [L_ i : \kappa (\xi )]$. Since $V_ j$ is nonempty for each $j$ and $m \geq [L : K]$ we conclude that $m = n$ and $[L_ i : \kappa (\xi )] = 1$ for all $i$. Then $V_ j \to U$ is an isomorphism in particular étale, hence $Y \times _ X U \to U$ is étale. By Descent, Lemma 35.23.29 we conclude that $Y \to X$ is étale over the image of $U \to X$ (an open neighbourhood of $x$).

Assume $L/K$ is Galois and (9) holds. Then $Y \to X$ is étale by Lemma 58.12.5. We omit the proof that (1) implies (9). $\square$

In the case of infinite Galois extensions of discrete valuation rings we can say a tiny bit more. To do so we introduce the following notation. A subset $S \subset \mathbf{N}$ of integers is multiplicativity directed if $1 \in S$ and for $n, m \in S$ there exists $k \in S$ with $n | k$ and $m | k$. Define a partial ordering on $S$ by the rule $n \geq _ S m$ if and only if $m | n$. Given a field $\kappa$ we obtain an inverse system of finite groups $\{ \mu _ n(\kappa )\} _{n \in S}$ with transition maps

$\mu _ n(\kappa ) \longrightarrow \mu _ m(\kappa ),\quad \zeta \longmapsto \zeta ^{n/m}$

for $n \geq _ S m$. Then we can form the profinite group

$\mathop{\mathrm{lim}}\nolimits _{n \in S} \mu _ n(\kappa )$

Observe that the limit is cofiltered (as $S$ is directed). The construction is functorial in $\kappa$. In particular $\text{Aut}(\kappa )$ acts on this profinite group. For example, if $S = \{ 1, n\}$, then this gives $\mu _ n(\kappa )$. If $S = \{ 1, \ell , \ell ^2, \ell ^3, \ldots \}$ for some prime $\ell$ different from the characteristic of $\kappa$ this produces $\mathop{\mathrm{lim}}\nolimits _ n \mu _{\ell ^ n}(\kappa )$ which is sometimes called the $\ell$-adic Tate module of the multiplicative group of $\kappa$ (compare with More on Algebra, Example 15.93.5).

Lemma 58.13.5. Let $A$ be a discrete valuation ring with fraction field $K$. Let $L/K$ be a (possibly infinite) Galois extension. Let $B$ be the integral closure of $A$ in $L$. Let $\mathfrak m$ be a maximal ideal of $B$. Let $G = \text{Gal}(L/K)$, $D = \{ \sigma \in G \mid \sigma (\mathfrak m) = \mathfrak m\}$, and $I = \{ \sigma \in D \mid \sigma \bmod \mathfrak m = \text{id}_{\kappa (\mathfrak m)}\}$. The decomposition group $D$ fits into a canonical exact sequence

$1 \to I \to D \to \text{Aut}(\kappa (\mathfrak m)/\kappa _ A) \to 1$

The inertia group $I$ fits into a canonical exact sequence

$1 \to P \to I \to I_ t \to 1$

such that

1. $P$ is a normal subgroup of $D$,

2. $P$ is a pro-$p$-group if the characteristic of $\kappa _ A$ is $p > 1$ and $P = \{ 1\}$ if the characteristic of $\kappa _ A$ is zero,

3. there is a multiplicatively directed $S \subset \mathbf{N}$ such that $\kappa (\mathfrak m)$ contains a primitive $n$th root of unity for each $n \in S$ (elements of $S$ are prime to $p$),

4. there exists a canonical surjective map

$\theta _{can} : I \to \mathop{\mathrm{lim}}\nolimits _{n \in S} \mu _ n(\kappa (\mathfrak m))$

whose kernel is $P$, which satisfies $\theta _{can}(\tau \sigma \tau ^{-1}) = \tau (\theta _{can}(\sigma ))$ for $\tau \in D$, $\sigma \in I$, and which induces an isomorphism $I_ t \to \mathop{\mathrm{lim}}\nolimits _{n \in S} \mu _ n(\kappa (\mathfrak m))$.

Proof. This is mostly a reformulation of the results on finite Galois extensions proved in More on Algebra, Section 15.112. The surjectivity of the map $D \to \text{Aut}(\kappa (\mathfrak m)/\kappa )$ is More on Algebra, Lemma 15.110.10. This gives the first exact sequence.

To construct the second short exact sequence let $\Lambda$ be the set of finite Galois subextensions, i.e., $\lambda \in \Lambda$ corresponds to $L/L_\lambda /K$. Set $G_\lambda = \text{Gal}(L_\lambda /K)$. Recall that $G_\lambda$ is an inverse system of finite groups with surjective transition maps and that $G = \mathop{\mathrm{lim}}\nolimits _{\lambda \in \Lambda } G_\lambda$, see Fields, Lemma 9.22.3. We let $B_\lambda$ be the integral closure of $A$ in $L_\lambda$. Then we set $\mathfrak m_\lambda = \mathfrak m \cap B_\lambda$ and we denote $P_\lambda , I_\lambda , D_\lambda$ the wild inertia, inertia, and decomposition group of $\mathfrak m_\lambda$, see More on Algebra, Lemma 15.112.5. For $\lambda \geq \lambda '$ the restriction defines a commutative diagram

$\xymatrix{ P_\lambda \ar[d] \ar[r] & I_\lambda \ar[d] \ar[r] & D_\lambda \ar[d] \ar[r] & G_\lambda \ar[d] \\ P_{\lambda '} \ar[r] & I_{\lambda '} \ar[r] & D_{\lambda '} \ar[r] & G_{\lambda '} }$

with surjective vertical maps, see More on Algebra, Lemma 15.112.10.

From the definitions it follows immediately that $I = \mathop{\mathrm{lim}}\nolimits I_\lambda$ and $D = \mathop{\mathrm{lim}}\nolimits D_\lambda$ under the isomorphism $G = \mathop{\mathrm{lim}}\nolimits G_\lambda$ above. Since $L = \mathop{\mathrm{colim}}\nolimits L_\lambda$ we have $B = \mathop{\mathrm{colim}}\nolimits B_\lambda$ and $\kappa (\mathfrak m) = \mathop{\mathrm{colim}}\nolimits \kappa (\mathfrak m_\lambda )$. Since the transition maps of the system $D_\lambda$ are compatible with the maps $D_\lambda \to \text{Aut}(\kappa (\mathfrak m_\lambda )/\kappa )$ (see More on Algebra, Lemma 15.112.10) we see that the map $D \to \text{Aut}(\kappa (\mathfrak m)/\kappa )$ is the limit of the maps $D_\lambda \to \text{Aut}(\kappa (\mathfrak m_\lambda )/\kappa )$.

There exist canonical maps

$\theta _{\lambda , can} : I_\lambda \longrightarrow \mu _{n_\lambda }(\kappa (\mathfrak m_\lambda ))$

where $n_\lambda = |I_\lambda |/|P_\lambda |$, where $\mu _{n_\lambda }(\kappa (\mathfrak m_\lambda ))$ has order $n_\lambda$, such that $\theta _{\lambda , can}(\tau \sigma \tau ^{-1}) = \tau (\theta _{\lambda , can}(\sigma ))$ for $\tau \in D_\lambda$ and $\sigma \in I_\lambda$, and such that we get commutative diagrams

$\xymatrix{ I_\lambda \ar[r]_-{\theta _{\lambda , can}} \ar[d] & \mu _{n_\lambda }(\kappa (\mathfrak m_\lambda )) \ar[d]^{(-)^{n_\lambda /n_{\lambda '}}} \\ I_{\lambda '} \ar[r]^-{\theta _{\lambda ', can}} & \mu _{n_{\lambda '}}(\kappa (\mathfrak m_{\lambda '})) }$

see More on Algebra, Remark 15.112.11.

Let $S \subset \mathbf{N}$ be the collection of integers $n_\lambda$. Since $\Lambda$ is directed, we see that $S$ is multiplicatively directed. By the displayed commutative diagrams above we can take the limits of the maps $\theta _{\lambda , can}$ to obtain

$\theta _{can} : I \to \mathop{\mathrm{lim}}\nolimits _{n \in S} \mu _ n(\kappa (\mathfrak m)).$

This map is continuous (small detail omitted). Since the transition maps of the system of $I_\lambda$ are surjective and $\Lambda$ is directed, the projections $I \to I_\lambda$ are surjective. For every $\lambda$ the diagram

$\xymatrix{ I \ar[d] \ar[r]_-{\theta _{can}} & \mathop{\mathrm{lim}}\nolimits _{n \in S} \mu _ n(\kappa (\mathfrak m)) \ar[d] \\ I_{\lambda } \ar[r]^-{\theta _{\lambda , can}} & \mu _{n_\lambda }(\kappa (\mathfrak m_\lambda )) }$

commutes. Hence the image of $\theta _{can}$ surjects onto the finite group $\mu _{n_\lambda }(\kappa (\mathfrak m)) = \mu _{n_\lambda }(\kappa (\mathfrak m_\lambda ))$ of order $n_\lambda$ (see above). It follows that the image of $\theta _{can}$ is dense. On the other hand $\theta _{can}$ is continuous and the source is a profinite group. Hence $\theta _{can}$ is surjective by a topological argument.

The property $\theta _{can}(\tau \sigma \tau ^{-1}) = \tau (\theta _{can}(\sigma ))$ for $\tau \in D$, $\sigma \in I$ follows from the corresponding properties of the maps $\theta _{\lambda , can}$ and the compatibility of the map $D \to \text{Aut}(\kappa (\mathfrak m))$ with the maps $D_\lambda \to \text{Aut}(\kappa (\mathfrak m_\lambda ))$. Setting $P = \mathop{\mathrm{Ker}}(\theta _{can})$ this implies that $P$ is a normal subgroup of $D$. Setting $I_ t = I/P$ we obtain the isomorphism $I_ t \to \mathop{\mathrm{lim}}\nolimits _{n \in S} \mu _ n(\kappa (\mathfrak m))$ from the surjectivity of $\theta _{can}$.

To finish the proof we show that $P = \mathop{\mathrm{lim}}\nolimits P_\lambda$ which proves that $P$ is a pro-$p$-group. Recall that the tame inertia group $I_{\lambda , t} = I_\lambda /P_\lambda$ has order $n_\lambda$. Since the transition maps $P_\lambda \to P_{\lambda '}$ are surjective and $\Lambda$ is directed, we obtain a short exact sequence

$1 \to \mathop{\mathrm{lim}}\nolimits P_\lambda \to I \to \mathop{\mathrm{lim}}\nolimits I_{\lambda , t} \to 1$

(details omitted). Since for each $\lambda$ the map $\theta _{\lambda , can}$ induces an isomorphism $I_{\lambda , t} \cong \mu _{n_\lambda }(\kappa (\mathfrak m))$ the desired result follows. $\square$

Lemma 58.13.6. Let $A$ be a discrete valuation ring with fraction field $K$. Let $K^{sep}$ be a separable closure of $K$. Let $A^{sep}$ be the integral closure of $A$ in $K^{sep}$. Let $\mathfrak m^{sep}$ be a maximal ideal of $A^{sep}$. Let $\mathfrak m = \mathfrak m^{sep} \cap A$, let $\kappa = A/\mathfrak m$, and let $\overline{\kappa } = A^{sep}/\mathfrak m^{sep}$. Then $\overline{\kappa }$ is an algebraic closure of $\kappa$. Let $G = \text{Gal}(K^{sep}/K)$, $D = \{ \sigma \in G \mid \sigma (\mathfrak m^{sep}) = \mathfrak m^{sep}\}$, and $I = \{ \sigma \in D \mid \sigma \bmod \mathfrak m^{sep} = \text{id}_{\kappa (\mathfrak m^{sep})}\}$. The decomposition group $D$ fits into a canonical exact sequence

$1 \to I \to D \to \text{Gal}(\kappa ^{sep}/\kappa ) \to 1$

where $\kappa ^{sep} \subset \overline{\kappa }$ is the separable closure of $\kappa$. The inertia group $I$ fits into a canonical exact sequence

$1 \to P \to I \to I_ t \to 1$

such that

1. $P$ is a normal subgroup of $D$,

2. $P$ is a pro-$p$-group if the characteristic of $\kappa _ A$ is $p > 1$ and $P = \{ 1\}$ if the characteristic of $\kappa _ A$ is zero,

3. there exists a canonical surjective map

$\theta _{can} : I \to \mathop{\mathrm{lim}}\nolimits _{n\text{ prime to }p} \mu _ n(\kappa ^{sep})$

whose kernel is $P$, which satisfies $\theta _{can}(\tau \sigma \tau ^{-1}) = \tau (\theta _{can}(\sigma ))$ for $\tau \in D$, $\sigma \in I$, and which induces an isomorphism $I_ t \to \mathop{\mathrm{lim}}\nolimits _{n\text{ prime to }p} \mu _ n(\kappa ^{sep})$.

Proof. The field $\overline{\kappa }$ is the algebraic closure of $\kappa$ by Lemma 58.12.1. Most of the statements immediately follow from the corresponding parts of Lemma 58.13.5. For example because $\text{Aut}(\overline{\kappa }/\kappa ) = \text{Gal}(\kappa ^{sep}/\kappa )$ we obtain the first sequence. Then the only other assertion that needs a proof is the fact that with $S$ as in Lemma 58.13.5 the limit $\mathop{\mathrm{lim}}\nolimits _{n \in S} \mu _ n(\overline{\kappa })$ is equal to $\mathop{\mathrm{lim}}\nolimits _{n\text{ prime to }p} \mu _ n(\kappa ^{sep})$. To see this it suffices to show that every integer $n$ prime to $p$ divides an element of $S$. Let $\pi \in A$ be a uniformizer and consider the splitting field $L$ of the polynomial $X^ n - \pi$. Since the polynomial is separable we see that $L$ is a finite Galois extension of $K$. Choose an embedding $L \to K^{sep}$. Observe that if $B$ is the integral closure of $A$ in $L$, then the ramification index of $A \to B_{\mathfrak m^{sep} \cap B}$ is divisible by $n$ (because $\pi$ has an $n$th root in $B$; in fact the ramification index equals $n$ but we do not need this). Then it follows from the construction of the $S$ in the proof of Lemma 58.13.5 that $n$ divides an element of $S$. $\square$

[1] This is possible because $\kappa (\mathfrak m^{sh})$ is a separable algebraic closure of $\kappa$ and $\kappa (\mathfrak m^{sep})$ is an algebraic closure of $\kappa$ by Lemma 58.12.1.

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