The Stacks project

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
\begin{equation} \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) } } \end{equation}

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 15.110.7. 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 15.110.8. 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 37.41.1 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 35.15.3 and 35.17.2 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 37.18.2 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 35.22.29 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 15.110.7. 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.

Comments (0)


Post a comment

Your email address will not be published. Required fields are marked.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.




In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 0BSD. Beware of the difference between the letter 'O' and the digit '0'.