Algebraic Number Theory — Chapter 6 Ramification theory

Chapter 6
Ramification theory

6.1. Semi-local theory

Notation 6.1.1. §

We often use a subscript $v$ to denote a valuation $| |_{v}$ on a field $K$ , even when that valuation is archimedean. When $| |_{v}$ is nonarchimedean, $v$ also denotes an additive valuation corresponding to $v$ , and when it is discrete, the additive valuation is chosen to have image $ℤ \cup {\infty}$ .

Notation 6.1.2. §

We let $K_{v}$ denote the completion of $K$ with respect to a valuation denoted $| |_{v}$ .

Remark 6.1.3. §

Let $A$ be a Dedekind domain and $𝔭$ a prime ideal of $A$ . Let $| |_{𝔭}$ be a valuation on the quotient field $K$ of $A$ such that $| a |_{𝔭} = c^{- v_{𝔭} (a)}$ for some $c \in ℝ_{> 1}$ and all $a \in K^{\times}$ . Then we may speak of the completion $K_{𝔭}$ of $K$ with respect to this valuation.

The following theorem explores extensions of valuations in the case that the ground field is not complete. In this case, uniqueness of the extension need not hold, but we can classify the distinct extensions.

Theorem 6.1.4. §

Let $K$ be a field and $v$ a valuation on $K$ . Let $\bar{v}$ be an extension of $v$ to a valuation on an algebraic closure $\bar{K_{v}}$ of $K_{v}$ . Let $L$ be an algebraic extension of $K$ . For any extension of $v$ to a valuation $w$ on $L$ , there exists an embedding $τ : L \to \bar{K_{v}}$ fixing $K$ such that $w = \bar{v} \circ τ$ , by which we mean that

| β |_{w} = | τ (β) |_{\bar{v}}

for all $β \in L$ . If $τ^{'} : L \to \bar{K_{v}}$ is another embedding fixing $K$ , then $w^{'} = \bar{v} \circ τ^{'}$ is equal to $w$ if and only if $τ$ and $τ^{'}$ are conjugate over $K_{v}$ : i.e., $τ^{'} = σ \circ τ$ for some automorphism $σ$ of $\bar{K_{v}}$ fixing $K_{v}$ .

Proof.

The valuation that $w$ induces on the completion $L_{w}$ can only be the unique valuation extending the valuation that $v$ induces on $K_{v}$ . If $τ : L_{w} \to \bar{K_{v}}$ is any field embedding fixing $K_{v}$ , then $\bar{v} \circ τ$ is a valuation on $L_{w}$ that extends $v$ on $K_{v}$ , and hence it must be $w$ .

For the second statement, suppose first that $τ^{'} = σ \circ τ$ with $σ \in {Aut}_{K_{v}} (\bar{K_{v}})$ . Again, $\bar{v}$ is the unique valuation on $\bar{K_{v}}$ extending the valuation on $K_{v}$ , so $\bar{v} \circ σ = \bar{v}$ . But then we have $\bar{v} \circ τ^{'} = \bar{v} \circ τ$ , and restricting to $L$ , this means that $w^{'} = w$ .

Conversely, suppose that $w^{'} = w$ . Note that $τ^{'} \circ τ^{- 1} : τ (L) \to τ^{'} (L)$ is an isomorphism fixing $K$ . Suppose that $L ∕ K$ is finite. As $K$ is dense in $K_{v}$ and $τ (L)$ is also a finite extension of $K$ , we have that $τ (L)$ is dense in $τ (L) \cdot K_{v}$ (and similarly for $τ^{'}$ ). Define

σ : τ (L) \cdot K_{v} \to τ^{'} (L) \cdot K_{v}

σ (α) = \lim_{n \to \infty} τ^{'} (α_{n})

for $α \in τ (L) \cdot K_{v}$ , choosing a sequence ${(α_{n})}_{n}$ of elements of $L$ such that ${(τ (α_{n}))}_{n}$ converges to $α$ . Note that this is independent of choice, as if $α_{n}^{'}$ is another such sequence, then the limits of ${(τ^{'} (α_{n}))}_{n}$ and ${(τ^{'} (α_{n}^{'}))}_{n}$ are the same by continuity of $τ^{'}$ . But then $τ^{'} = σ \circ τ$ , and $σ$ is the unique isomorphism fixing $K_{v}$ with this property. We then extend $σ$ to an element of ${Aut}_{K_{v}} (\bar{K_{v}})$ , obtaining the desired map.

In general, $L$ is a union of finite extensions $E$ of $K$ . For each such $E$ , we have defined a unique isomorphism $σ_{E} : τ (E) K_{v} \to τ^{'} (E) K_{v}$ such that $τ^{'} |_{E} = σ_{E} \circ τ |_{E}$ . If $E$ and $E^{'}$ are two finite extensions of $K$ in $L$ , then $σ_{E}$ and $σ_{E^{'}}$ agree on $τ (E \cap E^{'}) K_{v} \subseteq τ (E) K_{v} \cap τ (E^{'}) K_{v}$ by uniqueness. Together, the collection of maps $σ_{E}$ defines an embedding of the compositum of the fields $τ (E) K_{v}$ into $\bar{K_{v}}$ . We then extend this embedding to an automorphism of $\bar{K_{v}}$ fixing $K_{v}$ , and by definition, it has the property that $τ^{'} = σ \circ τ$ . □

Notation 6.1.5. §

If $L ∕ K$ is an extension of fields, $v$ is a valuation on $K$ , and $w$ is a valuation on $L$ , then we write $w ∣ v$ to denote that $w$ is equivalent to an extension of $v$ . The set of $w ∣ v$ will mean a set of representatives of the equivalence classes of the extensions of $v$ to $L$ .

The following is in essence a consequence of Proposition 1.1.1.

Proposition 6.1.6. §

Let $L ∕ K$ be a finite separable extension of fields and $v$ a valuation on $K$ . Then there is an isomorphism

κ : L \otimes_{K} K_{v} \to_{}^{\sim} \prod_{w ∣ v} L_{w}

such that $κ (β \otimes 1) = {(ι_{w} β)}_{w}$ for all $β \in L$ , where $ι_{w} : L \to L_{w}$ is the canonical embedding of a field in its completion.

Proof.

As $L ∕ K$ is finite and separable, there exists an element $𝜃 \in L$ such that $L = K (𝜃)$ . Let $f$ be the minimal polynomial of $𝜃$ over $K$ , and let $f = \prod_{i = 1}^{m} f_{i}$ over $K_{v}$ , where the $f_{i}$ are irreducible (and necessarily distinct by separability of $f$ ). Choose a root $𝜃_{i}$ of $f_{i} (x)$ for each $i$ inside a fixed algebraic closure $\bar{K_{v}}$ of $K$ . Proposition 1.1.1 provides an isomorphism

L \otimes_{K} K_{v} \to_{}^{\sim} \prod_{i = 1}^{m} K_{v} (𝜃_{i})

such that $𝜃 \otimes 1$ is sent to $𝜃_{i}$ in the $i$ th coordinate. By Theorem 5.4.7, the field $K_{v} (𝜃_{i})$ is necessarily complete with respect to a valuation $w_{i}$ extending $v$ . The embedding $τ_{i} : L \to K_{v} (𝜃_{i})$ sending $𝜃$ to $𝜃_{i}$ and fixing $K$ has dense image, so $K_{v} (𝜃_{i})$ is isomorphic to the completion of $L$ with respect to $w_{i}$ .

If $w$ is any valuation on $L$ extending $v$ , then Theorem 6.1.4 yields an embedding $τ$ of $L$ into $\bar{K_{v}}$ such that $w = \bar{v} \circ τ$ , where $\bar{v}$ is the unique extension of $v$ from $K_{v}$ to $\bar{K_{v}}$ . Then $τ (𝜃)$ is the root of $f_{i}$ for some $i$ , so there exists an automorphism $σ$ over $\bar{K_{v}}$ fixing $K_{v}$ such that $τ = σ \circ τ_{i}$ and therefore $w = \bar{v} \circ σ \circ τ_{i} = \bar{v} \circ τ_{i} = w_{i}$ . □

Corollary 6.1.7. §

Let $L ∕ K$ be a finite separable extension of fields and $v$ a valuation on $K$ . Then we have

[L : K] = \sum_{w ∣ v} [L_{w} : K_{v}] .

We can get more out of Proposition 6.1.6, which we will use later.

Definition 6.1.8. §

Let $L ∕ K$ be a finite separable extension of fields and $v$ a valuation on $K$ .

a.

The norm map for $L ∕ K$ at $v$ is the map

N_{L ∕ K}^{v} : \prod_{w ∣ v} L_{w} \to K_{v}

given by

N_{L ∕ K}^{v} ({(β_{w})}_{w}) = \prod_{w ∣ v} N_{L_{w} ∕ K_{v}} (β_{w}) .

b.

The trace map for $L ∕ K$ at $v$ is the map

{Tr}_{L ∕ K}^{v} : \prod_{w ∣ v} L_{w} \to K_{v}

given by

{Tr}_{L ∕ K}^{v} ({(β_{w})}_{w}) = \sum_{w ∣ v} {Tr}_{L_{w} ∕ K_{v}} (β_{w}) .

When $v$ is understood, these are denoted more simply by $N_{L ∕ K}$ and ${Tr}_{L ∕ K}$ .

We have the following, which says that $N_{L ∕ K}^{v}$ and $N_{L ∕ K}$ coincide on $L$ (using its natural embedding in each $L_{w}$ ), and similarly for trace maps.

Proposition 6.1.9. §

Let $L ∕ K$ be a finite separable extension of fields, and let $v$ be a valuation on $K$ . For $β \in L$ , we may view it as an element of $L_{w}$ for each $w ∣ v$ , and as elements of $K_{v}$ , we have

\begin{array}{rcl} N_{L ∕ K} (β) = \prod_{w ∣ v} N_{L_{w} ∕ K_{v}} (β) & and & {Tr}_{L ∕ K} (β) = \sum_{w ∣ v} {Tr}_{L_{w} ∕ K_{v}} (β) . \end{array}

Proof.

Let $m_{β} : L \to L$ be left multiplication. This indues a $K_{v}$ -linear transformation $m_{β} \otimes {id}_{K_{v}}$ on $K_{v}$ , and the characteristic polynomials of $m_{β}$ and $m_{β} \otimes {id}_{K_{v}}$ agree. Noting that the isomorphism of Proposition 6.1.6 is one of $L$ -vector spaces, the characteristic polynomial of $m_{β} \otimes {id}_{K_{v}}$ coincides with the product of the characteristic polynomials of multiplication by $β$ on the $L_{w}$ for $w ∣ v$ . The result is then a consequence of Proposition 1.3.3. □

The following result on valuation rings will later be useful to us.

Proposition 6.1.10. §

Let $L ∕ K$ be a finite separable extension of fields and $v$ a discrete valuation on $K$ . Suppose that $(β_{1}, \dots, β_{n})$ is an ordered basis for $L ∕ K$ such that $| β_{i} |_{w} \leq 1$ for all $1 \leq i \leq n$ and places $w$ of $L$ lying over $v$ , and $| D (β_{1}, \dots, β_{n}) |_{v} = 1$ . Then the isomorphism

κ : L \otimes_{K} K_{v} \to_{}^{\sim} \prod_{w ∣ v} L_{w}

of Proposition 6.1.6 restricts to an isomorphism

κ^{'} : ⨁_{i = 1}^{n} 𝒪_{v} (β_{i} \otimes 1) \to_{}^{\sim} \prod_{w ∣ v} 𝒪_{w},

where $𝒪_{v}$ $($ resp., $𝒪_{w})$ denotes the valuation ring of $K_{v}$ $($ resp., $L_{w})$ .

Proof.

First, we note that both the domain and codomain of $κ^{'}$ are free $𝒪_{v}$ -modules of rank $n = [L : K]$ . Moreover, $κ$ is injective, so it suffices to show that $κ^{'}$ is surjective. For this, note that the trace pairing $ψ : L \times L \to K$ given by

ψ (α, β) = {Tr}_{L ∕ K} (𝛼𝛽)

extends to a unique $K_{v}$ -bilinear pairing

ψ_{v} : (L \otimes_{K} K_{v}) \times (L \otimes_{K} K_{v}) \to K_{v},

which is given on simple tensors by the equation

ψ_{v} (α \otimes a, β \otimes b) = 𝑎𝑏 {Tr}_{L ∕ K} (𝛼𝛽) .

Let $w_{1}, \dots, w_{g}$ be the places of $L$ lying over $v$ . Since $𝒪_{v}$ is a PID, each valuation ring $𝒪_{w_{i}}$ is a free $𝒪_{v}$ -module of rank $n_{i} = [L_{w_{i}} : K_{v}]$ . So, for $1 \leq i \leq g$ , let

m_{i} = \sum_{j = 1}^{i - 1} n_{i},

and let $(λ_{m_{i} + 1}, \dots, λ_{m_{i + 1}})$ be an ordered basis of $𝒪_{w_{i}}$ as a free $𝒪_{v}$ -module. We view each $λ_{i}$ as sitting in the product $\prod_{i = 1}^{g} 𝒪_{w_{i}}$ by taking the other coordinates to be zero. Let $A = (a_{i, j}) \in {GL}_{n} (K_{v})$ be the matrix such that

λ_{i} = \sum_{j = 1}^{n} κ (β_{j} \otimes a_{i, j}),

(6.1.1)

for each $1 \leq i \leq n$ . We must show that $A$ has entries in $𝒪_{v}$ .

Via $κ$ , the pairing $ψ_{v}$ gives rise to a pairing

\tilde{ψ} : \prod_{i = 1}^{g} L_{w_{i}} \times \prod_{i = 1}^{g} L_{w_{i}} \to K_{v}

that is given on the basis $(λ_{1}, \dots, λ_{n})$ by

\begin{array}{l} \tilde{ψ} (λ_{i}, λ_{j}) & = ψ_{v} (κ^{- 1} (λ_{i}), κ^{- 1} (λ_{j})) \\ = \sum_{k = 1}^{n} \sum_{l = 1}^{n} ψ_{v} (β_{k} \otimes a_{i, k}, β_{l} \otimes a_{j, l}) \\ = \sum_{k = 1}^{n} a_{i, k} \sum_{l = 1}^{n} a_{j, l} {Tr}_{L ∕ K} (β_{k} β_{l}) \\ = \sum_{k = 1}^{n} a_{i, k} \sum_{l = 1}^{n} a_{j, l} \sum_{h = 1}^{g} {Tr}_{L_{w_{h}} ∕ K_{v}} (β_{k} β_{l}) \\ = {Tr}_{L ∕ K}^{v} (λ_{i} λ_{j}) . \end{array}

Here, of course, we have applied Proposition 6.1.9.

Note that $λ_{i} λ_{j}$ has nontrivial component in $L_{w_{h}}$ if and only if $m_{h} + 1 \leq i, j \leq m_{h + 1}$ . If this is the case, then

\tilde{ψ} (λ_{i}, λ_{j}) = {Tr}_{L_{w_{h}} ∕ K_{v}} (λ_{i} λ_{j}),

and otherwise $\tilde{ψ} (λ_{i}, λ_{j}) = 0$ . It follows that the matrix $M = (\tilde{ψ} (λ_{i}, λ_{j}))$ is block-diagonal with determinant

\prod_{i = 1}^{g} D (λ_{m_{i} + 1}, \dots, λ_{m_{i + 1}}),

and this is exactly the discriminant of $\tilde{ψ}$ relative to the basis $(λ_{1}, \dots, λ_{n})$ .

On the other hand, the discriminant of $\tilde{ψ}$ relative to the basis $(β_{1}, \dots, β_{n})$ is $D (β_{1}, \dots, β_{n})$ by definition, and we have by (6.1.1) and Lemma 1.4.6 that

D (β_{1}, \dots, β_{n}) = \det {(A)}^{2} \det (M) = \prod_{i = 1}^{g} D (λ_{m_{i} + 1}, \dots, λ_{m_{i + 1}}) .

Since $D (λ_{m_{i} + 1}, \dots, λ_{m_{i + 1}}) \in 𝒪_{v}^{\times}$ for each $i$ and $D (β_{1}, \dots, β_{n}) \in 𝒪_{v}^{\times}$ as well, we therefore have $\det (A) \in 𝒪_{v}^{\times}$ . Since the inverse of $A$ has coefficients in $𝒪_{v}$ , we therefore have that $A$ does as well. □

Finally, let us treat the special case of valuations on global fields and prove a product formula that generalizes the cases of $ℚ$ and $𝔽_{p} (t)$ for prime numbers $p$ . The following modification of the complex absolute value is necessary to account for the fact that a complex embedding and its complex conjugate have the same valuation.

Notation 6.1.11. §

Let $K$ be a global field and $v$ a place $K$ . We set

∥ ∥_{v} : K_{v}^{\times} \to ℝ_{\geq 0}

by $∥ ∥_{v} = | |_{v}$ if $v$ is not complex, and $∥ ∥_{v} = | |_{v}^{2}$ if $v$ is complex.

We have the following consequence of Proposition 6.1.9.

Lemma 6.1.12. §

Let $L ∕ K$ be a finite separable extension of global fields, and let $v$ be a place of $K$ . For $β \in L$ , we have

∥ N_{L ∕ K} (β) ∥_{v} = \prod_{w ∣ v} ∥ β ∥_{w} .

Proof.

Remark 5.4.2 and the definition of $∥ ∥_{v}$ tell us that for a place $w$ of $L$ over $v$ , we have

∥ α ∥_{w} = ∥ α ∥_{v}^{[L_{w} : K_{v}]}

for all $α \in K$ . Noting Proposition 6.1.9, we then have

∥ N_{L ∕ K} (β) ∥_{v} = \prod_{w ∣ v} ∥ N_{L_{w} ∕ K_{v}} (β) ∥_{v} = \prod_{w ∣ v} ∥ N_{L_{w} ∕ K_{v}} (β) ∥_{w}^{{[L_{w} : K_{v}]}^{- 1}} = \prod_{w ∣ v} ∥ β ∥_{w} .

□

Theorem 6.1.13 (Product formula). §

Let $K$ be a global field, and let $α \in K^{\times}$ . Then

\prod_{v \in V_{K}} ∥ α ∥_{v} = 1 .

Proof.

By Propositions 5.2.34 and 5.2.35, we have the result for $ℚ$ and $𝔽_{p} (t)$ for all primes $p$ . The field $K$ is in the general case a finite extension of exactly one of these fields, which we denote by $F$ . By Lemma 6.1.12, we have

\prod_{v \in V_{K}} ∥ α ∥_{v} = \prod_{u \in V_{F}} \prod_{v ∣ u} ∥ α ∥_{v} = \prod_{u \in V_{F}} ∥ N_{K ∕ F} (α) ∥_{u} = 1 .

□

We make the following definitions.

Definition 6.1.14. §

Suppose that $L ∕ K$ is a finite separable extension of complete discrete valuation fields.

a.: The ramification index $e_{L ∕ K}$ of $L ∕ K$ is the additive valuation on $L$ of a uniformizer of $K$ .
b.: The residue degree $f_{L ∕ K}$ of $L ∕ K$ is the degree of its residue field of $L$ over the residue field of $K$ .

Remark 6.1.15. §

In the case that $L ∕ K$ is a finite separable extension of complete discrete valuation fields for which $K = ℚ_{p}$ or $K = 𝔽_{p} ((t))$ , we denote $e_{L ∕ K}$ and $f_{L ∕ K}$ by $e_{L}$ and $f_{L}$ , respectively.

Remark 6.1.16. §

The latter definitions agree with those previously given. That is, let $L ∕ K$ be a finite extension of complete discrete valuation fields. Let $𝔪_{L}$ be the maximal ideal of $L$ and $𝔪_{K}$ that of $K$ . We then have $e_{L ∕ K} = e_{𝔪_{L} ∕ 𝔪_{K}}$ and $f_{L ∕ K} = f_{𝔪_{L} ∕ 𝔪_{K}}$ .

For complete discrete valuation fields, the degree formula is rather simpler than before.

Lemma 6.1.17. §

Let $L ∕ K$ be a finite separable extension of complete discrete valuation fields. We have $[L : K] = e_{L ∕ K} f_{L ∕ K}$ .

Proof.

There is only one nonzero prime ideal in the valuation ring of $L$ , and it lies over the maximal ideal of the valuation ring of $K$ . Theorem 2.5.11 then yields the result. □

Definition 6.1.18. §

We say that a finite separable extension $L ∕ K$ of complete discrete valuation fields is unramified, ramified, or totally ramified if the maximal ideal of the valuation ring of $L ∕ K$ is inert ( $e_{L ∕ K} = 1$ ), ramified ( $e_{L ∕ K} > 1$ ), or totally ramified ( $e_{L ∕ K} = [L : K]$ ) in the extension, respectively.

We compare these invariants with those defined previously.

Proposition 6.1.19. §

Let $A$ be a Dedekind domain with quotient field $K$ , and let $B$ be the integral closure of $A$ in a finite, separable extension $L$ of $K$ . Let $𝔭$ be a nonzero prime ideal of $A$ , and let $𝔓$ be a prime ideal of $B$ lying over $𝔭$ . Then $e_{𝔓 ∕ 𝔭} = e_{L_{𝔓} ∕ K_{𝔭}}$ and $f_{𝔓 ∕ 𝔭} = f_{L_{𝔓} ∕ K_{𝔭}}$ .

Proof.

We know that the residue field of $K$ for $𝔭$ is isomorphic to the residue field of $K_{𝔭}$ , and similarly for $L$ and $L_{𝔓}$ . Therefore, the second equality holds. As for the first, note that the valuation on $K_{𝔭}$ (resp., $L_{𝔓}$ ) is just the unique extension of that on $K$ (resp., $L$ ). If $π_{K}$ (resp., $π_{L}$ ) is a uniformizer of $K_{𝔭}$ (resp., $L_{𝔓}$ ), and $𝒪$ is the valuation ring of $L_{𝔓}$ then we have

π_{K} 𝒪 = 𝔭 𝒪 = 𝔓^{e_{𝔓 ∕ 𝔭}} 𝒪 = (π_{L}^{e_{𝔓 ∕ 𝔭}}),

so the ramification index of $L_{𝔓} ∕ K_{𝔭}$ is $v_{𝔓} (π_{K}) = e_{𝔓 ∕ 𝔭}$ . □

Definition 6.1.20. §

Let $K$ be a field, let $L$ be a Galois extension of $K$ , and let $w$ be a valuation on $L$ . For $σ \in Gal (L ∕ K)$ , the conjugate valuation $σ (w)$ is defined by

| β |_{σ (w)} = | σ^{- 1} (β) |_{w}

for $β \in L$ .

Remark 6.1.21. §

Definition 6.1.20 provides an action of the Galois group of $L ∕ K$ on the set of valuations of $L$ .

Remark 6.1.22. §

Suppose that $L ∕ K$ is an extension of global fields. Let $v$ be a valuation on $K$ , and let $w$ be a valuation on $L$ lying over it (i.e., such that $| |_{w}$ extends $| |_{v}^{[L_{w} : K_{v}]}$ ).

a.: If $v$ is the $𝔭$ -adic valuation of a finite prime $𝔭$ , then $w$ is the $𝔓$ -adic valuation of a prime $𝔓$ lying over it, and $σ (w)$ is just the $σ (𝔓)$ -adic valuation. If $v$ is an infinite prime of a finite extension of $𝔽_{p} (t)$ , then it arises as a prime ideal of the integral closure of $𝔽_{p} [t^{- 1}]$ in $K$ , and we have the analogous description.
b.: If $v$ is an archimedean prime, then $v$ arises from a real or complex embedding $τ_{v}$ of $K$ , and $w$ arises from an embedding $τ_{w}$ extending it. We then have $| β |_{σ (w)} = | τ_{w} \circ σ^{- 1} (β) |$ , and if $τ_{w}$ is complex, the complex conjugate embedding ${\bar{τ}}_{w}$ yields the same absolute value.

Given all this, we may speak of decomposition and inertia groups almost as before.

Definition 6.1.23. §

Let $L ∕ K$ be a Galois extension of fields with Galois group $G$ , and let $w$ be a valuation on $L$ .

a.: The decomposition group $G_{w}$ of $w$ is the set of $σ \in G$ fixing $w$ .
b.: The inertia group $I_{w}$ of $w$ is the set of $σ \in G_{w}$ such that $| σ (β) - β |_{w} < 1$ for all $β \in L$ with $| β |_{w} \leq 1$ if $w$ is nonarchimedean and the decomposition group if $w$ is archimedean.

We leave the following simple check to the reader.

Lemma 6.1.24. §

Let $L ∕ K$ be a Galois extension of fields with Galois group $G$ , and let $w$ be a valuation on $L$ . The inertia group $I_{w}$ is a normal subgroup of the decomposition group $G_{w}$ .

The following is an immediate consequence of Proposition 2.6.14 in the case of discrete valuation fields, but note that the analogous proof goes through in general.

Proposition 6.1.25. §

If $L ∕ K$ is a Galois extension of complete nonarchimedean valuation fields, then the decomposition group of the prime of $K$ is all of $G = Gal (L ∕ K)$ , and we have an exact sequence

1 \to I \to G \to Gal (κ (L) ∕ κ (K)) \to 1,

where $I$ is the inertia group in $G$ and $κ (L)$ (resp., $κ (K)$ ) is the residue field of $L$ (resp., $K$ ).

For the valuations attached to nonzero prime ideals in Dedekind domains, the decomposition and inertia groups agree with Definitions 2.6.5 and 2.6.15, as seen from the following.

Proposition 6.1.26. §

Let $L ∕ K$ be a Galois extension of fields, and let $w$ be a valuation on $L$ extending a valuation $v$ of $K$ . Then the restriction map

Gal (L_{w} ∕ K_{v}) \to Gal (L ∕ K), σ \mapsto σ |_{L}

is an injection with image the decomposition group of $w$ . If $w$ is nonarchimedean, the image of the inertia subgroup of $Gal (L_{w} ∕ K_{v})$ under this map is the inertia group of $w$ .

Proof.

Note that any $σ \in Gal (L_{w} ∕ K_{v})$ acts continuously on $L_{w}$ , since $w = σ (w)$ , as there is a unique valuation on $L_{w}$ extending the restriction of $w$ to $K_{v}$ . If $σ (β) = β$ for all $β \in L$ , then continuity forces $σ (β) = β$ for all $β \in L_{w}$ , since $L$ is dense in $L_{w}$ . So, the restriction map is injective, and its image is by definition contained in the decomposition group $G_{w}$ of $G = Gal (L ∕ K)$ . Any $τ$ in the decomposition group of $w$ in $G$ satisfies $w = τ (w)$ on $L$ , so on $L_{w}$ as well by continuity. Therefore, $τ : L \to L$ is continuous, and its composite with the natural inclusion $L ↪ L_{w}$ extends to a unique element of $Gal (L_{w} ∕ K_{v})$ by continuity, as in Proposition 5.3.12. That is, the image of restriction is $G_{w}$ .

For $w$ nonarchimedean, it remains to show that the image of the inertia group in $Gal (L_{w} ∕ K_{v})$ is the inertia group $I_{w}$ in $G$ . By definition, the image is contained in this group. Let $τ \in I_{w}$ , and take $β \in 𝒪_{w}$ . As $L_{w}$ is the completion of $L$ , we have $| β - b |_{w} < 1$ for some $b \in L$ with $| b |_{w} \leq 1$ . Set $β^{'} = β - b$ . Since $w$ is nonarchimedean and $| τ (b) - b |_{w} < 1$ , we have $| τ (β) - β |_{w} < 1$ if and only if $| τ (β^{'}) - β^{'} |_{w} < 1$ . But $| τ (β^{'}) - β^{'} |_{w} \leq | β^{'} |_{w} < 1$ as $| τ (β^{'}) |_{w} = | β^{'} |_{w}$ . Therefore, the extension of $τ$ to an element of $Gal (L_{w} ∕ K_{v})$ lies in the inertia subgroup. □

6.2. Differents and discriminants

Lemma 6.2.1. §

Let $A$ be a integrally closed domain with quotient field $K$ , let $L$ be a finite separable extension of $K$ , and let $B$ be the integral closure of $A$ in $L$ . Let

ℭ = {α \in L ∣ {Tr}_{L ∕ K} (𝛼𝛽) \in A for all β \in B} .

Then $ℭ$ is a fractional ideal of $B$ .

Proof.

Let $α_{1}, \dots, α_{n}$ be a basis of $L$ as a $K$ -vector space that consists of elements of $B$ . Let $d = D (α_{1}, \dots, α_{n})$ . By Lemma 1.4.19, we have that $𝑑ℭ \subseteq B$ , so $ℭ$ is a fractional ideal of $B$ . □

Lemma 6.2.1 allows us to make this following definition.

Definition 6.2.2. §

Let $A$ be a Dedekind domain with quotient field $K$ , let $L$ be a finite separable extension of $K$ , and let $B$ be the integral closure of $A$ in $L$ . The different $𝔇_{B ∕ A}$ of $B$ over $A$ is the inverse of the fractional ideal

{α \in L ∣ {Tr}_{L ∕ K} (𝛼𝛽) \in A for all β \in B} .

Remark 6.2.3. §

Since $L ∕ K$ is finite separable in Definition 6.2.2, the trace pairing of Example 1.4.4 is nonnegenerate by Proposition 1.4.13. In particular, $𝔇_{B ∕ A}$ is well-defined since it is the inverse of a submodule of $L$ .

Remark 6.2.4. §

The inverse different $𝔇_{B ∕ A}$ of Definition 6.2.2 is the smallest nonzero ideal of $B$ such that

{Tr}_{L ∕ K} (𝔇_{B ∕ A}^{- 1}) \subseteq A,

and it is a nonzero ideal of $B$ since $𝔇_{B ∕ A}^{- 1}$ contains $B$ by definition.

Lemma 6.2.5. §

Let $A$ be a Dedekind domain with quotient field $K$ . Let $L ∕ K$ and $M ∕ L$ be finite separable extensions, let $B$ be the integral closure of $A$ in $L$ , and let $C$ be the integral closure of $A$ in $M$ . We then have

𝔇_{C ∕ A} = 𝔇_{C ∕ B} \cdot 𝔇_{B ∕ A} .

Proof.

We have

{Tr}_{M ∕ K} (𝔇_{C ∕ B}^{- 1} 𝔇_{B ∕ A}^{- 1}) = {Tr}_{L ∕ K} (𝔇_{B ∕ A}^{- 1} {Tr}_{M ∕ L} (𝔇_{C ∕ B}^{- 1})) \subseteq {Tr}_{L ∕ K} (𝔇_{B ∕ A}^{- 1}) \subseteq A,

which implies that $𝔇_{C ∕ A} \subseteq 𝔇_{C ∕ B} 𝔇_{B ∕ A}$ . On the other hand, we may compute

{Tr}_{L ∕ K} ({Tr}_{M ∕ L} (𝔇_{C ∕ A}^{- 1})) = {Tr}_{M ∕ K} (𝔇_{C ∕ A}^{- 1}) \subseteq A,

so ${Tr}_{M ∕ L} (𝔇_{C ∕ A}^{- 1}) \subset 𝔇_{B ∕ A}^{- 1}$ . We therefore have that

{Tr}_{M ∕ L} (𝔇_{B ∕ A} 𝔇_{C ∕ A}^{- 1}) = 𝔇_{B ∕ A} {Tr}_{M ∕ L} (𝔇_{C ∕ A}^{- 1}) \subseteq B,

so $𝔇_{C ∕ B} \subseteq 𝔇_{B ∕ A}^{- 1} 𝔇_{C ∕ A}$ , which is to say that $𝔇_{C ∕ B} 𝔇_{B ∕ A} \subseteq 𝔇_{C ∕ A}$ . □

Lemma 6.2.6. §

Let $A$ be a Dedekind domain with quotient field $K$ , let $L$ be a finite separable extension of $K$ , let $B$ be the integral closure of $A$ in $L$ . For any multiplicatively closed subset $S$ of $A$ , one has

𝔇_{S^{- 1} B ∕ S^{- 1} A} = S^{- 1} 𝔇_{B ∕ A} .

Proof.

Note that

{Tr}_{L ∕ K} (S^{- 1} 𝔇_{B ∕ A}^{- 1}) = S^{- 1} {Tr}_{L ∕ K} (𝔇_{B ∕ A}^{- 1}) \subseteq S^{- 1} B,

so $𝔇_{S^{- 1} B ∕ S^{- 1} A} \subseteq S^{- 1} 𝔇_{B ∕ A}$ . On the other hand, we have

{Tr}_{L ∕ K} (𝔇_{S^{- 1} B ∕ S^{- 1} A}) \subseteq S^{- 1} B,

so for each $α \in 𝔇_{S^{- 1} B ∕ S^{- 1} A}^{- 1}$ , there exists $s \in S$ such that ${Tr}_{L ∕ K} (𝑠𝛼) = s {Tr}_{L ∕ K} (α) \in B$ , so $𝑠𝛼 \in 𝔇_{B ∕ A}^{- 1}$ . Therefore, we have the other containment. □

Somewhat more involved is the following.

Lemma 6.2.7. §

Let $A$ be a Dedekind domain with quotient field $K$ , let $L$ be a finite separable extension of $K$ , let $B$ be the integral closure of $A$ in $L$ . Let $𝔭$ be a prime ideal of $A$ , and let $𝔓$ be a prime ideal of $B$ lying over it. Let $𝒪_{𝔓}$ (resp., $𝒪_{𝔭}$ ) denote the valuation ring of $L_{𝔓}$ (resp., $K_{𝔭}$ ). Then

𝔇_{B ∕ A} 𝒪_{𝔓} = 𝔇_{𝒪_{𝔓} ∕ 𝒪_{𝔭}} .

Proof.

Let $α \in 𝔇_{B ∕ A}^{- 1}$ , and let $β \in 𝒪_{𝔓}$ . We claim that ${Tr}_{L_{𝔓} ∕ K_{𝔭}} (𝛼𝛽) \in 𝒪_{𝔭}$ . Let $𝔓_{1}, \dots, 𝔓_{g}$ be the prime ideals of $B$ lying over $𝔭$ , taking $𝔓_{1} = 𝔓$ . For this, let ${(β_{n})}_{n}$ be a sequence in $B$ with limit $β$ in the $𝔓$ -adic topology and with limit $0$ in the $𝔓_{i}$ -adic topology for $i \geq 2$ , which exists by the Chinese remainder theorem. Since the trace map is continuous, we have

{Tr}_{L_{𝔓} ∕ K_{𝔭}} (𝛼𝛽) = \lim_{n \to \infty} {Tr}_{L_{𝔓} ∕ K_{𝔭}} (α β_{n}) .

Moreover, we have by Proposition 6.1.9 that

{Tr}_{L_{𝔓} ∕ K_{𝔭}} (α β_{n}) = {Tr}_{L ∕ K} (α β_{n}) - \sum_{i = 2}^{g} {Tr}_{L_{𝔓_{i}} ∕ K_{𝔭}} (α β_{n}) .

Note that ${Tr}_{L ∕ K} (α β_{n}) \in A$ and that, for $i \geq 2$ , the sequence of elements ${Tr}_{L_{𝔓_{i}} ∕ K_{𝔭}} (α β_{n}) \in K_{𝔭}$ tends to $0$ in the $𝔭$ -adic topology, again by continuity of the trace map. Therefore, for sufficiently large $n$ , the element ${Tr}_{L_{𝔓} ∕ K_{𝔭}} (α β_{n})$ lies in $𝒪_{𝔭}$ , proving the claim. In particular, we have $𝔇_{B ∕ A}^{- 1} \subseteq 𝔇_{𝒪_{𝔓} ∕ 𝒪_{𝔭}}^{- 1}$ .

On the other hand, if $α \in 𝔇_{𝒪_{𝔓} ∕ 𝒪_{𝔭}}^{- 1}$ , then we may write $α$ as the limit of a sequence ${(α_{n})}_{n}$ in $L$ that has limit $0$ in $L_{𝔓_{i}}$ for $i \geq 2$ . For $β \in B$ , we have

\lim_{n \to \infty} {Tr}_{L ∕ K} (α_{n} β) = \sum_{i = 1}^{g} \lim_{n \to \infty} {Tr}_{L_{𝔓_{i}} ∕ K_{𝔭}} (α_{n} β) = {Tr}_{L_{𝔓} ∕ K_{𝔭}} (𝛼𝛽) \in 𝒪_{𝔭} .

Since $K \cap 𝒪_{𝔭} = A_{𝔭}$ , we have that ${Tr}_{L ∕ K} (α_{n} β) \in A_{𝔭}$ for sufficiently large $n$ . By Lemma 6.2.6, we therefore have $α_{n} \in S^{- 1} 𝔇_{B ∕ A}^{- 1}$ for all such $n$ , where $S$ is the complement of $𝔭$ in $A$ . But then $α \in 𝔇_{B ∕ A}^{- 1} 𝒪_{𝔓}$ as the limit of these elements. We therefore have the reverse containment $𝔇_{𝒪_{𝔓} ∕ 𝒪_{𝔭}}^{- 1} \subseteq 𝔇_{B ∕ A}^{- 1} 𝒪_{𝔓}$ and therefore equality. Since the these fractional ideals in $𝒪_{𝔓}$ agree, so do their inverses, proving the lemma. □

The following is an immediate corollary.

Corollary 6.2.8. §

Let $A$ be a Dedekind domain with quotient field $K$ , let $L$ be a finite separable extension of $K$ , let $B$ be the integral closure of $A$ in $L$ . For any prime ideal $𝔓$ of $B$ , we will use $𝔭$ to denote $𝔓 \cap A$ and $𝔇_{𝔓 ∕ 𝔭}$ to denote the intersection with $B$ of the local different $𝔇_{𝒪_{𝔓} ∕ 𝒪_{𝔭}}$ , where $𝒪_{𝔓}$ (resp., $𝒪_{𝔭}$ ) is the valuation ring of $L_{𝔓}$ (resp., $K_{𝔭}$ ). We then have

𝔇_{B ∕ A} = \prod_{𝔓} 𝔇_{𝔓 ∕ 𝔭},

with the product taken over the nonzero prime ideals of $B$ .

In the case that $B ∕ A$ is an extension Dedekind domains such that $B$ is generated by a single element as an $A$ -algebra, we have the following explicit recipe for the different.

Proposition 6.2.9. §

Let $A$ be a Dedekind domain with quotient field $K$ , let $L$ be a finite separable extension of $K$ . Let $β \in L$ be integral over $A$ , let $f \in A [x]$ be the minimal polynomial of $β$ , and let $f^{'} \in A [x]$ be the formal derivative of $f$ . Then $f^{'} {(β)}^{- 1}$ generates the $A [β]$ -module

{α \in L ∣ {Tr}_{L ∕ K} (𝛼𝐴 [β]) \subseteq A} .

Proof.

Write $f = \sum_{i = 0}^{n} a_{i} x^{i}$ for some $a_{i} \in A$ with $a_{n} = 1$ . Let $β_{1}, \dots, β_{n}$ be the roots of $f$ in an algebraic closure of $L$ . We claim that for any nonnegative integer $k < n$ , we have

\sum_{i = 1}^{n} \frac{f}{x - β_{i}} \frac{β_{i}^{k}}{f^{'} (β_{i})} = x^{k} .

(6.2.1)

To see this, note that the two sides of (6.2.1) are equal upon evaluation at each $β_{i}$ , yet both sides are polynomials of degree less than $n$ , so their difference is identically zero.

We have

\frac{f}{x - β} = \sum_{j = 0}^{n - 1} b_{j} x^{j}

for some $b_{j} \in A [β]$ , so

\sum_{j = 0}^{n - 1} {Tr}_{L ∕ K} (β^{k} \frac{b_{j}}{f^{'} (β)}) x^{j} = x^{k},

and therefore

{Tr}_{L ∕ K} (β^{k} \frac{b_{j}}{f^{'} (β)}) = δ_{j, k} .

Now, the elements $\frac{b_{j}}{f^{'} (β)}$ span ${α \in L ∣ {Tr}_{L ∕ K} (𝛼𝐴 [β]) \subseteq A}$ as an $A$ -module. We therefore need only show that the $b_{j}$ span $A [β]$ . For this, note that

(x - β) \sum_{i = 0}^{n - 1} b_{i} x^{i} = \sum_{i = 0}^{n} a_{i} x^{i},

so $b_{n - 1} = 1$ and $b_{j} - β b_{j + 1} = a_{j + 1}$ for each $0 \leq j \leq n - 1$ . Solving for the $b_{j}$ , we obtain

b_{j} = \sum_{i = 0}^{n - j - 1} a_{i + j + 1} β^{i},

for each $j$ . Since the coefficients of the powers of $β$ in the latter expression $b_{j}$ form the columns of a unipotent matrix in $A$ , each power $β^{i}$ of $β$ with $0 \leq i \leq n - 1$ may be written as an $A$ -linear combination of the $b_{j}$ . Thus, the $b_{i}$ span $A [β]$ . □

Corollary 6.2.10. §

Let $A$ be a Dedekind domain with quotient field $K$ , let $L$ be a finite extension of $K$ . Suppose that the integral closure of $A$ in $L$ equals $A [β]$ for some $β \in L$ . Let $f \in A [x]$ be the minimal polynomial of $β$ , and let $f^{'} \in A [x]$ be the formal derivative of $f$ . Then $𝒟_{B ∕ A} = (f^{'} (β))$ .

We will require the following.

Lemma 6.2.11. §

Let $K$ be a complete discrete valuation field with valuation ring $𝒪_{K}$ , and let $L$ be a finite extension of $K$ , with valuation ring $𝒪_{L}$ . Suppose that the corresponding extension $κ (L) ∕ κ (K)$ of residue fields is separable. Then there exists $β \in 𝒪_{L}$ such that $𝒪_{L} = 𝒪_{K} [β]$ . Moreover, any $β^{'} \in 𝒪_{L}$ sufficiently close to $β$ in the topology of $L$ also satisfies $𝒪_{L} = 𝒪_{K} [β^{'}]$ .

Proof.

Since $κ (L) ∕ κ (K)$ is separable, there exists $\bar{β} \in κ (L)$ such that $κ (L) = κ (K) (\bar{β})$ . Let $\bar{f} \in κ (K) [x]$ be the minimal polynomial of $\bar{β}$ , and let $f \in 𝒪_{K} [x]$ be any lift of $\bar{f}$ . We claim that there exists a lift $β \in 𝒪_{L}$ of $\bar{β}$ to an element with $f (β)$ a uniformizer of $L$ . For any lift $α$ , we must have at least that the valuation of $f (α)$ is positive, since $\bar{f} (\bar{β}) = 0$ . If it is not $1$ , then $α + π_{L}$ , where $π_{L}$ is a uniformizer of $L$ is another lift with

f (α + π_{L}) \equiv f (α) + f^{'} (α) π_{L} mod {(π_{L})}^{2} .

Since $\bar{f}$ is separable, we have that $f^{'} (α) \in 𝒪_{L}^{\times}$ . Therefore, we $f (α + π_{L})$ does indeed have valuation $1$ .

Now, with $β$ chosen, we set $π_{L} = f (β)$ and claim that the $β^{i} π_{L}^{j} = β^{i} f {(β)}^{j}$ with $0 \leq i \leq f_{L ∕ K} - 1$ and $0 \leq j \leq e_{L ∕ K} - 1$ form an $𝒪_{K}$ -basis of $𝒪_{L}$ , which will finish the proof, aside from the final statement. Given a uniformizer $π_{K}$ of $K$ , it suffices to show that the elements $β^{i} π_{L}^{j}$ are a basis of $𝒪_{L} ∕ π_{K} 𝒪_{L}$ as an $κ (K)$ -vector space. To see this, fix a set $S_{K}$ of representatives of $κ (K)$ in $𝒪_{K}$ , and note that the set $S_{L}$ of elements

\sum_{i = 0}^{f_{L ∕ K} - 1} c_{i} β^{i}

with $c_{i} \in S_{K}$ is a set of representatives of $κ (L)$ . While $π_{L}^{e_{L ∕ K}}$ is a multiple of $π_{K}$ , the elements

\sum_{i = 0}^{e_{L ∕ K} - 1} a_{i} π_{L}^{i}

with $a_{i} \in S_{L}$ clearly have distinct image in the quotient, which has dimension $e_{L ∕ K} f_{L ∕ K}$ . Hence, we have the claim.

Finally, note that the proof that $𝒪_{L} = 𝒪_{K} [β]$ depended only on the facts that $β$ lifts $\bar{β}$ and that $f (β)$ is a uniformizer. Since this holds true for any element in the congruence class of $β$ modulo $π_{L}^{2}$ , we are done. □

The latter lemma helps to extend the recipe of Corollary 6.2.10 to the general case. We omit the proof of the following theorem.

Theorem 6.2.12. §

Let $A$ be a Dedekind domain with quotient field $K$ , let $L$ be a finite separable extension of $K$ , and let $B$ be the integral closure of $A$ in $L$ . The $𝔇_{B ∕ A}$ is the ideal generated by the elements $f^{'} (β)$ with $β \in B$ such that $L = K (β)$ , for $f \in A [x]$ the minimal polynomial of $β$ .

containment, we consider the completion $𝒪_{𝔓}$ of $B$ with respect to a nonzero prime ideal

We may now show that the different detects ramification of primes.

Theorem 6.2.13. §

Let $A$ be a Dedekind domain with quotient field $K$ , let $L$ be a finite extension of $K$ , and let $B$ be the integral closure of $A$ in $L$ . Let $𝔓$ be a nonzero prime ideal of $B$ , let $𝔭 = A \cap 𝔓$ , and suppose that the corresponding extension of residue fields is separable. Then $𝔓$ is ramified over $A$ if and only if it divides $𝔇_{B ∕ A}$ .

Proof.

By Lemma 6.2.7, we may replace $B$ by its completion at $𝔓$ and $A$ by its completion at $𝔭$ . Therefore, we assume that $A$ is a complete discrete valuation ring, as is $B$ . By Lemma 6.2.11, we have that $B = A [β]$ for some $β \in B$ . Let $f \in A [x]$ be the minimal polynomial of $β$ . As $𝔇_{B ∕ A} = (f^{'} (β))$ , the prime $𝔓$ does not divide $𝔇_{B ∕ A}$ if and only if $f^{'} (β)$ is a unit, which is to say if and only if the image $\bar{β} \in B ∕ 𝔓$ of $β$ is a simple root of the image $\bar{f}$ of $f$ in $(A ∕ 𝔭) [x]$ .

If $𝔓$ is unramified, then $B ∕ 𝔓 = (A ∕ 𝔭) [\bar{β}]$ is of degree $[L : K]$ over $A ∕ 𝔭$ , so $\bar{f}$ is irreducible. Since the extension of residue fields is separable, $\bar{f}$ is itself separable, so $f^{'} (β)$ is a unit.

Conversely, suppose that $f^{'} (β)$ is a unit. Then the minimal polynomial $\bar{g}$ of $\bar{β}$ is relatively prime to $\bar{f} {\bar{g}}^{- 1}$ . By Theorem 5.3.33, there is a lift $g \in A [x]$ of $\bar{g}$ that divides $f$ and has the same degree as $\bar{g}$ . As $f$ is irreducible, this forces $\deg \bar{g} = [L : K]$ , which means that $\bar{f} = \bar{g}$ is irreducible. Thus, $𝔓$ is unramified. □

The different is closely is closely related to the discriminant, which we now define in greater generality than before, though with slightly less specificity in the already defined case that the ground ring is $ℤ$ , since the different we now consider is an ideal, not an integer.

Definition 6.2.14. §

Let $A$ be a Dedekind domain with quotient field $K$ , let $L$ be a finite separable extension of $K$ , let $B$ be the integral closure of $A$ in $L$ . The discriminant $𝔡_{B ∕ A}$ of $B ∕ A$ is the ideal of $A$ generated by all discriminants $D (α_{1}, \dots, α_{n})$ of ordered bases $(α_{1}, \dots, α_{n})$ of $L$ over $K$ that are contained in $B$ .

Proposition 6.2.15. §

Let $A$ be a Dedekind domain with quotient field $K$ , let $L$ be a finite separable extension of $K$ , let $B$ be the integral closure of $A$ in $L$ . Then

𝔡_{B ∕ A} = N_{L ∕ K} (𝔇_{B ∕ A}) .

Proof.

Let $𝔭$ be a prime ideal of $A$ that divides $𝔡_{B ∕ A}$ . (By Proposition 2.5.15, the ideals of $A$ that ramify in $B$ , hence lie below primes dividing $𝔇_{B ∕ A}$ , divide $𝔡_{B ∕ A}$ , so this suffices.) Let $S = A - 𝔭$ , and consider the localizations $S^{- 1} 𝔇_{B ∕ A}$ and $S^{- 1} 𝔡_{B ∕ A}$ . We know that $S^{- 1} 𝔇_{B ∕ A} = 𝔇_{S^{- 1} B ∕ S^{- 1} A}$ and that $S^{- 1} 𝔡_{B ∕ A} = 𝔡_{S^{- 1} B ∕ S^{- 1} A}$ follows directly from the definition of the discriminant (since $S$ is contained in $A$ and the discriminant function $D$ is $K$ -multilinear). Therefore, we may assume that $A$ is a DVR, from which it follows that $B$ is a PID (as a Dedekind domain with only finitely many nonzero prime ideals).

Since $B$ is a torsion-free $A$ -module of finite rank, it admits an $A$ -basis $(α_{1}, \dots, α_{n})$ , and we have $𝔡_{B ∕ A} = D (α_{1}, \dots, α_{n})$ . Let $(β_{1}, \dots, β_{n}) \in L^{n}$ be the dual basis to $(α_{1}, \dots, α_{n})$ for which ${Tr}_{L ∕ K} (α_{i} β_{j}) = δ_{i, j}$ for $1 \leq i, j \leq n$ . Then $(β_{1}, \dots, β_{n})$ is a free $A$ -module basis of $𝔇_{B ∕ A}^{- 1}$ . Let $γ \in B$ be such that $(γ) = 𝔇_{B ∕ A}$ , and note that $(γ^{- 1} α_{1}, \dots, γ^{- 1} α_{n})$ is also an ordered basis of $𝔇_{B ∕ A}^{- 1}$ as an $A$ -module. We therefore have

(D (β_{1}, \dots, β_{n})) = (D (γ^{- 1} α_{1}, \dots, γ^{- 1} α_{n})) = {(N_{L ∕ K} (γ))}^{- 2} (D (α_{1}, \dots, α_{n})) .

(6.2.2)

Let $σ_{1}, \dots, σ_{n}$ be the $K$ -linear embeddings of $L$ in an algebraic closure $\bar{K}$ of $K$ . Note that the product of the transpose of the matrix ${(σ_{i} α_{j})}_{i, j}$ and the matrix ${(σ_{i} β_{j})}_{i, j}$ has $(i, j)$ -entry

\sum_{k = 1}^{n} σ_{k} (α_{i}) σ_{k} (β_{j}) = {Tr}_{L ∕ K} (α_{i} β_{j}),

so is the identity matrix. Therefore, we have that

D (β_{1}, \dots, β_{n}) = \pm D {(α_{1}, \dots, α_{n})}^{- 1} .

Combining this with (6.2.2), we have

{(D (α_{1}, \dots, α_{n}))}^{2} = {(N_{L ∕ K} (γ))}^{2},

so we obtain $𝔡_{B ∕ A} = N_{L ∕ K} (𝔇_{B ∕ A})$ . □

We derive a few corollaries.

Corollary 6.2.16. §

Let $A$ be a Dedekind domain with quotient field $K$ , let $L$ be a finite separable extension of $K$ , let $B$ be the integral closure of $A$ in $L$ . A prime ideal of $A$ ramifies in $B$ if and only if it divides $𝔡_{B ∕ A}$ .

Proof.

This is immediate from Theorem 6.2.13 and Proposition 6.2.15. □

Remark 6.2.17. §

Corollary 6.2.16 tells us that the ideal $𝑝ℤ$ generated by a prime $p$ ramifies in $K ∕ ℚ$ for a number field $K$ if and only if $p$ divides $disc (K)$ .

Corollary 6.2.18. §

Let $A$ be a Dedekind domain with quotient field $K$ . Let $L ∕ K$ and $M ∕ L$ be finite extensions, let $B$ be the integral closure of $K$ in $L$ , and let $C$ be the integral closure of $K$ in $M$ . We then have

𝔡_{C ∕ A} = 𝔡_{B ∕ A}^{[M : L]} N_{L ∕ K} (𝔡_{C ∕ B}) .

Proof.

By Lemma 6.2.5 and Proposition 6.2.15, we have

𝔡_{C ∕ A} = N_{M ∕ K} (𝔇_{C ∕ A}) = N_{M ∕ K} (𝔇_{C ∕ B}) N_{L ∕ K} {(𝔇_{B ∕ A})}^{[M : L]} = N_{L ∕ K} (𝔡_{C ∕ B}) 𝔡_{B ∕ A}^{[M : L]},

as desired. □

Corollary 6.2.19. §

Let $A$ be a Dedekind domain with quotient field $K$ , let $L$ be a finite separable extension of $K$ , let $B$ be the integral closure of $A$ in $L$ . For any prime ideal $𝔓$ of $B$ , we will use $𝔭$ to denote $𝔓 \cap A$ and $𝔡_{𝔓 ∕ 𝔭}$ to denote the intersection with $A$ of the local discriminant $𝔡_{𝒪_{𝔓} ∕ 𝒪_{𝔭}}$ , where $𝒪_{𝔓}$ (resp., $𝒪_{𝔭}$ ) is the valuation ring of $L_{𝔓}$ (resp., $K_{𝔭}$ ). We then have

𝔡_{B ∕ A} = \prod_{𝔓} 𝔡_{𝔓 ∕ 𝔭},

with the product taken over the nonzero prime ideals of $B$ .

Proof.

By Lemma 6.1.9, Proposition 6.2.15, and Lemma 6.2.7, we have

𝔡_{B ∕ A} 𝒪_{𝔭} = N_{L ∕ K} (𝔇_{B ∕ A}) 𝒪_{𝔭} = \prod_{𝔓 ∣ 𝔭} N_{L_{𝔓} ∕ K_{𝔭}} (𝔇_{B ∕ A} 𝒪_{𝔓}) = \prod_{𝔓 ∣ 𝔭} N_{L_{𝔓} ∕ K_{𝔭}} (𝔇_{𝒪_{𝔓} ∕ 𝒪_{𝔭}}) = \prod_{𝔓 ∣ 𝔭} 𝔡_{𝒪_{𝔓} ∕ 𝒪_{𝔭}},

hence the result by intersection with $A$ . □

In the case of global or complete discrete valuation fields, which come equipped with canonical subrings, the ring of integers and valuation ring in the respective cases, we can use the field as the subscript in the definition of the different and discriminant, which we will typically do below.

Definition 6.2.20. §

a.: The different $𝔇_{L ∕ K}$ (resp., discriminant $𝔡_{L ∕ K}$ ) of an extension $L ∕ K$ of global fields is the different (resp., discriminant) of the corresponding extension $𝒪_{L} ∕ 𝒪_{K}$ of rings of integers.
b.: The different $𝔇_{L ∕ K}$ (resp., discriminant $𝔡_{L ∕ K}$ ) of an extension $L ∕ K$ of complete discrete valuation fields is the different (resp., discriminant) of the corresponding extension of valuation rings.

Combining the above results with Minkowski theory would allow us to derive the following fascinating result, which we state without proof.

Theorem 6.2.21. §

Let $K$ be a number field and $S$ a finite set of prime ideals of $K$ . For each $n \geq 1$ , there exist only finitely many extensions $L ∕ K$ of degree $n$ in which the prime ideals of $K$ that ramify in $L$ are all contained in $S$ .

As a consequence of Theorem 6.2.21 and Corollary 4.3.6, one has the following.

Theorem 6.2.22. §

For any $N \geq 1$ , there exist only finitely many number fields $K$ with $| disc (K) | \leq N$ .

Definition 6.2.23. §

A separable algebraic extension $L$ of a global field $K$ is unramified if every place of $K$ is unramified in $L$ .

We have the following corollary of Theorem 4.3.6.

Corollary 6.2.24. §

The field $ℚ$ has no nontrivial extension that is unramified at all finite primes.

Proof.

Corollary 4.3.6 tells us that if $[K : ℚ] \geq 2$ , then

| disc (K) | \geq \frac{2^{2}}{2!} \cdot (\frac{π}{4})^{2} = \frac{π^{2}}{8} > 1,

so $K$ is not unramified. □

6.3. Multiplicative groups of local fields

In this section, we study the structure of multiplicative groups of local fields. For the rest of this chapter, “local field” should be taken to mean “nonarchimedean local field”. Let us make the following definition.

Definition 6.3.1. §

Let $K$ be a complete discrete valuation field with valuation ring $𝒪$ and maximal ideal $𝔪$ . Then the $i$ th unit group $U_{i} = U_{i} (K)$ of $K$ is defined by $U_{0} = 𝒪^{\times}$ for $i = 0$ and $U_{i} = 1 + 𝔪^{i}$ for $i \geq 1$ .

Notation 6.3.2. §

In this section, we let $K$ be a local field. We let $p$ be the characteristic of its residue field, $𝒪$ its valuation ring, $𝔪$ its maximal ideal, $π$ a fixed uniformizer, $κ$ the residue field, and $q$ the order of $κ$ . We let $e = e_{K}$ and $f = f_{K}$ .

Lemma 6.3.3. §

The set of roots of unity of order prime to $p$ in a local field $K$ has order $q - 1$ , where $q$ is the order of the residue field of $K$ .

Proof.

The polynomial $x^{q} - x$ splits completely over the residue field $κ$ of $K$ , so Hensel’s Lemma tells us that $μ_{q - 1} (K)$ has order $q$ and maps isomorphically onto $κ {(K)}^{\times}$ . □

Proposition 6.3.4. §

Let $K$ be a local field with residue field of order $q$ . The canonical map

⟨ π ⟩ \times μ_{q - 1} (K) \times U_{1} (K) \to_{}^{\sim} K^{\times}

is an isomorphism.

Proof.

Since $K$ is a discrete valuation field with valuation we denote $v$ , we may write any $a \in K^{\times}$ uniquely as $a = π^{v (a)} b$ for some $b \in U_{0}$ . By Lemma 6.3.3, each $b \in U_{0}$ may then be written uniquely as $b = ξ \cdot u$ with $ξ \in μ_{q - 1}$ and $u \in U_{1}$ . □

Lemma 6.3.5. §

Let $K$ be a local field. For $a \in 𝒪$ , let $\bar{a}$ denote its image in $κ$ . We have isomorphisms of groups

U_{0} ∕ U_{1} \to_{}^{\sim} κ^{\times}, u U_{1} \mapsto \bar{u},

and

U_{i} ∕ U_{i + 1} \to_{}^{\sim} κ, (1 + π^{i} a) U_{i + 1} \mapsto \bar{a}

for $i \geq 1$ .

Proof.

The first statement follows from Proposition 6.3.4. The bijectivity of the second map is clear, and that it is a homomorphism is simply that

(1 + π^{i} a) (1 + π^{i} b) \equiv 1 + π^{i} (a + b) mod 𝔪^{i + 1} .

□

Lemma 6.3.6. §

Let $K$ be a local field of residue characteristic $p$ . We have an isomorphism of $ℤ_{p}$ -modules

U_{1} \to_{}^{\sim} \underset{i}{lim_{\leftarrow}} U_{1} ∕ U_{i}

via the map induced by the universal property of the inverse limit.

Proof.

Since $⋂_{i} U_{i} = {1}$ , the map in question is injective. Any sequence ${(a_{i})}_{i}$ in $U_{1}$ with $a_{i + 1} a_{i}^{- 1} \in U_{i}$ for each $i \geq 1$ is the image of the limit of the sequence. □

Note also the following.

Lemma 6.3.7. §

Let $K$ be a $p$ -adic field. Let $e = v (p)$ . For $i \geq 1$ and $a \in 𝒪^{\times}$ , we have

{(1 + π^{i} a)}^{p} \equiv 1 + p π^{i} a + π^{𝑖𝑝} a^{p} mod 𝔪^{2 i + e} .

In particular, we have

{(1 + π^{i} a)}^{p} \equiv {\begin{matrix} 1 + π^{𝑖𝑝} a^{p} mod 𝔪^{i + e} & if i < \frac{e}{p - 1}, \\ 1 + p π^{i} a mod 𝔪^{i + e + 1} & if i > \frac{e}{p - 1} . \end{matrix}

Proof.

The first statement is an easy consequence of the binomial expansion

{(1 + π^{i} a)}^{p} = \sum_{k = 0}^{p} (\binom{p}{k}) π^{𝑖𝑘} a^{k},

since $p$ exactly divides $(\binom{p}{k})$ for $0 < k < p$ . The second follows from the fact that $i > e ∕ (p - 1)$ if and only if

v (p π^{i}) = i + e < 𝑖𝑝 = v (π^{𝑖𝑝}) .

□

Lemma 6.3.8. §

The $p$ th power map is an isomorphism $U_{i} \to_{}^{\sim} U_{i + e}$ for $i > \frac{e}{p - 1}$ .

Proof.

For any $α \in U_{i + e}$ and $k \geq 1$ , suppose by induction that we have found $β_{k} \in U_{i}$ with

β_{k}^{p} α^{- 1} = 1 + p π^{i + k - 1} a_{k} mod 𝔪^{i + e + k}

for some $a_{k} \in 𝒪$ . Set $β_{k + 1} = β_{k} (1 + π^{i + k - 1} a_{k})$ and

β = \lim_{k \to \infty} β_{k} .

Then $β \in U_{i}$ with $β^{p} = α$ , and the $p$ th power map $U_{i} \to U_{i + e}$ is surjective.

Note that $p ℤ_{p} [μ_{p}] = {(1 - ζ_{p})}^{p - 1}$ for a primitive $p$ th root of unity $ζ_{p}$ by Lemma 3.1.13. So, if $ζ_{p} \in K$ , then $v (ζ_{p} - 1) = e ∕ (p - 1)$ . That is, $ζ_{p} \notin U_{e ∕ (p - 1) + 1}$ , so the map is injective as well. □

As a pro- $p$ group, $U_{1} (K)$ for $K$ a local field of residue characteristic $p$ is generated by any lift of a set of generators of $U_{1} ∕ U_{1}^{p}$ .

Proposition 6.3.9. §

Let $K$ be a $p$ -adic field, let $q$ be the order of the residue field of $K$ , let $π_{K}$ be a uniformizer of $K$ , and let $p^{n}$ be the number of $p$ -power roots of unity in $K$ . Then we have an isomorphism

U_{1} (K) ≅ ℤ_{p}^{[K : ℚ_{p}]} \times ℤ ∕ p^{n} ℤ,

of finitely generated $ℤ_{p}$ -modules. In particular, there are isomorphisms of topological groups

K^{\times} ≅ ⟨ π_{K} ⟩ \times 𝒪_{K}^{\times} ≅ ⟨ π_{K} ⟩ \times μ_{q - 1} (K) \times U_{1} (K) ≅ ℤ \times ℤ ∕ (q - 1) ℤ \times ℤ ∕ p^{n} ℤ \times ℤ_{p}^{[K : ℚ_{p}]},

where $K^{\times}$ has the subspace topology from $K$ , and the direct products are all given the product topology, with $⟨ π_{K} ⟩$ , $ℤ$ , and the finite groups involved given the discrete topology.

Proof.

Lemma 6.3.8 tells us that $U_{1} ∕ U_{1}^{p}$ is finite, so $U_{1}$ is a finitely generated $ℤ_{p}$ -module. Since the $p$ -power torsion in $U_{1}$ is the group of $p$ -power roots of unity in $K$ , which is cyclic of order $p^{n}$ , we have $U_{1} ≅ ℤ_{p}^{r} \times ℤ ∕ p^{n} ℤ$ for some $r \geq 0$ , and this is a topological isomorphism.

No nontrivial $p$ -power root of unity $ζ$ lies in $U_{j}$ for any integer $j > \frac{e}{p - 1}$ , where $e = v (p)$ . Since $U_{j}$ has finite index in $U_{1}$ , we therefore have that $U_{j} ≅ ℤ_{p}^{r}$ . By Lemma 6.3.8, we know that $U_{j}^{p} = U_{j + e}$ , so

r = \dim_{𝔽_{p}} U_{j} ∕ U_{j + e} = \sum_{k = j}^{j + e - 1} \dim_{𝔽_{p}} U_{k} ∕ U_{k + 1} = 𝑒𝑓 = [K : ℚ_{p}],

noting that $U_{k} ∕ U_{k + 1} ≅ 𝔽_{p^{f}}$ for all $k \geq 1$ , where $f$ is the residue degree of $K ∕ ℚ_{p}$ .

Finally, note that the $U_{i}$ form a basis of open neighborhoods of $1$ in $K^{\times}$ under both the subspace topology and topology induced by the product topology in the isomorphism with $K^{\times}$ of the theorem. Therefore, these isomorphisms are of topological groups. □

Let us also mention the case of finite characteristic. We provide an outline of the proof.

Proposition 6.3.10. §

Let $K = 𝔽_{q} ((t))$ . Then there exists a continuous $ℤ_{p}$ -linear isomorphism from $U_{1} (K)$ to a countable direct product of copies of $ℤ_{p}$ .

Proof.

Let $B$ be a basis of $𝔽_{q}$ as an $𝔽_{p}$ -vector space. Let $I$ be the countable set

I = {(c, i) ∣ c \in B, i \geq 1, p ∤ i} .

Define a homomorphism

κ : \prod_{(c, i) \in I} ℤ_{p} \to U_{1} (K)

κ ({(a_{c, i})}_{(c, i)}) = \prod_{(c, i) \in I} {(1 + c t^{i})}^{a_{c, i}} .

This is easily seen to be well-defined, and one may check that every element of $U_{1}$ has a unique expansion of this form.

The inverse image of $U_{j}$ under $κ$ contains the open neighborhood of $1$ that is the direct product of $p^{n_{i}} ℤ_{p}$ is the $(c, i)$ -coordinate for each $(c, i) \in I$ , where each $n_{i} \geq 0$ is minimal satisfying $p^{n_{i}} i > j$ . On the other hand, the image of an open neighborhood

\prod_{c, i \in I} p^{n_{c, i}} ℤ_{p}

with $n_{c, i} = 0$ for $i$ sufficiently large contains $U_{j}$ for $j$ with

j = \max {p^{n_{c, i}} i ∣ (c, i) \in I, n_{c, i} \geq 1},

which we leave to the reader to check. Therefore, $κ$ is in fact a topological isomorphism. □

6.4. Tamely ramified extensions

Before studying the larger class of tamely ramified extensions of a local field, let us first consider unramified extensions.

Lemma 6.4.1. §

Let $K$ be a local field. For each positive integer $n$ , there exists a unique unramified extension of $K$ of degree $n$ , equal to $K (μ_{q^{n} - 1})$ , where $q$ is the order of the residue field of $K$ .

Proof.

Let $L ∕ K$ be an unramified extension of degree $n$ . Then $κ (L)$ is a degree $n$ extension of $κ (K)$ by the degree formula. That $L$ contains $K (μ_{q^{n} - 1})$ is then simply Lemma 6.3.3. Moreover, $K (μ_{q^{n} - 1})$ is then by definition of degree $n$ over $K$ , so equals $L$ . □

Definition 6.4.2. §

We say that an algebraic extension $L$ of a local field $K$ is unramified if it is separable and every finite degree subextension of $K$ in $L$ is unramified.

Similarly, we have the following.

Definition 6.4.3. §

We say that an algebraic extension $L$ of a local field $K$ is totally ramified if it is separable and every finite degree subextension of $K$ in $L$ is totally ramified.

Definition 6.4.4. §

A Frobenius automorphism in a Galois extension $L ∕ K$ , with $K$ a local field, is any lift of the Frobenius automorphism of the extension of residue fields to $L$ .

Remark 6.4.5. §

If $L ∕ K$ is unramified in Definition 6.4.4, then there is a unique Frobenius automorphism in $Gal (L ∕ K)$ .

Proposition 6.4.6. §

Let $L$ be an separable extension of a local field $K$ . Then there is a unique maximal unramified extension $E$ of $K$ in $L$ , and $Gal (E ∕ K)$ is topologically generated by its Frobenius automorphism.

Proof.

It suffices to consider the case that $L ∕ K$ is finite. Let $q$ be the order of the residue field of $L$ . Set $E = K (μ_{q - 1})$ , which is unramified over $K$ . Any unramified extension of $K$ in $L$ is generated prime-to- $p$ roots of unity, of which there are only $q$ in $L$ . □

The following definition makes sense as the union of all finite unramified extensions of a local field (in a fixed separable closure).

Definition 6.4.7. §

The maximal unramified extension $K^{ur}$ of a local field is the unique largest unramified extension of $K$ inside a given separable closure of $K$ .

Proposition 6.4.8. §

The maximal unramified extension $K^{ur}$ of a local field $K$ is given by adjoining all prime-to- $p$ roots of unity in a separable closure of $K$ . Its Galois group $Gal (K^{ur} ∕ K)$ is isomorphic to $\hat{ℤ}$ via the map that takes the Frobenius automorphism to $1$ .

Proof.

By definition, $K^{ur}$ is the union of the finite unramified subextensions of $K$ in $K^{sep}$ , which is to say the fields $K_{n} = K (μ_{q^{n} - 1})$ . Since any prime-to- $p$ integer $m$ divides $q^{n} - 1$ for some $n$ , we have that $K^{ur}$ is given by adjoining all prime-to- $p$ roots of $1$ . Recall that $Gal (K_{n} ∕ K) ≅ ℤ ∕ 𝑛ℤ$ via the map that takes the Frobenius automorphism to $1$ . Therefore, the isomorphism in question is the composite of the canonical maps

Gal (K^{ur} ∕ K) ≅ \underset{n}{lim_{\leftarrow}} Gal (K_{n} ∕ K) ≅ \underset{n}{lim_{\leftarrow}} Gal (𝔽_{q^{n}} ∕ 𝔽_{q}) ≅ \underset{n}{lim_{\leftarrow}} ℤ ∕ 𝑛ℤ ≅ \hat{ℤ} .

□

Definition 6.4.9. §

Let $K$ be a local field and $L$ a Galois extension of $K$ . The inertia subgroup of $Gal (L ∕ K)$ is the subgroup of elements fixing the maximal unramified extension of $K$ in $L$ .

The following is analogous to Remark 6.1.25 and almost immediate from Proposition 6.4.8.

Proposition 6.4.10. §

Let $K$ be a local field, and let $I_{K}$ denote the inertia subgroup of $G_{K}$ . Then there is an exact sequence

1 \to I_{K} \to G_{K} \to G_{κ (K)} \to 1,

where $κ (K)$ is the residue field of $K$ , and the map $G_{K} \to G_{κ (K)}$ is the composition of restriction $G_{K} \to Gal (K^{ur} ∕ K)$ with the continuous isomorphism that takes a Frobenius element of $Gal (K^{ur} ∕ K)$ to the Frobenius element of $G_{κ (K)}$ .

Definition 6.4.11. §

a.: A separable extension $L ∕ K$ of local fields of residue characteristic $p$ is tamely ramified if $p ∤ e_{L ∕ K}$ .
b.: A separable extension $L$ of a local field $K$ is tamely ramified if every finite extension $E$ of $K$ in $L$ is tamely ramified.

Example 6.4.12. §

Let $K$ be a local field, $π$ a uniformizer, and $e$ an integer not divisible by the residue characteristic $p$ of $K$ . Then $K (π^{1 ∕ e}) ∕ K$ is totally and tamely ramified, with ramification index $e$ .

In fact, every tamely ramified extension is given in essentially this way.

Proposition 6.4.13. §

Let $L ∕ K$ be a tamely ramified extension of local fields of ramification index $e$ . Then there exists a finite unramified extension $E$ of $K$ and a unifomizer $λ$ of $E$ such that $L = E (μ)$ for an $e$ th root $μ$ of $λ$ .

Proof.

Let $E$ be the maximal unramified subextension of $K$ in $L$ . Then $L ∕ E$ is a totally ramified extension of degree $e$ . Let $π_{L}$ be a uniformizer of $L$ and $π_{K}$ a uniformizer of $K$ . A simple check of valuations tells us that $π_{L}^{e} = π_{K} α$ for some $α \in 𝒪_{L}^{\times}$ . In turn, Proposition 6.3.4 allows us to write $α = ξ \cdot u$ with $ξ \in L$ a root of unity of order prime-to- $p$ , where $p$ is the residue characteristic of $K$ , and $u \in U_{1} (L)$ . Since $L ∕ E$ is totally ramified, we actually have that $ξ \in E$ . Since $p ∤ e$ and $U_{1} (L)$ is an abelian pro- $p$ group, the element $u$ has an $e$ th root $β$ in $L$ . Set $λ = π_{K} ξ$ , which is a uniformizer in $E$ . We have ${(π_{L} β^{- 1})}^{e} = λ$ , so $L$ contains an $e$ th root $μ$ of $λ$ . Since the degree of $E (μ) ∕ E$ is $e$ , we have $L = E (μ)$ . □

Example 6.4.14. §

The extension $ℚ_{p} (μ_{p}) ∕ ℚ_{p}$ is totally ramified of degree $p - 1$ , hence tamely ramified. In fact, we have that $ℚ_{p} (μ_{p}) = ℚ_{p} ({(- p)}^{\frac{1}{p - 1}})$ . To see this, note that

p = \prod_{i = 1}^{p - 1} (1 - ζ_{p}^{i}) = {(1 - ζ_{p})}^{p - 1} \prod_{i = 1}^{p - 1} (1 + ζ_{p} + \dots + ζ_{p}^{i - 1}),

where $ζ_{p}$ is a primitive $p$ th root of $1$ . We have

\prod_{i = 1}^{p - 1} (1 + ζ_{p} + \dots + ζ_{p}^{i - 1}) \equiv \prod_{i = 1}^{p - 1} i \equiv (p - 1)! \equiv - 1 mod (1 - ζ_{p}),

so $- p = {(1 - ζ_{p})}^{p - 1} u$ for some $u \in U_{1} (ℚ_{p} (μ_{p}))$ . Since any such $u$ is a $(p - 1)$ st power, we have the claim.

If a separable extension of a local field is not tamely ramified, we say it is wildly ramified.

Definition 6.4.15. §

A separable extension $L$ of a local field $K$ of residue characteristic $p$ is wildly ramified if $p$ divides the ramification index of some finite extension $E$ of $K$ in $L$ .

Examples 6.4.16. §

Let $p$ be a prime number.

a.: The extension $ℚ_{p} (p^{1 ∕ p}) ∕ ℚ_{p}$ is wildly ramified.
b.: The extension $ℚ_{p} (μ_{p^{2}}) ∕ ℚ_{p}$ is wildly ramified, since its ramification index is $p (p - 1)$ .

The compositum of any collection of tamely ramified extensions is tamely ramified, so we may speak of the maximal tamely ramified extension of a local field inside a separable algebraic closure.

Remark 6.4.17. §

We can make sense of exponentiation on $ℤ_{p}^{\times}$ by elements $\hat{ℤ}$ as follows: if $u \in ℤ_{p}$ and $a \in \hat{ℤ}$ , then we set

u^{a} = \lim_{n \to \infty} u^{a_{n}},

where ${(a_{n})}_{n}$ is any sequence of integers with limit $a$ . We leave it to the reader to check that the limit converges independently of the choice of sequence.

Proposition 6.4.18. §

The maximal tamely ramified extension $L$ of a local field $K$ of residue characteristic $p$ is the Galois extension given by adjoining to $K^{ur}$ the roots $π^{1 ∕ m}$ of a uniformizer $π$ of $K$ for all $m \geq 1$ with $p ∤ m$ . The Galois group $Gal (L ∕ K^{ur})$ is isomorphic to the direct product

{\hat{ℤ}}^{(p)} = \prod_{ℓ \neq p} ℤ_{ℓ},

where $ℓ$ ranges over the prime numbers other than $p$ . The Galois group $Gal (L ∕ K)$ is the semi-direct product of $Gal (L ∕ K^{ur})$ with $Gal (K^{ur} ∕ K) ≅ \hat{ℤ}$ under the map $\hat{ℤ} \to Aut ({\hat{ℤ}}^{(p)})$ taking $a \in \hat{ℤ}$ to multiplication by $q^{a}$ in each coordinate of ${\hat{ℤ}}^{(p)}$ , where $q = | κ (K) |$ .

Remark 6.4.19. §

The Galois group of any Galois extension of the maximal tamely ramified extension of a local field is a pro- $p$ group, where $p$ is the residue characteristic. Any nontrivial such Galois extension is by necessity wildly ramified with no nontrivial tamely ramified subextension.

6.5. Ramification groups

Ramification groups provide a measure of the level of ramification in a Galois extension of a local field.

Notation 6.5.1. §

In this section, for a local field $L$ , we use $v_{L}$ to denote its valuation, $𝒪_{L}$ its valuation ring, and $π_{L}$ a uniformizer. If $𝔞$ is a fractional ideal of $𝒪_{L}$ , it is generated by a power of $π_{L}$ , and $v_{L} (𝔞)$ denotes that power.

Definition 6.5.2. §

Let $L ∕ K$ be a Galois extension of local fields with Galois group $G$ . For an integer $i \geq - 1$ , the $i$ th (higher) ramification group of $L ∕ K$ is the subgroup of $G$ given by

G_{i} = {τ \in G ∣ v_{L} (τ (α) - α) \geq i + 1 for all α \in 𝒪_{L}} .

Remark 6.5.3. §

We have by definition that $G_{- 1} = G$ and $G_{0}$ is the inertia group of $G$ . In particular, $G ∕ G_{0}$ is the Galois group of the maximal unramified subextension of $L ∕ K$ .

Remark 6.5.4. §

Let $L ∕ K$ be a Galois extension of local fields with Galois group $G$ . By Lemma 6.2.11, we have that $𝒪_{L} = 𝒪_{K} [β]$ for some $β \in 𝒪_{L}$ , so for any $i \geq - 1$ , we have

G_{i} = {τ \in G ∣ v_{L} (τ (β) - β) \geq i + 1} .

Lemma 6.5.5. §

Let $L ∕ K$ be a Galois extension of local fields with Galois group $G$ . The ramification groups $G_{i}$ are normal subgroups of $G$ .

Proof.

This is easy: let $σ \in G$ and $τ \in G_{i}$ . Then for any $α \in 𝒪_{L}$ , we have

τ (σ^{- 1} (α)) \equiv σ^{- 1} (α) mod π_{L}^{i + 1},

𝜎𝜏 σ^{- 1} (α) \equiv α mod π_{L}^{i + 1} .

□

Lemma 6.5.6. §

Let $L ∕ K$ be a Galois extension of local fields with Galois group $G$ . Let $σ \in G_{0}$ . Then $σ \in G_{i}$ for $i \geq 0$ if and only if $\frac{σ (π_{L})}{π_{L}} \in U_{i} (L)$ .

Proof.

First, we note that for $σ \in G_{i}$ , we have $σ (π_{L}) \equiv π_{L} mod π_{L}^{i + 1}$ , so we have $\frac{σ (π_{L})}{π_{L}} \in U_{i} (L)$ . Conversely, if $\frac{σ (π_{L})}{π_{L}} \in U_{i} (L)$ , then for any $a \in 𝒪_{L}$ , we may write

a = \sum_{k = 0}^{\infty} c_{k} π_{L}^{k}

for some $c_{k} \in μ_{q - 1} (L) \cup {0}$ , where $q = | κ (L) |$ . Since $σ \in G_{0}$ , it fixes elements $μ_{q - 1} (L)$ , so

σ (a) = \sum_{k = 0}^{\infty} c_{k} σ {(π_{L})}^{k} \equiv a mod π_{L}^{i + 1},

the last step as $σ (π_{L}) \equiv π_{L} mod π_{L}^{i + 1}$ . That is, $σ$ is an element of $G_{i}$ . □

Lemma 6.5.7. §

Let $L ∕ K$ be a Galois extension of local fields with Galois group $G$ . For each $i \geq 0$ , we have an injection

ρ_{i} : G_{i} ∕ G_{i + 1} \to U_{i} (L) ∕ U_{i + 1} (L)

given by

ρ_{i} (σ \cdot G_{i + 1}) = \frac{σ (π_{L})}{π_{L}} U_{i + 1} (L) .

Proof.

Lemma 6.5.6 tells us immediately that $ρ_{i}$ is well-defined and sends only the coset $G_{i + 1}$ to the coset $U_{i + 1} (L)$ . Therefore, it remains only to see that $ρ_{i}$ is a homomorphism. For this, let $σ, τ \in G_{i}$ , and note that

\frac{𝜎𝜏 (π_{L})}{π_{L}} = \frac{σ (π_{L})}{π_{L}} \cdot σ (\frac{τ (π_{L})}{π_{L}}) \equiv \frac{σ (π_{L})}{π_{L}} \cdot \frac{τ (π_{L})}{π_{L}} mod π_{L}^{i + 1},

the last step following from $σ \in G_{0}$ since $τ (π_{L}) π_{L}^{- 1} \in U_{i} (L)$ . □

Since the quotients $G_{i} ∕ G_{i + 1}$ are abelian for all $i \geq - 1$ , we have the following corollary.

Corollary 6.5.8. §

The Galois group of any Galois extension of local fields is solvable.

The following lemma is immediate from the definition of ramification groups.

Lemma 6.5.9. §

Let $L ∕ K$ be a Galois extension of local fields with Galois group $G$ , and let $H$ be a subgroup of $G$ . Then $H_{i} = H \cap G_{i}$ for all $i \geq - 1$ .

Lemma 6.5.10. §

Let $L ∕ K$ be a Galois extension of local fields with Galois group $G$ . Then $G ∕ G_{1}$ is the Galois group of the maximal tamely ramified subextension of $L ∕ K$ .

Proof.

If $L ∕ K$ is tamely ramified, then each quotient $G_{i} ∕ G_{i + 1}$ for $i \geq 1$ must be trivial. By Lemma 6.5.7, each quotient $G_{i} ∕ G_{i + 1}$ for $i \geq 1$ is a $p$ -group since since $U_{i} (L) ∕ U_{i + 1} (L)$ is one. Thus, $G_{1} = Gal (L ∕ L^{G_{1}})$ is a $p$ -group. As $G_{1}$ is contained in the inertia subgroup $G_{0}$ of $G$ , the field $L^{G_{1}}$ is totally ramified over $L^{G_{0}}$ and contains the maximal tamely ramified subextension of $L ∕ K$ . On the other hand, $G_{0} ∕ G_{1}$ injects into $U_{0} (L) ∕ U_{1} (L)$ , which has prime-to- $p$ order, so $L^{G_{1}} ∕ L^{G_{0}}$ is in fact tamely ramified. As $L^{G_{0}} ∕ K$ is unramified, $L^{G_{1}} ∕ K$ is then tamely ramified. □

Definition 6.5.11. §

Let $L ∕ K$ be a Galois extension of local fields with Galois group $G$ . The subgroup of wild inertia for $L ∕ K$ is $G_{1}$ .

Proposition 6.5.12. §

Let $p$ be a prime and $n \geq 1$ . Then

Gal {(ℚ_{p} (μ_{p^{n}}) ∕ ℚ_{p})}_{i} = {\begin{matrix} Gal (ℚ_{p} (μ_{p^{n}}) ∕ ℚ_{p}) & if - 1 \leq i \leq 0, \\ Gal (ℚ_{p} (μ_{p^{n}}) ∕ ℚ_{p} (μ_{p^{k}})) & if p^{k - 1} \leq i \leq p^{k} - 1 with 1 \leq k \leq n - 1 \\ 1 & if i \geq p^{n - 1} . \end{matrix}

Proof.

Let $F_{0} = ℚ_{p}$ and $F_{k} = ℚ_{p} (μ_{p^{k}})$ for $k \geq 1$ . Let $G = Gal (F_{n} ∕ ℚ_{p})$ . Fix a primitive $p^{k}$ th root of unity $ζ_{p^{k}}$ for each $k \geq 1$ such that $ζ_{p^{k + 1}}^{p} = ζ_{p^{k}}$ for each $k$ . Let $σ \in G$ be nontrivial, let $i \in ℤ$ be such that $σ (ζ_{p^{n}}) = ζ_{p^{n}}^{i}$ , and let $k \geq 0$ be maximal such that $i \equiv 1 mod p^{k}$ . Set $c = (i - 1) ∕ p^{k}$ . We then have

σ (ζ_{p^{n}}) - ζ_{p^{n}} = ζ_{p^{n}}^{i} - ζ_{p^{n}} = ζ_{p^{n}}^{1 + c p^{k}} - ζ_{p^{n}} = ζ_{p^{n}} (ζ_{p^{n - k}}^{c} - 1),

Since $p ∤ c$ , this has valuation $p^{k}$ in $F_{n}$ . As the valuation ring of $F_{n}$ is $ℤ_{p} [ζ_{p^{n}}]$ , we then have that $σ \in G_{p^{k} - 1} - G_{p^{k}}$ . On the other hand, the fact that $p^{k}$ exactly divides $i - 1$ (for $k < n$ ) says that $σ$ is an element of $Gal (F_{n} ∕ F_{k})$ but not $Gal (F_{n} ∕ F_{k + 1})$ . The result follows. □

Ramification groups have the following interesting property, which we state without proof.

Proposition 6.5.13. §

Let $L ∕ K$ be a Galois extension of local fields with Galois group $G$ , and let $i$ and $j$ be positive integers. For any $σ \in G_{i}$ and $τ \in G_{j}$ , the commutator $[σ, τ] = 𝜎𝜏 σ^{- 1} τ^{- 1}$ is an element of $G_{i + j + 1}$ .

Let us make the following useful definition.

Definition 6.5.14. §

Let $L ∕ K$ be a Galois extension of local fields with Galois group $G$ . Then we define a function $i_{L ∕ K} : G \to ℤ_{\geq 0} \cup {\infty}$ by

i_{L ∕ K} (σ) = \min {v_{L} (σ (α) - α) ∣ α \in 𝒪_{L}}

for $σ \in G$ .

Remark 6.5.15. §

In Definition 6.5.14, we have $σ \in G_{i}$ if and only if $i_{L ∕ K} (σ) \geq i + 1$ .

We now show how the ramification filtration determines the different of a Galois extension of local fields.

Proposition 6.5.16. §

Let $L ∕ K$ be a Galois extension of local fields with Galois group $G$ . Then we have

v_{L} (𝔇_{L ∕ K}) = \sum_{σ \in G - {1}} i_{L ∕ K} (σ) = \sum_{i = 0}^{\infty} (| G_{i} | - 1) .

Proof.

Let $β \in 𝒪_{L}$ be such that $𝒪_{L} = 𝒪_{K} [β]$ . We let $f \in 𝒪_{K} [x]$ be the minimal polynomial of $β$ . Then

f^{'} (β) = \prod_{σ \in G - {1}} (β - σ (β))

generates $𝔇_{L ∕ K}$ by Corollary 6.2.10, and so

v_{L} (f^{'} (β)) = \sum_{σ \in G - {1}} i_{L ∕ K} (σ) = \sum_{i = 0}^{\infty} \sum_{\begin{array}{c} σ \in G - {1} \\ i_{L ∕ K} (σ) = i \end{array}} i = \sum_{i = 0}^{\infty} i (| G_{i - 1} | - | G_{i} |) = \sum_{i = 0}^{\infty} (| G_{i} | - 1) .

□

Corollary 6.5.17. §

Let $L ∕ K$ be a Galois extension of local fields with Galois group $G$ . Let $H$ be a subgroup and $E = L^{H}$ its fixed field. Then

v_{E} (𝔇_{E ∕ K}) = \frac{1}{e_{L ∕ E}} \sum_{σ \in G - H} i_{L ∕ K} (σ) .

Proof.

Note first that

v_{E} (𝔇_{E ∕ K}) = \frac{1}{e_{L ∕ E}} v_{L} (𝔇_{E ∕ K}) .

By Lemma 6.2.5 and Proposition 6.5.16, we have

v_{L} (𝔇_{E ∕ K}) = v_{L} (𝔇_{L ∕ K}) - v_{L} (𝔇_{L ∕ E}) = \sum_{σ \in G - {1}} i_{L ∕ K} (σ) - \sum_{τ \in H - {1}} i_{L ∕ E} (τ) = \sum_{σ \in G - H} i_{L ∕ K} (σ)

noting for the last step that $i_{L ∕ E} (τ) = i_{L ∕ K} (τ)$ for $τ \in H$ by definition. □

Let us extend the definition of the lower ramification groups to all real numbers in the interval $[- 1, \infty)$ .

Definition 6.5.18. §

Let $L ∕ K$ be a Galois extension of local fields with Galois group $G$ . Let $t \in [- 1, \infty)$ . The $t$ th ramification group $G_{t}$ of $L ∕ K$ in the lower numbering is defined to be equal to the ramification group $G_{⌈ t ⌉}$ , where $⌈ \cdot ⌉$ is the ceiling function.

We now define a function from $[- 1, \infty)$ to $[- 1, \infty)$ as follows.

Definition 6.5.19. §

Let $L ∕ K$ be a Galois extension of local fields with Galois group $G$ . We define

ϕ_{L ∕ K} : [- 1, \infty) \to [- 1, \infty)

by $ϕ_{L ∕ K} (t) = t$ for $t < 0$ and

ϕ_{L ∕ K} (t) = \int_{0}^{t} {[G_{0} : G_{x}]}^{- 1} 𝑑𝑥

for $t \geq 0$ .

Remarks 6.5.20. §

a.: Definition 6.5.19 for an integer $k \geq - 1$ may be written as
$ϕ_{L ∕ K} (k) = \frac{1}{| G_{0} |} \sum_{i = 1}^{k} | G_{i} | .$
b.: The function $ϕ$ is continuous, piecewise linear, increasing, and concave down. Its slope between $k - 1$ and $k$ for an integer $k \geq 0$ is $| G_{k} | ∕ | G_{0} |$ , which is nonincreasing in $k$ .

Example 6.5.21. §

Let $F_{n} = ℚ_{p} (μ_{p^{n}})$ for a prime $p$ and $n \geq 1$ . By Proposition 6.5.12, we have

ϕ_{F_{n} ∕ ℚ_{p}} (t) = {\begin{matrix} t & if - 1 \leq i \leq 0, \\ \frac{t - p^{k - 1} + 1}{p^{k - 1} (p - 1)} + k - 1 & if p^{k - 1} - 1 \leq t \leq p^{k} - 1 with 1 \leq k \leq n - 1, \\ \frac{t - p^{n} + 1}{p^{n - 1} (p - 1)} + n - 1 & if t \geq p^{n - 1} - 1 . \end{matrix}

Note that $ϕ_{F_{n} ∕ ℚ_{p}} (p^{k} - 1) = k$ for each $0 \leq k \leq n$ .

We intend to show that $ϕ_{L ∕ K}$ behaves well under composition in towers of extension, and to investigate the behavior of ramification groups in quotients. For this, we first require several lemmas.

Lemma 6.5.22. §

Let $L ∕ K$ be a Galois extension of local fields with Galois group $G$ , and let $E ∕ K$ be a normal subextension. Set $N = Gal (L ∕ E)$ . For $δ \in G ∕ N$ , we have

i_{E ∕ K} (δ) = \frac{1}{e_{L ∕ E}} \sum_{\begin{array}{c} σ \in G \\ σ |_{E} = δ \end{array}} i_{L ∕ K} (σ) .

Proof.

Write $𝒪_{L} = 𝒪_{K} [β]$ for some $β \in 𝒪_{L}$ and $𝒪_{E} = 𝒪_{K} [α]$ for some $α \in 𝒪_{E}$ . Let $g \in 𝒪_{E} [x]$ be the minimal polynomial of $β$ over $E$ . We have

g = \prod_{τ \in N} (x - τ (β)),

so letting $g^{σ} \in 𝒪_{E} [x]$ denote the polynomial obtained by letting $σ \in G$ act on the coefficients of $g$ , we have

g^{σ} = \prod_{τ \in N} (x - 𝜎𝜏 (β)) .

Each coefficient of $g^{σ} - g$ is the difference of values of a symmetric polynomial in $| N |$ -variables on the elements $τ (β)$ and the elements $𝜎𝜏 (β)$ . Since each coefficient of $g$ lies in $𝒪_{E}$ , it is a polynomial in $α$ , and the coefficients of $g^{σ}$ are the same polynomials in $σ (α)$ , so $σ (α) - α$ divides the coefficient of $g^{σ} - g$ . In particular, $σ (α) - α$ divides $g^{σ} (β) - g (β) = g^{σ} (β)$ .

Now, write $α = f (β)$ for some $f \in 𝒪_{K} [x]$ . Note that $f (x) - α$ has $β$ as a root, so

f (x) - α = g (x) h (x)

for some $h \in 𝒪_{E} [x]$ . Then

f (x) - σ (α) = g^{σ} (x) h^{σ} (x) .

Plugging in $β$ , we obtain

α - σ (α) = g^{σ} (β) \cdot h^{σ} (β),

so $g^{σ} (β)$ divides $σ (α) - α$ . In particular, they have the same valuation.

Now, let $δ \in G ∕ N$ , which we may assume is not $1$ , since for $δ = 1$ the result is obvious, with both sides of the equation in the statement being infinite. As we have seen, $δ (α) - α$ and $g^{δ} (β)$ have the same valuation. Thus, we have

e_{L ∕ E} i_{E ∕ K} (δ) = v_{L} (δ (α) - α) = v_{L} (g^{δ} (α)) = v_{L} (\prod_{\begin{array}{c} σ \in G \\ σ |_{E} = δ \end{array}} (β - σ (β))) = \sum_{\begin{array}{c} σ \in G \\ σ |_{E} = δ \end{array}} i_{L ∕ K} (σ) .

□

Lemma 6.5.23. §

Let $L ∕ K$ be a Galois extension of local fields with Galois group $G$ . For $t \geq - 1$ , we have

ϕ_{L ∕ K} (t) + 1 = \frac{1}{| G_{0} |} \sum_{σ \in G} \min {i_{L ∕ K} (σ), t + 1} .

Proof.

Note that both sides of the equation in the statement are equal to $1$ for $t = 0$ . Also, both sides are piecewise linear and continuous, with slopes for any non-integral $t$ equal to

ϕ_{L ∕ K}^{'} (t) = {[G_{0} : G_{⌈ t ⌉}]}^{- 1} = \frac{| G_{⌈ t ⌉} |}{| G_{0} |}

and

\frac{1}{| G_{0} |} \sum_{\begin{array}{c} σ \in G \\ i_{L ∕ K} (σ) \geq ⌈ t ⌉ + 1 \end{array}} 1 = \frac{| G_{⌈ t ⌉} |}{| G_{0} |} .

Hence, we have the result. □

Lemma 6.5.24. §

Let $L ∕ K$ be a Galois extension of fields with Galois group $G$ , and let $E ∕ K$ be a Galois subextension. For every $t \geq - 1$ and $δ \in Gal (E ∕ K)$ , we have

i_{E ∕ K} (δ) - 1 = \max {ϕ_{L ∕ E} (i_{L ∕ K} (σ) - 1) ∣ σ \in G, σ |_{E} = δ} .

Proof.

Let $N = Gal (L ∕ E)$ . Let $δ \in G ∕ N$ . Let $σ \in G$ with $σ |_{E} = δ$ be such that $i = i_{L ∕ K} (σ) - 1$ is maximal. Let $β \in 𝒪_{L}$ be such that $𝒪_{L} = 𝒪_{K} [β]$ . For any $τ \in N$ , we have

i_{L ∕ K} (𝜎𝜏) = v_{L} (𝜎𝜏 (β) - β) \geq \min {v_{L} (𝜎𝜏 (β) - σ (β)), v_{L} (σ (β) - β)} = \min {i_{L ∕ K} (τ), i_{L ∕ K} (σ)}

with equality if $i_{L ∕ K} (τ) \neq i_{L ∕ K} (σ)$ . In particular, if $τ \notin N_{i}$ , then $i_{L ∕ K} (𝜎𝜏) = i_{L ∕ K} (τ)$ , and if $τ \in N_{i}$ , then $i_{L ∕ K} (𝜎𝜏) \geq i + 1$ and then in fact equality so by maximality of $i$ . We therefore have

i_{L ∕ K} (δ) = \frac{1}{| N_{0} |} \sum_{τ \in N} i_{L ∕ K} (𝜎𝜏) = \frac{1}{| N_{0} |} \sum_{τ \in N} \min {i_{L ∕ K} (τ), i + 1} = ϕ_{L ∕ E} (i) + 1,

the first step by Lemma 6.5.22 and the last step by Lemma 6.5.23. □

The latter lemma has the following corollary.

Theorem 6.5.25 (Herbrand’s theorem). §

Let $L ∕ K$ be a Galois extension of fields with Galois group $G$ , let $E ∕ K$ be a Galois subextension, and set $N = Gal (L ∕ E)$ . For any $t \geq - 1$ , one has

{(G ∕ N)}_{ϕ_{L ∕ E} (t)} = G_{t} N ∕ N .

Proof.

Note that $δ \in G ∕ N$ lies in ${(G ∕ N)}_{ϕ_{L ∕ E} (t)}$ if and only if $i_{E ∕ K} (δ) - 1 \geq ϕ_{L ∕ E} (t)$ . This occurs by Lemma 6.5.24 if and only if there exists $σ \in G$ that restricts to $δ$ such that

ϕ_{L ∕ E} (i_{L ∕ K} (σ) - 1) \geq ϕ_{L ∕ E} (t)

and so if and only if $i_{L ∕ K} (σ) - 1 \geq t$ for some such $σ$ . In turn, this is exactly to say that $σ \in G_{t}$ for some lift $σ \in G$ of $δ$ , or in other words that $δ \in G_{t} N ∕ N$ . □

Proposition 6.5.26. §

Let $L ∕ K$ be a Galois extension of local fields and $E$ a normal subextension of $K$ in $L$ . Then

ϕ_{L ∕ K} = ϕ_{E ∕ K} \circ ϕ_{L ∕ E} .

Proof.

Note first that both sides agree at $- 1$ and then that it suffices to consider the slope of both sides at non-integral values $t > - 1$ . Let $G = Gal (L ∕ K)$ and $N = Gal (L ∕ E)$ . The derivative of the right-hand side at $t$ is

ϕ_{E ∕ K}^{'} (ϕ_{L ∕ E} (t)) \cdot ϕ_{L ∕ E}^{'} (t) = {[{(G ∕ N)}_{0} : {(G ∕ N)}_{ϕ_{L ∕ E} (t)}]}^{- 1} {[N_{0} : N_{t}]}^{- 1}

and that of the left is ${[G_{0} : G_{⌈ t ⌉}]}^{- 1}$ , so it suffices to note by Lemma 6.5.9 and Theorem 6.5.25 that

[G_{s} : N_{s}] = [G_{s} : G_{s} \cap N] = [G_{s} N : N] = | {(G ∕ N)}_{ϕ_{L ∕ E} (s)} |

for $s = 0$ and $s = t$ . □