The abstract reciprocity map

Section 5.2 The abstract reciprocity map

We next define the reciprocity map in abstract class field theory (Definition 5.2.6). As a bonus, this definition will give an explicit recipe for computing the reciprocity map in local class field theory, but we will not expand on this point.

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Subsection Construction of the reciprocity map

In order to define a candidate

r : Gal (L / K) \to A_{K} / {Norm}_{L / K} A_{L}

for the reciprocity map, we must first give a partial definition using a different domain. See Remark 5.2.9 for motivation.

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Definition 5.2.1.

Under Hypothesis 5.1.12, let

L / K

be a Galois extension of finite subextensions of

\overset{―}{k} / k .

Let

H

be the semigroup of

g \in Gal (L^{unr} / K)

such that

d_{K} (g)

is a positive integer. Define the map

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r^{'} : H \to A_{K} / {Norm}_{L / K} A_{L}

as follows. For

g \in Gal (L^{unr} / K),

let

M

be the fixed field of

g,

so that

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e (M / K) = e ((M \cap L) / K), f (M / K) = d_{K} (g) .

We may set

r^{'} (g) := {Norm}_{M / K} (π_{M})

for some uniformizer

π_{M}

for

M,

once we check that this doesn’t depend on the choice of

π_{M} .

To wit, if

π_{M}^{'}

is another uniformizer for

M,

then

π_{M} / π_{M}^{'} \in U_{L}

belongs to

{Norm}_{L^{unr} / L} U_{L^{unr}}

by Corollary 5.1.15, so

{Norm}_{M / K} (π_{M} / π_{M}^{'})

belongs to

{Norm}_{L^{unr} / K} U_{L^{unr}} \subseteq {Norm}_{L / K} U_{L} .

So at least

r^{'}

is now a well-defined map, if not yet a semigroup homomorphism.

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Remark 5.2.2.

Before getting into the weeds, let’s make some other observations about Definition 5.2.1.

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First,

r^{'}

is invariant under conjugation: if we replace

g

h^{- 1} g h,

then its fixed field

M

is replaced by

M^{h}

and we can take the uniformizer

π_{M}^{h} .

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Next, if

g \in H

is actually in

Gal (L^{unr} / L),

then

r^{'} (g) \in {Norm}_{L / K} A_{L} .

In that case,

M

contains

L,

r^{'} (g) = {Norm}_{M / K} (π_{M})

can be rewritten as

{Norm}_{L / K} {Norm}_{M / L} (π_{M}) \in {Norm}_{L / K} A_{L} .

That is, if

r^{'}

were known to be multiplicative, it would induce a group homomorphism from

Gal (L / K)

A_{K} / {Norm}_{L / K} A_{L} .

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The remaining difficulty is to check that

r^{'}

is multiplicative. The issue here is that the definition of

r^{'} (g)

involves taking the norm from a field extension that depends on

g,

so it is hopeless to directly compare different values of

g .

Instead, we rewrite the definition in a more uniform manner.

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Proposition 5.2.3.

With notation as in Definition 5.2.1, put

n := d_{K} (g)

and choose

ϕ \in H

with

d_{K} (ϕ) = 1 .

Then for

x \in A_{M},

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{Norm}_{M / K} (x) = {Norm}_{L^{unr} / K^{unr}} (x x^{ϕ} \dots x^{ϕ^{n - 1}}) .

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Proof.

Put

U = M \cap K^{unr},

so that

{Norm}_{M / K} = {Norm}_{U / K} \circ {Norm}_{M / U} .

The group

Gal (U / K)

is of order

n

generated by

ϕ,

so for

y \in A_{U}

we have

{Norm}_{U / K} (y) = y y^{ϕ} \dots y^{ϕ^{n - 1}} .

Meanwhile, on

A_{M}

we can view

{Norm}_{M / U}

as the restriction of

{Norm}_{L^{unr} / K^{unr}},

{Norm}_{M / U} (x) = {Norm}_{L^{unr} / K^{unr}} (x) .

By taking

y = {Norm}_{M / U} (x)

and rewriting

y^{ϕ^{i}}

{Norm}_{L^{unr} / K^{unr}} (x)^{ϕ^{i}} = {Norm}_{L^{unr} / K^{unr}} (x^{ϕ^{i}}),

we deduce the claim.

Using the formula from Proposition 5.2.3, we can now establish multiplicativity, following [38], Proposition IV.5.5.

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Lemma 5.2.4.

With notation as in Definition 5.2.1, put

G := Gal (L^{unr} / K^{unr})

and choose

ϕ \in H

with

d_{K} (ϕ) = 1 .

x \in H_{0} (G, U_{M})

is fixed by

ϕ,

then

{Norm}_{G} (x)

vanishes in

H^{0} (G, U_{M}) .

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Proof.

By hypothesis,

x

is the class of some

u \in U_{M}

for which there exist

u_{1}, \dots, u_{r} \in U_{M}

and

τ_{1}, \dots, τ_{r} \in G

such that

u^{ϕ = 1} = \prod_{i = 1}^{r} u_{i}^{τ_{i} - 1} .

Put

n := [M : K]

and

σ := ϕ^{n},

and let

Σ

be the fixed field of

σ,

which contains

M .

Let

Σ_{n}

be the unramified extension of

Σ

of degree

n;

it is the fixed field of

σ^{n} .

Since

H_{T}^{0} (Gal (Σ_{n} / Σ), U_{Σ_{n}})

vanishes by Proposition 5.1.14, we can find

\tilde{u}, {\tilde{u}}_{i} \in U_{Σ_{n}}

such that

\begin{aligned} u & = {Norm}_{Σ_{n} / Σ} (\tilde{u}) = \tilde{u} {\tilde{u}}^{σ} \dots {\tilde{u}}^{σ^{n - 1}} \\ u_{i} & = {Norm}_{Σ_{n} / Σ} ({\tilde{u}}_{i}) = {\tilde{u}}_{i} {\tilde{u}}_{i}^{σ} \dots {\tilde{u}}_{i}^{σ^{n - 1}} . \end{aligned}

Since

H_{T}^{- 1} (Gal (Σ_{n} / Σ), U_{Σ_{n}})

vanishes by Proposition 5.1.14 again, we can find

\tilde{y} \in U_{Σ_{n}}

with

{\tilde{u}}^{ϕ - 1} / \prod_{i = 1}^{r} {\tilde{u}}_{i}^{τ_{i} - 1} = {\tilde{y}}^{σ - 1},

and so (because

σ = ϕ^{n}

)

{\tilde{u}}^{ϕ - 1} = (\tilde{y} {\tilde{y}}^{ϕ} \dots {\tilde{y}}^{ϕ^{n - 1}})^{ϕ - 1} \prod_{i = 1}^{r} {\tilde{u}}_{i}^{τ_{i} - 1} .

Applying

{Norm}_{G}

yields

{Norm}_{G} (\tilde{u})^{ϕ - 1} = {Norm}_{G} (\tilde{y} {\tilde{y}}^{ϕ} \dots {\tilde{y}}^{ϕ^{n - 1}})^{ϕ - 1} .

That is, if we set

z := {Norm}_{G} (\tilde{u}) / {Norm}_{G} (\tilde{y} {\tilde{y}}^{ϕ} \dots {\tilde{y}}^{ϕ^{n - 1}}),

we have

z^{ϕ - 1} = 1

and so

z \in U_{K} .

Put

y := \tilde{y} {\tilde{y}}^{σ} \dots {\tilde{y}}^{σ^{n} - 1} = {Norm}_{Σ_{n} / Σ} (\tilde{y}) \in U_{Σ};

using Proposition 5.2.3 we obtain

\begin{aligned} {Norm}_{G} (u) & = {Norm}_{G} (\tilde{u} {\tilde{u}}^{σ} \dots {\tilde{u}}^{σ^{n - 1}}) \\ = {Norm}_{G} (y y^{ϕ} \dots y^{ϕ^{n - 1}}) z^{n} \\ = {Norm}_{Σ / K} (y) {Norm}_{M / K} (u) \in {Norm}_{M / K} U_{M}, \end{aligned}

proving the claim. (Compare [38], Lemma IV.5.4.)

Lemma 5.2.5.

The map

r^{'} : H \to A_{K} / {Norm}_{L / K} A_{L}

exhibited in Definition 5.2.1 is a homomorphism of semigroups.

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Proof.

Let

g_{1}, g_{2} \in H

be arbitrary and put

g_{3} := g_{1} g_{2} .

Let

M_{i}

be the fixed field of

g_{i},

let

π_{i} \in A_{M_{i}}

be a uniformizer of

M_{i},

and put

ρ_{i} = r^{'} (g_{i}) = {Norm}_{M_{i} / K} (π_{i}) \in A_{K} .

Put

ρ := ρ_{1} ρ_{2} / ρ_{3};

note that

v_{K} (ρ_{i}) = f (M_{i} / K) v_{M_{i}} (π_{i}) = f (M_{i} / K) = d_{K} (g_{i}),

which implies that

v_{K} (ρ) = 0

and hence

ρ \in U_{K} .

Our goal is to check that

ρ \in {Norm}_{L / K} A_{L};

our plan is to rephrase this as a relation among units, to which Lemma 5.2.4 will apply.

We first make an adjustment at the level of group elements. Put

G := Gal (L^{unr} / K^{unr}) .

Choose

ϕ \in H

such that

d_{K} (ϕ) = 1 .

Put

d_{i} := d_{K} (g_{i})

and

τ_{i} := g_{i}^{- 1} ϕ^{d_{i}} \in G;

then

τ_{3} = g_{2}^{- 1} g_{1}^{- 1} ϕ^{d_{1} + d_{2}} = g_{2}^{- 1} ϕ^{d_{2}} (ϕ^{- d_{2}} g_{1} ϕ^{d_{2}})^{- 1} ϕ^{d_{1}} .

It will be convenient to replace

g_{1}

and

τ_{1}

with

g_{1}^{'} := ϕ^{- d_{2}} g_{1} ϕ^{d_{2}}, τ_{1}^{'} = (g_{1}^{'})^{- 1} ϕ^{g_{1}},

so that

τ_{1}^{'} τ_{2} = τ_{3} .

We correspondingly define

M_{1}^{'}

to be the fixed field of

g_{1}^{'}

and set

π_{1}^{'} := π_{1}^{ϕ^{n_{2}}} \in A_{M_{1}^{'}},

noting that

{Norm}_{M_{1}^{'} / K} (π_{1}^{'}) = {Norm}_{M_{1} / K} (π_{1}) = ρ_{1} .

Let

N

be a finite subextension of

L^{unr} / L

containing

M_{1}, M_{2}, M_{3}, M_{1}^{'} .

Set

σ_{i} := π_{i} π_{i}^{ϕ} \dots π_{i}^{ϕ^{d_{i} - 1}}, σ_{1}^{'} := (π_{1}^{'}) (π_{1}^{'})^{ϕ} \dots (π_{1}^{'})^{ϕ^{d_{1} - 1}}

and

u = σ_{1}^{'} σ_{2} / σ_{3} \in U_{N};

by Proposition 5.2.3 we have

ρ = {Norm}_{G} (u) .

By defining

u_{1} := (π_{1}^{'})^{1 - τ_{2}}, u_{2} := π_{2} / π_{1}^{'}, u_{3} := π_{3} / π_{1}^{'} \in U_{N}

and using the equality

τ_{1}^{'} τ_{2} = τ_{3},

we compute that

u^{ϕ - 1} = (π_{1}^{'})^{τ_{1}^{'} - 1} π_{2}^{τ_{2} - 1} / π_{3}^{τ_{3} - 1} = u_{1}^{τ_{1}^{'} - 1} u_{2}^{τ_{2} - 1} / u_{3}^{τ_{3} - 1}

vanishes in

H_{0} (G, U_{N});

by Lemma 5.2.4 we obtain

{Norm}_{G} (u) \in {Norm}_{N / K} U_{N},

proving the claim.

Definition 5.2.6.

With notation as in Definition 5.2.1, by combining Definition 5.2.1, Remark 5.2.2, and Lemma 5.2.5, we obtain a semigroup homomorphism

r^{'} : H \to A_{K} / {Norm}_{L / K} A_{L}

which kills

Gal (L^{unr} / L)

and thus induces a homomorphism

r_{L / K} : Gal (L / K) \to A_{K} / {Norm}_{L / K} A_{L} .

We call

r_{L / K}

the reciprocity map.

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Here are some consequences of the construction whose proofs are left to the reader.

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Proposition 5.2.7.

Under Hypothesis 5.1.12, if

L / K

and

L^{'} / K^{'}

are Galois extensions of finite subextensions of

\overset{―}{k} / k

such that

K \subseteq K^{'}

and

L \subseteq L^{'},

then the diagram in Figure 5.2.8 commutes.

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Proof.

See Exercise 1.

Remark 5.2.9.

Note that Proposition 5.2.7 dictates the form of Definition 5.2.1: we want to be able to compute the map

r : Gal (L / K) \to A_{K} / {Norm}_{L / K} A_{L}

by first computing the map

r : Gal (M L / M) \to A_{M} / {Norm}_{M L / M} A_{M L},

which should take Frobenius to a uniformizer (as

M L / M

is an unramified extension), and then applying

{Norm}_{M / L} .

We will use this picture again in Lemma 5.3.2.

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Proposition 5.2.10.

Under Hypothesis 5.1.12, if

L / K

is a Galois extension of finite subextensions of

\overset{―}{k} / k

and

K^{'}

is an intermediate field, then the diagram in Figure 5.2.11 commutes. Here

V : Gal (L / K)^{ab} \to Gal (L / K^{'})^{ab}

denotes the transfer map for the inclusion

Gal (L / K^{'}) \subseteq Gal (L / K)

(Theorem 2.3.8).

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Proof.

See Exercise 2.

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Exercises Exercises

1.

Prove Proposition 5.2.7.

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Hint.

See [38], Proposition IV.5.8.

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2.

Prove Proposition 5.2.10.

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Hint.

See [38], Proposition IV.5.9.

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3.

Under Hypothesis 5.1.12, choose

c \in {\hat{Z}}^{*}

and let

s_{L / K}

be the reciprocity map defined using

c d

instead of

d .

Show that

s_{L / K} = c^{- 1} r_{L / K} .

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4.

Under Hypothesis 5.1.12, choose

c \in {\hat{Z}}^{*}

which acts on

im (v),

and let

s_{L / K}

be the reciprocity map defined using

c d, c v

instead of

d, v .

Show that

s_{L / K} = r_{L / K} .

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