NUMBER THEORETIC BACKGROUND

5

be the map given by cup product with aE/F. Since CE, -

C%al(E/E/)

is a bijection for

F c E' c E, the property (W3) above implies, via an abstract cohomological

theorem (combine the corollary of p. 184 of [AT] with Theorem 12, p. 154, of [SI]),

that ctn(E/F) is an isomorphism for every n. Moreover, the canonical classes are

interrelated by

(1.2.4) infl (xE'/F = [E:E']aE/F and rtsaE/F = ocE/E

(for the first, use Theorem 6 on p. 188 of [AT]; the second is obvious). Thus, implicit

in the existence of Weil groups is all the cohomology of class field theory.

For example, taking n = - 1 in (1.2.3) we find

Hl(Gsd(E/F),

CE) = 0. Taking

n = 0, we find

H2(Ga\(E/F),

CE) is cyclic of order [E:F], generated by aE/F. Taking

n = - 2 we find an isomorphism

GEh/F

**

CF/NE/FCE

which, by (W^, is that given

by the reciprocity law. For E/F cyclic, this isomorphism determines aE/F,

a n

d it

follows that aE/F i s t n e "canonical" or "fundamental" class of class field theory.

The same is true for arbitrary E/F as one sees by taking a cyclic EJF of the same de-

gree as E/F, and inflating aE/F a n d ccEl/F to EEX/F, where they are equal by (1.2.4).

Conversely, if we are given classes aE/E' satisfying (1.2.4) and such that the maps

(1.2.3) are isomorphisms, then we can construct a Weil group WF as the projective

limit of group extensions WE/F made with these classes. This construction is ab-

stracted and carried out in great detail in Chapter XIV of [AT]. The existence of

such classes aE/E is proved in [AT] and [CF].

Thus, a Weil group exists for every F; to what extent is it unique?

(1.3) Unicity. A Weil group for F/F is unique up to isomorphism. More precisely:

(1.3.1)

PROPOSITION.

Let WF and WF be two Weil groups for F/F. There exists an

isomorphism 0: WF ^* WF such that the diagrams

WF Wf

wF,

(wEyh

are commutative.

For each finite Galois E/F, let 1(E) denote the set of isomorphisms/such that the

following diagram is commutative

0 CE • WE/F Gol(E/F) • 0

id / id

0 CE WE/F Gal(E/F) 0

Since the two group extensions WE/F&nd WE/F each have the same class, namely