Lemma 2.16. [Z1] (5, Lemma 10)
Any two pro nite groups Cj, j = 1, 2, which satisfy the following conditions are iso-
morphic:
(1) Cj contains a subgroup Dj of index 2 isomorphic to Gal(K(2)/k(i)),
(2) the 2-Sylow subgroup ˜Cj of Cj is isomorphic to Gal(K(2)/K),
(3) the symplectic Z /q[F ]-spaces ˜Dj/ ˜D (1,q)
j and K(i)
∗/(K(i)∗)q are isomorphic.
Remark 2.17. If a pro nite group C satis es condition (1) and (2) it is obvious that the 2-Sylow subgroup ˜D of D is isomorphic to Gal(K(2)/K(i)) and we identify the cyclic group C/ ˜D of order 2f with F . We can now consider the abelian group ˜D/ ˜D(1,q) as an F -module. The unique relation connecting the generators of ˜D/ ˜D(2,q) induces a
symplectic Z /q[F ]-structure on ˜D/ ˜D(1,q) (this works analogue to the symplectic space structure on Ak, which we explained in the section before).
Let again Gk be the pro nite group with m + 2 generators x1, . . . , xm+2, subject to a
single relation which depends on k, and k ∈ L or Nα. We now use the above lemma to
prove the following theorem:
Theorem 2.18. The groups Gk and Gal(K(2)/k) are isomorphic.
Proof. The two Theorems 2.12, 2.13 and the Lemma 2.15, prove that the group Gk
satis es the conditions (1) to (3) in Lemma 2.16. If k ∈ L, this is due to Zelvenskii and we proved the case k ∈ Nα. Furthermore, the Hasse norm residue symbol (class eld theory)
gives an F -isomorphism from K(i)∗/(K(i)∗)q to Gal(K(2)/K(i))/ Gal(K(2)/K(i))(1,q)
and thus also the group Gal(K(2)/k) satis es the three conditions.
The next step is to choose a co nal subset W in the set of all odd natural numbers consisting of numbers congruent to 1 mod 2κ, where κ = 2 if k ∈ L and κ = 2(α + 1)
if k ∈ Nα. Let S = {K ⊇ k | K/k unrami ed, [K : k] ∈ W }. Since the maximal
extension of k without simple rami cation is equal to ∪K∈SK(2), we get a description
of its Galois group as an inverse limit of the groups de ned in terms of generators and relations at the beginning of section 2.3.
Theorem 2.19. [Z1] (Theorem 4)
Let k be an extension of the eld of 2-adic numbers having odd degree m. The Galois group of the maximal extension without simple rami cation of the eld k is isomorphic to the pro nite group with m + 2 generators x1, . . . , xm+2 subject to the relation
x21x42[x1, x2] . . . [xm+1, xm+2] = 1
and the relation x∆(2) = 1 on the normal subgroup generated by the elements
x1x22, x3, . . . , xm+2.
Theorem 2.20. Let k be an extension of the eld of 2-adic numbers having even degree m, such that the maximal unrami ed extension of k does not contain a primitive 4-th root of unity. Further, let k0 denote the intersection of k with the extension Q
2∞ of the
eld Q2, and let q ≥ 4 denote the largest power of 2 such that the q-th roots of unity
belong to k(i).
If k0 is not contained in the real sub eld of Q
2∞, then q ≥ 8, and the Galois group of
the maximal extension without simple rami cation of the eld k is isomorphic to the pro nite group with m + 2 generators x1, . . . , xm+2 subject to the relation
x2+q/21 [x1, x2] . . . [xm+1, xm+2] = 1
and the relation x∆(2) = 1 on the normal subgroup generated by the elements
x1, x2, x4, x5, . . . , xm+2.
Remark 2.21. Zelvenskii also claims that one has the following description for the Galois group of the maximal extension without simple rami cation of the eld k if k ∈ Mα:
extension without simple rami cation of the eld k is isomorphic to the pro nite group with m + 2 generators x1, . . . , xm+2 subject to the relation
x21[x1, x2]x3q[x3, x4] . . . [xm+1, xm+2] = 1
and the relation x∆(2) = 1 on the normal subgroup generated by the elements
x1x q/2
3 , x2, x4, . . . , xm+2.
Zelvenskii wrote that one could prove the claim above in an analogous manner as for k ∈ L. But all our attempts to calculate the de ning relations of Hk and ˜Gk in the
case k ∈ Mα needed in the proofs of theorem (2.12) and (2.13) failed. In principle this
problem is tractable with the methods of computer algebra. But even the simplest case is too complex to be solved by a computer program such as GAP. Indeed, nding such a presentation for Hk and ˜Gk is essentially an isomorphism test. According to computer
algebra developer Derek Holt, there is no known algorithm for isomorphism testing of nite groups that has complexity better than O(|G|d(G)), where d(G) is the number of
3 Conjecture on the absolute Galois group of a local
eld with residue characteristic 2
Let k be a 2-adic number eld over Q2 having even degree n, such that the maximal
tamely rami ed extension of k does not contain a primitive 4-th root of unity. As mentioned in the introduction, we can not use the same method for describing the absolute Galois group of k as in the p 6= 2 case and it turns out to be very di cult to nd a dierent approach. But after working several years on this topic, we have an idea what the absolute Galois group of a eld k ∈ Nα might look like.
So we also require that k ∈ Nα. Then q = 2α+1 denotes the largest power of 2 such that
the q-th roots of unity belong to the eld k(i).
Furthermore, let T be the maximal tamely rami ed extension of k and 2s, s ∈ N, the
number of elements in the residue eld of k. Hasse and Iwasawa have shown that the Galois group G = Gal(T/k) has two generators σ and τ with the de ning relation
στ σ−1 = τ2s.
The group µT of roots of unity of 2-power order in T is isomorphic to Z /2 Z and the
operation of G on µT is trivial.
Conjecture 3.1. The absolute Galois group Gal(ksep/k) is isomorphic to the pro nite
group X with n + 3 generators σ, τ, x0, . . . , xn and the following de ning conditions:
(i) the normal subgroup generated by x0, . . . , xn is a pro-2-group,
(ii) the elements σ and τ ful ll the relation στσ−1 = τ2s
, (iii) the generators satisfy the relation
x2+
q 2
0 [x0, x1]x−σ2 (x2, τ )[x3, x4] . . . [xn−1, xn] = 1,
The pro nite group X is constructed as in (1.3). Thus X = X(G , n) = F (n + 1, G )/(r),
where (r) is the (closed) normal subgroup generated by the following element r: r = x2+
q 2
0 [x0, x1]x−σ2 (x2, τ )[x3, x4] . . . [xn−1, xn].
P is the normal subgroup generated by x0, . . . , xn, so
r ≡ (x2, τ ) ≡ (τ )π ≡ 1 mod P,
since the order of τ is prime to 2.
To describe the absolute Galois group of k it is enough to describe the Galois groups Gal(L(2)/k)of all tamely rami ed extensions L/k. Let L ⊂ T be a nite tamely rami ed Galois extension of k with degree ef0. Then e is odd (e divides 2s− 1) and the Galois
group Gal(L/k) is generated by σ and τ with the relations (see [I]) σf0 = 1 , τe= 1 , στ σ−1 = τ2s,
where f0 is the inertia degree and e the rami cation index of L/k. Then Gal(ksep/T ) is
a pro-2 group and ksep is equal to S
k⊂L⊂T L(2).
For each L we consider the diagram
Loo 2b K l f OO f0=2bf `` k ef0 OO e 77 Q2 n OO
where the integers denote the degree of the extensions and l is the xed eld of the subgroup of Gal(L/k) generated by σ. Therefore Gal(L/l) is cyclic of order f0 = 2bf
with b ≥ 0 and f odd. Additionally K/l is an unrami ed extension and since e is odd, l(i)/l is rami ed and q = 2α+1 is still the largest power of 2 such that the q-th roots of unity belong to l(i). The degree of l over Q2 is even and Gal(Q2∞/k ∩ Q2∞) =
Gal(Q2∞/l ∩ Q2∞)(again since f is odd), thus l ∈ Nα.
The next step is to de ne a special subgroup Xe of X. Let H be the normal subgroup of
G generated by σ and τeand let X
e = φ−1(H)denote the preimage of H in X (φ: X → G
is the surjection de ned by φ(σ) = σ, φ(τ) = τ, φ(xi) = 1 (see 1.3)).
The idea to prove the conjecture above is to start by showing that Xe/(τe) is iso-
morphic to the maximal extension without simple rami cation of l. To obtain this isomorphism we can follow Zelvenskiis argument, which we explained in the previous section. So if Xe,f denotes the factor group Xe/(τe) of Xe with one further relation,
namely the relation x∆(2) = 1 on the normal subgroup generated by σf, it is enough
to show that Xe,f is isomorphic to Gal(K(2)/l) = Gal(L(2)/l). For this it su ces that
both groups Xe,f and Gal(K(2)/l) ful ll the three conditions of the following lemma:
Lemma 3.2. [Z1] (5, Lemma 10)
Any two pro nite groups Cj, j = 1, 2, which satisfy the following conditions are iso-
morphic:
(1) Cj contains a subgroup Dj of index 2 isomorphic to Gal(K(2)/l(i)),
(2) the 2-Sylow subgroup ˜Cj of Cj is isomorphic to Gal(K(2)/K),
(3) the symplectic Z /q[F ]-spaces ˜Dj/ ˜Dj(1,q) and K(i)∗/(K(i)∗)q are isomorphic.
We already know that Gal(K(2)/l) satis es all three conditions. We are able to show the second requirement for Xe,f:
Proof. As in Theorem 2.13 we show that both groups Xge,f and Gal(K(2)/K) are De-
mu²kin pro-2 groups with the same two invariants n(Xge,f) = n(Gal(K(2)/K)) = ef n + 2 and Im(χ) = h−1 + 2αi. For Gal(K(2)/K) this is clear, hence we will prove the claim
forXge,f: In the setting of (1.3), let H be the normal subgroup of G generated by τe and σf, G = G /H and U the preimage of H in F (n + 1, G ). Denote by X
H the maximal
pro-2 factor group of XH. Then Xge,f = XH and if fσf denotes the image of σf under
the projection from U to Xge,f , Xge,f is generated as normal subgroup by the elements
x0, . . . , xn, fσf (the order of τ is prime to 2).
Let λ = 1 e
e−1
P
i=0
τi, then we obtain the following equivalences (?) modulo [P, U]:
(x2, τ ) ≡ (x2τ )π ≡ (x2τ )e π e ≡ (x 2τ x2τ−1τ2x2. . . x2τe−1x2τ−(e−1)τe) π e ≡ (x2xτ −1 2 x τ−2 2 . . . x τ−(e−1) 2 τ e )πe ≡ (xeλ 2 τ e )πe ≡ (xλ2)π ≡ xλ2 mod [P, U ].
The last equivalences hold since yπ = y for any y ∈ P and τe ∈ U. As a result, we have
r ≡ x2+20 αx−σ2 (x2, τ ) mod [P, U ]hhhhhhhhhhhhhh(??)
≡ (x1+20 α)2x−σ2 xλ2 mod [P, U ]. The statements in [JW] 2.4-2.7 show that
dim H2( gXe,f, F2) = 1,
dim H1( gXe,f, F2) = (G : H)m + 2 = efn + 2,
Tor(Xge,f ab
) ∼= Z /2 Z as G-module. It is left to show that the cup product H1( gX
e,f, F2) × H1( gXe,f, F2) → H2( gXe,f, F2) is a
nondegenerate bilinear form:
As [P/N, XH] ⊂ XH(1,2), it follows from (??) that
xσ2 ≡ xλ
2 mod ˜X
(1,2)
Let I be the closed two-sided ideal in Z2JG K generated by 2 and τ
e− 1, then
σλ ≡ λσ mod I and τ λ ≡ λ mod I.
Therefore we get from (3.0.1) xτ2 ≡ xλ 2 ≡ x σ 2 ≡ x2 mod ˜X (1,2) e,f (3.0.2)
and from that
[xaρ2 , xbρ2 0] ≡ mod ˜Xe,f(2,2), (3.0.3) for all a, b ∈ Z2 and ρ, ρ0 ∈ G . Consequently, the equivalences (?) also hold modulo
˜
Xe,f(2,2) since τee = 1, i.e.
^ (x2, τ ) ≡ xλ2 mod ˜X (2,2) e,f . Let κ =f −1P i=0
σi. Then in the group ring Z 2JG K κλ(λ − σ) ≡ κ(1 − σ)λ
≡ (1 − σf)λ mod I,
and for a v ∈ Z2JG K it follows from (3.0.2) that x2v2 ≡ x2a
2 mod ˜X
(2,2) e,f
with a ∈ Z2.
As a result from the above, we get the equivalence (x−σ2 (xg2, τ ))eλκ ≡ x κλ(λ−σ)e 2 ≡ (xλ2)(1−σf)ex2a2 ≡ [x2, fσf]ex2a2 mod ˜X (2,2) e,f .
Finally, by applying eλκ to the relation r we obtain
The efn + 2 elements ˜σf, x
2, xρi, i = 0, 1, 3, . . . , n, where ρ ∈ G runs through a set
of representatives of G /H, form a minimal system of generators of ˜Xe,f. Analogously
as in the proof of Theorem 2.13 we now can show that the cup product for ˜Xe,f is
nondegenerate and calculate the invariant Im(χ). For the calculation of the latter, we write reλκ= r0g with g ∈ ˜X(2,2)
e,f and de ne χ: ˜Xe,f → U2 by
χ(xρ1) = (−1 − 2α)−1, χ( ˜σf) = χ(x
2) = χ(x ρ
i) = 1 for ρ ∈ G, i = 0, 3, . . . , n,
which allows the lifting argument from the proof of Theorem 2.13. Then for any crossed homomorphism D
D(r) = D(r0) + χ(r0)D(g) = D(r0) =X
ρ∈G
(2α+ χ(xρ1)−1+ 1)D(xρ0) = 0.) Consequently, ˜Xe is a Demu²kin group with the required invariants.
The idea how to prove the rst condition, i.e. that the normal subgroup H of Xe,f
generated by the elements x0, x21, x2, . . . , xn, σis isomorphic to the group Gal((K(2)/l(i)),
would be to show that the normal subgroup of X generated by x0, x21, x2, . . . , xn, σ, τ is a
so called Demu²kin formation over the maximal tamely rami ed extension of k(i) de ned by Diekert in [D] with methods similar to the ones he used. However, the computations are complicated and we have not succeeded so far.
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