Cohomology groups of radical extensions

COHOMOLOGY GROUPS OF RADICAL EXTENSIONS

Eunmi Choi

Reprinted from the

Journal of the Korean Mathematical Society Vol. 44, No. 1, January 2007

c

°2007 The Korean Mathematical Society

J. Korean Math. Soc. 44 (2007), No. 1, pp. 151–167

COHOMOLOGY GROUPS OF RADICAL EXTENSIONS

Eunmi Choi

Abstract. If k is a subfield of Q(ε

) then the cohomology group H

(k (ε

)/k) is isomorphic to H

(k(ε

)/k) with gcd(m, n

) = 1. This enables us to reduce a cyclotomic k-algebra over k(ε

) to the one over k(ε

).

A radical extension in projective Schur algebra theory is regarded as an analog of cyclotomic extension in Schur algebra theory. We will study a reduction of cohomology group of radical extension and show that a Galois cohomology group of a radical extension is isomorphic to that of a certain subextension of radical extension. We then draw a cohomological characterization of radical group.

1. Introduction

Let k be a field, k ^∗ be the multiplicative subgroup of k and µ(k) be the group of roots of unity in k. For a Galois extension L of k with Galois group G = G(L/k) and for a 2-cocycle α ∈ Z ² (G, L ^∗ ) = Z ² (L/k, L ^∗ ), a crossed product algebra (L/k, α) = P

σ∈G Lu σ with u σ u τ = α(σ, τ )u στ and u σ x = σ(x)u σ

(x ∈ L, σ, τ ∈ G) is called a cyclotomic algebra if L is a cyclotomic extension of k and α has values in µ(L) (i.e., α ∈ Z ² (L/k, µ(L))). Let H _ι ² (L/k) be the image of a canonical homomorphism ι of H ² (L/k, µ(L)) into H ² (L/k, L ^∗ ) induced by the inclusion µ(L) ,→ L ^∗ . Since µ(L) is a subgroup of the torsion group of L ^∗ , ι is injective ([7, p.91]), so we may identify H ² (L/k, µ(L)) = H _ι ² (L/k) ≤ H ² (L/k, L ^∗ ).

Suppose k is a subfield of the cyclotomic extension Q(ε m ) (Q: the rational number field, ε m : a primitive m-th root of unity). Let L = k(ε n ). Due to [6], m and n are assumed to be either odd or divisible by 4. Then the Galois cohomology group H _ι ² (L/k) is isomorphic to H _ι ² (K/k) where K = k(ε n

) is a subextension of L such that n ⁰ is a certain divisor of n which is prime to m [13, (7.12)]. Employing this result, Janusz’s reduction theorem on cyclotomic algebras in [6] ([13, (7.9)]) follows that, a cyclotomic algebra (L/k, α) with α ∈ Z ² (L/k, µ(L)) can be reduced to the case gcd(m, n) = 1, i.e., to (K/k, β) where β is a 2-cocycle in Z ² (K/k, µ(K)) defined over the smaller group G(K/k).

Received August 9, 2005.

2000 Mathematics Subject Classification. Primary 16K50, 20G10.

Key words and phrases. cohomology group, radical extension.

This study was supported by Hannam University Research Fund 2005.

c

°2007 The Korean Mathematical Society

151

It is well known that every Schur k-algebra (a central simple k-algebra which is a homomorphic image of a group algebra kG for a finite group G) is similar to a cyclotomic k-algebra [13, (3.10)]. The idea of Schur algebra has been generalized to a projective Schur algebra in [9] by replacing group algebra by twisted group algebra; a projective Schur k-algebra is a central simple algebra that is a homomorphic image of a twisted group algebra k ^α G for a finite group G and α ∈ Z ² (G, k ^∗ ).

The analogue of cyclotomic algebra in the theory of projective Schur algebra is the radical algebra ([1]). A radical k-algebra is a crossed product algebra (L/k, α) where L = k(Ω) is a finite Galois radical extension of k, Ω is a subgroup of L ^∗ which is finite modulo k ^∗ (i.e., Ω/k ^∗ finite), and α ∈ Z ² (L/k, L ^∗ ) is represented by a 2-cocycle with values in Ω.

In this paper we study radical extensions and radical algebras, and obtain a corresponding result to Janusz’s reduction theorem on radical extensions. We derive a reduction of Galois cohomology groups over radical extension fields, indeed prove that for a radical extension L of k, there exists a Galois radi- cal extension K of k in L such that the cohomology group of G(L/k) is iso- morphic to that of G(K/k) (in Theorem 10). We then verify a cohomologi- cal characterization of radical groups that a homomorphism of radical groups R(K/k) → R(L/k) commutes with certain homomorphisms of cohomology groups (in Theorem 14).

All notations are standard. H ² (L/k, M ) is the 2-dimensional cohomology group H ² (G, M ) where G = G(L/k) and M is a G-module, while Z ² (L/k, M ) is the 2-cocycle group. If M = L ^∗ , we write H ² (L/k, L ^∗ ) = H ² (L/k). Let ε d

(d > 0) denote a primitive d-th root of unity, a|b denote the division of b by a, while a ^t ||b denote the highest power t of a to be a ^t |b.

2. Preliminaries

Lemma 1. ([13, 7.10]) Let H be a cyclic normal subgroup of G and M be a fi- nite G-module. Let N H = Q

h∈H h. If N H (M ) = M ^H then inf : H ² (G/H, M ^H )

→ H ² (G, M ) is an isomorphism, where inf is the inflation map from G/H to G and M ^H is the subset of M consisting of elements fixed by H.

For finite Galois extensions K and L of k with K < L, the norm N L/K : L → K, x 7→ ( Q

σ∈G(L/K) σ)(x) is a homomorphism for x ∈ L. If H = G(L/K) is normal in G = G(L/k), then N _L/K corresponds to N H in Lemma 1. In particular it is clear that N Q(ε

)/Q(ε

) hε p

i = hε p

i = hε p

i ^H for a prime p and i > 0, thus the following theorem is due to Lemma 1.

Theorem 2. ([13, (7.12)]) Let k ≤ Q(ε m ) and L = Q(ε m , ε n ). Let n ⁰ = 4 ^δ p 1 · · · p s where p i are distinct odd prime divisors of n not dividing m, and δ = 1 if 4|n, 4 6 |m; δ = 0 otherwise. Let K = Q(ε m , ε n

). Then H _ι ² (K/k) ∼ = H _ι ² (L/k).

As a consequence of Theorem 2, Janusz proved the next theorem on algebras.

Theorem 3. ([6], [13, (7.9)]) If k ≤ Q(ε m ) then any cyclotomic algebra over k is similar to the cyclotomic algebra (Q(ε m , ε t )/k, α) with t = 4 ^δ p 1 · · · p s ; δ = 0 if 4|m and δ = 1 otherwise, where all p i are distinct odd primes not dividing m.

Theorem 2 and 3 can be generalized to any cyclotomic extension field L containing finitely many roots of unity in [4].

Theorem 4. Let k ≤ Q(ε m ) and L = Q(ε m , ε n

, . . . , ε n

). Let

n ⁰ ₁ = 4 ^δ

p 1 · · · p s with distinct odd primes p j |n 1 , p j - m (1 ≤ j ≤ s) and δ 1 = 1 if 4|n 1 , 4 - m; δ 1 = 0 otherwise. And for 1 < i ≤ w, let

n ⁰ _i = 4 ^δ

p i1 · · · p iu

with p ij |n i , p ij - m(1 ≤ j ≤ u i ), and p ij - n v (1 ≤ v < i) where p ij are distinct odd primes, and δ i = 1 if 4|n i , 4 - m, 4 - n v (1 ≤ v < i); δ i = 0 otherwise.

Then H _ι ² (L/k) ∼ = H ι ² (K/k) where K = Q(ε m , ε n

, . . . , ε n

). Furthermore, a crossed product algebra (L/k, α) with α ∈ Z ² (L/k, µ(L)) is similar to (K/k, β) where β is a 2-cocycle having values in µ(K).

Proof. We will prove this when L = Q(ε m , ε n

, ε n

). Write n ⁰ ₁ = 4 ^δ

p 1 · · · p s

and n ⁰ ₂ = 4 ^δ

q 1 · · · q u where p i [resp. q j ] are distinct odd prime divisors of n 1

[resp. n 2 ] with p i - m, q j - m and q j - n 1 for 1 ≤ i ≤ s, 1 ≤ j ≤ u. And δ 1 = 1 if 4|n 1 , 4 - m; δ 2 = 1 if 4|n 2 , 4 - m, 4 - n 1 ; and δ 1 = δ 2 = 0 otherwise.

Let K = Q(ε m , ε _n

, ε _n

) and E = Q(ε m , ε _n

, ε n

). Then k < K = Q(ε mn

, ε n

) ≤ E = Q(ε mn

, ε n

)

for, gcd(m, n ⁰ _i ) = 1(i = 1, 2). Since each q j in the factorization of n ⁰ ₂ satisfies q j - mn ⁰ ₁ ; δ 2 = 1 if 4|n 2 and 4 - mn ⁰ ₁ ; and δ 2 = 0 otherwise, we are able to use Theorem 2 on K and E to get H _ι ² (K/k) ∼ = H ι ² (E/k).

Moreover with l = lcm(m, n 2 ), we have

E = Q(ε m , ε n

)(ε _n

) = Q(ε l , ε _n

) ≤ L = Q(ε m , ε n

)(ε n

) = Q(ε l , ε n

) and each p i in the factorization of n ⁰ ₁ satisfies

p i - l; δ 1 = 1 if 4|n 1 and 4 - l; and δ 1 = 0 otherwise.

Thus by applying Theorem 2 to L and E, we have H _ι ² (E/k) ∼ = H _ι ² (L/k).

Now the isomorphism (L/k, α) ∼ = (K/k, β) of crossed product algebra with

α ∈ Z ² (L/k, µ(L)), β ∈ Z ² (K/k, µ(K)) follows immediately by Theorem 3. ¤

3. Norm on radical extension fields

A field L is a (finite) radical extension of k if there is a subgroup Ω of L ^∗ such that L = k(Ω) and Ω/k ^∗ is a (finite) torsion group. We may exhibit L by k(

√

a 1 , . . . ,

√

a w ) with a i ∈ k ^∗ and n i > 0 (1 ≤ i ≤ w). Moreover if L = k(Ω) is a Galois extension over k, then Ω is G(L/k)-invariant, so k(Ω) contains enough roots of unity, i.e., L = k({ε n

,

√

a i |1 ≤ i ≤ w}) with primitive n i -th roots of unity ε n

. Clearly Ω is not determined uniquely, while Ω/k ^∗ is unique and finite.

The most interesting case of radical extension is L = k(λ) with λ ∈ L ^∗ and λ ⁿ ∈ k. If L/k is Galois radical then we may regard L = k(ε _n , λ). In particular if ε n ∈ k then L/k is a cyclic extension of degree dividing n(see [8, Theorem 14.4]). The radical extension L = k(λ) is said to be irreducible if degree [L : k] = n. Thus L/k is irreducible radical if and only if L = k(λ) and λ ∈ L ^∗ is a root of an irreducible polynomial X ⁿ − a ∈ k[X], and, if and only if the order of λk ^∗ in L ^∗ /k ^∗ is equal to the degree of λ over k.

Remark that, in case of a cyclotomic extension L = Q(ε m , ε n

, . . . , ε n

), the reduction in Theorem 4 would follow immediately by taking n = lcm(n 1 , . . . , n w ) and applying Theorem 2 to m and n. However when L = k(

√

a 1 , . . . ,

√ a w ) is a Galois radical extension, to get a kind of reduction we need to know how to choose n ⁰ _i from n i each other explicitly, not just taking n ⁰ from n =lcm of n i ’s.

Lemma 5. Let L = k(Ω) be a finite Galois radical extension of k. Then Ω/k ^∗ is a finite G-module where G = G(L/k).

Proof. The finite group Ω/k ^∗ has a G-module structure by defining an action σλ = σ(λ ⁰ )k ^∗ where σ ∈ G and λ ∈ Ω/k ^∗ such that λ = λ ⁰ k ^∗ for λ ⁰ ∈ Ω. ¤

We begin with a radical extension containing an n-th root of 1 6= a ∈ k ^∗ . Theorem 6. Let k = Q(ε m ) and K = k( √

a) be a Galois radical extension of k, where m and n are positive odd integers. For a prime p dividing n, let E = k(

√

a) and F = k(ε pn , √

a) be Galois radical extensions of k. Then (i) E = F , or E/F is cyclic of degree p and N _E/F hε pn ,

√

ai = hε n , √

ai.

(ii) Furthermore we assume p ^z ||m and p ^b ||n.

a) If 0 ≤ z ≤ b then F/K is cyclic of degree p, and N F/K hε pn , √

ai

= hε n , √

a ^p i < hε n , √

ai.

b) If b < z then E = K, or F = K so N _E/K hε pn ,

√

ai = hε n , √

ai.

Proof. The Galois radical extensions K and E form k < K = k(ε n , √

a) < F = k(ε pn , √

a) < E = k(ε pn ,

√ a).

Since E = k(

√

a) = F ( √

λ) where λ = √

a ∈ F and √

λ is a root of X ^p − √

a ∈

F [X], and since ε pn ∈ F , E/F is a cyclic extension of degree dividing p [8,

Theorem 14.4]. Hence E = F or [E : F ] = p.

If [E : F ] = p, any σ ∈ G(E/F ) maps

√

a to a zero of X ^pn − a, say σ(

√

a) = ε ^l _pn

√

a for 0 ≤ l < pn. But

√

a ^p is a zero of X ⁿ −a, so

√

a ^p ∈ F ^∗ ,

pn

√

a ^p = σ(

√

a ^p ) = ε ^pl _pn

√

a ^p and ε ^pl _pn = 1. Thus l = tn for some t = 0, . . . , p−1.

Moreover since each σ ∈ G(E/F ) leaves ε pn fixed, we have σ(

√

a)ε ^tn _pn

√

a, σ ² (

√

a) = ε ^2tn _pn

√

a, . . . , σ ^p−1 (

√

a) = ε ^(p−1)tn _pn

√ a, hence it follows that

N E/F (

√

a) = Y

σ∈G(E/F )

σ(

√

a) = ε tn(1+···+(p−1))

pn

√ a ^p =

√ a ^p .

Thus N E/F h

√

ai = h

√

a ^p i = h √

ai and N E/F hε pn i = hε ^p _pn i = hε n i.

Suppose that m = p ^z m ⁰ and n = p ^b n ⁰ with p - m ⁰ n ⁰ . If 0 ≤ z ≤ b then lcm(m, pn) = p ^z m ⁰ p ^b+1 n ⁰

p ^z gcd(m ⁰ , p ^b+1−z n ⁰ ) = m ⁰ p ^b+1 n ⁰

gcd(m ⁰ , n ⁰ ) = p ^b+1 lcm(m ⁰ , n ⁰ ) and similarly lcm(m, n) = p ^b lcm(m ⁰ , n ⁰ ). Set lcm(m ⁰ , n ⁰ ) = l. Then Q(ε m , ε pn )

= Q(ε _p

l ) and Q(ε m , ε n ) = Q(ε _p

l ) because p - l, hence [Q(ε m , ε pn ) : Q(ε m , ε n )] = φ(p ^b+1 l)

φ(p ^b l) = φ(p ^b+1 ) φ(p ^b ) = p (where φ is the Euler phi function), which shows that [F : K] = p.

Now each τ ∈ G(F/K) maps ε pn ∈ F to ε ^l _pn for some 0 ≤ l < pn. Because ε ^p _pn ∈ K ^∗ , we have ε ^p _pn = τ (ε ^p _pn ) = ε ^pl _pn , hence p(l − 1) ≡ 0 (mod pn) and l = tn + 1 with 0 ≤ t < p. Indeed, the cyclic group G(F/K) of order p consists of automorphisms τ t such that τ t (ε pn ) = ε ^tn+1 _pn for 0 ≤ t < p. Thus we have

N _F/K (ε pn ) = ε

P

t=0

(tn+1)

pn = ε ^p+pn pn

= ε ^p _pn ,

so N F/K hε pn i = hε ^p _pn i ≤ hε n i. Comparing the orders, we get N F/K hε pn i= hε n i.

And N F/K h √

ai = h Q

τ ∈G(F/K) τ ( √

a)i = h √

a ^p i < hε n , √

ai, this is (ii-a).

In case of m = p ^z m ⁰ , n = p ^b n ⁰ with 0 < b < z, gcd(m, pn) = pgcd(m, n), lcm(m, pn) = lcm(m, n), so Q(ε m , ε pn ) = Q(ε m , ε n ) and [F : K] = 1. Hence E = K, or [E : K] = p and N E/K hε pn ,

√

ai = N E/F hε pn ,

√

ai = hε n , √

a i.

¤ We observe that the assumption in (ii-b), i.e., m = p ^z m ⁰ , n = p ^b n ⁰ with b < z implies that k = Q(ε m ) already contains enough p-th roots of unity.

Corollary 7. Assume the same context as in (ii-b) Theorem 6 for radical extensions K = k( √

a) = k(Ω K ) and E = k(

√

a) = k(Ω E ) with finite Ω K /k ^∗ and Ω E /k ^∗ . If H = G(E/K) 6= 1 then N H (Ω E /k ^∗ ) = Ω K /k ^∗ .

Proof. E = k(ε pn ,

√

a) is a cyclic extension of K = k(ε n , √

a) of degree p with cyclic Galois group H. The mapping N H = Q

σ∈H σ in Lemma 1 determines N H hε pn ,

√

ai = hε n , √

ai by Theorem 6. By the action in Lemma 5, we have

N H (Ω E /k ^∗ ) = N H hε pn k ^∗ ,

√

ak ^∗ i = hε n k ^∗ , √

ak ^∗ i = Ω K /k ^∗ .

¤ We shall denote N H (Ω E /k ^∗ ) by the same notation N _E/K (Ω E /k ^∗ ) which is the norm map. As a generalization of Theorem 6, we have the following theorem.

Theorem 8. Let k = Q(ε m ) and L = k( √

a) = k(Ω) be a Galois radical exten- sion of k. Assume a prime p with p ^z ||m and p ^b ||n. Let F = k(

√

a) = k(Ω F ) be a Galois radical extension with L 6= F . If 0 < b ≤ z then N L/F (Ω/k ^∗ ) = Ω F /k ^∗ .

Proof. Let E = k(ε n ,

√

a) be a Galois radical extension. Then k < F = k(ε _n/p ,

√

a) < E = k(ε n ,

√

a) < L = k(ε n , √

a), and L = E( √

λ) where λ =

√

a ∈ E and √

λ is a root of a polynomial X ^p −

√

a in E[X]. Since ε p ∈ E, L/E is a cyclic radical extension of degree dividing p. So L = E or [L : E] = p. Because b ≤ z, k contains enough p-th roots of unity so F = E. But since L 6= F , we have [L : F ] = p and

N _L/F hε n , √

ai = hε ^p _n , √

a ^p i = hε _n/p ,

√ ai.

Similar to Corollary 7, we consequently have N _L/F (Ω/k ^∗ ) = Ω _F /k ^∗ . ¤ 4. Cohomology group on radical extension fields

We shall discuss cohomology groups over Galois radical extension fields, and have a reduction of cohomology group that is an analog of Theorem 2. Let H be a normal subgroup of G and M be a G-module. The inflation-restriction sequence on cohomology group

(1) 1 → H ^r (G/H, M ^H ) inf

→ H ^r (G, M ) res

→ H ^r (H, M )

is exact if r = 1. When r > 1, the sequence is exact if H ⁱ (H, M ) = 1 for all 1 ≤ i ≤ r − 1 (refer to [12, (3.4.2), (3.2.3)]).

Theorem 9. Let n = p ^b n 0 and m = p ^z m 0 with an odd prime p - n 0 m 0 . Let k = Q(ε m ), and L = k( √

a) = k(Ω) and L 0 = k(

√

a) = k(Ω 0 ) be Galois radical extensions of k with L 6= L 0 . If 0 < b ≤ z then the inflation map is an isomorphism on cohomology groups

H ² (L 0 /k, Ω 0 /k ^∗ ) inf

∼ = H ² (L/k, Ω 0 /k ^∗ ).

Proof. Obviously L = k(ε n , √

a) and L 0 = k(ε n

,

√

a). Since b ≤ z, ε _p

is contained in k < L 0 so ε n ∈ L 0 . Thus L = L 0 (

√

λ) where λ =

√

a and

√ λ is a root of X ^p

−

√

a in L 0 [X]. So L/L 0 is cyclic of degree p ^c for some c ≤ b.

Due to Lemma 5, Ω/k ^∗ is a G(L/k)-module with module action that, for any τ ∈ G(L/k) and √

ak ^∗ ∈ Ω/k ^∗ , τ ( √

ak ^∗ ) = τ ( √

a)k ^∗ = ε ^l _n √

ak ^∗ ∈ Ω/k ^∗

for some l > 0. Similarly Ω 0 /k ^∗ is a G(L 0 /k)-module. Moreover it is easy to see that Ω ₀ /k ^∗ is also a G(L/k)-module by regarding

√

a as ( √

a) ^p

. Write H = G(L/L 0 ). From the sequence of groups

1 → G(L/L 0 ) → G(L/k) → G(L 0 /k) → 1, we consider the inflation-restriction sequence

H ² (L 0 /k, (Ω 0 /k ^∗ ) ^H ) inf → H ² (L/k, Ω 0 /k ^∗ ) res → H ² (L/L 0 , Ω 0 /k ^∗ ).

Since H is cyclic, it follows from [12, (1.5.6)] and [12, (3.2.1)] that H ² (L/L 0 , Ω 0 /k ^∗ ) = H ⁰ (H, Ω 0 /k ^∗ ) = (Ω 0 /k ^∗ ) ^H

N _L/L

(Ω 0 /k ^∗ ) .

Every σ ∈ H leaves all elements in Ω 0 /k ^∗ fixed, so (Ω 0 /k ^∗ ) ^H = Ω 0 /k ^∗ . More- over due to Theorem 8, we compute the norm N _L/L

directly to get

N L/L

(Ω 0 /k ^∗ ) = hε ^p _n

k ^∗ ,

√

a ^p

k ^∗ i = hε n

k ^∗ ,

√

ak ^∗ i = Ω 0 /k ^∗ for gcd(p ^c , n 0 ) = 1. Hence H ² (L/L 0 , Ω 0 /k ^∗ ) = 1.

Again since H is finite cyclic and Ω 0 /k ^∗ is a finite H-module, we use the Herbrand’s quotient of Ω 0 /k ^∗ (refer to [3, (23.2)]) that

1 = h 2/1 (Ω 0 /k ^∗ ) = |H ² (L/L 0 , Ω 0 /k ^∗ )|

|H ¹ (L/L 0 , Ω 0 /k ^∗ )|

hence it follows that H ¹ (L/L 0 , Ω 0 /k ^∗ ) = H ² (L/L 0 , Ω 0 /k ^∗ ) = 1. Therefore we can conclude from (1) that the sequence

1 → H ² (L 0 /k, (Ω 0 /k ^∗ ) ^H ) inf → H ² (L/k, Ω 0 /k ^∗ ) res → H ² (L/L 0 , Ω 0 /k ^∗ ) = 1 is exact, so have an isomorphism H ² (L 0 /k, Ω 0 /k ^∗ ) ∼ = H ² (L/k, Ω 0 /k ^∗ ). ¤

For a given finite radical extension L over k, we observed in Theorem 9 that the cohomology group over G(L/k) can be decreased down to that over G(L 0 /k), where L 0 is the Galois radical extension of k smaller than L by ‘one’

prime factor power p ^b . We can strengthen this observation with the following theorem which will go to reduction of Galois cohomology groups.

Theorem 10. Let k = Q(ε m ) and L = k( √

a) be the same fields as in Theorem 9. Assume n = p ^b ₁

· · · p ^b _u

with distinct primes p i and b i > 0. For each p i

(1 ≤ i ≤ u), write m = p ^z _i

m i (with p i - m i and z i ≥ 0). Suppose b i ≤ z i for some 1 ≤ i ≤ u, and after appropriate renumbering, we assume that b i ≤ z i for all s + 1 ≤ i ≤ u. Let n 0 = p ^b ₁

· · · p ^b _s

, and let the Galois extensions be

L 0 = k(

√

a) = k(Ω 0 ), L j = k(

√

a) = k(Ω j ) for 1 ≤ j ≤ u − s.

Then H ² (L j /k, Ω j−1 /k ^∗ ) ∼ = H ² (L j−1 /k, Ω j−1 /k ^∗ ) for all 1 ≤ j ≤ u − s, so

H ² (L/k, Ω 0 /k ^∗ ) and H ² (L 0 /k, Ω 0 /k ^∗ ) are isomorphic.

Proof. Since L 0 , L j (1 ≤ j ≤ u − s) are all Galois radical extensions of k, and since n = p ^b ₁

· · · p ^b _s

· p ^b _s+1

· · · p ^b _u

= n 0 · p ^b _s+1

· · · p ^b _u

, we have L u−s = L and

k < L 0 = k(ε n

,

√

a) < L 1 = k(ε _n

0

p

,

√

a) < · · ·

< L j = k(ε _n

0

p

···p

,

√

a) < · · · < L u−s = L.

Since m = p ^z _i

m i and 0 < b i ≤ z i for s + 1 ≤ i ≤ u, each ε _p

i

∈ k. Together with ε n

∈ L 0 , ε _n

0

p

···p

belong to L 0 for all 1 ≤ j ≤ u − s. Hence L j = L j−1 (λ j ), where λ j ∈ L j and λ ^p

bs+j s+j

j ∈ L j−1 (1 ≤ j ≤ u − s), and L j /L j−1 is cyclic of degree p ^c _s+j

for some c j ≤ b s+j . Thus we have the isomorphism on cohomology groups

H ² (L 1 /k, Ω 0 /k ^∗ ) ∼ = H ² (L 0 /k, Ω 0 /k ^∗ ) due to Theorem 9, and proceeding inductively we get

(2) H ² (L j /k, Ω j−1 /k ^∗ ) ∼ = H ² (L j−1 /k, Ω j−1 /k ^∗ ) for each 1 ≤ j ≤ u − s.

Let H = G(L 2 /L 0 ), and consider the inflation-restriction sequence (3) H ² (L 0 /k, (Ω 0 /k ^∗ ) ^H ) inf

→ H ² (L 2 /k, Ω 0 /k ^∗ ) res

→ H ² (L 2 /L 0 , Ω 0 /k ^∗ ).

Since L 2 = L 0 (ξ) with ξ ^p

^p

∈ L 0 , L 2 /L 0 is a cyclic radical extension of degree dividing p ^b _s+1

p ^b _s+2

, and in fact [L 2 : L 0 ] = p ^c _s+1

p ^c _s+2

. Thus

H = G(L 2 /L 0 ) ∼ = Z _p

s+1

p

∼ = Z _p

s+1

× Z _p

s+2

∼ = G(L 2 /L 1 ) × G(L 1 /L 0 ), and by [7, (2.3.14)], we have

H ² (L 2 /L 0 , Ω 0 /k ^∗ ) ∼ = H ² (L 2 /L 1 , Ω 0 /k ^∗ ) × H ² (L 1 /L 0 , Ω 0 /k ^∗ ).

But since p i - n 0 for s + 1 ≤ i ≤ u, we obtain N L

/L

(Ω 0 /k ^∗ ) = hε n

k ^∗ ,

√

ak ^∗ i ^p

= Ω 0 /k ^∗ = (Ω 0 /k ^∗ ) ^L

^/L

and similarly, N L

/L

(Ω 0 /k ^∗ ) = (Ω 0 /k ^∗ ) ^p

= Ω 0 /k ^∗ = (Ω 0 /k ^∗ ) ^L

^/L

.

By the proof of Theorem 9, both H ² (L 1 /L 0 , Ω 0 /k ^∗ ) and H ² (L 2 /L 1 , Ω 0 /k ^∗ ) are trivial groups, so that H ² (L 2 /L 0 , Ω 0 /k ^∗ ) = 1. Then again the Herbrand quotient of Ω 0 /k ^∗ is equal to 1, i.e.,

1 = |H ² (L 2 /L 0 , Ω 0 /k ^∗ )|/|H ¹ (L 2 /L 0 , Ω 0 /k ^∗ )|.

So H ¹ (L 2 /L 0 , Ω 0 /k ^∗ ) is trivial, and the inflation-restriction sequence (3) is exact. We thus obtain the isomorphism on cohomology groups

H ² (L 0 /k, Ω 0 /k ^∗ ) inf

∼ = H ² (L 2 /k, Ω 0 /k ^∗ ).

And together with (2), it follows that

H ² (L 2 /k, Ω 0 /k ^∗ ) ∼ = H ² (L 1 /k, Ω 0 /k ^∗ ) ∼ = H ² (L 0 /k, Ω 0 /k ^∗ ).

Applying this process to the cyclic group H = G(L 3 /L 0 ) of order p ^c _s+1

p ^c _s+2

p ^c _s+3

and to the sequence

H ² (L 0 /k, (Ω 0 /k ^∗ ) ^H ) inf

→ H ² (L 3 /k, Ω 0 /k ^∗ ) res

→ H ² (L 3 /L 0 , Ω 0 /k ^∗ ), we also get H ² (L 3 /L 0 , Ω 0 /k ^∗ ) = 1, for G(L 3 /L 0 ) ∼ = G(L 3 /L 2 ) × G(L 2 /L 0 ) and H ² (L 3 /L 2 , Ω 0 /k ^∗ ) = 1 = H ² (L 2 /L 0 , Ω 0 /k ^∗ ).

The exactness of the inflation-restriction sequence guarantees the isomor- phism H ² (L 3 /k, Ω 0 /k ^∗ ) ∼ = H ² (L 0 /k, Ω 0 /k ^∗ ). Continuing, we eventually get

H ² (L/k, Ω 0 /k ^∗ ) = H ² (L u−s /k, Ω 0 /k ^∗ ) ∼ = H ² (L 0 /k, Ω 0 /k ^∗ ).

¤ Remark 1. In Theorem 10, we assume a = 1, i.e., √

a = ε n for n = p ^b ₁

· · · p ^b _u

(odd primes). Suppose m = p ^z _i

m i with z i < b i for 1 ≤ i ≤ s, and b i ≤ z i

for s + 1 ≤ i ≤ u. Let n 0 = p ^b ₁

· · · p ^b _s

, L = Q(ε m , ε n ) and L 0 = Q(ε m , ε n

).

Then L = L 0 and H ² (L/k, Ω L /k ^∗ ) ∼ = H ² (L 0 /k, Ω L

/k ^∗ ) (this is Theorem 10).

Moreover, by rearrangement if necessary, we assume z ₁ = · · · = z _t = 0 for t ≤ s, i.e., p i 6 |m for 1 ≤ i ≤ t. By letting n ⁰ ₀ = p 1 · · · p t , we can further reduce the cohomology group by Janusz theorem that

H ² (L/k, µ(L)) = H ² (L 0 /k, µ(L 0 )) ∼ = H ² (k(ε n

)/k, µ(k(ε n

))).

Remark 2. It is natural to ask whether the Remark 1 is true for a 6= 1 with the same n and n ⁰ ₀ , i.e., is H ² (k( √

a)/k, h √

ai/k ^∗ ) ∼ = H ² (k(

√

a)/k, h

√ ai/k ^∗ )?

The following proposition provides a partial answer.

Proposition 11. Let L = k( √

a) be a Galois radical extension of k = Q(ε m ).

Let n = p ^b ₁

· · · p ^b _u

(b i > 0, odd primes p i ) and m = p ^z _i

m i (z i ≥ 0, p i 6 |m i ) for all i. By rearrangement, assume z i ≤ b i for 1 ≤ i ≤ s(≤ u), z i > b i for s + 1 < i ≤ u, and moreover z j = 0 for 1 ≤ j ≤ t(≤ s). Set n 0 = p ^b ₁

· · · p ^b _s

and n ⁰ ₀ = p 1 · · · p t . Let K = k(

√

a) = k(Ω), and F v = k(

√ a) and B v = k(ε _n

0

/p

,

√

a) be Galois radical extensions of k for t + 1 ≤ v ≤ s.

Then H ² (L/k, Ω/k ^∗ ) ∼ = H ² (F v /k, Ω/k ^∗ ), H ² (K/k, Ω/k ^∗ ) ∼ = H ² (B v /k, Ω/k ^∗ ), but H ² (F v /k, Ω/k ^∗ ) 6∼ = H ² (B v /k, Ω/k ^∗ ).

Proof. Let L ₀ = k(

√

a) = k(Ω _L

) be a Galois radical extension of k. With the integers n = p ^b ₁

· · · p ^b _u

, n 0 = p ^b ₁

· · · p ^b _s

and n ⁰ ₀ = p 1 · · · p t for t ≤ s ≤ u, it is clear that k < K < L 0 < L and H ² (L/k, Ω L

/k ^∗ ) ∼ = H ² (L 0 /k, Ω L

/k ^∗ ) due to Theorem 10. Hence it is follows that

H ² (L/k, Ω/k ^∗ ) ∼ = H ² (L 0 /k, Ω/k ^∗ ).

Let p v be one of p t+1 , . . . , p s . Since m = p ^z _v

m v with 0 < z v ≤ b v , the Galois radical extensions of k are

K = k(ε n

,

√

a) < F v = k(ε _n

0

/p

,

√

a) < L 0 = k(ε n

,

√

a).

Now let K, F v and L 0 denote the cyclotomic extensions K = Q(ε m , ε _n

0

) < F v = Q(ε m , ε _n

0

/p

) < L 0 = Q(ε m , ε n

).

Owing to p ^z _v

|| _p

ⁿ

v

and ε _p

∈ K, we have p v - [F v : K] and H ² (F v /K, µ(F v )) = H ² (L 0 /F v , µ(L 0 )) = 1. Thus due to Janusz theorem we have

H ² (K/k, µ(K)) ∼ = H ² (F _v /k, µ(F _v )) ∼ = H ² (L ₀ /k, µ(L ₀ )).

On the other hand, from the tower of fields k < K = K(

√

a) < F _v = F _v (

√

a) < L ₀ = L ₀ (

√ a), consider a field A v = k(ε n

,

√

a). We now observe the followings.

(i) K < B v < F v < A v < L 0

(ii) L 0 = A v (

√

λ) where λ =

√

a and

√

λ is a root of X ^p

− λ ∈ A v [X]. Since ε _p

∈ A v , L 0 /A v is a cyclic extension of order p ^w _v

with w v ≤ z v .

(iii) Similar to (ii), it can be seen that F v = B v (

√

θ) where θ =

n00

√

a ∈ B v and

√

θ is a root of X ⁿ

^/n

^p

− θ ∈ B v [X]. Since ε _n

0

/n

p

belongs to B v = Q(ε m , ε _n

0

/p

,

√

a), F v /B v is cyclic of degree dividing p ^b ₁

⁻¹ · · · p ^b _t

⁻¹ · p ^b _t+1

· · · p ^z _v

· · · p ^b _s

.

(iv) A v = F v (ε n

) = F v (ε _p

v

), so G(A v /F v ) ∼ = G(L 0 /F v ).

(v) B v = K(ε _n

0

/p

), so G(B v /K) ∼ = G(F v /K).

L 0 = k(ε n

) L 0 = k(ε n

,

√

a) = L 0 (

√ a) A v = k(ε n

,

√

a) F v = k(ε _n

0

/p

) F v = k(

√

a) = F v (

√ a) B v = k(ε _n

0

/p

,

√ a) K = k(ε n

) K = k(

√

a) = k(Ω) = K(

√ a) Thus from (ii), L 0 /A v is cyclic of order p ^w _v

with w v ≤ z v , so

N L

/A

(Ω/k ^∗ ) = N L

/A

hε m , ε n

,

√

aik ^∗ = hε m k ^∗ , ε ^p _n

k ^∗ ,

√

a ^p

k ^∗ i, and this is equal to Ω/k ^∗ because gcd(n ⁰ ₀ , p v ) = 1 for t + 1 ≤ v ≤ s. Hence

H ² (L 0 /A v , Ω/k ^∗ ) = (Ω/k ^∗ ) ^G(L

^/A

⁾ N _L

_/A

(Ω/k ^∗ ) = 1

= H ⁰ (L 0 /A v , Ω/k ^∗ ) = H ¹ (L 0 /A v , Ω/k ^∗ ).

So the exact sequence

H ² (A v /k, Ω/k ^∗ ) → H ² (L 0 /k, Ω/k ^∗ ) → H ² (L 0 /A v , Ω/k ^∗ )

yields the isomorphism H ² (A v /k, Ω/k ^∗ ) ∼ = H ² (L 0 /k, Ω/k ^∗ ).

From (iv), A v = F v (ε _p

v

) and G(A v /F v ) ∼ = G(L 0 /F v ) cyclic, so the in- variant set Ω/k ^∗ by G(A v /F v ) corresponds to µ(K)/k ^∗ by G(L 0 /F v ). Thus

H ² (A v /F v , Ω/k ^∗ ) ∼ = H ² (L 0 /F v , µ(K)/k ^∗ ).

Since H ² (L 0 /F v , µ(K)) ,→ H ² (L 0 /F v , µ(L 0 )) = 1 by Janusz theorem, we have H ² (L 0 /F v , µ(K)) = 1. Moreover since H ² (L 0 /F v , µ(K)) → H ² (L 0 /F v , µ(K)/

k ^∗ ), we have

H ² (L 0 /F v , µ(K)/k ^∗ ) = 1 = H ² (A v /F v , Ω/k ^∗ ).

So we obtain the isomorphism H ² (F v /k, Ω/k ^∗ ) ∼ = H ² (A/k, Ω/k ^∗ ) from the exact sequence H ² (F v /k, Ω/k ^∗ ) → H ² (A v /k, Ω/k ^∗ ) → H ² (A v /F v , Ω/k ^∗ ) = 1.

We therefore have the isomorphisms

H ² (L/k, Ω/k ^∗ ) ∼ = H ² (L 0 /k, Ω/k ^∗ ) ∼ = H ² (A v /k, Ω/k ^∗ )

∼ = H ² (F v /k, Ω/k ^∗ ).

Now from (v), B v = K(ε _n

0

/p

) and G(B v /K) ∼ = G(F v /K) cyclic. As above,

H ² (B v /K, Ω/k ^∗ ) ∼ = H ² (F v /K, µ(K)/k ^∗ ) = 1.

Thus H ² (K/k, Ω/k ^∗ ) → H ² (B v /k, Ω/k ^∗ ) → H ² (B v /K, Ω/k ^∗ ) = 1 is exact, so the isomorphism H ² (K/k, Ω/k ^∗ ) ∼ = H ² (B v /k, Ω/k ^∗ ) follows.

However we observe that H ² (B v /k, Ω/k ^∗ ) is not isomorphic to H ² ( F v /k, Ω/

k ^∗ ). In fact, F v /B v is cyclic of degree d dividing p ^b ₁

⁻¹ · · · p ^b _t

⁻¹ · p ^b _t+1

· · · p ^z _v

· · · p ^b _s

by (iii). Thus the N _F

_/B

(Ω/k ^∗ ) = N _F

_/B

hε m k ^∗ , ε _n

k ^∗ ,

√

ak ^∗ i = hε m k ^∗ , ε ^d _n

0

k ^∗ ,

√

a ^d k ^∗ i 6= Ω/k ^∗ , because gcd(d, n ⁰ ₀ ) need not be 1. ¤ The exact correspondence of Theorem 2 with respect to radical extension is to show H ² (K/k, Ω K /k ^∗ ) ∼ = H ² (L/k, Ω L /k ^∗ ) where K = k(Ω K ) < L = k(Ω L ). Instead of this, we proved in Theorem 10 that H ² (K/k, Ω K /k ^∗ ) ∼ = H ² (L/k, Ω _K /k ^∗ ) which is a subgroup of H ² (L/k, Ω _L /k ^∗ ). We have discussed a radical extension field with one n-th root of an element in k. The next theorem is about a radical extension having more than one n-th root.

Theorem 12. Let k = Q(ε m ). Write m = p ^z m ⁰ and n i = p ^b

n ⁰ _i (i = 1, 2) with an odd prime p 6 |m ⁰ n ⁰ ₁ n ⁰ ₂ , and z, b i ≥ 0. Let L = k(

√

a 1 ,

√

a 2 ) = k(Ω L ), F = k(

√

a 1 ,

√

a 2 ) = k(Ω F ), and K = k(

√ a 1 ,

√

a 2 ) = k(Ω) be Galois radical extensions of k. Assume b i ≤ z for i = 1, 2. Then

(i) N F/K (Ω F /k ^∗ ) = Ω/k ^∗ = N L/K (Ω L /k ^∗ ).

(ii) Moreover, H ² (L/k, Ω/k ^∗ ) ∼ = H ² (F/k, Ω/k ^∗ ) ∼ = H ² (K/k, Ω/k ^∗ ).

Proof. We may write the Galois radical extensions of k by K = k(ε n

,

√

a 1 ,

√

a 2 ) < F = k(ε n

, ε n

,

√ a 1 ,

√

a 2 )

< L = k(ε n

,

√ a 1 ,

√

a 2 )