The following theorem of Kummer is known (see [1], p. 11):

VOL. LXVIII 1995 FASC. 2

CRITERION FOR A FIELD TO BE ABELIAN

BY

J. W ´ O J C I K

The following theorem of Kummer is known (see [1], p. 11):

Let α ∈ P

, p a prime, P

= Q(ζ

), ζ

= e

. Assume that α is of order p with respect to (P

)

. Let σ = (ζ

→ ζ

), where % is a primitive root mod p. The extension P

( √

α)/Q is abelian if and only if the number α

is a pth power in P

.

H. Hasse gives in [1], p. 11, a more general result:

Let F , M be algebraic number fields such that F ⊆ M and the extension M/F is cyclic. Assume that ζ

∈ M . Let σ denote a generator of G(M/F ), ζ

= ζ

. Let α ∈ M . Assume that α is of order n with respect to M

. Put L = M ( √

α). The extension L/F is abelian if and only if the number α

is an nth power in M .

The aim of this paper is to prove a similar theorem which contains the above result. Namely, we have the following:

Theorem. Let F be a field and n a positive integer not divisible by the characteristic of F . Let M/F be an abelian extension of finite degree and L = M ( √

α) for some α ∈ M

. Further , let σ

, . . . , σ

be a basis of G(M (ζ

)/F ) with ζ

= ζ

, a

∈ Z. The extension L/F is abelian if and only if there exist A

, . . . , A

∈ M

such that

(1) α

= A

(1 ≤ j ≤ r), (2) A

= A

(1 ≤ i, j ≤ r).

Corollary 1. Let F be a field and n a positive integer not divisible by the characteristic of F . Let the extension M (ζ

)/F be cyclic and L = M ( √

α) for some α ∈ M

. Further , let σ be a generator of G(M (ζ

)/F ) with ζ

= ζ

, a ∈ Z. The extension L/F is abelian if and only if the number α

is an nth power in M .

R e m a r k 1. Corollary 1 contains Hasse’s result quoted above.

1991 Mathematics Subject Classification: Primary 11R20.

[187]

Corollary 2 (A. Schinzel [3]). Let F be a field and n a positive integer not divisible by the characteristic of F . A binomial x

− α has an abelian Galois group over F if and only if α

= γ

, where γ ∈ F and w

is the number of nth roots of unity contained in F .

P r o o f o f T h e o r e m. Let α = β

, L = M (β) and L = L(ζ

).

Necessity. Assume that the extension L/F is abelian. Then L/F and L/M are also abelian. Put G = G(L/F ) and H = G(L/M ). Let σ

∈ G with σ

= σ

on M (ζ

), and τ ∈ H. We have β

= ζ

β, β

= β

= ζ

β

and

(3) β

= β

=: A

∈ M

. Hence α

= A

. Thus (1) holds.

By (3), A

= A

= β

= β

= A

. Thus (2) holds.

Sufficiency. Assume that conditions (1) and (2) hold. We shall prove that the extension L/F is abelian. It is enough to prove that L/F is abelian.

We have F ⊆ M ⊆ L ⊆ L. Since M/F , L/M and L/L are separable, so is L/F .

Let σ be an arbitrary isomorphism of L over F with σ = σ on M (ζ

), σ ∈ G(M (ζ

)/F ). We have

(4) M = F (γ)

and

(5) L = F (β, γ, ζ

).

Since the extension M/F is normal,

(6) γ

∈ M ⊆ L.

Obviously

(7) ζ

∈ L.

We have

(8) σ = σ

. . . σ

, t

∈ Z, 0 ≤ t

< h

, h

= ord σ

. Put

(9) A

:=

r

Y

j=1

A

t1

1 ...a^tj−1_j−1 σ^tj+1_j+1 ...σ_r^tr

σtj j −atj

j σj −aj

j

.

Obviously A

∈ M

. We now show that

(10) α

= A

, where a = a

. . . a

.

We have

a

. . . a

σ

. . . σ

(σ

− a

) + a

. . . a

σ

. . . σ

(σ

− a

)

= a

. . . a

σ

. . . σ

− a

a

σ

. . . σ

for 1 ≤ j ≤ r − 1.

Hence

(11) σ − a =

r

X

j=1

a

. . . a

σ

. . . σ

(σ

− a

).

By (11), (1) and (9), α

=

r

Y

j=1

α

tj−1

j−1 σ^tj+1_j+1 ...σ^tr_r

σtj j −atj

j σj −aj

=

 Y

j=1

A

a^t1₁ ...a^tj−1_j−1 σ^tj+1_j+1 ...σ^tr_r

σtj j −atj

j σj −aj

j



= A

. Thus (10) holds.

By (10), β

= α

= α

= α

A

= (β

A

)

. Hence (12) β

= ζ

β

A

∈ L.

Since the extension L/F is separable and, by (5)–(7) and (12), normal, it is a Galois extension and σ is an automorphism.

Let τ be any automorphism of L over F with τ = τ on M (ζ

), τ ∈ G(M (ζ

)/F ). Since the extension M/F is abelian we have, by (4),

(13) γ

= γ

.

Obviously

(14) ζ

= ζ

.

We have

(15) τ = σ

. . . σ

, u

∈ Z, 0 ≤ u

< h

, h

= ord σ

. We now show that

(16) A

= A

, where b = a

. . . a

. By (15) and (11),

(17) τ − b =

r

X

i=1

a

. . . a

σ

. . . σ

(σ

− a

).

By (15) and (9),

(18) A

=

r

Y

i=1

A

u1

1 ...a^ui−1_i−1 σ_i+1^ui+1...σ^ur_r ^σ

ui i −aui

i σi−ai

i

.

By (2), (9), (17), (18) and (11),

A

=

r

Y

j=1 r

Y

i=1

A

t1

1 ...a^tj−1_j−1 σ_j+1^tj+1...σ^tr_r

σtj j −atj

j

σj −aj a^u1₁ ...a^ui−1_i−1 σ^ui+1_i+1 ...σ^ur_r ^σ

ui i −aui

i σi−ai

j

=

r

Y

i=1 r

Y

j=1

A

u1

1 ...a^ui−1_i−1 σ^ui+1_i+1 ...σ_r^urσ

ui i −aui

i

σi−ai a^t1₁ ...a^tj−1_j−1 σ_j+1^tj+1...σ^tr_r

σtj j −atj

j σj −aj

i

= A

. Thus (16) holds.

By (12),

(19) β

= ζ

β

A

.

By (8),

(20) ζ

= ζ

= ζ

tr

n 1

= ζ

. Similarly,

(21) ζ

= ζ

.

By (12) and (19)–(21),

(22) β

= ζ

β

A

A

, (23) β

= ζ

β

A

A

. By (16), A

A

= A

A

. By (22) and (23),

(24) β

= β

.

By (5), (24), (13) and (14) the extension L/F is abelian.

P r o o f o f C o r o l l a r y 2. We put M = F in the Theorem. It is enough to prove that α

= A

and A

= A

(A

, A

∈ F ) ⇔ α

= γ

(γ ∈ F ).

By Galois theory w

= (1 − a

, . . . , 1 − a

, n). Hence α

= A

⇔ α

= γ

. It is enough to prove that α

= A

⇒ A

= A

. Assume that α

= A

. Then

α

= α

= γ

= A

. Hence A

= ζ

γ

and

A

= γ

= A

.

R e m a r k 2. In special cases conditions (1) and (2) in the Theorem can be replaced just by (1). We have such a situation in Corollaries 1 and 2.

In general we cannot drop (2). This is shown by the following example:

F = Q, M = P

, n = 8, α = −4, L = P

( √

−4). Put σ

= (ζ

→ ζ

), σ

= (ζ

→ ζ

), a

= −1, a

= 5. Then (1) is satisfied:

α

= (−4)

= A

, α

= (−4)

= A

,

where A

= ζ

(1 − ζ

), A

= ζ

(1 + ζ

)

, A

, A

∈ P

, i, j are arbitrary rational integers. However, the extension L/F is not abelian. Otherwise by Corollary 2 we would have α

= 16 = γ

with γ ∈ Q, which is impossible.

The condition (2) is not satisfied. Indeed, A

= −1/4, A

= 1/4.

R e m a r k 3. In the case F = Q, M = P

, where P

= Q(ζ

) and m(n − 1) is even, there is a simple criterion for abelianity. Namely, the extension L/F is abelian if and only if α is of the form

α = ζτ (χ)

γ

,

where ζ, γ ∈ P

, ζ is a root of unity, χ is some proper character with conductor f and of order k such that (m, f ) = 1 or 2, k | (n, m) and τ (χ) is the normalized proper Gaussian sum corresponding to χ, with τ (χ)

∈ P

. This follows from Kronecker–Weber’s theorem and from the Theorem in [4].

R e m a r k 4. Below we give a new proof of Corollary 2 connected with the proof of the Theorem (in fact, with the proof of necessity). This proof is much shorter than other known proofs of Corollary 2 (see [3], [5] and [2], p. 435).

P r o o f. Sufficiency. Assume that α

= γ

, γ ∈ F . Put α = β

, γ = β

. We have β

∈ F

(ζ

∈ F ) and β = ζ

β

∈ F

. Thus the extension F (β, ζ

)/F is abelian.

Necessity. Assume that the Galois group of x

− α is abelian. Put α = β

, G = G(F (β, ζ

)/F ), H = G(F (ζ

)/F ) and σ

= (ζ

→ ζ

). Let σ, τ ∈ G with σ = σ

on F (ζ

). We have β

= ζ

β, β

= β

= ζ

β

and β

= β

= A

∈ F . Hence α

= A

. By Galois theory w

= g.c.d.

({1 − a}, n). Hence α

= γ

, γ ∈ F .

REFERENCES

[1] H. H a s s e, Invariante Kennzeichnung relativ-abelscher Zahlk¨ orper mit vorgegebener Galoisgruppe ¨ uber einem Teilk¨ orper des Grundk¨ orpers, Abh. Deutsch. Akad. Wiss. 8 (1947), 5–56.

[2] G. K a r p i l o w s k y, Topics in Field Theory , North-Holland, Amsterdam, 1989.

[3] A. S c h i n z e l, Abelian binomials, power residues and exponential congruences, Acta Arith. 32 (1977), 245–274.

[4] J. W ´ o j c i k, Powers of cyclotomic numbers, Comment. Math. 32 (1992), 213–223.

[5] W. Y. Y e l e z, On normal binomials, ibid. 36 (1980), 113–124.

Re¸ cu par la R´ edaction le 19.12.1993