### 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}

^{∗}

### , p a prime, P

p### = Q(ζ

p### ), ζ

p### = e

^{2πi/p}

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

_{p}

^{∗}

### )

^{p}

### . Let σ = (ζ

p### → ζ

_{p}

^{%}

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

p### ( √

^{p}

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

^{σ−%}

### is a pth power in P

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

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

_{n}

^{σ}

### = ζ

_{n}

^{%}

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

^{n}

### . Put L = M ( √

^{n}

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

^{σ−%}

### 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 ( √

^{n}

### α) for some α ∈ M

^{∗}

### . Further , let σ

1### , . . . , σ

i### be a basis of G(M (ζ

n### )/F ) with ζ

n^{σ}

^{j}

### = ζ

n^{a}

^{j}

### , a

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

1### , . . . , A

r### ∈ M

^{∗}

### such that

### (1) α

^{σ}

^{j}

^{−a}

^{j}

### = A

^{n}

_{j}

### (1 ≤ j ≤ r), (2) A

^{σ}

_{j}

^{i}

^{−a}

^{i}

### = A

^{σ}

_{i}

^{j}

^{−a}

^{j}

### (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

n### )/F be cyclic and L = M ( √

^{n}

### α) for some α ∈ M

^{∗}

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

n### )/F ) with ζ

_{n}

^{σ}

### = ζ

_{n}

^{a}

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

^{σ−a}

### 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

^{n}

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

^{w}

^{n}

### = γ

^{n}

### , where γ ∈ F and w

n### 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

^{n}

### , L = M (β) and L = L(ζ

n### ).

### 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

j### ∈ G with σ

j### = σ

j### on M (ζ

n### ), and τ ∈ H. We have β

^{τ}

### = ζ

_{n}

^{k}

### β, β

^{σ}

^{j}

^{τ}

### = β

^{τ σ}

^{j}

### = ζ

n^{a}

^{j}

^{k}

### β

^{σ}

^{j}

### and

### (3) β

^{(σ}

^{j}

^{−a}

^{j}

^{)τ}

### = β

^{σ}

^{j}

^{−a}

^{j}

### =: A

j### ∈ M

^{∗}

### . Hence α

^{σ}

^{j}

^{−a}

^{j}

### = A

^{n}

_{j}

### . Thus (1) holds.

### By (3), A

^{σ}

_{j}

^{i}

^{−a}

^{i}

### = A

^{σ}

_{j}

^{i}

^{−a}

^{i}

### = β

^{(σ}

^{j}

^{−a}

^{j}

^{)(σ}

^{i}

^{−a}

^{i}

^{)}

### = β

^{(σ}

^{i}

^{−a}

^{i}

^{)(σ}

^{j}

^{−a}

^{j}

^{)}

### = A

^{σ}

_{i}

^{j}

^{−a}

^{j}

### . 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

n### ), σ ∈ G(M (ζ

n### )/F ). We have

### (4) M = F (γ)

### and

### (5) L = F (β, γ, ζ

n### ).

### Since the extension M/F is normal,

### (6) γ

^{σ}

### ∈ M ⊆ L.

### Obviously

### (7) ζ

_{n}

^{σ}

### ∈ L.

### We have

### (8) σ = σ

_{1}

^{t}

^{1}

### . . . σ

_{r}

^{t}

^{r}

### , t

j### ∈ Z, 0 ≤ t

j### < h

j### , h

j### = ord σ

j### . Put

### (9) A

σ### :=

r

### Y

j=1

### A

^{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

σ### ∈ M

^{∗}

### . We now show that

### (10) α

^{σ−a}

### = A

^{n}

_{σ}

### , where a = a

^{t}

_{1}

^{1}

### . . . a

^{t}

_{r}

^{r}

### .

### We have

### a

^{t}

_{1}

^{1}

### . . . a

^{t}

_{j−1}

^{j−1}

### σ

^{t}

_{j+1}

^{j+1}

### . . . σ

_{r}

^{t}

^{r}

### (σ

^{t}

_{j}

^{j}

### − a

^{t}

_{j}

^{j}

### ) + a

^{t}

_{1}

^{1}

### . . . a

^{t}

_{j}

^{j}

### σ

_{j+2}

^{t}

^{j+2}

### . . . σ

^{t}

_{r}

^{r}

### (σ

_{j+1}

^{t}

^{j+1}

### − a

^{t}

_{j+1}

^{j+1}

### )

### = a

^{t}

_{1}

^{1}

### . . . a

^{t}

_{j−1}

^{j−1}

### σ

_{j}

^{t}

^{j}

### . . . σ

^{t}

_{r}

^{r}

### − a

^{t}

_{1}

^{1}

### a

^{t}

_{j+1}

^{j+1}

### σ

_{j+2}

^{t}

^{j+2}

### . . . σ

^{t}

_{r}

^{r}

### for 1 ≤ j ≤ r − 1.

### Hence

### (11) σ − a =

r

### X

j=1

### a

^{t}

_{1}

^{1}

### . . . a

^{t}

_{j−1}

^{j−1}

### σ

^{t}

_{j+1}

^{j+1}

### . . . σ

_{r}

^{t}

^{r}

### (σ

^{t}

_{j}

^{j}

### − a

^{t}

_{j}

^{j}

### ).

### By (11), (1) and (9), α

^{σ−a}

### =

r

### Y

j=1

### α

^{(σ}

^{j}

^{−a}

^{j}

^{)a}

^{t1}

^{1}

^{...a}

tj−1

j−1 σ^{tj+1}_{j+1} ...σ^{tr}_{r}

σtj j −atj

j σj −aj

### =

### Y

^{r}

j=1

### A

a^{t1}_{1} ...a^{tj−1}_{j−1} σ^{tj+1}_{j+1} ...σ^{tr}_{r}

σtj j −atj

j σj −aj

j

n### = A

^{n}

_{σ}

### . Thus (10) holds.

### By (10), β

^{σn}

### = α

^{σ}

### = α

^{σ}

### = α

^{a}

### A

^{n}

_{σ}

### = (β

^{a}

### A

σ### )

^{n}

### . Hence (12) β

^{σ}

### = ζ

_{n}

^{u}

### β

^{a}

### A

σ### ∈ 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

n### ), τ ∈ G(M (ζ

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

### (13) γ

^{σ τ}

### = γ

^{τ σ}

### .

### Obviously

### (14) ζ

_{n}

^{σ τ}

### = ζ

_{n}

^{τ σ}

### .

### We have

### (15) τ = σ

_{1}

^{u}

^{1}

### . . . σ

^{u}

_{r}

^{r}

### , u

i### ∈ Z, 0 ≤ u

i### < h

i### , h

i### = ord σ

i### . We now show that

### (16) A

^{τ −b}

_{σ}

### = A

^{σ−a}

_{τ}

### , where b = a

^{u}

_{1}

^{1}

### . . . a

^{u}

_{r}

^{r}

### . By (15) and (11),

### (17) τ − b =

r

### X

i=1

### a

^{u}

_{1}

^{1}

### . . . a

^{u}

_{i−1}

^{i−1}

### σ

^{u}

_{i+1}

^{i+1}

### . . . σ

^{u}

_{r}

^{r}

### (σ

_{i}

^{u}

^{i}

### − a

^{u}

_{i}

^{i}

### ).

### By (15) and (9),

### (18) A

τ### =

r

### Y

i=1

### A

^{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

^{τ −b}

_{σ}

### =

r

### Y

j=1 r

### Y

i=1

### A

^{(σ}

^{i}

^{−a}

^{i}

^{)a}

t1

1 ...a^{tj−1}_{j−1} σ_{j+1}^{tj+1}...σ^{tr}_{r}

σtj j −atj

j

σj −aj a^{u1}_{1} ...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

^{(σ}

^{j}

^{−a}

^{j}

^{)a}

u1

1 ...a^{ui−1}_{i−1} σ^{ui+1}_{i+1} ...σ_{r}^{ur}^{σ}

ui i −aui

i

σi−ai a^{t1}_{1} ...a^{tj−1}_{j−1} σ_{j+1}^{tj+1}...σ^{tr}_{r}

σtj j −atj

j σj −aj

i

### = A

^{σ−a}

_{τ}

### . Thus (16) holds.

### By (12),

### (19) β

^{τ}

### = ζ

_{n}

^{v}

### β

^{b}

### A

τ### .

### By (8),

### (20) ζ

_{n}

^{σ}

### = ζ

_{n}

^{σ}

### = ζ

^{a}

^{t1}

^{1}

^{...a}

tr

n 1

### = ζ

_{n}

^{a}

### . Similarly,

### (21) ζ

_{n}

^{τ}

### = ζ

_{n}

^{b}

### .

### By (12) and (19)–(21),

### (22) β

^{σ τ}

### = ζ

_{n}

^{ub+va}

### β

^{ab}

### A

^{a}

_{τ}

### A

^{r}

_{σ}

### , (23) β

^{τ σ}

### = ζ

_{n}

^{ub+va}

### β

^{ab}

### A

^{b}

_{σ}

### A

^{σ}

_{τ}

### . By (16), A

^{a}

_{τ}

### A

^{τ}

_{σ}

### = A

^{b}

_{σ}

### A

^{σ}

_{τ}

### . By (22) and (23),

### (24) β

^{σ τ}

### = β

^{τ σ}

### .

### 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

^{1−a}

^{j}

### = A

^{n}

_{j}

### and A

^{1−a}

_{j}

^{i}

### = A

^{1−a}

_{i}

^{j}

### (A

i### , A

j### ∈ F ) ⇔ α

^{w}

^{n}

### = γ

^{n}

### (γ ∈ F ).

### By Galois theory w

n### = (1 − a

1### , . . . , 1 − a

r### , n). Hence α

^{1−a}

^{j}

### = A

^{n}

_{j}

### ⇔ α

^{w}

^{n}

### = γ

^{n}

### . It is enough to prove that α

^{1−a}

^{j}

### = A

^{n}

_{j}

### ⇒ A

^{1−a}

_{j}

^{i}

### = A

^{1−a}

_{i}

^{j}

### . Assume that α

^{1−a}

^{j}

### = A

^{n}

_{j}

### . Then

### α

^{1−a}

^{j}

### = α

^{w}

^{n}

^{(1−a}

^{j}

^{)/w}

^{n}

### = γ

^{n(1−a}

^{j}

^{)/w}

^{n}

### = A

^{n}

_{j}

### . Hence A

j### = ζ

w^{x}

^{j}n

### γ

^{(1−a}

^{j}

^{)/w}

^{n}

### and

### A

^{1−a}

_{j}

^{i}

### = γ

^{(1−a}

^{j}

^{)(1−a}

^{i}

^{)/w}

^{n}

### = A

^{1−a}

_{i}

^{j}

### .

### 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

4### , n = 8, α = −4, L = P

4### ( √

^{8}

### −4). Put σ

_{1}

### = (ζ

8### → ζ

_{8}

^{−1}

### ), σ

2### = (ζ

8### → ζ

_{8}

^{5}

### ), a

1### = −1, a

2### = 5. Then (1) is satisfied:

### α

^{σ}

^{1}

^{−a}

^{1}

### = (−4)

^{2}

### = A

^{8}

_{1}

### , α

^{σ}

^{2}

^{−a}

^{2}

### = (−4)

^{−4}

### = A

^{8}

_{2}

### ,

### where A

1### = ζ

_{4}

^{i}

### (1 − ζ

4### ), A

2### = ζ

_{4}

^{j}

### (1 + ζ

4### )

^{−2}

### , A

1### , A

2### ∈ P

_{4}

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

^{2}

### = 16 = γ

^{8}

### with γ ∈ Q, which is impossible.

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

^{σ}

_{1}

^{2}

^{−a}

^{2}

### = −1/4, A

^{σ}

_{2}

^{1}

^{−a}

^{1}

### = 1/4.

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

m### , where P

m### = Q(ζ

m### ) 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)

^{n}

### γ

^{n}

### ,

### where ζ, γ ∈ P

m### , 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)

^{n}

### ∈ P

_{m}

### . 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

^{w}

^{n}

### = γ

^{n}

### , γ ∈ F . Put α = β

^{n}

### , γ = β

_{1}

^{w}

^{n}

### . We have β

1### ∈ F

^{ab}

### (ζ

wn### ∈ F ) and β = ζ

_{nw}

^{a}

_{n}

### β

1### ∈ F

^{ab}

### . Thus the extension F (β, ζ

n### )/F is abelian.

### Necessity. Assume that the Galois group of x

^{n}

### − α is abelian. Put α = β

^{n}

### , G = G(F (β, ζ

n### )/F ), H = G(F (ζ

n### )/F ) and σ

a### = (ζ

n### → ζ

_{n}

^{a}

### ). Let σ, τ ∈ G with σ = σ

a### on F (ζ

n### ). We have β

^{τ}

### = ζ

_{n}

^{j}

### β, β

^{στ}

### = β

^{τ σ}

### = ζ

_{n}

^{aj}

### β

^{σ}

### and β

^{(σ−a)τ}

### = β

^{σ−a}

### = A

a### ∈ F . Hence α

^{1−a}

### = A

^{n}

_{a}

### . By Galois theory w

n### = g.c.d.

_{σ}

_{a}

_{∈H}

### ({1 − a}, n). Hence α

^{w}

^{n}

### = γ

^{n}