The Galois module of a twisted element in the p m -th cyclotomic field

LXI.4 (1992)

The Galois module of a twisted element in the p ^m -th cyclotomic field

by

Kurt Girstmair (Innsbruck)

1. Introduction and results. For a natural number n let ξ n ∈ C denote the primitive nth root of unity: ξ n = e ^2πi/n . Then Q n = Q[ξ n ] is the nth cyclotomic field and G n = Gal(Q n /Q) is its Galois group over Q. The field Q n is a module over the group ring Q[G n ] of G n , and, by the normal basis theorem, it is isomorphic to Q[G n ] itself. In what follows let n = p ^m , where p is a prime number and m a nonnegative integer. Let q = p ^e , e ≥ 1, be another power of p. For x ∈ Q n we consider

x nq = ξ nq x ∈ Q nq .

We call x nq the twisted element of x, because it arises from x by means of the rotation of the plane C through 2π/(nq). It is almost obvious that the Galois module Q[G nq ]x nq is contained in Q[G nq ]ξ nq (cf. proof of Theorem 1 below). Suppose that d is the number of divisors of p − 1. Then Q[G nq ]ξ nq

is the direct sum of d simple Q[G nq ]-submodules if p ≥ 3, and it is the direct sum of two simple submodules for p = 2 and nq ≥ 4. Hence Q[G ^nq ]ξ nq has 2 ^d different submodules for p ≥ 3, and four different submodules for p = 2, nq ≥ 4. We consider the case q 6= 2 first. Here Q[G nq ]x nq is always one of the two trivial submodules of Q[G nq ]ξ nq ; indeed, we show

Theorem 1. Let p be a prime, n = p ^m , q = p ^e , m ≥ 0, e ≥ 1. In addition, if p = 2, let e ≥ 2. For each element x ∈ Q n , x 6= 0,

Q[G nq ]x nq = Q[G nq ]ξ nq .

For q = 2 the result is different: Let M 1 , M 2 be the simple submodules of Q[G 2n ]ξ 2n , n = 2 ^m ≥ 4. For k = 1, 2, let

V k = {x ∈ Q n ; Q[G 2n ]x 2n = M k } ∪ {0} .

Theorem 2. With the above notations, V k is a Q-subspace of Q n of dimension dim V k = n/4, k = 1, 2. Moreover ,

Q n = V 1 ⊕ V ₂ .

2. Proofs. We adopt the above notations. Most of the representation theory of Q[G n ] we use in the sequel can be found in [1], Section 1, and in [4].

Consider the map

Z \ pZ → G n , k 7→ σ k ,

where σ k (ξ n ) = ξ ^k _n . This map is surjective and multiplicative. It is used in order to identify the character group

X n = {χ : G n → C ^× ; χ a group homomorphism}

of G n with the group of Dirichlet characters modulo n. Indeed, put χ(k) = n χ(σ k ) if p - k,

0 otherwise.

For an element α = P{a _σ σ ; σ ∈ G n } in Q[G n ] let χ(α) = P a _σ χ(σ)

∈ C. Let Y ⊆ X n be a conjugacy class of characters, i.e., all characters χ, χ ⁰ in Y generate the same group hχi = hχ ⁰ i. Then the group ring Q[G n ] splits into the simple submodules

Q[G n ] Y = {α ∈ Q[G n ] ; χ(α) 6= 0 only if χ ∈ Y } , of Q-dimension dim Q[G n ] Y = |Y |.

Next fix χ ∈ X n . According to [3], [1] there is a map y(χ|−) : Q n → C with the following properties:

(i) y(χ|−) is χ-linear, i.e., for all α ∈ Q[G n ] and all x ∈ Q n , y(χ|αx) = χ(α)y(χ|x).

(ii) Let Y be the conjugacy class of χ. Then

Q n,Y = {x ∈ Q n ; y(χ ⁰ |x) 6= 0 only if χ ⁰ ∈ Y }

is the uniquely determined Q[G n ]-submodule of Q n that is isomorphic to Q[G n ] Y .

The map y(χ|−) is uniquely determined by χ up to factors in C ^× ; this means that the maps c · y(χ|−), c ∈ C ^× , are the only ones having properties (i), (ii), too.

Consider the numbers y(χ|ξ n ) ∈ C, χ ∈ X n . Then y(χ|ξ n ) 6= 0 iff χ is a primitive character modulo n (cf. [4]). Hence

Q[G n ]ξ n = M Q n,Y ,

where Y runs through the conjugacy classes of primitive characters mod- ulo n. We obtain

Lemma 1. Let n be as above, and let x ∈ Q ⁿ . Then

(1) Q[G n ]x ⊆ Q[G n ]ξ n iff y(χ|x) = 0 for all imprimitive characters

χ mod n;

(2) Q[G n ]x = Q[G n ]ξ n iff {χ∈X n ; y(χ|x) 6= 0} = {χ∈X n ; χ primitive}.

For the proof of Theorem 1 we need two additional lemmas.

Lemma 2. Let p be a prime, n = p ^m , q = p ^e , m ≥ 0, e ≥ 1. For p = 2 let e ≥ 2.

(1) For each k ∈ {1, . . . , n} there is a uniquely determined number j ∈ {1, . . . , n} such that 1 + qk ≡ (1 + q) ^j mod nq.

(2) The map {1, . . . , n} → {1, . . . , n} : k 7→ j is bijective.

(3) Let k, k ⁰ ∈ {1, . . . , n} and let j, j ⁰ be their images under the above map. Let 0 ≤ l ≤ m. Then k ≡ k ⁰ mod p ^l iff j ≡ j ⁰ mod p ^l . Furthermore, k ≡ 0 mod p ^l iff j ≡ 0 mod p ^l .

The proof of Lemma 2 consists, essentially, in the observation that the subgroups {1 + qk ; k = 1, . . . , n} and h 1 + q i of (Z/nqZ) ^× coincide and have order n (cf. [2], p. 72 ff.). Note, however, that the map Z/nZ → Z/nZ : k 7→ j is not a group homomorphism in general.

Lemma 3. Let p be a prime, m ≥ 2, and n = p ^m . Let β = P{b j σ j ; j ∈ {1, . . . , n}, p - j} ∈ Q[G n ] be such that βξ n = 0. Then b j = b j

for all j, j ⁰ with j ≡ j ⁰ mod n/p.

P r o o f. Put M = {α ∈ Q[G n ] ; αξ n = 0}. According to Lemma 1, the element β = P b j σ j is in M iff χ(β) = 0 for each primitive character χ mod n. From this we conclude that

dim M = |{χ ∈ X n ; χ is imprimitive}| = ϕ(n/p) ,

where ϕ is Euler’s function. Observe now that Z ^p − ξ _n ^p = Z ^p − ξ _n/p is the minimal polynomial of ξ n over Q n/p . This means that the trace

T (ξ ^j _n ) = X

{σ _j

ξ n ; j ⁰ ≡ j mod n/p}

vanishes for each j, j ∈ {1, . . . , n}, p - j. Hence the elements α j = X

{σ j

; j ⁰ ∈ {1, . . . , n}, j ⁰ ≡ j mod n/p} ,

j ∈ {1, . . . , n/p}, p - j, are in M . It is obvious that the α j are Q-linearly independent. So, for dimensional reasons, they form a Q-basis of M . When the element β ∈ M is expressed in terms of this basis, the assertion follows.

P r o o f o f T h e o r e m 1. Let n, q be as in Theorem 1. Let x =

n

X

k=1

a k ξ _n ^k ∈ Q n .

Then x nq = P{a k ξ _nq ^1+qk ; k = 1, . . . , n} is a linear combination of primitive

nqth roots of unity. Therefore x nq ∈ Q[G nq ]ξ nq and Q[G nq ]x nq ⊆ Q[G nq ]ξ nq .

Next suppose that χ is primitive mod nq. Since y(χ|−) is determined up to factors in C ^× only, we may assume that y(χ|ξ nq ) = 1. Suppose that y(χ|x nq ) = 0. We show that x = 0, which proves the theorem, by Lemma 1.

We use induction with respect to the exponent m. Let m = 0, i.e., n = 1 and x ∈ Q. Then 0 = y(χ|x q ) = y(χ|xξ q ) = x · y(χ|xξ q ) = x .

Now let m > 0, which means p | n, and let n ⁰ = n/p. The induction hypothesis is as follows: Let q ⁰ = p ^e

, e ⁰ ≥ 1 (e ⁰ ≥ 2 for p = 2), x ⁰ ∈ Q n

, and χ ⁰ a primitive character mod n ⁰ q ⁰ ; if y(χ ⁰ |x ⁰ _n

q

) = 0 then x ⁰ = 0.

Take x as above. Then y(χ|x nq ) =

n

X

k=1

a k χ(1 + qk) = 0 .

For each j ∈ {1, . . . , n} we put b j = a k , where k is the uniquely determined number in {1, . . . , n} with (1 + q) ^j ≡ 1 + qk mod nq (Lemma 2). Observe that η = χ(1 + q) is a primitive nth root of unity (use [2], p. 212, and Lemma 2). Now

y(χ|x nq ) =

n

X

j=1

b j η ^j = 0 .

Consider the case n = p first. Because 1 + Z + . . . + Z ⁿ⁻¹ is the minimal polynomial of η over Q, all the coefficients b ^j are equal. Therefore a 1 = . . . = a n and x = 0. Suppose now that n = p ^m , m ≥ 2. Put

x ⁰ = X

{a _k ξ _n ^k ; k ∈ {1, . . . , n}, p | k}

and x ⁰⁰ = x − x ⁰ . The “trace argument” in the proof of Lemma 3 shows that T (η ^j ) = 0, for all j ∈ {1, . . . , n}, p - j (T is the trace of Q n over Q n

). Therefore y(χ|x ⁰⁰ _nq ) = 0, which implies that y(χ|x ⁰ _nq ) = y(χ|x nq ) − y(χ|x ⁰⁰ _nq ) = 0. However, x ⁰ _nq is the same as x ⁰ _n

q

, with n ⁰ = n/p, q ⁰ = qp.

The induction hypothesis yields x ⁰ = 0. Let β = X

{b _j σ j ; j ∈ {1, . . . , n}, p - j} ∈ Q[G n ] .

Then βη = y(χ|x ⁰⁰ _nq ) = 0. By Lemma 3, the coefficients of β fulfill: b j = b j

for all j, j ⁰ ∈ {1, . . . , n}, p - j, j ⁰ , j ≡ j ⁰ mod n ⁰ . Then a k = a k

for all k, k ⁰ ∈ {1, . . . , n}, p - k, k ⁰ , k ≡ k ⁰ mod n ⁰ . We obtain

x ⁰⁰ =

n

X

k=1 p - k

a k n

X

k

=1 k

≡k mod n

ξ _n ^k

=

n

X

k=1 p - k

a k T (ξ ^k _n ) .

But the traces in the last sum vanish, whence x ⁰⁰ = 0 and x = x ⁰ + x ⁰⁰ = 0 follows.

P r o o f o f T h e o r e m 2. Let n = 2 ^m , m ≥ 2. There are exactly two

conjugacy classes of primitive characters mod 2n, viz. the set of even and

the set of odd primitive characters (cf. [2], p. 212). Choose an arbitrary odd character χ 1 and an arbitrary even character χ 2 , both of them primitive.

Then

M 1 = {z ∈ Q[G 2n ]ξ 2n ; y(χ 1 |z) = 0} = Q[G 2n ](ξ 2n + ξ _2n ⁻¹ ) , M 2 = {z ∈ Q[G 2n ]ξ 2n ; y(χ 2 |z) = 0} = Q[G 2n ](ξ 2n − ξ _2n ⁻¹ ) ,

are the simple submodules of Q[G 2n ]ξ 2n . For each σ ∈ G 2n , χ k (σ) is an (n/2)th root of unity, k = 1, 2, which shows that the Q-linear map

g k : Q n → Q n/2 , x 7→ g k (x) = y(χ k |ξ _2n x) , is well defined. Let V k denote the kernel of g k , k = 1, 2. Then

V k = {x ∈ Q ⁿ ; ξ 2n x ∈ M k } = {x ∈ Q ⁿ ; Q[G ²ⁿ ]x 2n = M k } ∪ {0} , since M k is simple. Moreover,

dim V k = ϕ(n) − dim g k (Q n ) ≥ ϕ(n) − ϕ(n/2) = n/4, k = 1, 2 . But V 1 ∩ V ₂ = {0}, so dim(V 1 ⊕ V ₂ ) ≥ n/2 = dim Q n . Thus V 1 ⊕ V ₂ = Q n

and dim V k = n/4, k = 1, 2.

Example. Let n = 2 ^m and m ≥ 2 be as above. Consider the elements x ⁺ = 1 + ξ _n ⁻¹ , x ⁻ = 1 − ξ _n ⁻¹

in Q n . Then ξ 2n x ⁺ = ξ 2n + ξ _2n ⁻¹ , Q[G 2n ]x ⁺ _2n = M 1 , ξ 2n x ⁻ = ξ 2n − ξ ⁻¹ _2n , Q[G 2n ]x ⁻ _2n = M 2 . Furthermore, Q[G n ]x ⁺ = Q[G n ]x ⁻ = Q ⊕ Q[G n ]ξ n . Hence V 1 and V 2 cannot be Q[G ⁿ ]-modules. Indeed, if they were, Q ⊕ Q[G n ]ξ n ⊆ V ₁ ∩ V ₂ would follow, which is impossible.

R e m a r k. Clearly the results of this note do not depend on the partic- ular choice ξ n = e ^2πi/n of a primitive nth root of unity. This choice was just made for reasons of convenience, e.g., for the sake of the simple relation ξ _nq ^q = ξ n .

References

[1] K. G i r s t m a i r, Character coordinates and annihilators of cyclotomic numbers, Ma- nuscripta Math. 59 (1987), 375–389.

[2] H. H a s s e, Vorlesungen ¨ uber Zahlentheorie, Springer, Berlin 1950.

[3] H. W. L e o p o l d t, ¨ Uber die Hauptordnung der ganzen Elemente eines abelschen Zahlk¨ orpers, J. Reine Angew. Math. 201 (1959), 119–149.

[4] G. L e t t l, The ring of integers of an abelian number field , ibid. 404 (1990), 162–170.

INSTITUT F ¨ UR MATHEMATIK UNIVERSIT ¨ AT INNSBRUCK TECHNIKERSTR. 25/7

A-6020 INNSBRUCK, ¨ OSTERREICH

Received on 8.10.1991 (2181)