EXPLICIT KUMMER GENERATORS FOR CYCLOTOMIC EXTENSIONS

A BSTRACT . If p is a prime number congruent to 1 modulo 3 , then we explicitly describe an element of the cyclotomic ﬁeld Q ( ζ 3 ) whose third root generates the cubic subextension of Q ( ζ 3 p ) / Q ( ζ 3 ) . Similarly, if p is a prime number congruent to 1 modulo 4 , then we explicitly describe an element of the cyclotomic ﬁeld Q ( ζ 4 ) whose fourth root generates the quartic cyclic subextension of Q ( ζ 4 p ) / Q ( ζ 4 ) . For further number ﬁelds we express generators of Kummer extensions inside cyclotomic ﬁelds in terms of Gauss sums.


INTRODUCTION
Consider the cyclotomic extensions of the form Q(ζ 3p )/Q(ζ 3 ), where p is a prime number congruent to 1 modulo 3. Since the base field contains the third roots of unity, the subextension of Q(ζ 3p )/Q(ζ 3 ) is a Kummer extension, generated by the third root of some element of Q(ζ 3 ). Similarly, the cyclotomic extensions Q(ζ 4p )/Q(ζ 4 ), where p is a prime number congruent to 1 modulo 4, have a quartic cyclic subextension which is a Kummer extension, generated by the fourth root of some element of Q(ζ 4 ). We describe generators of those Kummer extensions: Theorem 1. Let p be a prime number such that p ≡ 1 (mod 3).
Theorem 2. Let p be a prime number such that p ≡ 1 (mod 4).
We also have: Theorem 3. Let p be a prime number such that p ≡ 1 (mod 4). The quadratic extension of Q( √ p) inside Q(ζ p ) is generated by the square roots of ω p := u √ p, where u is a fundamental unit of Q( √ p) such that for any embedding σ : Q( √ p) → R the sign of σ(u √ p) is positive if and only if p ≡ 1 (mod 8). Moreover, we have where γ p is as in Theorem 2.
Example 4. The cyclotomic field Q(ζ 5 ) is a quartic cyclic extension of Q, which is generated by ζ 5 . By considering the sine and cosine of 2π 5 we get the expression is the quadratic extension generated by the square root of − 5+ √ 5 8 . Up to a square this element is − √ 5+1 2 · √ 5 and we may easily check with [6] that the first factor is a fundamental unit.
In the last section we consider further number fields and present generators for Kummer extensions inside cyclotomic fields by making use of Gauss sums. Notice that the results in this note may be in part well-known to the experts however they are useful for applications and hence we have provided a comprehensive reference. We thank the referee for suggesting a short proof of Lemma 5.
Notation. For an integer n 1 we denote by ζ n a primitive n-th root of unity, and by µ n the multiplicative group generated by ζ n . Recall that a prime number p ≡ 1 (mod n) decomposes fully (i.e. it splits completely) in Q(ζ n ). If K is a number field, then after choosing an embedding of K in C we write x for the complex conjugate of an element x ∈ K. Given a prime p of K, we denote by K p the completion of K at p with respect to the p-adic valuation of K. If L is a local field of characteristic 0, we denote by v L its valuation and by O L its valuation ring. Finally, if we write a congruence modulo an algebraic integer in K, then we mean the congruence modulo the integral ideal generated by that element in the ring of integers.
Since 1 is the only root of unity in µ 6 which is congruent to 1 modulo 3, we deduce that π · π = p. As for the unicity, we could replace π by π, but to mantain the property π ≡ 1 (mod 3) we cannot multiply π by a root of unity in O distinct from 1, see Lemma 5. Lemma 6. Let K be the unique unramified extension of Q 3 (ζ 3 ) of degree 3. If α ∈ K is a unit such that v K (1 − α) 4, then α is a third power.
and denote by U (n) the image of U (n) in (K ) × /(K ) ×3 . We have to show that U (4) is trivial. We know that U (5) is trivial by applying Hensel's Lemma to the polynomial x 3 − a with value . Consider the map raising to the third power from where u is the unit such that 3 = uξ 2 . Thus the map is the multiplication by u, which is surjective. We have found that any element of U (4) is an element of U (5) up to third powers, and hence U (4) = U (5) is trivial.
Among the elements uα with u ∈ µ 6 , only α and −α have this ramification property.
. It suffices to see that α acquires a third root over the unique unramified cubic extension of local fields K /K (ξ) . We will consider elements in K as power series in ξ but with coefficients in a fixed residue system of F 27 in the unique unramified extension of Q 3 of degree 3.
Noticing that ζ 2 3 = −1 − ζ 3 and hence ξ 2 = −3ζ 3 , and considering that −ζ 3 ≡ −1 (mod ξ), we can write for some x ∈ F 27 (notice that we can work with the residues in F 27 because two representatives for x differ by an element of (ξ)). The ratio α/(1 + x ξ) 3 , where x is a lift of x, thus lies in U (4) Thus α has a third root in K up to multiplication by some element of U (4) K , but all those elements are third powers in K by Lemma 6.
As for the second assertion, notice that Proof of Theorem 1 (2). By Kummer theory there is some β p ∈ Q(ζ 3 ) such that 3 β p generates the cubic extension inside Q(ζ 3p ), which is unramified at ξ because (3) = (ξ 2 ). Since this extension is invariant by complex conjugation, we are allowed to replace β p by its complex conjugate. Up to multiplying β p by a cube we may suppose that each prime ideal appearing in the factorization of the fractional ideal (β p ) has exponent either 1 or 2. Such prime ideals ramify in Q(ζ 3p ) by [3, Lemma C.1.7 and its proof]. Since only the ideals over p ramify in the extension Q(ζ 3p )/Q(ζ 3 ), we deduce that where u ∈ µ 6 is a unit and where x, y ∈ {0, 1, 2}. The exponents x and y cannot be both 0 by [3, Lemma C.1.7 and its proof] because p is totally ramified in Q(ζ p )/Q.
Since the extension Q(ζ 3p ) is abelian over Q and invariant under complex conjugation, the fields Q(ζ 3 , 3 β p ) and Q(ζ 3 , 3 β p ) must be equal. By Kummer theory we then have, up to cubes, either β p = β p or β 2 p = β p . Thus in (1) it cannot be that exactly one among x and y is 0.
We can also exclude the cases x = y = 1 and x = y = 2 because then β p = u·p or β p = u·p 2 , which would imply that 3 √ p is contained in a cyclotomic extension of Q. So we have found that {x, y} = {1, 2}, and, up to working with the complex conjugate of β p instead, we have for some u ∈ µ 6 .
We have also found that any element of U (7) is an element of U (8) up to fourth powers, and hence U (7) = U (8) is trivial.
Notice that when y = 0, then α is already a square (by reasoning as in Lemma 8).
For the second assertion consider that if K( 4 √ α) and K( 4 ζ n 4 α) are both unramified at (ξ), then the same holds for K( 4 ζ n 4 ) and hence ζ n 4 = 1. For the first assertion it suffices to see that α acquires a fourth root in K . We will work with a fixed representative system of the residue field taken in the unique unramified extension of Q 2 , so that the difference of any two lifts lies in (2) = (ξ 2 ). Notice that 2 = ξ 2 − ξ 3 hence ±3 ≡ 1 (mod ξ 2 ), so we have: but x 2 + x = y with y ∈ {0, 1} is solvable in F 4 and hence also x 4 + x 2 = (x 2 + x) 2 = y. We then have α ≡ (1 + xξ) 4 (mod ξ 6 ) with x ∈ O K /(ξ) F 4 . So α/(1 + xξ) 4 lies in U (6) This means that α has a fourth root in K up to multiplication by an element of U (6) K , which is of the form 1 + wξ 6 (mod ξ 7 ), for some w ∈ K . We can write (1 + zξ 2 ) 4 ≡ 1 + (z + z 2 )ξ 6 (mod ξ 7 ) , but z + z 2 = w with w ∈ F 4 is solvable in F 16 , hence α has a fourth root in K up to an element of U (7) K consists of fourth powers by Lemma 8. Lemma 10. Let π ∈ Q(ζ 4 ) be such that π · π = p holds. The element π can be chosen, by multiplying it with a root of unity, in such a way that π · π 3 ≡ 1 + (x + xξ)ξ 4 (mod ξ 6 ) holds for some x ∈ {0, 1}. In particular, we have π · π 3 ≡ 1 (mod 4). The element π is then unique up to sign and up to complex conjugation.
Proof. Write pO = (π) · (π). Let γ p ∈ Q(ζ 4 ) × be such that 4 √ γ p generates the cyclic quartic subextension of Q(ζ 4p ) over Q(ζ 4 ). Up to multiplying this element by a fourth power in Q(ζ 4 ), we may suppose that each prime ideal appearing in the factorization of the fractional ideal (γ p ) has exponent 1, 2 or 3. Such ideals must ramify in Q(ζ 4p ) by [3, Lemma C.1.7 and its proof]. Since only the ideals over p ramify in the extension Q(ζ 4p )/Q(ζ 4 ), we deduce that (γ p ) = (π) x · (π) y with x, y ∈ {0, 1, 2, 3}. We cannot have x = y = 0 because the Kummer extension needs to be ramified at p, as p is totally ramified in Q(ζ p ). We cannot have x = y = 1 or x = y = 3 because otherwise 4 √ p or 4 p 3 would be contained in a cyclotomic extension of Q. Since the quadratic subextension of Q(ζ 4p )/Q(ζ 4 ) is Q(ζ 4 , √ p), we must have that γ p equals p times a square in Q(ζ 4 ), so that {x, y} = {1, 3}. Thus we have Notice that γ p and its third power generate the same Kummer extension. Thus, up to replacing γ p with its third power, we may suppose that (γ p ) = (π) · (π) 3 and hence that for some root of unity ζ ∈ µ 4 . By Lemma 10 and Proposition 9 there is exactly one choice of ζ such that the field generated by 4 √ γ p is unramified at 2 (as it is the case for the considered subextension of Q(ζ 4p )/Q(ζ 4 )), and this is exactly the choice that makes ζ · π · π 3 congruent to 1 modulo 4.
Proof of Theorem 2. The result is a consequence of Lemma 10 and Proposition 11.
Proof of Theorem 3. Let ω p ∈ Q( √ p) be such that √ ω p generates the quadratic extension of Since p is the only prime number ramifying in Q(ζ p ) and it is totally ramified, by [3, Lemma C.1.7 and its proof] the prime ( √ p) above p is the only prime ideal appearing in the factorization of the fractional ideal (ω p ) with odd valuation. So we can write (ω p ) = ( √ p)I 2 for some ideal I, and hence I 2 is principal. Since the class number of Q( √ p) is odd (see for example [5,Example 2.9]), I is principal. By working up to squares of elements in K, we may then suppose that (ω p ) = ( √ p). So we can write ω p = u √ p where u is a unit.
Since 4 √ p is not contained in a cyclotomic extension, the element u cannot be a root of unity.
Up to squares of a fundamental unit, we may then suppose that u is a fundamental unit.
Considering the tower of quadratic extensions in Q(ζ p ) we deduce that only the largest field is not fixed by complex conjugation, since this automorphism generates a subgroup of order 2.
The quadratic extension of Q( √ p) is then totally imaginary if and only if it is that largest field, which means that p ≡ 1 (mod 8). We have a totally imaginary field by adding if and only if the sign of σ(u √ p) is negative for any embedding σ : Q( √ p) → R.

GENERATORS FOR KUMMER EXTENSIONS IN TERMS OF GAUSS SUMS
Let q and p be prime numbers such that p ≡ 1 mod q n holds for some n 1. We call K = Q(ζ q n ) and L = Q(ζ q n p ), and denote by K the cyclic subextension of L/K of degree q n . We call O K the ring of integers of K: notice that, if ℘ ⊂ O K is a prime ideal over p, then we have a canonical isomorphism O K /℘ ∼ = Z/pZ. Theorem 12. We keep the above notation.
(3) If ℘ corresponds to χ, then we have the prime factorization where σ x denotes the image of x under the isomorphism (Z/q n Z) * ∼ = Gal(K/Q).