A Normal Relative Integral Basis for the Normal Closure of a Pure Cubic Field over Q( 3)

Transcription

1 Int. J. Contemp. Math. Sciences, Vol. 3, 2008, no. 29, A Normal Relative Integral Basis for the Normal Closure of a Pure Cuic Field over Q Blair K. Spearman Department of Mathematics and Statistics University of British Columia Okanagan Kelowna, BC, Canada V1V 1V7 lair.spearman@uc.ca Kenneth S. Williams School of Mathematics and Statistics Carleton University, Ottawa, ON, Canada K1S 5B6 kwilliam@connect.carleton.ca Astract An explicit normal relative integral asis is given for the normal closure of a pure cuic field over Q. This asis is shown to e unique up to permutation and units. 1 Introduction In [1] Carter proved that k = Q is a Hilert-Speiser field of type C 3. This means that if E is a tamely ramified normal extension of k with GalE/k = C 3 then E has a normal relative integral asis over k. Let K e a pure cuic field. Let L = K so that L is the normal closure of K. By Carter s theorem we know that if L/k is tamely ramified then L/k possesses a normal relative integral asis NRIB. We prove that in this case the converse holds, that is, if L/k possesses a NRIB then L/k is tamely ramified. When L/k is tamely ramified we use the relative integral asis RIB given in [2] to give explicitly a NRIB for L/k. Further we show that this NRIB is unique up to permutation and units of k. We prove 3 Theorem 1.1. Let K e a pure cuic field so that K = Q a 2 for some coprime squarefree integers a and. Let L e the normal closure of K. Let

2 1428 B. K. Spearman and K. S. Williams k = Q. i The extension L/k is tamely ramified if and only if 3 a, 3, 9 a ii A NRIB exists for L/k if and only if 1.1 holds. iii If 1.1 holds then { a 2 1/3 + ω a 2 1/3 + ω 2 a 2 1/3 + a 2 1/3, ω 2 a 2 1/3, } ω a 2 1/3 is a NRIB for L/k, where ω = 1, and for m Z the Legendre- 2 Jacoi-Kronecker symol is given y m = m +1, if m 1 mod 3, 1, if m 2 mod 3, 0, if m 0 mod 3. iv The NRIB given in iii is unique up to permutation and units. 2 Proof of Theorem 1.1 We egin with a simple lemma. Lemma 2.1. Let Q E F e a tower of fields with F/E normal. Suppose that {θ 1,θ 2,...,θ n } is a normal relative integral asis for F/E. Then θ 1 +θ θ n is a unit in O E, the ring of integers of E. Proof. Let t = θ 1 + θ θ n O F. As θ 1,θ 2,...,θ n are conjugates over E, we have t O E. Then 1= 1 t θ t θ t θ n.

3 Normal relative integral asis 1429 But {θ 1,θ 2,...,θ n } is a relative integral asis for F/E so 1 t O E. Hence t is a unit of O E. We are now ready to prove Theorem 1.1. Proof of Theorem 1.1. i By [2, eq. 2.6, p. 1624] we have { a dl/k = 2 2, if 3 a, 3, 9 a 2 2, 9a 2 2, otherwise. If 3 a, 3, 9 a 2 2, is not ramified in L/k so that L/k is a tamely ramified extension. Otherwise, as =P 3 for some prime ideal P, L/k is wildly ramified. ii, iii We egin with the case 3 a, 3. In this case the integers of L are of the form [2, Tale 3.1i, p. 1624] α + βa 2 1/3 + γ a2 1/3, 2.1 where α, β, γ O k. Suppose that {θ 1,θ 2,θ 3 } is a NRIB for L/k. Then we see from 2.1 that θ 1 + θ 2 + θ 3 =3α. By Lemma 2.1, 3α is a unit of O k. This is impossile. Hence L/k does not possess a NRIB. The cases 3 a, 3 and 3 a, 3, 9 a 2 2 follow in exactly the same way using [2, Tale 3.1iiiii, p. 1624]. Again L/k does not possess a NRIB in oth cases. In the remaining case 3 a, 3, 9 a 2 2, we claim that {r 1,r 2,r 3 } is a NRIB for L/k, where r 1 = 1 a 2 1/3 + a 2 1/3, ωa 2 1/3 + ω 2 a 2 1/3, r 2 = 1 3 r 3 = 1 3 a a ω 2 a 2 1/3 + ωa 2 1/3. It is clear from [2, Tale 3.1, p. 1624] that each r i i {1, 2, 3} is an integer of L. Further a simple calculation shows that det R 2 = a 2 2 = dl/k,

4 1430 B. K. Spearman and K. S. Williams y [2, eq. 2.6, p. 1624], where R = r 1 r 2 r 3 r 2 r 3 r 1 r 3 r 1 r 2. Hence {r 1,r 2,r 3 } is a NRIB for L/k. iv Suppose that {s 1,s 2,s 3 } is another NRIB for L/k, where 3 a, 3, 9 a 2 2. Then there exist A, B, C O k such that s 1 = Ar 1 + Br 2 + Cr 3, s 2 = Cr 1 + Ar 2 + Br 3, s 3 = Br 1 + Cr 2 + Ar 3. Let Then S = s 1 s 2 s 3 s 2 s 3 s 1 s 3 s 1 s 2. det S 2 =A + B + C 2 A 2 + B 2 + C 2 AB BC CA 2 det R 2. Hence A + B + C 2 A 2 + B 2 + C 2 AB BC CA 2 is a unit of O k. As A + B + C O k and A 2 + B 2 + C 2 AB BC CA O k, each of A + B + C and A 2 + B 2 + C 2 AB BC CA is a unit of O k. But the units of O k are {±1, ±ω, ±ω 2 } so that there exist m, n Z such that and Then A + B + C = ±ω m 2.2 A 2 + B 2 + C 2 AB BC CA = ±ω n. A + B + C 2 3AB + BC + CA=±ω n so that AB + BC + CA = 1 ω 2m ω n. 3 As AB + BC + CA O k we must have 1 3 ω2m 1 3 ωn O k.

5 Normal relative integral asis 1431 But {1,ω} is an integral asis for k so 2m n mod nd the minus sign holds. Hence Then so AB + BC + CA = AB +A + B±ω m A + B = 0 A 2 +B ω m A + BB ω m = Hence the quadratic polynomial x 2 +B ω m x + BB ω m O k [x] has a root A in O k. Thus its discriminant must e a square in O k, that is B ω m 2 4BB ω m =H 2 for some H O k, that is 3B ω m + H 3B ω m H =4ω 2m. As O k is a unique factorization domain and 2 is a prime in O k, we have 3B ω m + H = ±2ω f, 3B ω m H = ±2ω 2m f, for some f Z. Then 3B ω m = ± ω f + ω 2m f so 3B = ±ω f ω 2m 2f ± ω m f. As 1 ± ω r + ω 2r 0 mod 3 in O k if and only if the plus sign holds, we see that the plus sign holds in 1 + ω 2m 2f ± ω m f.thus 3B = ±3ω f or 0, that is B is a unit of O k or 0. From 2.4 and then 2.nd 2.2, we deduce that exactly one of A, B, C is a unit and the others are 0. This proves that s 1,s 2,s 3 is a unit multiple of a permutation of r 1,r 2,r 3. 3 Acknowledgement The research of oth authors was supported y grants from the Natural Sciences and Engineering Research Council of Canada.

6 1432 B. K. Spearman and K. S. Williams References [1] J. E. Carter, Normal integral ases in quadratic and cyclic cuic extensions of quadratic fields, Archiv der Mathematik, vol. 81, pp , [2] B. K. Spearman and K. S. Williams, A relative integral asis over Q for the normal closure of a pure cuic field, International Journal of Mathematics and Mathematical Sciences, vol. 2003, pp , Received: April 10, 2008