A 4 -SEXTIC FIELDS WITH A POWER BASIS. Daniel Eloff, Blair K. Spearman, and Kenneth S. Williams
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1 A 4 -SEXTIC FIELDS WITH A POWER BASIS Daniel Eloff, Blair K. Spearman, and Kenneth S. Williams Abstract. An infinite family of monogenic sextic fields with Galois group A 4 is exhibited. 1. Introduction. Let K be an algebraic number field of degree n. Let O K denote the ring of integers of K. The field K is said to possess a power basis if there exists an element θ O K such that O K = Z+Zθ+ +Zθ n 1. A field having a power basis is called monogenic. For an extended history of monogenic number fields the reader should consult [3]. For recent work on this topic see [4,, 6, 8]. In this paper we exhibit infinitely many monogenic sextic fields with Galois group A 4. We prove the following result. Theorem. Let d Z. Set f d (x) := x 6 + (2d + 2)x 4 + (2d 1)x 2 1 Z[x]. (1.1) Let θ d C be a root of f d (x). Set K d = Q(θ d ). Then [K d : Q] = 6, Gal(f d ) A 4, and the fields K d (d Z) are distinct. Moreover K d is monogenic with ring of integers Z[θ d ] for infinitely many values of d. Remark. We prove that K d is monogenic whenever 4d 2 + 2d + 7 is squarefree, which occurs for infinitely many values of d by a result of Nagel [7]. We remark that Anai and Kondo [1] have stated without proof that Gal(x 6 + (2a + 2)x 4 + (2a 1)x 2 1) A 4 for all a Q for which x 6 + (2a + 2)x 4 + (2a 1)x 2 1 is irreducible in Q[x]. We show in Section 2 that f d (x) is, in fact, irreducible in Z[x] for all d Z (Lemma 2.2) and that Gal(f d ) A 4 for all d Z (Lemma 2.4). In Section 3 we complete the proof of the theorem. 2. Irreducibility and Galois Group of f d. Throughout this section d Z, f d (x) is given by (1.1), θ d C is a root of f d (x), and K d = Q(θ d ). Clearly [K d : Q] 6. We denote the splitting field of f d (x) by L d. 1
2 Lemma 2.1. (a) K d contains a unique cubic subfield C d. (b) C d is cyclic over Q. Proof. (a) Let g d (x) := x 3 + (2d + 2)x 2 + (2d 1)x 1 Z[x] so that g d (x 2 ) = f d (x). Suppose that g d (x) is reducible in Q[x]. As deg(g d (x)) = 3, g d (x) has a rational root r. As g d (x) Z[x], we have r Z. Then r g d (0)(= 1) so r = ±1. As g d ( 1) = 1 we have r 1 so r = 1. Hence, 4d + 1 = g d (1) = 0, contradicting d Z. Thus, g d (x) is irreducible in Q[x] so that Q(θd 2) is a cubic subfield of K d. Suppose C d = Q(θd 2) and F are two distinct cubic subfields of K d. Then the compositum field C d F is a subfield of K d of degree 9 over Q. Hence, 9 [K d : Q], contradicting [K d : Q] 6. We have shown that C d = Q(θd 2) is the unique cubic subfield of K d. (b) As disc(g d (x)) = (4d 2 + 2d + 7) 2, C d is a cyclic field. Lemma 2.2. f d (x) is irreducible in Q[x]. Proof. Suppose that f d (x) is reducible over Q so that [K d : Q] < 6. As 3 = [C d : Q] [K d : Q] we have [K d : Q] = 3. Thus, θ d is a root of an irreducible cubic in Z[x], say, h(x) = x 3 + ax 2 + bx + c. Clearly θ d is a root of h( x) and h( x) h(x). As θ d and θ d are roots of f d (x) it follows that f d (x) = h(x)h( x) = x 6 + ( a 2 + 2b)x 4 + (b 2 2ac)x 2 c 2. Equating constant terms of f d (x), we deduce that c = ±1. Next, equating the coefficients of x 4 and x 2, we obtain 2d + 2 = a 2 + 2b, 2d 1 = b 2 2ac. Eliminating d, we have (a c) 2 + (b 1) 2 = 1, 2
3 a contradiction, proving that f d (x) is irreducible in Q[x]. Lemma 2.3. [L d : Q] 12. Proof. Let ±φ 1, ±φ 2, ±φ 3 be the roots of f d (x) with φ 1 = θ d. Then 1 = f d (0) = φ 2 1φ 2 2φ 2 3 so φ 1 φ 2 φ 3 = ±1. Then L d = Q(φ 1, φ 2, φ 3 ) = Q(φ 1, φ 2 ) = C d (φ 1, φ 2 ) as C d = Q(θ 2 d ) = Q(φ2 1 ). Hence, [L d : C d ] 2 2 = 4 and so [L d : Q] = [L d : C d ][C d : Q] 4 3 = 12 by Lemma 2.1(a). Lemma 2.4. Gal(f d ) A 4. Proof. The field K d satisfies [K d : Q] = 6 (Lemma 2.2), K d contains a cubic subfield (Lemma 2.1(a)), and disc(f d ) = 2 6 (4d 2 + 2d + 7) 4 Z 2. Hence, from Cohen [2], we see that Gal(f d ) A 4 or S 4. By Lemma 2.3 we deduce that Gal(f d ) A 4. Lemma 2.. C d is the only subfield ( Q, K d ) of K d. Proof. This follows immediately from Lemmas 2.1 and Proof of Theorem. We begin by recalling the following result [8]. Lemma 3.1. Let f(x) = x n + a n 1 x n a 1 x + a 0 Z[x] be irreducible. Suppose that θ C is a root of f(x) and K = Q(θ). If p is a prime number such that p a 0 and p a 1, then the ideal < p > ramifies in K. We now prove our theorem. It is assumed throughout this section that d Z is such that 4d 2 + 2d + 7 is squarefree. If d 2 (mod 3), say d = 3m + 2 (m Z), then 4d 2 + 2d + 7 = 36m 2 + 4m (mod 9), a contradiction. Hence, d 2 (mod 3). As d Z we have (4d + 1) 2 1. Therefore, 4d 2 + 2d + 7 = (4d + 1) = 7. Also, (4d 2 + 2d + 7, 4d + 1) = (d + 7, 4d + 1) = (d + 7, 27) = 1 (3.1) 3
4 as d+7 0 (mod 3). Let p be a prime dividing 4d 2 +2d+7. As 4d 2 +2d+7 is assumed to be squarefree we have p 4d 2 + 2d + 7. Then from (3.1), we deduce that p 4d + 1. Let ( ) x 2d 2 k d (x) = 3 3 g d = x 3 3(4d 2 +2d+7)x+(4d+1)(4d 2 +2d+7). 3 Recall from the proof of Lemma 2.1(a) that g d (x) is irreducible in Q[x]. Hence, k d (x) is irreducible in Q[x]. A root of k d (x) is λ = 3θd 2 + (2d + 2) and Q(λ) = Q(3θd 2 + (2d + 2)) = Q(θ2 d ) = C d. Hence, by Lemma 3.1, the ideal < p > ramifies in C d so by Dedekind s Theorem p d(c d ). Thus, we have shown that every prime dividing 4d 2 + 2d + 7 divides d(c d ). As (4d 2 + 2d + 7) 2 = disc(g d ) = m 2 d(c d ) (3.2) for some m N, we see that every prime p dividing d(c d ) divides 4d 2 + 2d + 7. Thus, 4d 2 + 2d + 7 and d(c d ) are divisible by exactly the same set of primes. As 4d 2 + 2d + 7 is squarefree, we deduce from (3.2) that m = 1 and d(c d ) = (4d 2 + 2d + 7) 2. (3.3) Next, by the conductor-discriminant formula, see for example [2], we deduce using Lemma 2.1(a) and (3.3) that 4d 2 + 2d + 7) 4 d(k d ), say for some t N. Hence, d(k d ) = t(4d 2 + 2d + 7) 4 (3.4) 2 6 (4d 2 + 2d + 7) 4 = disc(f d ) = l 2 d(k d ) = l 2 t(4d 2 + 2d + 7) 4 (3.) for some l N. From (3.) we see that t = u 2 for some u N and lu = 2 3. Thus, u = 1, 2, 2 2 or 2 3 and d(k d ) = 2 α (4d 2 + 2d + 7) 4, α {0, 2, 4, 6}. We wish to show that α = 6. To do this it suffices to prove that none of the 32 elements { a0 + a 1 θ d + a 2 θ 2 d + a 3θ 3 d + a 4θ 4 d + θ d 2 } a 0, a 1, a 2, a 3, a 4 {0, 1} 4
5 is an algebraic integer. We just illustrate this with the element θ d + θ 3 d + θ4 d + θ d 2 as the remaining 31 elements can be treated in a similar manner. Let γ = θ d + θd 3 + θ4 d + θ d. We show first that γ does not belong to a proper subfield of K d. Suppose on the contrary that γ F, where F is a subfield of K d with F K d. Then by Lemma 2., γ C d. As γ = θ d (1 + θd 2 + θ4 d ) + θ4 d and θ2 d, θ4 d C d, we deduce that θ d C d, a contradiction. Hence, the minimal polynomial of γ over Q is of degree 6. Using MAPLE we find that the minimal polynomial of γ over Q is x 6 + a 1 x + a 2 x 4 + a 3 x 3 + a 4 x 2 + a x + a 6, where a 1 = (8d 2 + 8d + 12), a 2 = 32d + 64d d d d + 74, a 3 = (64d + 64d d d d + 76), a 4 = 64d + 208d 3 116d d 208, a = 32d d d d + 144, a 6 = (32d d d d + 104). Clearly, a = (4d4 + d d 2 + 9d + 13) d 3 + d (mod 2). Hence, 2 6 a 6 so that γ/2 is not an integer of K d. We treat the remaining 31 elements in a similar manner obtaining the same conclusion each time. Thus, we have proved that α = 6 and d(k d ) = 2 6 (4d 2 + 2d + 7) 4 = disc(f d ). Hence, { 1, θ d, θ 2 d, θ3 d, θ4 d, θ d} is an integral basis for Kd. As 4d 2 + 2d + 7 = 4d d (d, d 1 Z) = d = d 1,
6 we deduce that the fields K d are distinct. 4. Other Power Bases. If 4d 2 + 2d + 7 is squarefree, we have shown that {1, θ d, θ 2 d, θ3 d, θ4 d, θ d } is a power basis for the ring O K d of integers of K d. As N(θ d ) = 1, { 1, 1, 1 θ d θd 2, 1 θd 3, 1 θd 4, 1 } θd is also a power basis for O Kd, that is {1, φ d, φ 2 d, φ3 d, φ4 d, φ d } is a power basis for O Kd with φ d = 1 θ d = (2d 1)θ d + (2d + 2)θ 3 d + θ d. With d = 0 we carried out a computer search using an index form corresponding to { 1, θ d, θ 2 d, θ3 d, θ4 d, θ d}, and found five more power bases, namely the bases { 1, ψ i, ψ 2 i, ψ3 i, ψ4 i, ψ i } (i = 1, 2, 3, 4, ) with ψ 1 = 2θd 3 + θd, ψ 2 = 2θ d + 2θd 3 + θd, ψ 3 = 2θ d + θd 3 + θ d, ψ 4 = 2θ d + 3θd 3 + θd, ψ = 2θ d + 3θd 2 + 3θd 3 + θd 4 + θd. None of these power bases is an integer translate or a Galois conjugate of θ d. This can easily be checked by finding the minimal polynomials of the above elements and observing that they are distinct, even under integer translation. We do not know if there are any other power bases for this sextic field. Acknowledgement. The second and third authors were supported by grants from the Natural Sciences and Engineering Research Council of Canada. 6
7 Ref erences 1. H. Anai and T. Kondo, A Family of Sextic Polynomials With Galois Group A -the Computation of Splitting Fields and Galois Groups, Studies in the Theory of Computer Algebra and Its Applications, Kyoto, (199), H. Cohen, A Course in Computational Algebraic Number Theory, Springer-Verlag, I. Gaál, Diophantine Equations and Power Integral Bases: New Computational Methods, Birkhäuser, Boston, M. J. Lavallee, B. K. Spearman, K. S. Williams, and Q. Yang, Dihedral Quintic Fields With a Power Basis, Math. J. Okayama Univ., 47 (200), I. Járási, Power Integral Bases in Sextic Fields With a Cubic Subfield, Acta Sci. Math. (Szeged), 69 (2003), L. Miller-Sims and L. Robertson, Power Integral Bases for Real Cyclotomic Fields, Bull. Austral. Math. Soc., 71 (200), T. Nagel, Zur Arithmetik der Polynome, Abh. Math. Sem. Hamburg, 1 (1922), B. K. Spearman, A. Watanabe, and K. S. Williams, PSL(2,) Sextic Fields With a Power Basis, Kodai Math. J., 29 (2006), 12. Mathematics Subject Classification (2000): 11R21 Daniel Eloff Department of Mathematics and Statistics University of British Columbia Okanagan Kelowna, B.C. Canada V1V 1V7 dan.eloff@gmail.com Blair K. Spearman Department of Mathematics and Statistics University of British Columbia Okanagan Kelowna, B.C. Canada V1V 1V7 blair.spearman@ubc.ca Kenneth S. Williams School of Mathematics and Statistics Carleton University Ottawa, Ontario Canada K1S B6 williams@math.carleton.ca 7
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