Integral bases and monogenity of pure fields

Integral bases and monogenity of pure fields Istv´an Ga´al∗ , and L´aszl´o Remete University of Debrecen, Mathematical Institute H–4002 Debrecen Pf.400., Hungary e–mail: gaal.istv[email protected]u, [email protected]u April 26, 2016 Abstract Let mbe a square-free integer (m6= 0,1). We show that the structure of the integral bases of the fields K=Q(n √m) are periodic in m. For 3 ≤n≤9 we show that the period length is n2. We explicitly describe the integral bases, and for n= 3,4,5,6,8 we explicitly calculate the index forms of K. This enables us in many cases to characterize the monogenity of these fields. Using the explicit form of the index forms yields a new technics that enables us to derive new results on monogenity and to get several former results as easy consequences. For n= 4,6,8 we give an almost complete characterization of the monogenity of pure fields. 1 Introduction Let mbe a square-free integer (m6= 0,1) and n≥2 a positive integer. There is an extensive literature of pure fields of type K=Q(n √m). (Describing the ∗Research supported in part by K115479 from the Hungarian National Foundation for Scientific Research 2010 Mathematics Subject Classification: Primary 11R04; Secondary 11Y50 Key words and phrases: pure fields, integral basis, power integral basis, monogenity following results on pure fields we use some basic concepts on monogenity and power integral bases that are detailed in Section 2.) B.K.Spearman and K.S.Williams [14] gave an explicit formula for the integral basis of pure cubic fields. B.K.Spearman, Y.Qiduan and J.Yoo [13] showed that if iis a cubefree positive integer then there exist infinitely many pure cubic fields with minimal index equal to i. I.Ga´al and T.Szab´o [11] studied the behaviour of the minimal indices of pure cubic fields in terms of the discriminant. L. El Fadil [12] gave conditions for the existence of power integral bases of pure cubic fields in terms of the index form equation. T.Funakura [7] studied the integral basis in pure quartic fields. I.Ga´al and L.Remete [9] calculated elements of index 1 (with coefficients <101000) in pure quartic fields K=Q(4 √m) for 1 < m < 107,m≡2,3 (mod 4). S.Ahmad, T.Nakahara and S.M.Husnine [3] showed that if m≡1 (mod 4), m6≡ ±1 (mod 9) then Q(6 √m) is not monogenic. On the other hand [4], if m≡2,3 (mod 4), m6≡ ±1 (mod 9) then Q(6 √m) is monogenic. A.Hameed and T.Nakahara [1] constructed integral bases of pure octic fields Q(8 √m). They proved [2] that if m≡1 (mod 4) then Q(8 √m) is not monogenic. On the other hand A.Hameed, T.Nakahara, S.M.Husnine and S.Ahmad [5] proved that if m≡2,3 (mod 4) then Q(8 √m) is monogenic. A.Hameed, T.Nakahara, S.M.Husnine and S.Ahmad [5] showed that if m≡2,3 (mod 4) then Q(2n √m) is monogenic, this involves the pure quartic and pure octic fields, as well. Moreover, they showed [5] that if all the prime factors of ndivide mthen Q(n √m) is monogenic. Our purpose is for 3 ≤n≤9 to give a general characterization of the integral basis of K=Q(n √m). We prove that the integral bases of K= Q(n √m) is periodic in m. For 3 ≤n≤9 the period length is n2. The knowledge of the integral bases makes possible also to compete the sporadic results on the monogenity of these fields. Our method applying the explicit form of the index forms yields a new technics that enables us to obtain new results on the monogenity of these fields and to obtain several former results as easy consequences. In our Theorems 4, 7, 8 we give an almost complete characterization of the monogenity pure quartic, sextic and octic fields, respectively. The cubic case is well-known and easy, much less is known about the quintic, septic and nonic cases. 2 2 Basic concepts about the monogenity of number fields We recall those concepts [8] that we use throughout. Let αbe a primitive integral element of the number field K(that is K=Q(α)) of degree nwith ring of integers ZK. The index of αis I(α) = (Z+ K:Z[α]+) = s D(α) DK =1 p|DK|Y 1≤i<j≤nα(i)−α(j), where DKis the discriminant of Kand α(i)denote the conjugates of α. The minimal index of Kis iK= min I(α) where αruns through the primitive integral elements of K. If B= (b1= 1, b2, . . . , bn) is an integral basis of K, then the index form corresponding to this integral basis is I(x2, . . . , xn) = 1 p|DK|Y 1≤i<j≤n(b(i) 2−b(j) 2)x2+. . . + (b(i) n−b(j) n)xn (where b(i) jdenote the conjugates of bj) which is a homogeneous polynomial with integral coefficients. For the integral element α=x1+b2x2+. . . +bnxn we have I(α) = |I(x2, . . . , xn)| independently of x1.αgenerates a power integral basis (1, α, . . . , αn−1) if and only if I(α) = 1 that is (x2, . . . , xn)∈Zn−1is a solution of the index form equation I(x2, . . . , xn) = ±1 in(x2, . . . , xn)∈Zn−1.(1) In this case ZK=Z[α] and Kis called monogenic. 3 3 Basic results Throughout we assume that mis a square-free integer with m6= 0,1 and n > 2 an integer. Let K=Q(n √m) and ϑ=n √m. Our first theorem is on the prime divisors of the denominators of the integral basis elements: Theorem 1. If (1, ϑ, . . . , ϑn−1)is not an integral basis in K, then for any element α=a0+a1ϑ+. . . +an−1ϑn−1 q(2) of the integral basis (with a0, . . . , an−1, q ∈Z,q6= 0) the denominator qcan only be divisible by primes dividing n, the prime factors of qdo not divide m. Proof The discriminant of ϑ=n √mis ±nnmn−1. If (1, ϑ, . . . , ϑn−1) is not an integral basis in Q(ϑ), then there must be a number qdividing nnmn−1and an element αof type (2) such that αis an algebraic integer and an element of (1, ϑ, . . . , ϑn−1) can be replaced by αto get a basis with smaller discriminant. Let pbe a prime divisor of q. Then obviously α0=q pα=e0+e1ϑ+. . . +en−1ϑn−1 p(3) is also an algebraic integer. We can also assume that 0 ≤ei< p (0 ≤i≤ n−1) by taking each eimodulo p. We show that pis a divisor of n. Assume on the contrary that p|m. The element α0ϑ=e0ϑ+e1ϑ2+. . . +en−2ϑn−1+en−1m p is obviously an algebraic integer. By p|mthe element e0ϑ+e1ϑ2+. . . +en−2ϑn−1 p 4 is also an algebraic integer. We proceed by multiplying this element by ϑ and omitting the analogous integral part. Finally we obtain that %=e0ϑn−1 p is an algebraic integer. The element %is the root of the polynomial f%(x) = pxn−en 0m pn−1 . This polynomial is irreducible over Qif and only if its reciprocal polynomial f1/%(x) = en 0m pn−1 xn−p. is irreducible. Here m/p is an integer, not divisible by pbecause mis squarefree. e0is also not divisible by p(otherwise we did not have e0in (3) and we had the same result with the first non-zero ei). Hence f1/%(x) is an Eisentein polynomial, therefore f%(x) is irreducible. Then f%(x) is the defining polynomial of %. This contradicts to %being an algebraic integer. 2 Remark. Theorem 1 implies Theorem 3.1. of A.Hameed, T.Nakahara, S.M.Husnine and S.Ahmad [5]: if all the prime factors of ndivide mthen Q(n √m) is monogenic. Next we show that the integral bases of K=Q(n √m) are periodic. First we prove this statement with a period length much larger than n2but this result is valid for any n. Theorem 2. Let n=ph1 1. . . phk kand n0=p[nh1/2] 1. . . p[nhk/2] kwhere [x]denotes the lower integer part of x. Let ϑ=n √mand γ=n √m+nn 0. Then the structure of the integral bases of the fields Q(ϑ)and Q(γ)is the same in terms of ϑand γ, respectively. Remark. Under the ”same structure” we mean that if the integral basis of Q(ϑ) has an element a0+a1ϑ+. . . +an−1ϑn−1 q, 5 then the integral basis of Q(γ) has an element a0+a1γ+. . . +an−1γn−1 q and vice versa. Proof Assume that (1, ϑ, . . . , ϑn−1) is not an integral basis in Q(ϑ). Then there must be integer elements of type α=a0+a1ϑ+. . . +an−1ϑn−1 q(4) which can replace elements of (1, ϑ, . . . , ϑn−1) to obtain an integral basis. We show that the existence of analogous algebraic integers of type (4) is equivalent in the fields generated by ϑ=n √mand by γ=n √m+nn 0. If we replace an element of the basis (1, ϑ, . . . , ϑn−1) by αof (4) then the discriminant of the basis decreases by a factor q2. Hence q2divides ±nnmn−1 (the discriminant of ϑ=n √mis ±nnmn−1). By Theorem 1 the prime divisors pof qdo not divide m. Hence q2|nn, which implies that qdivides n0. Denote the conjugates of αby α(j), j = 1, . . . , n. The defining polynomial of αis n Y j=1 (x−α(j)) = 1 qn n Y j=1 (qx −a0−a1ϑ(j)−. . . −an−1(ϑ(j))n−1). The product is a symmetrical polynomial of ϑ(1), . . . , ϑ(n), hence its coefficients can be expressed as polynomials (with integer coefficients) of the defining polynomial of ϑ, that is xn−m. Hence there exist polynomials P0, . . . , Pn−1∈Z[x] such that n Y j=1 (x−α(j)) = 1 qn((qx)n+Pn−1(m)(qx)n−1+. . . +P1(m)(qx) + P0(m)). Therefore the element αis an algebraic integer if and only if qn|qjPj(m) that is qn−j|Pj(m) (j= 0,1, . . . , n −1).(5) 6 Replace now mby m0=m+nn 0and consider integral bases in the field Q(γ) = Q(n √m+nn 0). In this field an element of type (4), that is δ=a0+a1γ+. . . +an−1γn−1 q(6) is an algebraic integer if and only if qn−j|Pj(m+nn 0) (j= 0,1...,n−1) (7) with the same polynomials Pj. By q|n0the conditions (5) are equivalent to (7). Therefore Q(ϑ) and Q(γ) contains the same type of integer elements. Elements of that type are linearly independent in the first case if and only if they are linearly independent in the second case. Therefore Q(ϑ) and Q(γ) admits the same type of integral bases. 2 4 The structure of the integral bases is periodic in mmodulo n2for 3≤n≤9 Theorem 2 implies that the integral bases of K=Q(n √m) are periodic modulo nn 0. This number is of magnitude nn2/2. For small values of nwe have a much sharper assertion. Theorem 3. For 3≤n≤9the integral bases of Q(n √m)are periodic in m modulo n2. Proof For n= 3,4,5 the nn 0is 27, 655536, 9765625, respectively. Calculating the integral bases of Q(n √m) for square-free mup to nn 0it is easily seen that the structure of the integral bases of Q(n √m) are periodic modulo n2. One can easily detect a few types of integral bases that are repeated for square-free values of m,m+n2,m+ 2n2etc. Let now n > 5. Then nn 0is far too large for the calculations described above. However for n= 6,7,8,9 we managed to prove the same assertion. 7 Let 1 < r < n2. If ris square-free, then set r0=r. If rhas a common square factor with n, then none of r+kn2is square-free, we omit r. If rhas no common square factor with nbut contains another square factor, then we set r0=r+n2or r0=r+ 2n2etc. which is already square-free. Let ϑ=n √r0, calculate the integral bases of Q(n √r0) and denote the basis elements by (b1= 1, b2, . . . , bn), where bjis of the form bj=aj0+aj1ϑ+. . . +aj,n−1ϑn−1 q (with aj0, aj1, . . . , aj,n−1∈Zand with a non-zero denominator qthe prime factors of which divide n). Let m=r+kn2be a square-free integer, γ=n √mand b0 j=aj0+aj1γ+. . . +aj,n−1γn−1 q. We wonder I. if the analogues of the elements bj, that is the elements b0 jremain algebraic integer for any square-free m=r+kn2, further II. if for some square-free m=r+kn2some of the basis elements (b0 1= 1, b0 2, . . . , b0 n) can be replaced by an integral element of type d=e1b0 1+. . . +enb0 n p(8) (where 0 ≤e1, . . . , en≤p−1 and pis a prime divisor of n) to obtain a basis with smaller discriminant. I. The defining polynomial of e1b0 1+. . . +enb0 nis G(x) = n Y j=1 (x−e1b0(j) 1−. . . −enb0(j) n). This polynomial is symmetrical in the conjugates of γ=n √m, hence its coefficients will be polynomials in m: G(x) = xn+Gn−1(m)xn−1+. . . +G1(m)x+G0(m). 8 Since there are denominators in the b0 i, the polynomials Gj(depending also on e1, . . . , en) are not necessarily of integer coefficients. Let us substitute m=r+kn2. We obtain G(x) = xn+Hn−1(k)xn−1+. . . +H1(k)x+H0(k). For all possible residues rwe have explicitly calculated these polynomials Hj(k) which also depend on e1, . . . , en. In all cases we found that these are polynomials in k, e1, . . . , enwith integer coefficients. Substituting ei= 1 and ej= 0, j = 1, . . . , n, j 6=ithis implies that b0 iis integer for any square-free m=r+kn2(1 ≤i≤n). II.Consider now the defining polynomial of d(8), that is P(x) = 1 pnG(px) = 1 pn(px)n+Hn−1(k)(px)n−1+. . . +H1(k)(px) + H0(k). This polynomial has integer coefficients if and only if pndivides pjHj(k) that is pn−j|Hj(k) = Hj(k, e1, . . . , en) (0 ≤j≤n−1).(9) Let now (e1, . . . , en)∈Znbe an arbitrary given fixed tuple. Obviously by (9) the validity of the statement if dis integral or not, depends only on the behaviour of kmodulo pnand not on the value of k. This allows us to test the fields Q(n √m) for square-free m=r+kn2where kruns though all residue classes modulo pn. These fields were tested directly, calculating their integral bases. We found that in all cases the fields had the same structure of integral basis in terms of γ=n √mlike the field Q(n √r0) in terms of ϑ=n √r0. This proves our assertion. 2 Remark The test described at the end of the above proof required to calculate the integral bases of 24(26+ 36) = 18239 sextic fields, 48 ·77= 39530064 septic fields, 48 ·28= 12288 octic fields and 72 ·39= 1417176 nonic fields. 9 Case 6.3. r= 10,19, m=r+ 36ksquare-free 1, x, x2, x3,1 + x2+x4 3,x+x3+x5 3, D = 2632m5 Calculating the index form it is easily seen that the index form equation is not solvable modulo 3, hence these fields are not monogenic. This case is not covered by S. Ahmad, T. Nakahara and S. M. Husnine [3], [4] since m≡1 (mod 9). Case 6.4. r= 26,35, m=r+ 36ksquare-free 1, x, x2, x3,1+2x2+x4 3,x+ 2x3+x5 3, D = 2632m5 This case is not covered by S. Ahmad, T. Nakahara and S. M. Husnine [3], [4] since m≡ −1 (mod 9). If m= 26+36kthen 4m|(f2−9f3). If Kis monogenic then for a solution of the index form equation these factors are equal to ±1. The possible values of f2−9f3are ±8,±10, hence the above divisibility can not hold. If m= 35 + 36kthen 4(35 + 36k)|(f2−9f3). If Kis monogenic then for a solution of the index form equation these factors are equal to ±1. The possible values of f2−9f3are ±8,±10, hence the above divisibility can only hold for k=−1, that is m=−1. It is easily seen that the relative index [10] of elements of K=Q(6 √−1) is divisible by 9, therefore this field is not monogenic, either. Case 6.5. r= 17, m=r+ 36ksquare-free 1, x, x2,1 + x3 2,4+3x+ 2x2+x4 6,4x+ 3x2+ 2x3+x5 6, D = 32m5 This case is not covered by S. Ahmad, T. Nakahara and S. M. Husnine [3], [4] since m≡ −1 ( mod 9). Calculating the index form we can easily see that the index form equation is not solvable modulo 6. Case 6.6. m= 1, m= 1 + 36ksquare-free 1, x, x2,1 + x3 2,4+3x+ 4x2+x4 6,3+4x+ 3x2+x3+x5 6, D = 32m5 16 This case is not covered by S. Ahmad, T. Nakahara and S. M. Husnine [3], [4] since m≡1 (mod 9). Calculating the index form we can easily see that the index form equation is not solvable modulo 3. Summarizing the above statements we have Theorem 7. For the following values of rlet m=r+ 36k(k∈Z) be a square-free integer. The field K=Q(6 √m)is monogenic for r= 2,3,6,7,11,14,15,22,23,30,31,34 and is not monogenic for r= 1,5,10,13,17,19,21,25,26,29,33,35. 6 Pure septic fields, K=Q(7 √m) Case 7.1. r= 2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,20,21,22,23,24, 25,26,27,28,29,32,33,34,35,36,37,38,39,40,41,42,43,44,45,46,47, m=r+ 49ksquare-free B={1, x, x2, x3, x4, x5, x6}, D =−m677 These fields are obviously monogenic. Case 7.2. r= 18, m= 18 + 49ksquare-free B={1, x, x2, x3, x4, x5,1+2x+ 4x2+x3+ 2x4+ 4x5+x6 7}, D =−m675 Case 7.3. r= 19, m= 19 + 49ksquare-free B={1, x, x2, x3, x4, x5,1+3x+ 2x2+ 6x3+ 4x4+ 5x5+x6 7}, D =−m675 17 Case 7.4. r= 30, m= 30 + 49ksquare-free B={1, x, x2, x3, x4, x5,1+4x+ 2x2+x3+ 4x4+ 2x5+x6 7}, D =−m675 Case 7.5. r= 31, m= 31 + 49ksquare-free B={1, x, x2, x3, x4, x5,1+5x+ 4x2+ 6x3+ 2x4+ 3x5+x6 7}, D =−m675 Case 7.6. r= 48, m= 48 + 49ksquare-free B={1, x, x2, x3, x4, x5,1+6x+x2+ 6x3+x4+ 6x5+x6 7}, D =−m675 Case 7.7. r= 1, m= 1 + 49ksquare-free B={1, x, x2, x3, x4, x5,1 + x+x2+x3+x4+x5+x6 7}, D =−m675 7 Pure octic fields, K=Q(8 √m) In all these cases the index form is the product of three factors of degrees 4,8,16, respectively. We shall denote these factors by f1, f2, f3. These depend on the parameter mand on the variables x2, . . . , x8. Case 8.1. r= 2,3,6,7,10,11,14,15,18,19,22,23,26,27,30,31,34,35,38, 39,42,43,46,47,50,51,54,55,58,59,62,63, m=r+ 64ksquare-free 1, x, x2, x3, x4, x5, x6, x7, x8, D =−88m7 18 These fields are obviously monogenic. This also follows from A.Hameed, T.Nakahara, S.M.Husnine and S.Ahmad [5]. Case 8.2. r= 1,17,33,49, m=r+ 64ksquare-free 1, x, x2, x3,1 + x4 2,x+x5 2,1 + x2+x4+x6 4, 1 + x+x2+x3+x4+x5+x6+x7 8, D =−210m7 These fields are not monogenic by the theorem of A.Hameed and T.Nakahara [2]. We conjecture that in these fields the minimal index is 128. Case 8.3. r= 5,13,21,29,37,45,53,61, these cases can be included by m= 5 + 8k, square-free 1, x, x2, x3,1 + x4 2,x+x5 2,x2+x6 2,x3+x7 2, D =−216m7 Calculating and factorizing f3−16f2 2we find that it is divisible by m. If there existed a power integral basis then for a solution of the index form equation we would have f1, f2, f3=±1, hence f3−16f2 2=±1−16 is either −15 or −17. The possible divisors are ±3,±5,±15,±17 but only m=−3,5 is of type m= 5 + 8k. For m=−3 then element (−1,−1,0,1,1,0,−1) has index one, hence K=Q(8 √−3) is monogenic. For m= 5 the least index we found in K=Q(8 √5) was 16. Note that A.Hameed, T.Nakahara, S.M.Husnine and S.Ahmad [2] assert that these fields are not monogenic, they certainly did not involve K= Q(8 √−3). Case 8.4. r= 9,25,41,57, these cases can be included by m= 9 + 16k, square-free 1, x, x2, x3,1 + x4 2,x+x5 2,1 + x2+x4+x6 4,x+x3+x5+x7 4, D =−212m7 19 Calculating and factorizing f2−4f2 1we find that it is divisible by m. If there existed a power integral basis then for a solution of the index form equation we would have f1, f2, f3=±1, hence f2−4f2 1=±1−4 is either −3 or −5. The possible divisors are ±3,±5 but none of them is of type m= 9 + 16k. Therefore these fields are not monogenic. This also follows from the theorem of A.Hameed, T.Nakahara, S.M.Husnine and S.Ahmad [2]. Summarizing the above statements we have Theorem 8. For the following values of rlet m=r+ 64k(k∈Z) be a square-free integer. The field K=Q(8 √m)is monogenic for r= 2,3,6,7,10,11,14,15,18,19,22,23,26,27,30,31,34,35,38,39,42,43,46, 47,50,51,54,55,58,59,62,63 and is not monogenic for r= 1,5,9,13,17,21,25,29,33,37,41,45,49,53,57,61,m6= 5, with the exception of K=Q(8 √−3) which is monogenic. Remark 9. We conjecture that the minimal index of K=Q(8 √5) is 16. Octic fields of this type will be considered in a forecoming paper. 8 Pure nonic fields, K=Q(9 √m) Case 9.1. r= 2,3,4,5,6,7,11,12,13,14,15,16,20,21,22,23,24,25,29,30, 31,32,33,34,38,39,40,41,42,43,47,48,49,50,51,52,56,57,58,59,60,61,65, 66,67,68,69,70,74,75,76,77,78,79, m=r+ 81k, square-free 1, x, x2, x3, x4, x5, x6, x7, x8, D = 318m8 These fields are obviously monogenic. Case 9.2. r= 1,28,55, m=r+ 81k, square-free 1, x, x2, x3, x4, x5,1 + x3+x6 3,x+x4+x7 3, 20 1 + x+x2+x3+x4+x5+x6+x7+x8 9, D = 310m8 Case 9.3. r= 8,17,35,44,62,71, m=r+ 81k, square-free 1, x, x2, x3, x4, x5,1+2x3+x6 3,x+ 2x4+x7 3,x2+ 2x5+x8 3, D = 312m8 Case 9.4. r= 10,19,37,46,64,73, m=r+ 81k, square-free 1, x, x2, x3, x4, x5,1 + x3+x6 3,x+x4+x7 3,x2+x5+x8 3, D = 312m8 Case 9.5. r= 26,53,80, m=r+ 81k, square-free 1, x, x2, x3, x4, x5,1+2x3+x6 3,x+ 2x4+x7 3, 1+2x+x2+ 8x3+ 7x4+ 8x5+x6+ 2x7+x8 9, D = 310m8 9 Computational remarks In all our calculations we used Maple [6] and most of our programs executed a couple of seconds or a few minutes on an average laptop. For n= 4,6,8 we needed a very careful calculation of the factors of the index forms, which may take extremely long otherwise. The tests corresponding to Theorem 3 took also a few minutes for n= 3,4,5,6,8. For n= 9 it executed 5 hours. For n= 7 we executed our Malpe program on a supercomputer with nodes having 24 CPU-s. 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