Ideal Composition in Quadratic Fields: from Bhargava to Gauss

Transcription

1 Ideal Composition in Quadratic Fields: from Bhargava to Gauss Duncan Buell Department of Computer Science and Engineering University of South Carolina Clemson, 10 Oct 010 DAB Page 1

2 Composition in Quadratic Fields (following Bhargava) 1 Quadratic Forms Everything is an integer (except ). Binary quadratic form f(x, y) = ax + bxy + cy = (a, b, c) of discriminant = b 4ac. Primitive if gcd(a, b, c) = 1; we usually consider only primitive forms. (For imprimitive forms we can usually just factor out the common term, do the theory on the resulting primitive form, and then put the factor back in.) Forms f 1 = (a 1, b 1, c 1 ) and f = (a, b, c ) are equivalent if there exists M = that we have a matrix equivalence M T a 1 b 1 / M = a b / b 1 / c 1 b / c Alternatively, M : f 1 (x 1, y 1 ) f (x, y ) under x 1 = α β x γ δ y 1 y α β SL (Z) such γ δ Clemson, 10 Oct 010 DAB Page

3 with explicit change of variables a = a 1 α + b 1 αγ + c 1 γ b = b 1 (αδ + βγ) + (a 1 αβ + c 1 γδ) (1) c = a 1 β + b 1 βδ + c 1 δ f = (a, b, c) represents an integer m if there exist x, y such that m = ax + bxy + cy. Primitive representation if gcd(x, y) = 1) Theorem 1.1. Primitive representation is equivalent to equivalence. Because if a = a 1 α + b 1 αγ + c 1 γ with gcd(α, γ) = 1, then we can find β, δ with αδ βγ = 1 and we have the matrix from the modular group for the equivalence (1). Theorem 1.. The classes of forms of a fixed discriminant form the finite abelian group that is the ideal class group of the quadratic order of discriminant in the quadratic field Q( ). Theorem 1.3. A primitive form can represent some integer relatively prime to any chosen integer. (In deep obscurity mode, this is the Cebotarev Density Theorem, but it can be proven directly in this simple case.) Clemson, 10 Oct 010 DAB Page 3

4 Special cases for the generators of the modular group: 1 β : (a, b, c) (a, b + aβ, aβ + bβ + c) : (a, b, c) (c, b, a) 1 0 Clemson, 10 Oct 010 DAB Page 4

5 Quadratic Fields Let Z, 0, 1 (mod 4) be the discriminant of a quadratic number field K = Q( ). We have the ring of integers O in K, ideals in O, and the class group of ideals modulo principal ideals. Essentially, ideals are Z-modules [ with some conditions added, for a, b Z. a, b + ] Essentially, meaning: narrow versus wide ideals invertible ideals nonfundamental discriminants, orders not the principal order etc. Clemson, 10 Oct 010 DAB Page 5

6 3 Forms and Ideals The form f = (a, b, c) of discriminant corresponds to the Z-module [ a, b+ For positive discriminant, classes of forms correspond to the narrow classes of ideals modulo principal ideals of positive norm (an extra to 1 mapping). For nonfundamental discriminants of orders, classes of primitive forms of that disc are crisply and simply defined, but modules and ideals require some extra conditions. I have always preferred forms, in part because having the third coefficient c explicitly presented is useful, especially in what now follows. ]. Clemson, 10 Oct 010 DAB Page 6

7 4 Composition of forms/multiplication of ideals In essence, and [ a 1, b 1 + (a 1, b 1, c 1 ) (a, b, c ) (fudge a 1 a,, ) ] [ a, b + ] [ fudge a 1 a, + The real substance of a composition algorithm is to be able precisely to determine the value of fudge. ] Clemson, 10 Oct 010 DAB Page 7

8 5 Dirichlet s United Forms Assume f 1 = (a 1, b 1, c 1 ) and f = (a, b, c ) of the same discriminant. If f 1 = (a 1, b 1, c 1 ) (a 1, b 1 + a 1 β 1, C 1 ) = (a 1, B, C 1 ) f = (a, b, c ) (a, b + a β, C ) = (a, B, C ) then necessarily we have f 1 (a 1, B, a C) () and the composition f 1 f (a 1 a, B, C). f (a, B, a 1 C) Dirichlet defined two forms to be united if gcd(a 1, a, b 1+b ) = 1, and a solution to () can be shown to be guaranteed if the forms are united. This is the simplest of the composition algorithms. For existential purposes, since we can always cite Theorem 1.3 and argue that we can find f 3 f with gcd(a 3, a 1 ) = 1, we can in theory always do composition using Dirichlet united forms. This is unsatisfactory because this isn t an algorithmically explicit composition. Clemson, 10 Oct 010 DAB Page 8

9 6 Arndt s Algorithm Dirichlet united forms is not algorithmic in that we have to specify that we find some form with a relatively prime lead coefficient. Arndt s method: Assume f 1 = (a 1, b 1, c 1 ) and f = (a, b, c ) of discriminant. Let β = b 1+b and n = gcd(a 1, a, β), and solve a 1 t + a u + βv = n for t, u, v. Then B b 1 (mod a 1 /n) has the simultaneous solution and the composition is βb n B b b 1b + n (mod a /n) (mod a 1 a /n ) B = a 1b t + a b 1 u + v(b 1 b + )/ n f 1 f (a 1 a /n, B, ). (3) United forms, fudged to allow for n 1: find the most nearly united B possible. Hold that thought! Clemson, 10 Oct 010 DAB Page 9

10 7 Bhargava s cubes Bhargava defines a cube a c e g b d f h and three pairs of matrices derived from the three orientations of faces of the cube: M 1 = a b N 1 = e f c d g h M = a c N = b d (4) e g f h M 3 = a e N 3 = c g b f d h From these we get forms Q i (x, y) = Det(M i x N i y) Clemson, 10 Oct 010 DAB Page 10

11 which explicitly are Q 1 = x ( ad + bc) + xy(ah bg cf + de) + y ( eh + fg) Q = x ( ag + ce) + xy(ah + bg cf de) + y ( bh + df) (5) Q 3 = x ( af + be) + xy(ah bg + cf de) + y ( ch + dg) all of discriminant = a h + b g + c f + d e + (abgh + cdef + acfh + bdeg + adeh + bcfg) + 4(adfg +bceh) We let SL (Z) SL (Z) SL (Z) act on these cubes: in the i-th factor, we have r s : (M i, N i ) (rm i + sn i, tm i + un i ) t u 1 1 : (M i, N i ) (M i + N i, N i ) 0 1 : (M i, N i ) ( N i, M i ) Clemson, 10 Oct 010 DAB Page 11

12 Bhargava (continued) The action of SL (Z) SL (Z) SL (Z) is the action of equivalence on the derived forms. Theorem 7.1. We have composition in the group of classes of forms as follows. Q 1 Q Q 3 Id. Bhargava s identity is or / ( + 3)/4 Clemson, 10 Oct 010 DAB Page 1

13 Bhargava (continued) Theorem 7.. Bhargava reduces to Dirichlet united forms. Proof. Bhargava defines projectively the cube a c e g b d f h so we may assume that gcd(a, b, c, d, e, f, g, h) = 1. Use this fact and invoke the extended Euclidean algorithm. That is, apply modular group transformations to get a = 1. Apply further transformations with the left face to get a 0 for the b location; apply transformations with the top face to get a 0 for for the c location; and apply transformations with the front face to get a 0 for for the e location. Clemson, 10 Oct 010 DAB Page 13

14 Bhargava (continued) This yields 0 f 1 0 g 0 d h Ba Da Bing! The three forms are now Q 1 = dx + hxy + fgy Q = gx + hxy + dfy Q 3 = fx + hxy + dgy Note that Gauss/Dirichlet/Arndt/Matthews/Buell would have written the third form as the equivalent form Q 3 = dgx hxy fy ; it is typical of doing the computations in Bhargava fashion that we get the product written this way. Clemson, 10 Oct 010 DAB Page 14

15 Bhargava (continued) In theory, then, Bhargava subsumes Gauss. But given two forms, how do we find a cube on which both live, and then which among the compounded forms in the class shows up as the third form from the cube? That is, how do we make this algorithmic instead of existential? In fact, Arndt s method works just fine. There are no doubt other methods, but with eight variables and three sets of quadratic equations, showing how, where, and why the old method works is a good start. Remember: Q 1 = x ( ad + bc) + xy(ah bg cf + de) + y ( eh + fg) Q = x ( ag + ce) + xy(ah + bg cf de) + y ( bh + df) Q 3 = x ( af + be) + xy(ah bg + cf de) + y ( ch + dg) and we are asking for ways to solve this if we are given explicit coefficients for Q 1 and Q. Theorem 7.3 (Unproven?). A given octuple a, b, c, d, e, f, g, h uniquely determines a set of three forms (forms, not classes) and conversely. If I could prove this, things would be a little simpler. Clemson, 10 Oct 010 DAB Page 15

16 8 Bhargava continued The old algorithm still works Necessary Conditions for the Algorithm To compound two forms (A 1, B 1, C 1 ) and (A, B, C ) by putting them on a cube: Let β = B 1+B and set n = gcd(a 1, A, β). Start by choosing relatively prime values of a 0 and c 0. Now solve a 0 d + c 0 b = A 1 /n = A 10 a 0 g + c 0 e = A /n = A 0 (6) a 0 h c 0 f = β/n = β 0 Clemson, 10 Oct 010 DAB Page 16

17 Bhargava continued (continued) We get general solutions a = n a 0 c = n c 0 b = b 0 + a 0 λ d = d 0 + c 0 λ e = e 0 + a 0 µ (7) g = g 0 + c 0 µ f = f 0 + a 0 ν h = h 0 + c 0 ν We now have three equations to solve, one for B 1, one for C 1, and one for C. (We get B by combining the formula for B 1 with the formula for β.) Clemson, 10 Oct 010 DAB Page 17

18 Bhargava continued (continued) First, B 1 = a(h 0 + c 0 ν) (b 0 + a 0 λ)(g 0 + c 0 µ) c(f 0 + a 0 ν) + (d 0 + c 0 λ)(e 0 + a 0 µ) = ah 0 b 0 g 0 cf 0 + d 0 e 0 + λ( a 0 g 0 + c 0 e 0 ) + µ(a 0 d 0 c 0 b 0 ) We have = ah 0 b 0 g 0 cf 0 + d 0 e 0 + λa 0 µa 10 B 1 (ah 0 b 0 g 0 cf 0 + d 0 e 0 ) = λa 0 µa 10 (8) or, written another way, B 1 B + b 0 g 0 d 0 e 0 = λa 0 µa 10 (9) We assume that we can solve (9); assuming this, then the solutions are λ = λ 0 + A 10 η µ = µ 0 + A 0 η (10) for yet another parameter η. Clemson, 10 Oct 010 DAB Page 18

19 Bhargava continued (continued) Next, C 1 = (e 0 + a 0 µ)(h 0 + c 0 ν) + (f 0 + a 0 ν)(g 0 + c 0 µ) which reduces to C 1 + e 0 h 0 f 0 g 0 = β 0 µ A 0 ν and then to C 1 + e 0 h 0 f 0 g 0 + β 0 µ 0 = A 0 [ β 0 η ν] (11) Finally, C = (b 0 + a 0 λ)(h 0 + c 0 ν) + (d 0 + c 0 λ)(f 0 + a 0 ν) which reduces to C + b 0 h 0 d 0 f 0 + β 0 λ 0 = A 10 [ β 0 η ν] (1) Clemson, 10 Oct 010 DAB Page 19

20 Bhargava continued (continued) The Algorithm Leaving aside a fair number of issues about greatest common divisors and the resulting ability to do the extended Euclidean algorithm, the process is: 1. Compute n, and then choose a 0 and c 0 relatively prime.. Solve equations (6) to get the general forms of the variables, as in (7). This we are guaranteed to be possible by our choice of a 0 and c Solve (9) to get λ and µ as in (10). This is not guaranteed. However, in the development of the composition algorithm, especially that of united forms, there is a role to be played by gcd(a 1, A ), so I am not overly concerned here. It is also possible here that one would need to use a good choice of a 0 and c 0 and not an arbitrary choice. 4. Solve (11) and (1) for ν and η. This is really one equation ( ) 1 β 0 η + ν = (C 1 + e 0 h 0 f 0 g 0 + β 0 µ 0 ) A ( 0 ) 1 = (C + b 0 h 0 d 0 f 0 + β 0 λ 0 ) A 10 Clemson, 10 Oct 010 DAB Page 0

21 9 A Specific Version of an Algorithm To compound two forms (A 1, B 1, C 1 ) and (A, B, C ) by putting them on a cube: Let β = B 1+B and set n = gcd(a 1, A, β). Start by choosing relatively prime values of a 0 and c 0. Specifically, choose a 0 = 1 c 0 = 0 (13) Now solve a 0 d + c 0 b = A 1 /n = A 10 a 0 g + c 0 e = A /n = A 0 (14) a 0 h c 0 f = β/n = β 0 Clemson, 10 Oct 010 DAB Page 1

22 We get general solutions a = n a 0 = n c = n c 0 = 0 b = b 0 + a 0 λ = 0 + λ d = d 0 + c 0 λ = A λ e = e 0 + a 0 µ = 0 + µ (15) g = g 0 + c 0 µ = A 0 + 0µ f = f 0 + a 0 ν = 0 + ν h = h 0 + c 0 ν = β 0 + 0ν We now have three equations to solve, one for B 1, one for C 1, and one for C. (We get B by combining the formula for B 1 with the formula for β.) First, B 1 = a(h 0 + c 0 ν) (b 0 + a 0 λ)(g 0 + c 0 µ) c(f 0 + a 0 ν) + (d 0 + c 0 λ)(e 0 + a 0 µ) = ah 0 b 0 g 0 cf 0 + d 0 e 0 + λ( a 0 g 0 + c 0 e 0 ) + µ(a 0 d 0 c 0 b 0 ) = ah 0 b 0 g 0 cf 0 + d 0 e 0 + λa 0 µa 10 = nβ 0 + λa 0 µa 10 Clemson, 10 Oct 010 DAB Page

23 That is, B 1 β = λa 0 µa 10 (16) or, written another way, B 1 B = λa 0 µa 10 (17) We know that we can solve (17). This is because we have ( ) ( ) ( ) B1 B B1 + B B1 B β 0 = = A 10 C 1 A 0 C. (18) n Any common factor of A 10 and A 0 on the right hand side cannot divide β 0 by the definition of n, so it must divide B 1 B. This is enough to guarantee that we can solve (17), and then the solutions must be of the form λ = λ 0 + ηa 10 /p µ = µ 0 + ηa 0 /p (19) for yet another parameter η, and where p = gcd(a 10, A 0 ). Clemson, 10 Oct 010 DAB Page 3

24 Next, C 1 = (e 0 + a 0 µ)(h 0 + c 0 ν) + (f 0 + a 0 ν)(g 0 + c 0 µ) which reduces to C 1 = β 0 µ A 0 ν (0) and then to C 1 + β 0 µ 0 = (A 0 /p)( β 0 η pν) (1) We know that we can solve (0) and (1). To see this, we refer again to (18). Any common factor of β 0 and A 0 must divide A 10 C 1 and must therefore divide C 1 by the definition of n. Finally, C = (b 0 + a 0 λ)(h 0 + c 0 ν) + (d 0 + c 0 λ)(f 0 + a 0 ν) which reduces to C = β 0 λ A 10 ν () and then to C + β 0 λ 0 = (A 10 /p)( β 0 η pν) (3) Clemson, 10 Oct 010 DAB Page 4

25 We know that we can solve () and (3) by an argument analogous to that for equation (0). We know that we can simultaneously solve (17), (0), and (). What we have is a matrix equation A 0 A 10 0 β 0 0 A 10 0 β 0 A 0 λ µ ν = B 1 B C C 1 which reduces to A 0 A β 0 A 10 A 10 A λ µ ν = B 1 B C A 0 + β 0 ( B1 B ) β 0 ( B1 B ) + C A 0 C 1 A 10 and then by applying (17) this becomes A 0 A β 0 A λ µ ν = B 1 B C 1 0 so our solutions to (17), (0), and () will in fact work. Clemson, 10 Oct 010 DAB Page 5

26 As a comment on proper composition, we note that Q 3 = (A 3, B 3, C 3 ) has C 3 = ch + dg = A 10 A 0 = A 1 A /n, just as is usually done with the standard composition algorithms. Of course, the usual algorithms produce a form with this as the lead coefficient, but the usual algorithms produce the compounded form, not its inverse. If we take the compounded form to be (A 3, B 3, C 3 ) (C 3, B 3, A 3 ), then this composition is essentially the same as ordinary composition. Clemson, 10 Oct 010 DAB Page 6

27 10 The Specific Algorithm Is the Standard Algorithm The specific algorithm above for putting two forms on a Bhargava cube is in fact the standard algorithm for composition. The standard algorithm, from my book, as done by Arndt and presented in Mathews, is as follows. (Arndt/Mathews do an algorithm for the Gauss forms with the middle coefficient required to be even; the algorithm in my book is the same algorithm but for Eisenstein forms with arbitrary middle coefficient.) Arndt s Method: To compound forms F 1 = (A 1, B 1, C 1 ) and F = (A, B, C ), we let β = B 1+B, we let n = gcd(a 1, A, β), and we solve for t, u, and v. We let A 1 t + A u + βv = n The composition is A 3 = A 1 A /n, B 3 = A 1B t + A B 1 u + v(b 1 B + )/. n F 1 F (A 3, B 3, ) Clemson, 10 Oct 010 DAB Page 7

28 with the third coefficient computed from the discriminant (or painfully from the formulas above). This works because we have and and B 3 B 1 (mod (A 1 /n)) (4) B 3 B (mod (A /n)) (5) βb 3 B 1B + (mod A 1 A /n) (6) and this is essentially the united forms method of Dirichlet, adjusted slightly for the case that n 1. Clemson, 10 Oct 010 DAB Page 8

29 Bhargava can be derived from Arndt As above, let a = n c = 0 d = A 10 (7) g = A 0 h = β 0 I have somewhat overcomplicated the notation by introducing b = λ and e = µ, but I will stick with that. If we now solve equation (16) B 1 β = λa 0 µa 10 this leads to B 1 + A 10 µ = B + A 0 λ. In other words, we have solved congruences (4) and (5) and the common value of this expression is our desired B 3. The final solution, solving (0) (or ()) for f in Bhargava s cube, is what gives us a value for B 3 that also satisfies congruence (6) in Arndt s method. Clemson, 10 Oct 010 DAB Page 9