Generalizing Gauss s Gem

Transcription

1 For n = 2, 3, all airs of nilotent matrices satisfy 1, but not all nilotent matrices of size 2 commute, eg, [ ] [ ] A =, and B = ACKNOWLEDGMENTS We than the referees for their useful remars This wor has been suorted in art by the Microsoft Research INRIA Joint Centre and by Université Paris Diderot Paris 7 REFERENCES 1 S H Weintraub, A Guide to Advanced Linear Algebra, The Dolciani Mathematical Exositions, Vol 44, MAA Guides, 6, Mathematical Association of America, Washington, DC, M Balaj, Romanian National Mathematical Olymiad, Problems and Solutions 11th grade, Aril 2011, available at htt://wwwonm2011isjbihorro/subiecte/finala11_soldf in Romanian Algorithms Project, Inria Paris-Rocquencourt, Le Chesnay, France AlinBostan@inriafr Observatoire de Paris, 77 Avenue Denfert-Rochereau, Paris, France Combot@imccefr Generalizing Gauss s Gem Ezra Brown and Marc Chamberland Abstract Gauss s Cyclotomic Formula is extended to a formula with variables, where is an odd rime This new formula involves the determinant of a circulant matrix An alication involving the Wendt determinant is given Gauss s Cyclotomic Formula [3, , 467] is a neglected mathematical wonder Theorem 1 Gauss Let be an odd rime and set = 1 1/2 Then there exist integer olynomials Rx, y and Sx, y such that 4x + y x + y = Rx, y 2 Sx, y 2 The goal of this note is to generalize this theorem Denote a circulant matrix as x 1 x 2 x 3 x x x 1 x 2 x 1 circx 1, x 2,, x = x 1 x x 1 x 2 x 2 x 3 x 4 x 1 htt://dxdoiorg/104169/amermathmonthly MSC: Primary 11C08 August Setember 2012] NOTES 597

2 Let j be the Legendre symbol, that is, for j 0 mod, j = 1 or 1 according as j is or is not a quadratic residue mod A multivariable generalization of Theorem 1 follows Theorem 1 is a secial case of Theorem 2 with x 3 = = x = 0 Theorem 2 Let be an odd rime and = 1 1/2 Then there exist integer olynomials Rx 1, x 2,, x and Sx 1, x 2,, x such that 4 detcircx 1, x 2,, x x 1 + x x = Rx 1, x 2,, x 2 Sx 1, x 2,, x 2 Secifically, one can tae Rx 1, x 2,, x = A + B and Sx 1, x 2,, x = A B/ where A = B = j j = 1 and ζ is a rimitive th root of unity Proof It is well-nown [4] that x 1 + ζ j x 2 + ζ 2 j x ζ 1 j x, x 1 + ζ j x 2 + ζ 2 j x ζ 1 j x, 1 detcircx 1, x 2,, x = x 1 + ζ j x 2 + ζ 2 j x ζ 1 j x 1 x 1 + x x j The choice of R and S given above then easily satisfy the desired equation, A B 2 R 2 S 2 = A + B 2 = 4AB 1 = 4 j i x i ζ j i 1 = 4 detcircx 1, x 2,, x x 1 + x x The challenge now is to show that both R and S are olynomials with integer coefficients Let be a rime > 3, let = 1 1/2, let ζ be a rimitive th root of unity, and let K = Qζ be the cyclotomic field of th roots of unity For any integer such that 1 1, define the maing σ on K by setting σ ζ = ζ and extending the ma linearly Then K is a Galois extension of degree 1 over the rational field Q with cyclic Galois grou G = {σ 1 1} G also acts on Qζ [x 1,, x ] by setting σ x i = x i and extending the action linearly; see [2, 596ff] for details and further information 598 c THE MATHEMATICAL ASSOCIATION OF AMERICA [Monthly 119

3 Let α = r/ ζ r and β = n/= 1 ζ n A bit of algebra shows that β = α 1 and αβ = 1 /4; thus, α = 1 ± /2 and β = 1 /2, for some choice of signs The set of maings H = {σ / = 1} is a subgrou of G of index 2, whose fixed field is the quadratic field Qα Note that both A and B are in Zζ [x 1,, x ] We now show that A + B Z[x 1,, x ] and that A B Zα[x 1,, x ] The roduct rule for the Legendre symbol states that if j and are relatively rime to then j j = Thus, if = 1, then relacing ζ by ζ in A and B ermutes the factors of A and the factors of B Similarly, if = 1, then relacing ζ by ζ in A and B exchanges the factors of A with the factors of B It follows that if = 1, then the action of σ on Qζ [x 1,, x ] fixes both A and B, while if = 1, then the action of σ on Qζ [x 1,, x ] interchanges A and B We conclude that σ A + B = A + B for all, so that A + B is invariant under the action of every element of the Galois grou G Thus, the coefficients of A + B lie in the fixed field of G, namely the rational field Q, and so A + B Q[x 1,, x ] But A + B Zζ [x 1,, x ], so it follows that R = A + B is a olynomial with integer coefficients We now turn to S = A B/ By revious results, the coefficients of A and B are in the field fixed by the index-2 subgrou H of the Galois grou G, namely Qα Since A, B Zζ [x 1,, x ], it follows that both A and B are in Zα[x 1,, x ] Hence, there exist olynomials f = f x 1,, x and g = gx 1,, x with integer coefficients such that A = f + gα Let n be a fixed quadratic nonresidue mod The nontrivial automorhism of Qα sends α to β As A is not fixed by σ n, we see that σ n α = β Hence, B = σ n A = σ n f + gα = f + gβ It follows that A B = gα β, where g has integer coefficients Then, by revious wor and a little more algebra, we see that α β = ± It follows that S = A B a olynomial with integer coefficients = ±g = ±g, In the case when 1 mod 4, the functions R and S given in Theorem 2 are not unique The Pell equation x 2 y 2 = 1 2 has infinitely many integer solutions for any rime see [1] Since x 2 1 y2 1 x 2 2 y2 2 = x 1x 2 + y 1 y 2 2 x 1 y 2 + x 2 y 1 2, any solution x, y to equation 2 may be used in conjunction with the solution R, S in Theorem 2 to roduce another air of olynomials August Setember 2012] NOTES 599

4 R = x R + ys, S = x S + y R which mae Theorem 2 wor Indeed, infinitely many such R and S exist The olynomials R and S raidly grow in size For = 5, one has and R = 2 x 1 2 x 2 x 5 x 5 x 3 x 2 x x 2 2 x 1 x 3 x 5 x 4 x 3 x 2 x 1 x 4 x 2 x x x 5 2 x 1 x 5 x 4 x x 4 2 S = x 2 x 4 x 1 x 4 + x 4 x 3 + x 5 x 4 x 5 x 3 + x 3 x 2 + x 1 x 5 x 1 x 3 + x 2 x 1 x 2 x 5 For = 7, R has 84 terms and S has 56 terms A simle alication of Theorem 2 involves a determinant considered by Wendt in conjunction with Fermat s Last Theorem The so-called Wendt determinant is defined by W n = det circ n, 0 n, 1 n n,, 2 n 1 E Lehmer claimed later roved by JS Frame [5, 128] that W n = 1 n 1 2 n 1u 2 for some u N Since n 1 n = 2 n 1, =0 if n is an odd rime, Theorem 2 imlies 2u 2 = R 2 S 2 for some integers u, R, and S This equation clearly has a trivial solution if S = 0 This situation occurs when 1 mod 4 since B = 1 + ζ j 1 j = 1 = j = = A The first few cases where S = 0 are j 1 + ζ j ζ j = , = , 600 c THE MATHEMATICAL ASSOCIATION OF AMERICA [Monthly 119

5 = and = REFERENCES 1 E Barbeau, Pell s Equation Sringer, New Yor, D S Dummit, R M Foote, Abstract Algebra, third edition John Wiley, Hoboen, NJ, C F Gauss, Untersuchungen über höhere Arithmeti Translated from German by H Maser Chelsea, New Yor, G Golub, C Van Loan, Matrix Comutations, third edition, John Hoins University Press, Baltimore, P Ribenboim, Fermat s Last Theorem For Amateurs Sringer, New Yor, 1999 Deartment of Mathematics, Virginia Tech, Blacsburg, VA 24061, ezbrown@mathvtedu Deartment of Mathematics and Statistics, Grinnell College, Grinnell, IA 50112, chamberl@mathgrinnelledu Norms as a Function of Are Linearly Indeendent in Finite Dimensions Greg Kuerberg Abstract We show that there are no non-trivial linear deendencies among -norms of vectors in finite dimensions that hold for all The roof is by comlex analytic continuation Theorem 1 Let v 1, v 2,, v n be non-zero vectors with v R d Suose that α 1 v 1 + α 2 v α n v n = 0 1 for all [a, b] with 1 a < b Then the equation is trivial in the following sense Call two of the vectors equivalent if they differ by adding zeros, ermuting or negating coordinates, and rescaling Then the terms of 1 in each equivalence class, with the given coefficients, already sum to zero The result was stated as a question by Steve Flammia on MathOverflow htt://dxdoiorg/104169/amermathmonthly MSC: Primary 46B07 August Setember 2012] NOTES 601