# New PDF release: Linear orderings By Joseph G Rosenstein

ISBN-10: 0125976801

ISBN-13: 9780125976800

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Extra info for Linear orderings

Example text

Thus equivalent forms have the same discriminant. 24 §2. LAGRANGE, LEGENDRE AND QUADRATIC FORMS The sign of the discriminant D has a strong effect on the behavior of the form. 4) 4a/(x,y) = {lax + by)2 - Dy2. 4). Note that all of these notions are invariant under equivalence. The discriminant D influences the form in one other way: since D = b2 — Aac, we have D = b2 mod 4, and it follows that the middle coefficient b is even (resp. odd) if and only if D = 0 (resp. 1) mod 4. 5. Let D = 0,1 mod 4 be an integer and m be an odd integer relatively prime to D.

D/p) for odd primes p not dividing D. Furthermore, { [-1 1 when D > 0 whenD<0. Proof. The proof will make extensive use of the Jacobi symbol. Given m > 0 odd and relatively prime to M, recall that the Jacobi symbol (M/m) is defined to be the product -)=n(m) f=\\Pi where m = p\ ■ • ■ pr is the prime factorization of m. 15) (MN\ _ / M \ /N \ m ) \m) \m M\ mn I [M\(M \m I \ n 16 §1. 10). 10). 18) (_l)(D-i)("-V/4 ( To compare these expressions, first note that the two Jacobi symbols are equal since m = n mod D.

I) The values in (Z/DZ)* represented by the principal form of discriminant D form a subgroup H C ker(x). (ii) The values in (Z/DZ)* represented by f(x,y) form a coset ofH in ker(x). Proof. We first show that if a number m is prime to D and is represented by a form of discriminant D, then [m] G ker(x). 1, we can write m = d2m', where m' is properly represented by f(x,y). Then x([ m l) = x([d2m']) = x([d])2x([m']) = x([>w']). , D = b2 — km for some b and k. 14) imply that and our claim is proved.