Hatcher-Topology-of-Numbers · Problem 5.3.2

Classification of Quadratic Forms

Mathematics — Algebraic Topology · class-number

Problem

In this extended exercise the goal will be to show that the only negative even discriminants with class number 11 are −4,−8,−12,−16-4, -8, -12, -16, and −28-28. (Of these only −4-4 and −8-8 are fundamental discriminants.) The strategy will be to exhibit an explicit reduced primitive form QQ different from the principal form x2+dy2x^2 + dy^2 for each discriminant −4d-4d with d>4d > 4 except d=7d = 7. This will be done by breaking the problem into several cases, where in each case a form QQ will be given and you are to show that this form has the desired properties, namely it is of discriminant −4d-4d, primitive, reduced, and different from the principal form. You should also check that the cases considered cover all possibilities. (a) Suppose dd is not a prime power. Then it can be factored as d=acd = ac where 1<a<c1 < a < c and aa and cc are coprime. In this case let QQ be the form ax2+cy2ax^2 + cy^2. (b) The form ax2+2xy+cy2ax^2 + 2xy + cy^2 will work provided that d+1d+1 factors as d+1=acd+1 = ac where aa and cc are coprime and 1<a<c1 < a < c. If dd is odd, for example a power of an odd prime, then d+1d+1 is even so it has such a factorization d+1=acd+1 = ac unless d+1=2nd+1 = 2^n. (c) If d=2nd = 2^n the cases we need to consider are n≥3n \geq 3 since d>4d > 4. When n=3n = 3 take QQ to be 3x2+2xy+3y23x^2 + 2xy + 3y^2 and when n≥4n \geq 4 take QQ to be 4x2+4xy+(2n−2+1)y24x^2 + 4xy + (2^{n-2}+1)y^2. (d) When d+1=2nd+1 = 2^n the cases of interest are n≥3n \geq 3. When n=3n = 3 we have d=7d = 7 which is one of the allowed exceptions with class number 11. When n=4n = 4 we have d=15d = 15 and 3x2+5y23x^2 + 5y^2 works as in part (a). When n=5n = 5 we have d=31d = 31 and we take the form 5x2+4xy+7y25x^2 + 4xy + 7y^2. When n≥6n \geq 6 we use the form 8x2+6xy+(2n−3+1)y28x^2 + 6xy + (2^{n-3}+1)y^2.

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