Matching Tag: quadratic-orders
The inert quadratic order conjecture. For each m ≥ 1 m\geq 1 m ≥ 1 ,
Fix d ≥ 4 d\geq4 d ≥ 4 , write Δ = ( d + 1 ) ( d − 3 ) = f 2 Δ 0 \Delta=(d+1)(d-3)=f^2\Delta_0 Δ = ( d + 1 ) ( d − 3 ) = f 2 Δ 0 with Δ 0 \Delta_0 Δ 0 a fundamental discriminant, and let f ′ ∣ f f'\mid f f ′ ∣ f . Let M \mathcal M M satisfy the extended order-to-multiplet conject…
Fix d ∈ Z > 0 d\in\mathbb Z_{>0} d ∈ Z > 0 , d ≠ 3 d\neq3 d = 3 , write Δ = Δ d = ( d + 1 ) ( d − 3 ) = f 2 Δ 0 \Delta=\Delta_d=(d+1)(d-3)=f^2\Delta_0 Δ = Δ d = ( d + 1 ) ( d − 3 ) = f 2 Δ 0 with Δ 0 \Delta_0 Δ 0 a fundamental discriminant, and let M ( f ′ ) \mathcal M(f') M ( f ′ ) be the multiplet indexed by…
Fix d ∈ Z > 0 d\in\mathbb Z_{>0} d ∈ Z > 0 , d ≠ 3 d\neq3 d = 3 , set Δ = Δ d = ( d + 1 ) ( d − 3 ) \Delta=\Delta_d=(d+1)(d-3) Δ = Δ d = ( d + 1 ) ( d − 3 ) and K = \Q ( Δ ) K=\Q(\sqrt{\Delta}) K = \Q ( Δ ) . Let O Δ \mathcal O_\Delta O Δ and O K \mathcal O_K O K denote the quadratic order of discriminan…
Fix d ∈ Z > 0 d\in\mathbb Z_{>0} d ∈ Z > 0 , d ≠ 3 d\neq3 d = 3 , let Δ = ( d + 1 ) ( d − 3 ) \Delta=(d+1)(d-3) Δ = ( d + 1 ) ( d − 3 ) , and let O Δ \mathcal O_\Delta O Δ be the quadratic order of discriminant Δ \Delta Δ . Geometric Count Conjecture. The number of…
Fix a positive integer d ≠ 3 d\neq 3 d = 3 , let Δ d = ( d + 1 ) ( d − 3 ) \Delta_d=(d+1)(d-3) Δ d = ( d + 1 ) ( d − 3 ) , and write Δ d = f 2 Δ 0 \Delta_d=f^2\Delta_0 Δ d = f 2 Δ 0 , where Δ 0 = disc ( O K d ) \Delta_0=\operatorname{disc}(\mathcal O_{K_d}) Δ 0 = disc ( O K d ) and K d = \Q ( Δ d ) K_d=\Q(\sqrt{\Delta_d}) K d = \Q ( Δ d ) .…
Let d 1 d_1 d 1 and d 2 d_2 d 2 be discriminants of quadratic imaginary orders, and fix an integer t t t . Let … Assume that the conductors of d 1 d_1 d 1 and d 2 d_2 d 2 , together with m m m , have no simultan…