Quaternary descent #
Let F be a field of characteristic not two and Q a regular quadratic form of dimension four
over F whose discriminant is the square class of a nonsquare d. Let E = F(√d), a quadratic
field extension of F. Then Q is isotropic over F exactly when its scalar extension to E is
isotropic (O'Meara 58:7).
This is what makes a quaternary form accessible to arguments over the discriminant field: over
E the discriminant of Q becomes a square, and the theorem descends isotropy back to F.
The hypotheses have the following roles.
- The dimension is exactly four. Over
F = ℚ(i), the binary form⟨1, 2⟩has discriminant[2]and is anisotropic, as-2is not a square inℚ(i), but it becomes isotropic overF(√2), where-2 = (i√2)². - The discriminant is the plain discriminant
TauCeti.RegularFormClass.discr; in dimension four the signed discriminant differs from it by the class of-1. Eis a quadratic extension ofFcontaining a square rootsof the nonsquared, which pins it down asF(s). A larger field containing a square root ofd, such asF(√d, √e), is not covered.
Main results #
QuadraticForm.not_anisotropic_of_not_anisotropic_baseChange_quaternary: quaternary descent of isotropy fromF(√d)toF.QuadraticForm.anisotropic_baseChange_iff_quaternary: a quaternary form of nonsquare discriminantdis anisotropic overF(√d)exactly when it is anisotropic overF.
References #
- O. T. O'Meara, Introduction to Quadratic Forms (1963), 58:7.
Quaternary descent (O'Meara 58:7). Let Q be a regular quadratic form of dimension four
over a field F of characteristic not two, whose discriminant is the square class of a nonsquare
d, and let E be a field of degree two over F containing a square root s of d, so that
E = F(s). If Q is isotropic over E, then it is isotropic over F.
Quaternary descent, as an equivalence: a regular quaternary form whose discriminant is the
square class of a nonsquare d is anisotropic over F(√d) exactly when it is anisotropic over
F.