Double and simple roots of quadratics modulo the maximal ideal #
Let R be a local ring with perfect residue field. If the discriminant a² + 4 b of the monic
quadratic T² + a T − b lies in the maximal ideal, then the reduction of the quadratic has a double
root in the residue field
(Polynomial.exists_quadratic_eq_zero_and_two_mul_add_eq_zero_of_discrim_eq_zero), and any lift
s ∈ R of it is a root of the quadratic and of its derivative modulo the maximal ideal. In
characteristic two the double root is a square root, which is where perfectness is needed.
This is the completing-the-square step of Tate's algorithm (Step 6), where it is applied twice to the coefficients of a Weierstrass equation over a discrete valuation ring.
Conversely, over any local ring, if the discriminant a² + 4 b is a unit then every root t of
T² + a T − b modulo the maximal ideal is simple: the derivative 2 t + a is a unit, since
(2 t + a)² = 4 (t² + a t − b) + (a² + 4 b).
Main results #
TauCeti.IsLocalRing.exists_add_two_mul_mem_maximalIdeal_and_sub_mul_sub_sq_mem_maximalIdeal: ifa² + 4 b ∈ 𝔪, someshasa + 2 s ∈ 𝔪andb − s a − s² ∈ 𝔪.TauCeti.IsLocalRing.isUnit_two_mul_add_of_isUnit_sq_add_four_mul: ifa² + 4 bis a unit andt² + a t − b ∈ 𝔪, then2 t + ais a unit.TauCeti.HenselianLocalRing.exists_quadratic_root_of_isUnit_discriminant: in a Henselian local ring, a root modulo the maximal ideal of a monic quadratic with unit discriminant lifts to a root with unit derivative.
A root of T² + a T − b modulo the maximal ideal is simple when the discriminant a² + 4 b is
a unit: if t² + a t − b ∈ 𝔪, then 2 t + a is a unit, since
(2 t + a)² = 4 (t² + a t − b) + (a² + 4 b).
A double root of T² + a T − b modulo the maximal ideal, lifted from the perfect residue
field: if the discriminant a² + 4 b lies in the maximal ideal, some s ∈ R has
a + 2 s ∈ 𝔪 and b − s a − s² ∈ 𝔪.
A root of T² + a T − b modulo the maximal ideal of a Henselian local ring lifts to a root
in the same residue class with unit derivative when a² + 4 b is a unit.