One point with intersection multiplicity 2

We consider the product of two conjugate conics with just oneFp-rational point

of intersection, whose multiplicity of intersection is 2. The result, since the procedure involves the solubility absolutely irreducible quartics, describes the solubility for primes greater or equal than 19. We move the singular point to P = [0 : 0 : 1]; since the

intersection multiplicity is at least 2 in P, the tangent lines at P to the two conics

coincide, we move them tox= 0. We can write this reduction as T = (Q(x, y) +xz)(σ(Q(x, y)) +xz),

whereQ(x, y)is a binary quadratic overFp2 but not on_Fp, in order to have a product of

irreducible conics the coecient ofy2 inQ should be not null. Moreover, since we want

8.3. Conjugate conics

x - Q−σ(Q) (this condition is equivalent to have multiplicity of intersection equal to

2) the coecientα of y2 has to be in Fp2 \_Fp. We refer to the probability of lifting the

point[0 : 0 : 1] asτ.

The rst necessary condition for lifting P are the divisibility ofX andY coordi-

nates by p. Then we can divide all the coecients by p, the new reduction will be just T =az4.This reduction is not soluble by primitivity unless the coecientaofz4 is zero,

which happens with probability1/p. In this case we can divide further by p ending up

with the reduction

T =z2(az2+bxz+x2+cyz) =z2γ.

Notice that the coecient ofx2z2 is equal to one since it is the same coecient of

the original reduction (since we multiplied it byp2 and divided it byp2). The coecients a, b and c are uniformly and independently distributed on Fp. If the conic dened by

γ = 0 is smooth we are done, since it has at least one smooth point with z-coordinate

dierent from 0. Looking at the determinant of the matrix associated toγ, the conic is

irreducible if and only if c6= 0, this happens with probability (p−1)/p. If we focus on

the singular cases (they happen with probability1/p) we already know that on the line z= 0there is a double point, it can be the intersection of two distinct lines dened over

Fp (this case happens with probability p₂−_p1, and we have solubility), or the intersection

of two conjugate lines over Fp2 (this case happens with probability p−1

2p , and we have

insolubility because the only Fp point has z-coordinate null and so it is not liftable by

primitivity) or we can have a double line with probability1/p which we move to x= 0.

A necessary condition now for liftability is that p divides the x-coordinate of

the solution, making the substitution and dividing by p we end up with the following

reduction

T =z2(az2+cyz+dy2),

where a, c and d are uniformly and independently distributed on Fp. If d is non-zero

(with probability (p−1)/p) then we have probability of solubility 1/2 by Proposition

8.1.1. If d= 0 (with probability 1/p) the reduction is z3(az +cy) and, if c 6= 0 which

happens with probability(p−1)/p, we have a line which contains smooth points, which

implies solubility. Otherwise, if c= 0 = d(with probability 1/p2_{) the reduction is just} z4a = 0 where, by primitivity, we have potential solubility if a = 0 (with probability 1/p).

The null quartic is the only case where we need to investigate further. After 115

8.3. Conjugate conics

division by pthese are the valuations of all the coecients of the quartic Z4 ≥0

≥0 ≥0 = 0 ≥0 ≥0

≥2 ≥1 ≥0 ≥0

X4 ≥4 ≥3 ≥2 ≥1 = 0 Y4

In particular, looking at the reductionT the coecient ofxy2_{zis the trace of the element} α cited at the beginning, the one ofy4 is norm of α and the one ofx2z2 is 1.

About this family of quartics we know that they have a singular point in[1 : 0 : 0]

with double tangent z = 0. By primitivity we cannot lift this point. By interpolation

on small primes we have the counts of the possible reduction of these quartics (in total there are p6 of them):

ˆ p6 −p4 are irreducible, therefore if the cardinality of the base eld is greater or

equal than 19, by Section 8.2, we have solubility 1.

ˆ p4 are products of conjugate conics, in particular theFp-rational intersection points

are:

(i) One point with intersection multiplicity 2 in (p4−p3)/2 cases;

(ii) Two points with intersection multiplicity 2 inp3 cases;

(iii) One point with intersection multiplicity 2 and two points with multiplicity of intersection 1 in (p4−p3)/2 cases.

Recalling that we cannot lift the point Q = [1 : 0 : 0] the solubility of the

undetermined cases above are as follows:

(i) The only singular point isQ, since we cannot lift any point;

(ii) This case has solubilityτ, since we can just lift the second point of interesection;

(iii) This case has solubility 2p+1

p2_+2p+1, which is the solubility when we have two points

with multiplicity 1. Wrapping up we nd τ = 1 p p−1 p + 1 p p−1 2p + 1 p p−1 2p + 1 p p−1 p + 1 p2 1 p6 p6−p4+p3τ+p 4_−p3 2 2p+ 1 p2_{+ 2p+ 1} , 116

8.3. Conjugate conics

which give us the probability of solubility

2p9+ 5p8+ 5p7+ 5p6+ 5p5+ 4p4+ 6p3+ 6p2+ 4p+ 1 2(p+ 1)2 _p8_+p7_+p6_+p5_+p4_+p3_+p2_{+p+ 1} .