First sets of lifts

LetC be a projective curve in PN defined over K by F ∈ OK[X0, . . . , XN] and let p be a principal prime of good reduction for C with generator π. We note that in generalCmay not be a complete intersection and hence may be defined by more than

N−1 polynomials. However, for any pointP on C (or indeed ¯P on ¯C) only N −1 of the polynomials are needed to define C at that point. Because C is smooth, the Jacobian of such a tuple of polynomials will have full rankN−1 when evaluated at a representative ofP. The choice of tuple of polynomials will vary with P. We will assume for simplicity that our curves are always complete intersections; in practice for each reduced point ¯P a suitable tuple of polynomials can be chosen.

Let ¯P be a rational point on the reduced curve ¯C. We may choose a repre- sentative for ¯P inFNp+1; there are N(p)−1 possible choices, one for each non-zero element ofFp. By choosing a lift of each coordinate fromFp toOK, we may choose an integral representatives for ¯P. s∈ ON_K+1 satisfies s≡P¯ mod p and F(x) ≡0 modp. sis p-primitive. If it were not then each coordinate of swould be inp and reducing s mod p would give the zero vector in FNp +1, which does not represent a point in PN(Fp). This choice of s is clearly not unique: the set of all integral representatives for ¯P is the set of all x∈ O_KN+1 that reduce mod p to some repre- sentative of ¯P inFNp+1. Ifsis some integral representative for ¯P, then every integral representative for ¯P is given by cs+πy for some c ∈ O_K satisfying c 6≡ 0 modp and somey ∈ O_KN+1.

Because ∇F(s) has full rank, we may employ Hensel lifting to construct a p-adic solution from s. By Proposition 5.5 there exists an s0 ∈ O_KN_p+1 such that

s≡s0 mod pand F(s0) = 0. Although s0 depends on the choice of representative

s, the sets we proceed to construct will be unique. We fix s0 to be a lift of s and recall that_bs0 will represent a truncation of s0.

Definition 24. Let Sk be the set of all pointsx∈ ONK+1 such that

x is a representative for P¯ mod p and

F(x)≡0 mod pk.

Lemma 5.6. We have S1= n c0bs0+πy c₀ ∈ O_K, c₀ 6≡0 mod p, y∈ ON_K+1 o ,

where _bs0 is any mod p approximation to s0.

Proof. If x ∈S1 then x is a representative for ¯P mod p and x may be written as

c0bs0+πy for somec06≡0 modp and some y∈ O N+1 K .

If x = c0bs0 +πy then F(x) ≡ F(c0sb0) modp. Because F(x) is a set of homogeneous polynomials,F(c0bs0)≡0 if and only ifF(bs0)≡0 mod p.

S1 is simply the set of all elements of ON_K+1 that reduce to ¯P modp. How- ever, the setsSi for i >1 have a more complicated structure.

Lemma 5.7. Lets0∈ ON_Kp+1 be as above. Then the Jacobian matrix forF evaluated

at s0 gives a surjective linear map ∇F(s0)(X) :O_KN_p+1 → ON_Kp−1 with s0 contained

in the kernel.

Proof. The fact that s0 lies in the kernel of ∇F(s0)(X) follows from the fact that ∇Fj(X)(X) =λjFj(X) for some scalarλj for eachFj, by Euler’s theorem on homo- geneous functions. ∇F(s0)(X) is linear, so we need only to prove that∇F(s0)(X) is surjective. The matrix ∇F(s0) has full rank modp, so it defines a surjective mapFNp+1 →FNp−1.

Let y ∈ ON_K−1. Because ∇F(s0)(X) is surjective mod p there exists an

x∈ ON_K+1 such that

∇F(s0)(x)≡y mod p.

By Lemma 5.1, the Jacobian of ∇F(s0)(X)−y is ∇F(s0) and this has full rank modp. Therefore by Hensel’s Lemma (Proposition 5.5) we may liftx, which satisfies

F(s0)(x)−y ≡ 0 mod p, to a p-adic solution x0 ∈ O_KN_p+1 such that ∇F(s0)(x0) =

y.

This use of Hensel’s Lemma proves the existence of solutions to equations of the form∇F(s0)(X)−y ≡ 0 mod pm for anyy and any m and is a key step in the construction of further sets of lifts. By considering thesi as lying in the p-adic completion at their construction, we do not have to fix ap-adic precision at the start or update our calculations at each stage but can usesbi to represent the truncation ofsi to the required precision.

The rank of the kernel of ∇F(s0)(X) is 2. Choose a vectors1∈ OKNp+1 that

O_Kp, by the same method as the proof of Lemma 5.7) such thats0 and s1 generate the kernel of∇F(s0)(X). The new vector s1 will be used in the construction of the second lattice of lifts.

Lemma 5.8. If x ∈S1 thenx ∈ S2 if and only if x≡ c0(bs0+πc1bs1) modp 2 _for

some c0, c1 ∈ OK with c0 6≡ 0 mod p. Here bs0 and bs1 are any approximations modp2 and mod p respectively.

Proof. Writex=c0bs0+πy∈S1and set degF−1 to be the multi-index (deg(Fi)−1)i. We perform a Taylor expansion based atc0sb0:

F(x) =F(c0sb0+πy)≡F(c0bs0) +∇F(c0bs0)(πy) mod p 2 ≡cdeg(₀ F)F(s_b0) +πcdeg(F) −1 0 ∇F(bs0)(y) mod p 2 ≡cdeg(₀ F)−1π∇F(_bs0)(y) mod p2, because bs0 is a mod p 2 _{approximation to s}

0. Since c0 6≡0 mod p, x ∈S2 if and only if∇F(_bs0)(y)≡0 modp. This is the case if and only ifyis a linear combination of_bs0 and bs1, sox∈S2 if and only if

x≡c0bs0+πa0sb0+πa1bs1 mod p 2_.

By relabelling constants, we have

x=c0(bs0+πc1bs1) +π 2_y

for some c0, c1 ∈ OK such that c0 6∈ p. As x is an arbitrary element of S1 and

S1⊃S2, S2= n c0(bs0+πc1bs1) +π 2_y c₀, c₁∈ O_K, c₀ 6≡0 mod p, y∈ ON_K+1 o .