In this section, we show how instances of Problem 1 can be reduced to instances of Problem 2. The main technical ingredient, stated in Lemma 8 below, generalizes to any s > 1 the one given for s = 1 by Zeh, Gentner, and Augot in [39, Proposition 3]. To prove it, we use the same steps as in [39]; we rely on the notion of Hasse derivatives, which allow us to write Taylor expansions in positive characteristic (see for instance Hasse [19] or Roth [31, pp. 87, 276]).
In what follows, comparison and addition of s-tuples of integers are defined com- ponentwise. For example, writing i 6 j is equivalent to ik 6 jk for k = 1, . . . , s,
and i−j denotes (i1−j1, . . . , is−js). Similarly, ify= (y1, . . . , ys) is in K[X]s then
Y −y= Y1−y1, . . . , Ys−ys. Finally, for products of binomial coefficients, we shall
write j i = j1 i1 · · · js is .
Note that this coefficient is zero when i 66 j; note also that the monomial Yi has
total degree degY(Yi) =|i|=i1+· · ·+is.
IfAis any commutative ring with unity andA[Y] denotes the ring of polynomials in Y1, . . . , Ys over A, then for a polynomial P(Y) = PjPjY
j in A[Y] and a multi-
by P[i]=X j>i j i PjYj−i.
The Hasse derivative satisfies the following property (Taylor expansion): for all a in As,
P(Y) =X
i
P[i](a)(Y −a)i.
The following lemma shows how Hasse derivatives can help rephrase the vanishing condition (iv) of Problem 1. Below, the polynomials G and R are as defined in the introduction.
Lemma 8. For any polynomial Q in K[X,Y], Q satisfies the vanishing condition (iv) of Problem 1 if and only if for all i in Ns such that |i|< m,
Q[i](X,R) = 0 modGm−|i|.
Proof. Since the xr’s defining G =Qnr=1(X−xr) are pairwise distinct, it suffices to
prove, for 1 6 r 6 n, the following equivalence for the point (xr,yr): Q(xr,yr) = 0
with order at least m if and only if for all i in Ns such that |i| < m, Q[i](X,R) = 0 mod (X−xr)m−|i|. Now, up to a shift one can assume that this point is 0∈Ks+1;
in other words, it suffices to show that for R(0) = 0 ∈ Ks, we have Q(0,0) = 0 with order at least m if and only if, for all i in Ns such that |i|< m, Xm−|i| divides
Q[i](X,R).
Assume first that 0 ∈Ks+1 is a root of Q of order at least m. Then, Q(X,Y) = P
jQjY
j has only monomials of total degree at least m, so that for j > i, each
nonzero QjYj−i has only monomials of total degree at least m− |i|. Now, R(0) =
0 ∈ Ks implies that X divides each component of R. Consequently, Xm−|i| divides
QjRj−i for each j >i, and thus Q[i](X,R) as well.
Conversely, let us assume that for all i in Ns such that |i| < m, Xm−|i| divides
Q[i](X,R), and show that Q has no monomial of total degree less than m. Writing the Taylor expansion of Q with A=K[X] and a=R, we obtain
Q(X,Y) = X
i
Q[i](X,R)(Y −R)i.
Each component of R being a multiple of X, we deduce that for the multi-indices i
at least m. Using our assumption, the same conclusion follows for the multi-indices such that |i|< m.
For i inNs, with |i|< m, define the polynomials
Pi=Gm−|i| as well as Fi= (Fi,j)j∈Γ, with Fi,j = j i Rj−i mod Pi = j1 i1 Rj1−i1 1 · · · js is Rjs−is s modPi.
Then, the previous lemma implies that Qsatisfies properties (ii)-(iv) if and only if it can be written as Q=Pj∈ΓQjYj, with deg(Qj)< Nj for all j and, for all i inNs
such that |i|< m,
X
j∈Γ
QjFi,j = 0 modPi.
The latter conditions express the problem of finding such a Q as an instance of Problem 2. In order to make the reduction completely explicit, define further
µ= s+m−1 s , ν =|Γ|,
and choose arbitrary orders on the set of indices {i ∈ Ns | |i| < m} and Γ, that is, bijections
φ:{0, . . . , µ−1} → {i∈Ns | |i|< m} and ψ :{0, . . . , ν−1} →Γ. To i in {0, . . . , µ− 1}, we can then associate M′
i = Mφ(i), and similarly to j in
{0, . . . , ν−1}we associate N′
j =Nψ(j); we also set Pi′ =Pφ(i) and Fi,j′ =Fφ(i),ψ(j).
Proposition 9. Let (s, ℓ, m, n, k, b) be parameters for Problem 1, and let the param- eters µ, ν,M′ = (M′
0, . . . , Mµ′−1),N′ = (N0′, . . . , Nν′−1) be as above.
Then, one can reduce an instance of Problem 1 with parameters (s, ℓ, m, n, k, b)
to an instance of Problem 2 with parameters (µ, ν,M′,N′) and input polyno-
mials (P′
i,Fi′) using O(rM(M) log(M)) operations in K, where we write r =
max( s+ms−1,|Γ|) = max(µ, ν).
Proof. The only thing left to do is the complexity analysis. First, we need to compute
namely G, . . . , Gm, so it can be done using O(mM(mn)) operations; this will be dominated by the cost of the third step below.
Then, we have to compute the interpolation polynomials R= (R1, . . . , Rs)(using
Lagrange Interpolation). Each of them can be computed inO(M(n) log(n)) operations inK, for a total ofO(sM(n) log(n)), which isO(sM(M) log(M)). We have|Γ|>s, so that this cost isO(rM(M) log(M)), except in the extreme case when Γ = {(0, . . . ,0)}, so that |Γ|= 1; in that case, however, we need not compute R and we can omit the cost of interpolation.
Finally, we compute Fi,j modPi for every i,j. This is done by fixing i and
computing all products Fi,j modPi incrementally, starting from R1, . . . , Rs. Each
product takes O(M(Mi)) operations in K. Summing over all j leads to a cost of
O(|Γ|M(Mi)) per index i. Summing over all i and using the super-linearity of M
leads to a total cost of O(|Γ|M(M)), which is O(rM(M)).
Thus, in order to prove Theorem 4, it is enough to prove Theorem 3: in view of the cost reported in Theorem 3 for solving the linear system, the cost of reduction given above will always be negligible.