# Lecture 2: Newton's Method and P-adic Lifting.

### Date: Apr. 5

### Background Material and Further Resources

- Algorithms for polynomial arithmetic. Modular algorithms.

### Reading

- Sec. 6.1-3 of the text.
- Paper by Loos.

### Topics

- Modular algorithm for univariate polynomial integral resultant computation
- res(A,B) mod p = res(A mod p, B mod p) provided p does not divide
ldcf(A) or ldcf(B).
- L(res(A,B)) is dominated by nL(d), where deg(A), deg(B) <= n and
|A|_1, |B|_1 <= d. This bound the number of mod p images, assuming
p is a "word-sized" prime.
- The time for a single evaluation of A mod p is dominated by n*L(d).
The total time spent on evaluation is nL(d)*nL(d).
- The modular resultant can be computed in time dominated by n^2,
using a modification of the Euclidean algorithm (e.g. subresultant
PRS). The total time spent computing modular resultants is
nL(d)*n^2 = n^3L(d).
- The time for applying the CRT to the resulting modular resultants
is (nL(d))^2.
- The total time is therefore dominated by n^2L(d)^2 + n^3L(d).

- Discrimants and Squarefree polynomials.
- A polynomial is squarefree if it does not have any multiple roots.
A(r) = 0, is of multiplicity k if (x-r)^k|A(x), but (x-r)^(k+1) does
not divide A(x).
- A(x) is squarefree iff gcd(A,A') is trivial.
- The gcd(A,A') is non-trivial iff res(A,A') = 0. Therefore a
polynomial is squarefree iff res(A,A') != 0. The res(A,A')/ldcf(A)
is equal to the discriminant of A. Let alpha_1,...,alpha_n, be
the roots of A, then
disc(A) = ldcf(A)^(2m-1) * Prod_{i < j} (alpha_i - alpha_j)^2.
Clearly, disc(A) = 0 iff A has roots of multiplicity greater than 1.

- Computing rational zeros be exhaustive search
- Let A(x) = sum_{i=0}^n a_i x^i. If A(s/t) = 0 and gcd(s,t) = 1,
then a_n *s^n + ... + a_1 * s * t^(n-1) + a_0*t^n = 0, which
implies, s|a_0 and t|a_n.
- Rational zeros can be found by testing if A(s/t) = 0 for
all divisors s of a_0 and t of a_n. See the maple worksheet for
details.
- Evaluation of A(r) can be performed by Horner's method in time
dominated by n^2L(r)^2 + n^2L(r)L(d). In this case, L(r) is
dominated by L(d) since s and t must divide a_0 and a_n respectively.
- The total time for finding rational zeros by exhastively testing
all possible divisors of a_0 and a_n is dominated by d^2*(n*L(d))^2.

- P-adic representation
- Computing rational zeros by p-adic lifting.

### Programs

- Maple worksheet illustrating varous algorithms for computing the
rational zeros of integral prolynomials
(ratzero.mws).
- SACLIB programs:

### Assignment

Created: Apr. 6, 2000 by
jjohnson@mcs.drexel.edu