By Madhu Sudan

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Since g0 I g I g and h0 I h I h we have g1; h1 2 I. De ne u = a g1 b h1 : Since g1 ; h1 2 I we also have u 2 I. We claim that g0 I g (1 + u) and h0 I h (1 u). For g0 we have g (1 + u) g (1 + a g1 b h1 ) = g + a g g1 b g h 1 g + (1 b h )g1 b g h1 = g + g1 b (h g1 + g h1 ) g0 b (h g1 + g0 h1 ) = g0 b (h g0 h g + g0 h0 g0 h ) g0 b (f f) = g0 : 2 2 2 2 2 2 LECTURE 7 34 The proof for h0 is analogous. When R = F x; y] is the bivariate polynomial ring and I = (yk ) the \uniqueness" of the solution can be stated in a more expressive way.

Thus the polynomial N(x) F(x)p(x) has at least n e roots, while its degree is at most e+k. Hence if e+k < n e, the polynomial will be identically zero implying p(x) = NF ((xx)) . But this condition is precisely e < n 2 k which is given to us. The algorithm is thus guaranteed to work. Remarks on Run-time of the Algorithm Solving the linear system F(xi)yi = N(xi) can be done in O(npoly(log n)) time using the special form of the system. 40 LECTURE 8 Often a much faster way to obtain the message polynomial p(x) is to compute the (at most e) roots of F(x) using a root- nding algorithm, and to then obtain p(x) by fast interpolation at (any of) the points xi which are not roots of F(x).

C ) every univariate polynomial always factors. We will use substitutions which result in bivariate polynomials. , x y2 . 2 The Theorem We use the following substitution rule: for f(x; y1 ; y2; : : :; yn) 2 F x; y1 ; y2; : : :; yn ] and a^; ^b 2 Fn we consider the bivariate polynomial f(x; a1 t + b1; a2t + b2; : : :; ant + bn ): If f is reducible, say f = p q, then with high probability p(x; a1t+b1 ; a2t+b2 ; : : :; ant+bn ) and q(x; a1t+b1 ; a2 t+b2 ; : : :; ant+bn ) are not constant, and f(x; a1 t + b1; a2t + b2 ; : : :; ant + bn) is reducible.

### Algebra and Computation by Madhu Sudan

by Jason

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