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Thus while solving the linear system in step 3(b) of the algorithm, we should obtain a g~ with small sized coe cients. It is not necessary that we obtain the g~ with the smallest coe cients, but one close enough would do. Let us look at the linear system once again. Suppose gk has degree d0 < n(= deg(f)). Let g~(x) = d X i=0 ci xi where d = n 1; ci 2 Z lk (x) = The linear system is then given by the equation: d X i=0 ci xi = dX d0 i=0 i gk (x)xi + d X i=0 dX d0 i=0 i xi; i 2 Z i pk xi; i 2 Zwhere ci ; i; i are the unknowns.

The goal of Step 2 is to lift this factorization to one modulo y2k . 1. For what follows it will also be useful to know that g0 is irreducible, so we assume that too. In the next lecture we will show how to get rid of these assumptions by some initial preprocessing in Step 1. ) Having lifted the factorization for su ciently large value of k, the main sub-steps of Step 3 of the algorithm are the following: Step 3(a). 2) Step 3(b). Find gcdx (f; g) (viewed as polynomials in F(y) x]) and if non-trivial, a non-trivial factorization of f in F x; y] may be found using Gauss's Lemma.

Clearly, for any g0 I g (1 + u), h0 I h (1 + u) (for u 2 I), we have g0 I g I g, h0 I h I h and g0 h0 I g h (1 u2) I f. 1). Let g1 = g0 g and h1 = h0 h . 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.

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