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Sample text

Here I = (pk )) For any f 2 R and for any factorisationf = gh( mod I) of f and there exist a; b such that ag + bh = 1( mod I), then there exist g ; h ; a ; b such that f = g h ( mod I 2 ) g = g( mod I) h = h( mod I) a g + b h = 1( mod I 2 ) and the solution g ; h is unique in the sense: If any other g0 ; h0 also satisy these properties, then g0 = g (1 + u)( mod I 2 ) h0 = h (1 u)( mod I 2 ) for some u 2 I . Coming to steps 3 and 4 of the algorithm, as before we have the following questions: How do we know that g~ and lk satisfying (1) exist?

We know from Claim 1 that the coe cients of any factor g of f are not very large. 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.

Proof Notice rst g satis es the degree conditions and the condition g 6= 0. It su ce to show that there exists lk such that g(x; y) = gk (x; y)lk (x; y) (mod y2k ). We nd lk by applying the Hensel lifting procedure to the factorization of g modulo y. This yields polynomials g~k and lk such that g(x; y) = g~k (x; y):lk (x; y) (mod y2k ), where g~k is monic in x, and satis es g~k = g0 (mod y). We claim g~k = gk . To do so we set ~hk (x; y) = lk (x; y)h(x; y) and then notice that f(x; y) = g(x; y)h(x; y) = g~k (x; y)lk (x; y)h(x; y) (mod y2k ) = g~k (x; y)h~ k (x; y) (mod y2k ): Additionally g~k is monic and equals g0 (mod y).

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