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The idea to x this is to consider \shift" y by a value 2 R { speci cally, consider f (x; y)def = f(x; y + ) for 2 R. Clearly, if we factor f (x; y) then we can recover the factors of f(x; y) as well. We hope to nd (and work with) a such that f (x; 0) (= f(x; )) has no repeated factors as a polynomial in R x]. 1 If jRj > 4d2, then for any polynomial f(x; y) 2 R x; y] with degx(f) d and degy (f) d, there exists a 2 R such that f (x; 0) 2 R x] has no repeated factors. Proof: Consider the discriminant discx(f) treating f as belonging to (R y]) x].

C B~ = B @ ... A 0 0 b2n : : : bmn Repeat this process till B is reduced to a matrix of the form CA O where A is a m0 m0 lower triangular matrix (2 Zm0 m0 ), C is a (n m0 ) m0 matrix (2 Z(n m0) m0 ) and O is the n (m m0 ) zero matrix. Clearly the basis B = CA also spans L(b1 ; : : : ; bm ) and it is also a linearly independent set. With this the problem of nding a small vector reduces to: Given a linearly independent set of vectors fb1 : : : ; bm g 2 Zn, nd a small vector in L(b1; : : :bn ) as: For any lattice L = L(b1 ; : : : ; bn) in Zn, the determinant of the lattice L (denoted as det(L)) is de ned 0j j 1 j det(L) = det @b1 b2 : : : bn A j j j Though this de nition seems to be dependent of the choice of the basis, we shall show that det(L) (modulo f 1g) is independent of the basis.

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.

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