By Madhu Sudan

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4 For any two bases B = (b1; : : : ; bn); B0 = (b01 ; : : : ; b0n) of a lattice L, det(B) = det(B 0). Proof As B is a basis for the lattice L and b0i 2 L, there exists a n n matrix A 2 Zn n such that B 0 = BA. Similarly there exists a A0 2 Zn n such that B = B 0 A0 . ( In fact the moves we adopt to do so leave the sign of the determinant unaltered) We move from one basis to another by a sequence of (i; j; q) moves. An (i; j; q) move from a basis (b1 ; : : : ; bn) to (b01; : : : ; b0n) is de ned as follows: b0i = bi qbj b0k = bk ; k 6= i We choose q such that the b0i is the smallest such vector.

If either u or v is 0, then by hypothesis kz k2 kak2. If both u = 6 0 and v = 6 0 Case (i) : u > v We rst observe that (a + b) (a + b) b b ) 2a b a a kz k22 = (u2a a + v2 b b + 2uva b) (u2a a + v2 b b uva a) (u(u v)a a + v2 b b) (u(u v)a a) kak22 Case (i) : u v We now observe that (a + b) (a + b) ) 2a b kz k22 a a b b (u2 a a + v(v u)b b) (u2 a a) kak22 For the case n > 2, we employ the LLL basis reduction to nd a small vector. The LLL basis reduction will be discussed in the next lecture. 1 Introduction In the last lecture, we saw how nding a short vector in a lattice plays an important part in a polynomial time algorithm for factoring polynomials with rational coe cients.

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).