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But since R0 is a nite eld extension of R, we can treat it as Rl for some l, and then it is easy to transform the above requirements into a linear system over R. Indeed, assume the original system is Ax = y with A being a r n matrix over R0, and we seek solutions x 2 R0n and y 2 Rr . We can rewrite this system as the homogeneous system Bz = 0 over R0 where B = Aj Ir ] is a r n + r matrix and we seek solutions z1; z2 ; : : :; zn+r 2 R0 n+r with zn+i 2 R for 1 i r. Since R0 is a nite eld extension of R, it can be viewed as the vector space Rl , with a basis f 1; 2; : : :; l g where i 2 R0 for 1 i l and 1 = 1(2 R).
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.
This yields the claim. Now that we know a solution exists, we can easily nd polynomials g~ and lk of the required degree just by solving a system of linear equations, where the unknown are the coe cients of g~ and lk . This answers the second question above. Finally, we need to see why computing the greatest common divisor of f and g~, as polynomials in F y](x), yields a non-trivial factorization. 2). Let 2k > 2d2 where d is the (total) degree of f . Then gcdx (f; g) is non-trivial. Proof Obviously gcd(f; g~) 6= f because degx g < degx f.