By Madhu Sudan

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Moroever, if you compute gcd(f; f 0d) we see that there are no multiple roots (hence S has at least qd roots). Speci cally, gcd(f; f 0 ) = gcd(xq x; 1) = 1. We know from a previous lecture that if any polynomial and it's derivative have a greatest common divisor of 1, then the polynomial does not have multiple roots. So, we have shown that S has exactly qd elements. 3. 8 There are at least (qd 1)=d monic irreducible polynomials of degree at most d over GF(q). Proof Consider the unique extension eld K of F where jK j = qd .

8) then f is reducible. Furthermore, from the factorization of f, one can easily construct the factorization of r. P We next present the construction. Let r(x; y1; : : :; yn) = ei=0 ri (y1 ; : : :; yn)xi where e is the degree of x in r, that is, re 6= 0. 7 The function f is a polynomial in F x; y1; : : :; yn] which is monic in x. Furthermore, if f is reducible then r is reducible. Proof By de nition of f it holds that f(x; y1 ; : : :; yn) = e X i=0 (re(y1 ; : : :; yn ))e 1 iri(y1 ; : : :; yn )xi = xe + e 1 X i=0 (re(y1 ; : : :; yn ))e 1 iri (y1 ; : : :; yn)xi : Thus, f is a polynomial in F x; y1 ; : : :; yn ] that is monic in x.

For the case n = 2, a sequence of such moves leads us to a basis (a; b) where kak kbk kb + qak for all q 2 Z. In fact for this case with L2 norm, a is indeed the smallest vector in the lattice. 3. 5 If a; b 2 Z2 and kak2 kbk2 kb + qak2 for all q 2 Z, then for all z 2 L(a; b) f(0; 0)g, kak2 kz k2 . Proof Let z = ua + vb for some u; v 2 Z (u; v not both 0). 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.