By Madhu Sudan

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3 A Warm-up Lemma In this section we prove that if with \high" probability f(x; b1 ; : : :; bn) is not square free then f is reducible. This lemma will be used in the sequel to assume that f(x; 0; : : :; 0) is square free. 1. This lemma requires that @f @x 6= 0, which is not really a restriction when the characteristic of F is zero, but is more problematic over nite elds. 8. We start with a simple observation. 2 If the polynomial f(x; y1 ; : : :; yn) is monic in x and g(x; y1 ; : : :; yn) divides f then g is monic in x.

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 . Now, for all non zero elements 0 ; 1 : : : ; l where l is chosen to be smallest element such that there exist 2 K we form the sequence P 0 ; 1; : : : ; l 2 F with i i = 0 and ( 0 ; 1 ; : : : l ) 6= (0; 0; : : : ; 0). We know that l d because K has dimension d. Now,Pwel can restrict l = 1 because the last term is always non-zero, and we can divide out.

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. 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.