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Proof Consider the splittingd eld of xqd x over GF(q). Call this splitting eld K. Now, consider the set S = f 2 K such that q = 0g. We claim S satis es the conditions of the theorem. We need to show that: 1. S is a eld. 2. jS j = qd To see that S is a eld, it su ces to check that for all ; 2 S, + ; straightforward: ; ; 1 2 S { which is fairly 1. ( + )qd = qd + qd + multiples of p = qd + qd = + . Therefore, + 2 S. 2. ( )q d = q d q d = : Therefore, 2 S. 3. ( )qd = 1qd qd = 1qd = : The last equality follows from the previous one because qd is odd whenever q is odd, and q is even only if the underlying prime p is 2, in which case, 1 = 1 for that eld.

Factoring polynomials over the rationals can be reduced to this case by a clearing of denominators. This factorisation can be further extended to factoring multivariate P polynomials as discussed in the previous 3 lectures. Given a polynomial f(x) = ni=0 ai xi 2 Zx];(an 6= 0), factoring f involves nding irreducible polynomials f1 ; f2; : : :fk 2 Zx] and c 2 Zsuch that f(x) = cf1 (x)f2 (x) : : :fk (x). As integer factorisation is hard, we will relax the uniqueness restrictions to just the following: each of the fi 's is an irreducible polynomial over Zof degree greater than 1 and c 2 Z.

If and F be a nite set, and f(x; y1 ; y2; : : :; yn) be a monic polynomial in x whose total @f 6= 0 @x Pr^ f(x; a1 t + b1; a2t + b2; : : :; ant + bn) is reducible ] a^;b O(d5) ; jS j where a^ and ^b are chosen uniformly and independently from S n , then f is reducible. The rest of this section is devoted to the proof of this theorem. 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.