By Madhu Sudan

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**Example text**

Consider the irreducible polynomial x61 + x2 over n ) = 0 for every i then the degree of each variable in each monomial is GF(2)). Nevertheless, if @r(x@x;:::;x i divisible by p { the characteristic of the eld. Thus, if r is a non-constant polynomial and F is a nite eld then r is a p-th power of some polynomial. ;:::;xn ) 6= 0 for some i. In this case if we covert r to a function f that We are left with the case that @r(x @x i n ) 6= 0. 7, then @f (y@x;:::;x i to check if r is reducible. 2 Black-box Factorization of Multivariate Polynomials The task of black-box factorization of multivariate polynomials is de ned as follows.

Moreover, g0 = g (1 yk qg ) + yk q(g g) I g (1 + u) where u = yk qg 2 I. 1 that f g0h0 (mod I 2 ) for h0 = h (1 u), ie. g0 divides f modulo I 2 . This proves that there exists a solution g0 such that g0 is monic and of degree degx g. It remains to prove that the solution is unique. Let g00 I g be such that f = g00h00 (mod I 2 ). We have g00(h00 h) I g00 h00 gh I f f = 0, ie. yk jg00(h00 h). Since g00 is monic in x, this is possible i yk j(h00 h), ie. h00 I h. 1 and get g00 = g0(1 + u) for some u.

We have shown that if f(x; b1; : : :; bn) is not square free then its discriminant is zero. We observe that the discriminant of f(x; b1; : : :; bn) is obtained from the discriminant of f(x; y1 ; : : :; yn) by substituting y1 = b1 ; : : :; yn = bn . Thus, the discriminant of f(x; y1 ; : : :; yn) is a polynomial that is zero for \many" substitutions and therefore must be identically zero. However, if the discriminant of f is zero then f is reducible. 1. We use Hensel's lifting lemma to nd a factor of the univariate polynomial f(x; a1 t; : : :; ant).