Algebra and Computation by Madhu Sudan

By Madhu Sudan

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Now given f 2 R x; y], we work with a nite extension R0 of R such that jR0j > 4d2 and obtain a non-trivial irreducible factor (if any) g0 2 R0 x; y] of f 2 R0 x; y] using the above corollary. We now need to recover a factor g 2 R x; y] using g0 2 R0 x; y]. 3 Let g0 ; f be as above. Let g 2 R x; y] be the minimal polynomial such that g0 (x; y)jg(x; y) (in R0 x; y]). Then g(x; y)jf(x; y). Proof: Since g0 is irreducible, so is the minimal polynomial g 2 R x; y] such that g0 jg. Now if g does not divide f, then 9 ux; vx 2 R x; y] and p 2 R x] (resp.

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.

See Lecture 4). 22nl, we have N 6= 0( mod p). Thus reducing the above equation modulo pk , we have N = u(x)f(x) + v(x)~g (x)( mod pk ) = u(x)gk (x)hk (x) + v(x)gk (x)lk (x)( mod pk ) = gk (x)(u(x)hk (x) + v(x)lk (x))( mod pk ) But this is a contradiction. As gk (x) is a polynomial of degree greater than 1. On the RHS, we have either the constant 0 or polynomial in x depending on whether u(x)hk (x) + v(x)lk (x) is identically equal to 0 or not. However on the LHS we have a non-zero constant N. Thus, we should have gcd(f; g~) 6= 1: Thus step 4 computes a non-trivial factor of f if the coe cients of g~ are not very large.

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