By Madhu Sudan

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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. Since g00 are both monic of degree degxg, it must be g00 = g0 . 1. Let f be the given bivariate polynomial. Our basic idea is to use the univariate factoring algorithm to nd a factorization f(x; y) = g0(x; y)h0 (x; y) (mod y). The goal of Step 2 is to lift this factorization to one modulo y2k . 1. For what follows it will also be useful to know that g0 is irreducible, so we assume that too.

3) Furthermore, gk;a^(x; t) = g0 (x) (mod t) and hk;a^(x; t) = h0 (x) (mod t). 8 If f is not monic in x then p might be reducible. For example, if f (x; y1 ; y2 ) = xy1 + y2 then p(x; t; y1 ) = t(xy1 + y2 ) is reducible. 4) (since f(x; 0; : : :; 0) = f(x; y1 ; : : :; yn) (mod y1 ; : : :; yn )). 5) gk (x; t; y1; : : :; yn)def = gk0 (x; y1t; : : :; yn t); and hk (x; t; y1; : : :; yn)def = h0k (x; y1t; : : :; yn t): It holds that p(x; t; y1; : : :; yn ) = f(x; y1 t; : : :; ynt) = gk0 (x; y1 t; : : :; ynt) h0k (x; y1 t; : : :; yn t) (mod (y1 ; : : :; yn)k ) = gk (x; t; y1; : : :; yn) hk (x; t; y1; : : :; yn ) (mod tk ) (the last equality holds since in any monomial in gk and hk the degree of t is the sum of the degrees of the yi 's in the monomial).

But the solution to the Hensel lifting is unique, implying gk = g~k . This yields the claim. Now that we know a solution exists, we can easily nd polynomials g~ and lk of the required degree just by solving a system of linear equations, where the unknown are the coe cients of g~ and lk . This answers the second question above. Finally, we need to see why computing the greatest common divisor of f and g~, as polynomials in F y](x), yields a non-trivial factorization. 2). Let 2k > 2d2 where d is the (total) degree of f .

### Algebra and Computation by Madhu Sudan

by Jason

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