By Madhu Sudan

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**Extra resources for Algebra and Computation**

**Example text**

The previous example is, in some sense, a reduction in the wrong direction since over the integers it is easier to check irreducibility then to check primality. However, we want to use this idea over arbitrary elds. The rst obstacle is that f(a1 ; a2; : : :; an) is always a product of two elements of the eld. To overcome this problem we use partial substitutions. That is, we substitute the values of the variables such that the result is a polynomial in few variables. A simple partial substitution is to substitute all the variables except for x1: given f(x1 ; x2; : : :; xn) we choose a2 ; : : :; an and consider the polynomial f(x1 ; a2; : : :; an) (this is a univariate polynomial in x1).

Multiples of a polynomial p. The set of bivariate polynomials in x; y with monomials xiyj such that i + j d. Ideals are important because we can de ne the quotient R=I, where p q (mod I) if p q 2 I, and operations on R=I are well de ned. The product of two ideals I; J is the set of all nite sums of products of elements of I and J: n X I J = f aibi : ai 2 I; bi 2 J g: i=1 For example, if, for some p 2 R, Ip is the ideal fpq : q 2 Rg (denoted (p)) then I 2 = I:I = (p2). d. f; g; 2 R and ideal I in R, we say that pseudo gcd(f; g) = 1 (mod I) to imply that there exist a; binR such that af + bg = 1 (mod 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. Since g00 are both monic of degree degxg, it must be g00 = g0 . 1. Let f be the given bivariate polynomial.

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