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By Madhu Sudan

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Example text

The discriminant of f is a polynomial in y1 ; : : :; yn which we denote (y1 ; : : :; yn). Clearly, (b1 ; : : :; bn) = ^b: By the conditions of the lemma, with probability greater than (2d2)=jS j over the ^b's, the discriminant (b1 ; : : :; bn) is zero. Thus, (y1 ; : : :; yn ) is a polynomial of the n variables y1 ; : : :; yn which has degree at most 2d2, and is zero with \high" probability. , gcd(f; @f @x ) 6= 1. 3 the polynomial f is reducible, or there exists some ^b such that f(x; b1 ; : : :; bn) is square free.

A b1n b2n : : : bmn 10 Lp norm for a vector x = (x1 ; : : : ; xn ) is given by kxkp = ( P xp) p i 1 52 LECTURE 10 P 1. (a) Compute g = gcd(b11; b21; : : : ; bm1 ) and integers a11; a21; : : : ; am1 such that mi=1 ai1bi1 = g (b) Construct a new basis B 0 = (b01; : : : ; b0m ) as follows b01 = b0k = bk m X ai1bi i=1 bk1 b0 ; k 6= 1 g 1 Note that B 0 also spans L(b1; : : : ; bm ) and that b011 = g and b0k1 = 0; k 6= 1] 2. Repeat step 1 for B~ given by 0 b0 : : : b0 1 22 m2 .. C B~ = B @ ... A 0 0 b2n : : : bmn Repeat this process till B is reduced to a matrix of the form CA O where A is a m0 m0 lower triangular matrix (2 Zm0 m0 ), C is a (n m0 ) m0 matrix (2 Z(n m0) m0 ) and O is the n (m m0 ) zero matrix.

Thus, we get a factorization of r. 1 We next claim that if r is reducible and has a \good" factorization then f is reducible. 7 and therefore omitted. 8 If r = g(x; y1 ; : : :; yn) h(x; y1; : : :; yn) for some polynomials g and h such that degx(g) > 0 and degx (h) > 0 then f is reducible. Therefore, we get an e cient procedure to check if a polynomial r(x; y1; : : :; yn) is irreducible. If gcd(r0(y1 ; : : :; yn); r1(y1 ; : : :; yn ); : : :; re(y1 ; : : :; yn )) 6= 1 then r is reducible. 14). Test, using Hilbert's irreducibility theorem, if f is reducible.

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