Algebra and Computation, Edition: version 25 Mar 1999 by Madhu Sudan

By Madhu Sudan

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Output: A non-trivial factorisation of f if f is reducible over Zx]. 1. Preprocessing of f: Firstly perform Square Free Factorisation and reduce the problem to the case of factoring a square free polynomial f. Then obtain a suitable \small" prime p such that an 6= 0 (mod p). ) 2. (a) Factor f(x) (mod p) as f(x) = g0(x)h0 (x) (mod p) where g0 is an irreducible monic polynomial relatively prime to h0. (Note such a g0 exists as f is square free) (b) Iteratively lift this factorisation using Hensel's Lemma to obtain f(x) = gk (x)hk (x) (mod pk ) 3.

2 Hensel's Lifting Lemma Hensel's Lemma tell us how to lift a factorization modulo y, to obtain a factorization modulo yk for any k. For example, given polynomial f(x; y) = x(x + 1) + y2 we obtain factorizations f = x(x + 1) (mod y) f = x(x + 1) (mod y2 ) f = (x + y2 )(x + 1 y2 ) (mod y4 ) .. One may think of this process as nding approximate roots of polynomial fy (x) = f(x; y), when y is \small". If y 0, than fy ( 1) 0 is an approximate solution and x + 1 is a factor modulo y. A better approximation is 1 + y2 which gives the factor x + 1 y2 modulo y4 .

JS j = qd To see that S is a eld, it su ces to check that for all ; 2 S, + ; straightforward: ; ; 1 2 S { which is fairly 1. ( + )qd = qd + qd + multiples of p = qd + qd = + . Therefore, + 2 S. 2. ( )q d = q d q d = : Therefore, 2 S. 3. ( )qd = 1qd qd = 1qd = : The last equality follows from the previous one because qd is odd whenever q is odd, and q is even only if the underlying prime p is 2, in which case, 1 = 1 for that eld. Therefore, 2 S. 4. ( 1)qd = ( qd ) 1 = 1. Therefore, 1 2 S. To compute the cardinality of S observe that S cannot have more than qd elements since xqd x has at most qd roots (since its degree is qd ).

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