By Madhu Sudan

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Thus, f(x; y1 ; : : :; yn) = p(x; 1; y1; : : :; yn) = (x; 1; y1; : : :; yn ) (x; 1; y1; : : :; yn): Since f is irreducible then either (x; 1; y1; : : :; yn ) is constant or (x; 1; y1; : : :; yn) is constant. , assume that (x; 1; y1; : : :; yn) 1. Since is non-constant, there is a non-zero polynomial 1(x; t; y1; : : :; yn ) such that (x; t; y1; : : :; yn) = 1 + (1 t) 1(x; t; y1; : : :; yn): This contradicts the fact that is monic in x. 1. We show how to use the factoring algorithm provided by Hensel's lifting lemma to prove that p is reducible.

We show how to use the factoring algorithm provided by Hensel's lifting lemma to prove that p is reducible. 3, we can assume that f(x; 0; : : :; 0) is square free. 1) It must hold that gcd(g0 (x) h0(x)) = 1 (since f(x; 0; : : :; 0) is square free), and we can choose g0 and h0 such that g0 is monic and irreducible. 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.

8 There are at least (qd 1)=d monic irreducible polynomials of degree at most d over GF(q). Proof Consider the unique extension eld K of F where jK j = qd . Now, for all non zero elements 0 ; 1 : : : ; l where l is chosen to be smallest element such that there exist 2 K we form the sequence P 0 ; 1; : : : ; l 2 F with i i = 0 and ( 0 ; 1 ; : : : l ) 6= (0; 0; : : : ; 0). We know that l d because K has dimension d. Now,Pwel can restrict l = 1 because the last term is always non-zero, and we can divide out.