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Moreover, g0 = g (1 yk qg ) + yk q(g g) I g (1 + u) where u = yk qg 2 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.

To simplify the notations, we \forget" the xed ^b and assume that f(x; 0; : : :; 0) is square free. 3 and then we show how we use this ideas for proving the the theorem. We have shown that if f(x; b1; : : :; bn) is not square free then its discriminant is zero. We observe that the discriminant of f(x; b1; : : :; bn) is obtained from the discriminant of f(x; y1 ; : : :; yn) by substituting y1 = b1 ; : : :; yn = bn . Thus, the discriminant of f(x; y1 ; : : :; yn) is a polynomial that is zero for \many" substitutions and therefore must be identically zero.

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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