By Radford E.M.

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If i = 1 , . . , n. then there exists an integer m > 0 and elements a . Q , . . a m _i £ F L such that X™ + om-lXrn-1 + ••• + aiXi + OQ = 0. TM Copyright n 2003 by Marcel Dekker, Inc. All Rights Reserved. Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved. F) such that rXdX^ cancels either with X™ or with some C,Xd' ' X{ ', where j ^ f < m. In the first case rXd = X™~j and in the second (C,Xd'}-l(rXd} = xf~j. But (CiXd'}-l(rXd} is the smallest term of a~,la,j € FL viewed as an element of f .

D COROLLARY 3. Let m = we have f1 in (5). Then for any index j X,6M € k. We shall now prove Theorem 5 using the argument of [AC]. Suppose that Ui = Vi + v'^Xi), 1 < i < n, where suppf,; n Zej = 0. /j C Ze^. By Corollary 3 there exists w' € f such that supp w' C supp Uj and (adgW^X] = u : i - X j X E j , where Aj is from Corollary 3. Hence we can replace each ut by ut — (adgw')Xt and assume that Uj = XjXj. Applying again Corollary 3 we deduce that ut — XtXf for any index t = I, . . , n. D TM Copyright n 2003 by Marcel Dekker, Inc.

Uin), 1 < i < m. Xf according to Theorem 1, where oti, . . ,an € fc* and e = ±1. Then e ii+-+in/ 7=1 U % TM Copyright n 2003 by Marcel Dekker, Inc. All Rights Reserved. Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved. //, = 0. Proof. By Theorem 6 for any index j = 1, . . , n we have • • • X? = v VjX . Hence dlXv = (v\l • ••v'^)Xv. If t = (*i, . . ,t n ) G Zr x (NUO) n ~ r , then by Proposition 1 t v v t (ad A"')*" = x x - x x = n 9/i )-( 11 fy i Ov^\ I TT *il (13) D THEOREM 10. Let k be a field of characteristic zero and n > 3.