By John Dauns

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U(S)k(S) + u(T)k(T)c T + u(T)k(T) there is a S Ic +... u(R)k(R)c +... C € K R such S Ic ; \ and rearrangement of terms shows that I a(S,S-l)-la(l,l)-l If B = a(S-l,S)-S(-l)a(l,~)-S(-l). is defined as the element B= -1 -1 -1 -1 u(S )a(S,S ) a(l,l) 0 ~ a-ca(l/cS) Note that = u(T)k(T) (l-cT IcS)+.. +u(R)k(R) (l-cR/cS). 1 - cT/cS ~ O. In the lemma below, the hypothesis that F c K is a normal separable extension which is assumed throughout this section is actually not needed. - F c K, and regard v as a finite dimensional vector space v = K F.

E. 4Z, Theorem 14J, is I F The hypothesis that c K K F[6J : is a simple finite + u(R)K : u(S)kS ku(S) +... 185, Theorem 4J). k : u(ST)a(S,T) E K; S,T,R E G zoJ). is a finite separable algebraic for the basis of a right K-vector space associativity among all products of three basis elements. 4. select symbols not in The order G (KI F)! 194, Theorem zJ). of the Galois group : u(S)K over F left elementwise fixed by every G, ca : c} : F. 44, Theorem 15; E A ,. S,T,R u(S)(u(TR)a(T,R» : u(STR)a(S,TR)a(T,R); R (u(S)u(T»u(t) : (u(ST)a(S,T»u(R) : u(STR)a(ST,R)a(s,T) .

Will 1 € G. as will be a division algebra. 2. 14), th~ cyclic algebra D fixed. be simply The set G It is I;s. (ko(p-l» 49], form the fields sP = t, Then Q[I;](t) = 1Q1(I;,t)c K = F[s]. elementwise times K be the group of all automorphisms of and transcendental p for a prime 1;, I;P = 1, a primitive 1Q1, ... ,R} -1 S(-l) = S , S -S '-S(k) cS = c , l/cS = c , l/cS(k) = c . 3. 60 - - - or S,T,R,a Automorphisms or endomorphisms will only act on K but not y. 7. the following holds (Alternatively, they may be R a(S,TR)a(T,R) : a(ST,R)a(S,T) .