By John Dauns

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The F[x] is separable F[yu] as follows: (x-yu)(x-uy). The two maximal subfields I by t 3 (b 2 is nonseparable because its derivative vanishes and it has Without loss of generality O. + u multiple roots in Suppose with 2 2 u -t = 0; 2 F[y], (y+t) = 0; 2 2 F[yu],(yu) + tyu + t = o. nonseparable: The minimal det s solution contains the following maximal subfields. a+bu+dt det s is independent of d 2) are all divisibly by separable: The determinant b = bt, c = ct must be zero, or -1 ;to, sED, c + du s<-> + and The hypotheses on D (b+c+du The only 2 ao = 0, dO = 0, a,b,c, Hence a + bu ad t Thus :11y<->II: Thus = 0,1.

The converse of the next proposition is also true (see has an inverse for which I. 10). and A PROPOSITION. 14. crossed product and valent to {b(S,T) {a(S,T)}, I If A S,T € (KI F, = G} a ( . , )) . is any factor is a set equi- then these two factor sets define isomorphic I 0 ~ ak-ka = u(S)c(k-kS) +... K ~ C (K), (vii) b(',') shows fhat it follows that . Conclusion (vii) K = C(K»)c A a : G x G > K* 4 C(K). Since is a maximal algebras subfield. 11. PROOF. 13. a(','), € Finally, a straightforward verification shows that if we a is simple.

E2) b(S,l) = 1. (3) b(S -1 ,S) = b(S,S -1 S ). = w(sfl=weS-l)b(S,S-l)-l . =w(S-l)b(S-l,S)-se-l). isomor-=u(S-l)a(S-l,s)-S(-l)a(l,l)-S(-l) = l-K B: 1 = wO) -1 -1' contain a distinguished = Moreover, the K, ek above isomorphism is the identity on this k under :: B It follows from b(l,l) = c(l)a(l,l) that the identity in -1 -1 -1 -1 is e = u(l)a(l,l) = u(l)c(l)c(l) a(l,l) = v(l)b(l,l) . 16. 12 but u (S) c( S) S Since 69 In practice and literature, division rings defined in terms > l-k, of "left" are encountered as well as ones in which everything is E K.