# A concrete approach to division rings by John Dauns

By John Dauns

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Example text

C(K;D) D . is a left D, > module: D, KeD but also the subfield K(a) S C(K) is contained Then K(a) c D is a commutative subfield of D <~: If is a maximal subfield < a-I E C(K). properly c E K. t : D -> be any division ring and Then K[a] S C(K), s ::;Lb @ c E D ~ K, As a consequence of the commutativity of D @ K D . F, and take any element K. Let K. ~ K, which is a sum of a finite number of terms D, k LEMMA. CK. 3. product V k is understood, the commutator of as follows: the D-commutator For any subset such as.

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) .

Kz b = aN (c) .. (ccr)c € F, ,b) isomorphic. If where C € K, ==-=> : . 22. y <==-=> ,cr,a). PROOF. m - b(',') and induces Set c = The last equation becomes c(cr). i = 1: 1 = i = 2: = i = ccrc c(cr(2» c(cr(2» = ccrc; c(cr(3» = ccr(2) c crc; an 1 c(cr(2»crc c( cr(3 » (K/F,cr,a)-> I (K/F,cr,b). c crc = K. i = m-l:! 23. 1 - c(cr(m-2»crc c(cr(m-l» 2 : F\{O}. 23. - <-: for some 0(0) 75 - - Conversely, now Define C <: K. 74 E it = = is oem) given 1 = b that <: = aN(c) Being moti- G. D be a division algebra with Let contains a maximal subfield D of whether 0 ~ i ~ m-1.