Division algebras over fields of formal power series

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and select p>m, p>n. Let C=AB+N, D=BA. Then L=LCP 09(CP), M=MDPEfD%(DP). We shall prove that A induces an isomorphism A' on LCp onto MDP. Since for each j, CiA =ADi, we have LCPA =LADPCMDP. But MDP=MDP+1=MB(AB)PA =MBCPA CLCPA. Thus A is on LCP onto MDP. We observe that for any r, Cr=(AB)r+(AB)r-lN+ * * * +Nr, hence for r=2p, C2P=(AB)2P + (AB) 2p-'N+ * * * + (AB)P+'NP-1 since NP =Nm = o. If xCPA = 0, then x(AB)PA =O, x(AB)P+1=x(AB)p+2 = * =0. Thus XC2p=O, xCp=O, which proves that A' is an isomorphism. We finally have CA'=A'D so that the contraction of C to LCP is similar to the contraction of D to MDP. Thus C and D have the same elementary divisors which do not have zero as a root. The theorem follows from Theorem 2.