By Richard M. Wilson
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Extra info for Graph Puzzles, Homotopy, and the Alternating Group
Y ( A such that xy = b is a r e g u l a r (3) A ®R Q is the full Proof. (1) = > is a r e g u l a r (1). If x is a r e g u l a r of A. element of A, then Ax ideal of R by y ~ A such that yx = b is a r e g u l a r of R. (1). of J. By Let J be a r e g u l a r (2) there J O R is a r e g u l a r (2) = > (3). of A, then X = z/x, (2) t h e r e ideal is an e l e m e n t H e n c e k = z y / x y = zy/b. t h e r e is an e l e m e n t 1/x = y/b. 2. ring of q u o t i e n t s element of A. b o t h b and y are r e g u l a r expression ring of q u o t i e n t s Let x be a r e g u l a r element y ¢ A and a r e g u l a r J A R is a r e g u l a r in the for k shows that of A.
17. Let A be a torsion d i v i s i b l e R-module. a sum of Artinian divisible Proof. Corollary R-modules. 3.
Be a chain of divisible c HomR(K,D2) c ... 7. by the proof of (3) ~ > H-module HomR(K,Dm) c of D. 6. Thus there is an index m 0 such that = HomR(K,Dmo ) for all m _> m O. 3, p r o v i n g Therefore D m = Dm0 for all that D has ACC on divisible submod- ules. (6) -~-> (I). 3 there is an R - h o m o m o r p h i s m Since K is Artinian submodule by (3) = > and hence f(K) c A. (7). This is a trivial (7) = > (5). It is a consequence R-modules. submodules, Suppose divisible We assert divisible submodule implication.