Download Geometric Transformations II by I. M. Yaglom, Allen Shields PDF

By I. M. Yaglom, Allen Shields

This ebook is the sequel to Geometric modifications I which seemed during this sequence in 1962. half I treas length-preserving ameliorations, this quantity treats shape-preserving modifications; and half III treats affine and protecting modifications. those periods of changes play a basic position within the group-theoretic method of geometry. As within the prior quantity, the remedy is direct and straightforward. The creation of every new concept is supplemented through difficulties whose suggestions hire the belief simply provided, and whose distinctive suggestions are given within the moment half the ebook.

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PROOF. We easily can see that F is an inductive set, and F has a maximal member by Zorn's Lemma. If J1 , J2 are ideals such that P C J1 , P C J2, then by the maximality of P, both J1 fl S and J2 fl S are nonempty. Let s E J1 fl S, s' E J2 n S. Then ss' E S, and ss' E Jl J2 ¢ P. 1, we see that P is a prime ideal. D. REMARK. 2, maximal members of F are called maximal ideals with respect to S. If S consists only of invertible elements, then a maximal ideal with respect to S is simply called a maximal ideal of R.

The proof is easy and we omit it. REMARK. In the theorem above, if M is also an R-module, then Hom(M, N) can have an R-module structure by defining (ao) (m) = d(am) . This multiplication coincides with the one in the theorem if 0 E Horn R(M, N). 4. Let M be a module D Mr of submodules of M over a ring R. A descending sequence Mo D is called a descending chain, and r is called the length of the chain. If there is a submodule M' such that Mi D M' D Mi+I , then we can have a longer chain by inserting M'.

PROOF. We start with (iii)-(v). Assume that R is a field and I is an ideal different from 101. Let a be a nonzero element of I. By the definition of a field, a has an inverse a-1 . Therefore, I D aR = R, which shows (iii) implies (v). It is obvious that (iv) is equivalent to (v). Assume that (v) holds. If 0 54 a E R , then aR = R , and for some b E R , we have ab = I. By the commutativity of R, we see that b is a-1 . Thus, (v) implies (iii). 4 and the equivalence between (iii) and (iv) proved above.

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