Let R be a ring with identity. Show that R is not a free module on any
Question:
Let R be a ring with identity. Show that R is not a free module on any set in the category of all R-modules (as defined in Exercise 2).
Data from in exercise 2
Let R be any ring (possibly without identity) and X a nonempty set. In this exercise an R-module F is called a free module on X if F is a free object on X in the category of a/I left R-modules. Thus, F is the free module on X if there is a function L: X → F such that for any left R-module A and function ∫: X → A there is a unique R-module homomorphism f̅: F → A with f̅I, = f
(a) Let {Xi| i ϵ I) be a collection of mutually disjoint sets and for each i ϵ I, suppose F, is a free module on Xi with Li : xi → Fi, Let X = and Fi, with ϕi: Fi → F the canonical injection. Define L : X → F by i(x) = ϕiLi(x) for x ϵ Xi;(L is well defined since the Xi are disjoint). Prove that F is a free module on X.
(b) Assume R has an identity. Let the abelian group Z be given the trivial R-module structure (rm = 0 for all r ϵ R, m ϵ Z), so that R⊕z is an R-module with r(r' ,m) = (rr', 0) for all r ,r' ϵ R, m ϵ Z. If X is any one element set, X = { t) , let L : X → R⊕Z be given by L(t) = (lR,1). Prove that R⊕Z is a free module on X.
(c) If R is an arbitrary ring and X is any set, then there exists a free module on X.
Step by Step Answer:
Algebra Graduate Texts In Mathematics 73
ISBN: 9780387905181
8th Edition
Authors: Thomas W. Hungerford