MATH 341 Final, Fall 2002

American University of Beirut · Modules and Rings I · 1 page

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MATH 341 Final, Fall 2002, page 1

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Mathematics 341 Final Exam Jan 24, 2003 Time: 2.5 hrs N. Nahlus problems of the following 10 problems. 1 details and omit all trivialities) Part 1:(80%) Answer any 8 (Include all essentia xtensions. (Yes: trivial) Show that an injective module has no proper essential ¢: divisible R-module is injective. 1. that, over a p.i.d R, every 2. Use Baer’s Lemma to show 3, Show that every R-module M has an injective resolution: a long exact sequence as where each Q, is an injective R-module. Z-modules, show that Z, @Q=0 4, (a) Over that M@R=M (b) Over (general R-modules), show ng of R-modules). Suppose Ris a commutative domain, A is torsion- 5. Let O=a@beA@B (tensori free and B is free. Show that a=0 or b-0. modules over any domain R. 6. Let M<, Q bean essential extension of R- If M is also divisible & Q is torsion-free, show that M=Q. (Hint: Let geQ-M) Mis also torsion-free, show that dule over a commutative domain. If zero elt. in). 7. Let M be a divisible R-mo Mis injective. (Hint: In applying Baer’s Lemma, start with any non- ry Decomposition Theorem ). 8. Let M bea torsion module over a piDR.Let M= @M , (Prima Let A be a submodule of M. Show that 4=@(ANM,) ofan Artinian module is an automorphism . 9. Show that every injective endomorphism dose not split (in Z-modules) 10. Show that the inclusion map Z ®ZOZB Part 2: (20%) module which is also projective must be the zero module. A) Every injective Z- lem using divisibility). (Hint: Restate the prob ar map of R-modules. Jf R is a commutat sim & f(i)=i'm' where j and i’ are non ive domain, I is an ideal of R, Show that_m=m!' B) Let f:12M be line -zero elts of I. Mis torsion free, f(@) C) Let RCS bean extension of rings such that 1, =1s. Show that S" =S.@ R" ective. D) LetRbea left noetherian ring. Show that any direct sum of injective R-modules is also inj a domain with the property that every fg R-module is projective. E) Let R be ule of a free R-module is free. Show that every submod