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College Algebra Exam Review 381

# College Algebra Exam Review 381 - J are linearly dependent...

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8.5. FINITELY GENERATED MODULES OVER A PID, PART II. 391 The elementary divisors of M are .x ± 2/ 4 ;.x ± 2/ 2 ;.x ± 1/ 2 ;.x ± 1/ , and .x 2 C 1/ 3 . Regrouping the direct summands gives: M Š ± Q ŒxŁ=. .x ± 2/ 4 / ˚ Q ŒxŁ=. .x ± 1/ 2 / ˚ Q ŒxŁ=. .x 2 C 1/ 3 / ² ˚ ± Q ŒxŁ=. .x ± 2/ 2 / ˚ Q ŒxŁ=. .x ± 1// ² Š Q ŒxŁ=. .x ± 2/ 4 .x ± 1/ 2 ± x 2 C 1 ² 2 / ˚ Q ŒxŁ=. .x ± 2/ 2 .x ± 1//: The invariant factors of M are .x ± 2/ 4 .x ± 1/ 2 ± x 2 C 1 ² 3 and .x ± 2/ 2 .x ± 1/ . Exercises 8.5 8.5.1. Let R be an integral domain, M an R –module and S a subset of R . Show that ann .S/ is an ideal of R and ann .S/ D ann .RS/ . 8.5.2. Let M be a module over an integral domain R . Show that M=M tor is torsion free 8.5.3. Let M be a module over an integral domain R . Suppose that M D A ˚ B , where A is a torsion submodule and B is free. Show that A D M tor . 8.5.4. Let R be an integral domain. Let B be a maximal linearly indepen- dent subset of an R –module M . Show that RB is free and that M=RB is a torsion module. 8.5.5. Let R be an integral domain with a non–principal ideal J . Show that J is torsion free as an R –module, that any two distinct elements of
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Unformatted text preview: J are linearly dependent over R , and that J is a not a free R –module. 8.5.6. Show that M D Q = Z is a torsion Z –module, that M is not ﬁnitely generated, and that ann .M/ D f g . 8.5.7. Let R be a principal ideal domain. The purpose of this exercise is to give another proof of the uniqueness of the invariant factor decomposition for ﬁnitely generated torsion R –modules. Let p be an irreducible of R . (a) Let a be a nonzero, nonunit element of R and consider M D R=.a/ . Show that for k ² 1 , p k ± 1 M=p k M Š R=.p/ if p k divides a and p k ± 1 M=p k M D f g otherwise. (b) Let M be a ﬁnitely generated torsion R –module, with a direct sum decomposition M D A 1 ˚ A 2 ˚ ³³³ ˚ A s ; where ´ for i ² 1 , A i Š R=.a i / , and...
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