College Algebra Exam Review 340

# College Algebra Exam Review 340 - 350 8 MODULES Example...

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Unformatted text preview: 350 8. MODULES Example 8.1.25. Let R be the ring of n–by–n matrices over a ﬁeld. Let M be the left R–module of n–by–s matrices over K . Let T be a ﬁxed s –by–s matrix over K . Then right multiplication by T is an R–module endomorphism of M . We will discuss module homomorphisms in detail in the next section. Direct Sums Deﬁnition 8.1.26. The direct sum of several R–modules M1 ; M2 ; : : : ; Mn is the Cartesian product endowed with the operations 0 0 0 0 0 0 .x1 ; x2 ; : : : ; xn / C .x1 ; x2 ; : : : ; xs / D .x1 C x1 ; x2 C x2 ; : : : ; xn C xn / and r.x1 ; x2 ; : : : ; xn / D .rx1 ; rx2 ; : : : ; rxn /: The direct sum of M1 ; M2 ; : : : ; Mn is denoted M1 ˚ M2 ˚ ˚ Mn . In a direct sum of R–modules M D M1 ˚ M2 ˚ ˚ Mn , the subset f M i D f0g ˚ ˚ Mi ˚ ˚ f0g is a submodule isomorphic (as R–modules) to to Mi . The sum of these submodules is equal to M . When is an R–module M isomorphic to the direct sum of several R– submodules A1 ; A2 ; : : : ; An ? The module M must be isomorphic to the direct product of the Ai , regarded as abelian groups. In fact, this sufﬁces: Proposition 8.1.27. Let M be an R–module with submodules A1 ; : : : As such that M D A1 C C As . Then the following conditions are equivalent: (a) .a1 ; : : : ; as / 7! a1 C C as is a group isomorphism of A1 As onto M . (b) .a1 ; : : : ; as / 7! a1 C C as is an R–module isomorphism of A1 ˚ ˚ As onto M . (c) Each element x 2 M can be expressed as a sum x D a1 C (d) C as ; with ai 2 Ai for all i , in exactly one way. If 0 D a1 C C as , with ai 2 Ai for all i , then ai D 0 for all i. Proof. The equivalence of (a), (c), and (d) is by Proposition 3.5.1. Clearly (b) implies (a). On the other hand, the map ' W .a1 ; : : : ; as / 7! a1 C Cas ...
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