College Algebra Exam Review 345

College Algebra Exam Review 345 - 8.2. HOMOMORPHISMS AND...

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Unformatted text preview: 8.2. HOMOMORPHISMS AND QUOTIENT MODULES ' M qqqqq qqqq qqq qqqqq 355 P qqqqqqqqqq qqq q ∼ =' Q qqqqq qqqqq qqqqqqq M=N Proof. The homomorphism theorem for groups (Theorem 2.7.6) gives us an isomorphism of abelian groups ' W M=N ! P satisfying ' ı D ' . Q Q We have only to verify that ' also respects the R actions. But this follows Q at once from the definition of the R action on M=N : '.r.m C N // D '.rm C N / D '.rm/ Q Q D r'.m/ D r '.m C N /: Q I Example 8.2.6. Let R be any ring, M any R–module, and x 2 R. Consider the cyclic R–submodule Rx . Then r 7! rx is an R–module homomorphism of R onto Rx . The kernel of this map is called the annihilator of x , ann.x/ D fr 2 R W rx D 0g: Note that ann.x/ is a submodule of R, that is a left ideal. By the homomorphism theorem, R=ann.x/ Š Rx . Proposition 8.2.7. (Correspondence Theorem) Let ' W M ! M be an R–module homomorphism of M onto M , and let N denote its kernel. Then A 7! ' 1 .A/ is a bijection between R–submodules of M and R– submodules of M containing N . Proof. By Proposition 2.7.12, A 7! ' 1 .A/ is a bijection between the subgroups of M and the subgroups of M containing N . It remains to check that this bijection carries submodules to submodules. This is left as an exercise. I Proposition 8.2.8. Let ' W M ! M be a surjective R–module homomorphism with kernel K . Let N be a submodule of M and let N D ' 1 .N /. Then m C N 7! '.m/ C N is an isomorphism of M=N onto M =N . Equivalently, M=N Š .M=K/=.N=K/. ...
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