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Thread: show existence of complementary submodule

  1. #1
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    show existence of complementary submodule

    I think i understand the first 2 parts of the hints but the last part of this question just confuses me =/ Prove that if N is a submodule of an R module M so that the quotientM/N is a free R module then there exists a complementary submodule to N in M. lecturer's hint: let X be a subset of M st (x + N| x in X) is basis of M/N define fn B: M/N --> M st B(x +N) = x for all x in X and show that im(B) is complementary to N
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  2. #2
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    Quote Originally Posted by gtkc View Post
    I think i understand the first 2 parts of the hints but the last part of this question just confuses me =/ Prove that if N is a submodule of an R module M so that the quotientM/N is a free R module then there exists a complementary submodule to N in M. lecturer's hint: let X be a subset of M st (x + N| x in X) is basis of M/N define fn B: M/N --> M st B(x +N) = x for all x in X and show that im(B) is complementary to N
    first note that $\displaystyle B$ is well-defined because $\displaystyle X$ is a basis for $\displaystyle M/N.$ now let $\displaystyle \text{Im}(B)=L$ and $\displaystyle z \in N \cap L.$ so $\displaystyle z=B(u),$ for some $\displaystyle u \in M/N.$ but $\displaystyle u=\sum c_j(x_j + N)=\sum c_j x_j + N,$ for some

    $\displaystyle x_j \in X, \ c_j \in R.$ thus $\displaystyle z=B(u)=\sum c_j x_j \in N.$ therefore $\displaystyle u=\sum c_j x_j + N=0.$ so we proved that $\displaystyle N \cap L = \{0 \}.$ we only now need to prove $\displaystyle M=N+L$: let $\displaystyle y \in M.$ then $\displaystyle y + N \in M/N$

    and hence $\displaystyle y+N=\sum r_jx_j + N,$ for some $\displaystyle x_j \in X, \ r_j \in R.$ thus $\displaystyle y - \sum r_j x_j = a \in N$ and so $\displaystyle y=a+ \sum r_j x_j = a+ B(\sum r_jx_j + N) \in N + L.$ hence $\displaystyle M \subseteq N + L.$ the other direction of

    the inclusion is trivial. this completes the proof of $\displaystyle M=N \oplus L.$
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