Thread:

1. 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

2. Originally Posted by gtkc
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 $B$ is well-defined because $X$ is a basis for $M/N.$ now let $\text{Im}(B)=L$ and $z \in N \cap L.$ so $z=B(u),$ for some $u \in M/N.$ but $u=\sum c_j(x_j + N)=\sum c_j x_j + N,$ for some

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

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

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