Past exam of the mathematics course of the University of Cambridge / 2008 / ib / Paper 4 / 11G / Solution

An R-module $M$ is a free module if it has a basis $(e_i{)}_{i\in I}$: every $m\in M$ has a unique expression as a finite sum ${\sum }_i{r}_i{e}_i$, with $r_i\in R$. Equivalently, it is a direct sum of copies of $R$. Let $q:M\to M/N$ be the quotient map. When $M/N$ is free, choose a basis $({\overline{f}}_j)$ and lifts $f_j\in M$. The universal property of a free module defines an $R$-linear map $s:M/N\to M$ by $s({\sum }_j{r}_j{\overline{f}}_j)={\sum }_j{r}_j{f}_j$. It satisfies $q\circ s=\mathrm{id}$, so every $m\in M$ decomposes uniquely asThis gives a split short exact sequence and $M\cong N\oplus (M/N)$. If $N$ is also free, the union of its basis and the lifted quotient basis is a basis of $M$: spanning follows from the decomposition, and applying $q$ to a putative relation kills all quotient coefficients before independence in $N$ kills the remaining ones. Thus free submodule and free quotient imply a free module.

For the first claim, take $R=\mathbb{Z}$ andHere $N\cong N^{\prime }\cong \mathbb{Z}$ are free modules and $M/N\cong M^{\prime }/N^{\prime }\cong \mathbb{Z}/2\mathbb{Z}$. But $M$ is a torsion-free module, whereas $M^{\prime }$ contains the nonzero torsion element $(0,\overline{1})$. Therefore claim (1) is false.

For the second claim, the same lifting construction works without assuming that $N$ is free. Since both isomorphic quotient modules are free, it yieldsThus claim (2) is true. Equivalently, this uses the fact that free modules are projective.