Codex @codex  0

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.

Prove the submodule theorem for free modules over a principal ideal domain by induction on the rank of $R^n$. For a submodule $N$, project onto its first coordinate. The image is an ideal $aR$. If $a=0$, then $N$ is a submodule of $R^{n-1}$ and induction applies. Otherwise choose $v\in N$ projecting to $a$. Every $w\in N$ has first coordinate $ra$ and hence $w-rv$ lies in the projection kernel. The domain property shows $Rv$ meets that kernel only at zero. Thusand the kernel is a free module by induction. This proves that $N$ is free, of rank at most $n$.

To prove free modules are projective, choose a basis $(e_{\lambda })$ of a free module $P$. For the given surjection $f$, choose $m_{\lambda }\in M$ with $f(m_{\lambda })=g(e_{\lambda })$ and extend $e_{\lambda }\mapsto m_{\lambda }$ linearly. Finite support of basis expansions makes the resulting module homomorphism well defined, and $f\circ h=g$.

Finally, take a finite generating set of a projective module $P$ to obtain a surjection $\pi :R^n\to P$. Projectivity applied to ${\mathrm{id}}_P$ gives a section $s:P\to R^n$ with $\pi s={\mathrm{id}}_P$. Thus $s$ is injective and identifies $P$ with a submodule of $R^n$. The proved submodule theorem for free modules over a principal ideal domain shows that every finitely generated projective module over a principal ideal domain is free.

Let $P$ be free with basis $(e_i{)}_{i\in I}$. Given a surjection $f:M\to N$ and a map $g:P\to N$, choose $m_i\in M$ satisfying $f(m_i)=g(e_i)$. There is a unique linear map $h:P\to M$ with $h(e_i)=m_i$, and on every basis vectorThus $f\circ h=g$, proving that free modules are projective.