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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.

A subset $S\subseteq M$ freely generates $M$ when every $m\in M$ has a unique expressionwith $r_s\in R$ and only finitely many nonzero coefficients. Equivalently, $S$ is a basis of a module and $M$ is the free module on $S$.

If $S$ freely generates $M$ and $f:S\to N$ is any function, defineUnique coordinates make $\varphi$ well defined; it is an $R$-module homomorphism and is the only extension of $f$. Conversely, apply the proposed universal property to the free module $F=R^{(S)}$ and the inclusion of $S$ into $M$. It gives maps $F\to M$ and $M\to F$ extending the corresponding functions on $S$. Uniqueness makes both composites identity maps, so $M\cong F$ and $S$ freely generates $M$. This is the universal property of a free module.

Now let $T$ generate the free module $M$ and $|T|=m$. Choose a maximal ideal $\mathfrak{m}$ of the nontrivial ring $R$. Then $k=R/\mathfrak{m}$ is a field, andis a $k$-vector space. The images of a basis $S$ of $M$ form a vector-space basis, while the images of $T$ span it. ThereforeIn particular $S$ is finite. Applying this result in both directions to two finite bases shows that they have equal cardinality, the rank of a free module $\mathrm{rk}M$.

A Euclidean domain is a principal ideal domain. By the submodule theorem for free modules over a principal ideal domain, every submodule $N$ of the finite-rank free module $M$ is free. Since a basis of $M$ generates it, the preceding inequality applied in the standard proof gives

The primary decomposition theorem for finitely generated modules over a principal ideal domain states thatwhere $p$ ranges over finitely many nonassociate irreducibles and the positive exponents are uniquely determined up to order. Equivalently, the torsion part decomposes into its primary cyclic summands.

Let $H$ be a finite subgroup of the multiplicative group of a field. It is a finite abelian group, hence the theorem over $\mathbb{Z}$ gives an invariant-factor decompositionIts exponent is $d_r$, so every element of $H$ is a root of $X^{d_r}-1$. A degree-$d_r$ polynomial over a field has at most $d_r$ roots, whenceBut $d_r\le |H|$, and equality forces all earlier factors to be trivial. Thus