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Induct on $n$ to prove the stronger assertion that any submodule $M\subseteq R^n$ is a free module of rank of a free module at most $n$. For $n=0$ this is immediate. Project onto the first coordinate by $\pi :R^n\to R$. Its image $\pi (M)$ is an ideal of the principal ideal domain, so $\pi (M)=(a)$.

If $a=0$, then $M$ embeds in the remaining $n-1$ coordinates and the induction hypothesis applies. Otherwise choose $v\in M$ with $\pi (v)=a$. Put $N=M\cap \ker \pi$, a submodule of $R^{n-1}$, which is free by induction. For every $m\in M$, there is $b\in R$ with $\pi (m)=ba$, so $m-bv\in N$. If $bv\in N$, then $ba=0$; the integral domain property and $a\ne 0$ give $b=0$. ThereforeAppending $v$ to a basis of $N$ proves that $M$ is a free module of rank of a free module at most $n$. This proof of the submodule theorem for free modules over a principal ideal domain does not assume in advance that $M$ is a finitely generated module.

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

If (i) holds, the submodule theorem for free modules over a principal ideal domain says that any submodule $N\subseteq R^n$ is free. Its rank is at most $n$, since tensoring the inclusion with $F$ embeds $N{\otimes }_{R}F$ into $F^n$. Hence $N$ has at most $n$ generators, proving (iv).

Conversely, apply (iv) with $n=1$. Every ideal is an $R$-submodule of $R$ and therefore has one generator, which is (i). Thus all four conditions are equivalent.

We prove by induction on $n$ that every submodule $N\subseteq R^n$ is free. The case $n=0$ is immediate. Let $\pi :R^n\to R$ be projection onto the first coordinate. Since $R$ is a principal ideal domain,for some $d$. If $d=0$, then $N\subseteq R^{n-1}$ and induction applies. Otherwise choose $x\in N$ with $\pi (x)=d$. For every $y\in N$, write $\pi (y)=rd$; then $y-rx\in \ker (\pi {|}_N)$. Also $Rx\cap \ker (\pi {|}_N)=0$ because $R$ is a domain and $d\ne 0$. ThereforeThe kernel is a submodule of $R^{n-1}$ and is free by induction, while $Rx\cong R$. Hence $N$ is free. This is the submodule theorem for free modules over a principal ideal domain.

If $P$ is finitely generated and projective, choose a surjection $f:R^n\to P$. Projectivity supplies $h:P\to R^n$ with $f\circ h={\mathrm{id}}_P$, soThus $P\cong h(P)$ is a submodule of the finitely generated free module $R^n$, and the result just proved shows that $P$ is free.