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Every degree-$n$ number field has a ring of integers that is a free abelian group of rank $n$. To prove this without assuming finite generation, scale a rational basis to algebraic integers ${\alpha }_1,\dots ,{\alpha }_n$ and set $M={\sum }_j\mathbb{Z}{\alpha }_j$. Their trace pairing has an invertible integer matrix $T$, since the trace form of a number field is nondegenerate. For $x\in {\mathcal{O}}_K$, all $\mathrm{Tr}(x{\alpha }_j)$ are integers. The adjugate matrix formula gives $M\subseteq {\mathcal{O}}_K\subseteq (\det T{)}^{-1}M$. The submodule theorem for free modules over a principal ideal domain makes the middle group free of rank $n$, and a group basis is an integral basis.

For a number field $K$ of degree of a field extension $n$, an integral basis is a basis ${\omega }_1,\dots ,{\omega }_n$ over $\mathbb{Q}$ such thatThus every algebraic integer in $K$ has a unique expression with integer coefficients in this basis. Merely requiring its entries to be algebraic integers would not suffice: for example, $1,2\sqrt{2}$ is a rational basis of $\mathbb{Q}(\sqrt{2})$ but misses the algebraic integer $\sqrt{2}$ in its integer span.

Here is a proof of the existence of an integral basis. Start with any rational basis $v_1,\dots ,v_n$. Each $v_j$ has a nonzero integer multiple ${\alpha }_j$ that is an algebraic integer. Indeed, ifchoose a positive integer $q$ clearing all the denominators. Then $qv$ satisfies the monic polynomialThe ${\alpha }_j$ are still a rational basis. Put $M={\sum }_j\mathbb{Z}{\alpha }_j$ and form the matrixIts entries are integers, because products and sums of algebraic integers are algebraic integers and a rational algebraic integer is an integer. The nondegeneracy of the trace form of a number field gives $D\ne 0$: if $x\ne 0$, then $\mathrm{Tr}(x{x}^{-1})=n\ne 0$.

For $x={\sum }_j{x}_j{\alpha }_j\in {\mathcal{O}}_K$, each $\mathrm{Tr}(x{\alpha }_i)$ is an integer, soThe adjugate matrix formula for $T^{-1}$ gives $D{x}_j\in \mathbb{Z}$. ConsequentlyThe submodule theorem for free modules over a principal ideal domain says that every subgroup of the finite-rank free abelian group $D^{-1}M$ is again a free abelian group of rank at most $n$. Since ${\mathcal{O}}_K$ contains $M$, its rank is exactly $n$. A basis of this free abelian group is the required integral basis.

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.

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.