Structure theorem for finitely generated modules over a principal ideal domain

Remarks

Since $R$ is Noetherian, $M$ has a finite presentation $R^m \xrightarrow{A} R^n \to M \to 0$. Row and column operations bring the matrix $A$ to Smith normal form, a diagonal matrix with entries $d_1 \mid d_2 \mid \cdots$; this gives part 1. Part 2 follows from part 1 by the Chinese remainder theorem, which splits $R/(d)$ into $\bigoplus_p R/(p^{e})$ over the prime powers $p^e$ exactly dividing $d$. The part $R/(d_1) \oplus \cdots \oplus R/(d_k)$ is the torsion submodule of $M$, and $r$ is the rank of the free quotient $M$ modulo torsion.

Example: over $\Z$, the module $\Z/2\Z \oplus \Z/6\Z$ has invariant factors $2 \mid 6$ and elementary divisors $2, 2, 3$, since $\Z/6\Z \cong \Z/2\Z \oplus \Z/3\Z$. For $R = \Z$ the theorem is the structure theorem for finitely generated abelian groups. For $R = k[x]$ and $M = k^n$ with $x$ acting by a matrix, it gives the rational canonical form and the Jordan normal form.

The principal ideal hypothesis is necessary. Over $R = \Z[x]$, the ideal $I = (2, x)$ is a finitely generated torsion-free module. Any two elements of $I$ are linearly dependent, so a basis would have one element, and $I$ would be principal. It is not principal, so $I$ is not free and not of the form above.

Going deeper

For a student

The theorem unifies two classical results: the classification of finitely generated abelian groups, for $R = \Z$, and the normal forms of matrices, for $R = k[x]$. In the second case, a matrix $A$ makes $k^n$ a $k[x]$-module, the invariant factors are the invariant polynomials of $A$, and $d_k$ is the minimal polynomial. The proof by Smith normal form is algorithmic when $R$ is Euclidean. Over a Dedekind domain that is not principal, the torsion part still decomposes, but the torsion-free part is a sum of ideals rather than a free module.