Structure Theorem for Modules over a PID

Invariant factors and elementary divisors classify modules over a PID.

The idea

Theorem (Structure theorem for modules over a PID).

Let $R$ be a principal ideal domain. Every finitely generated $R$-module is isomorphic to $R^{r} \oplus R/(d_{1}) \oplus R/(d_{2}) \oplus \cdots \oplus R/(d_{k}), \qquad d_{1} \mid d_{2} \mid \cdots \mid d_{k},$ and the rank $r$ together with the list $d_{1}, \ldots, d_{k}$ is uniquely determined by the module. Two such modules are isomorphic if and only if those data agree.

A complete classification is a rare thing to have. Modules over a general ring are hopeless; over a principal ideal domain, where every ideal is generated by a single element, only two kinds of thing can occur — a free part $R^{r}$, and cyclic torsion pieces $R/(d)$ — and the divisibility chain $d_{1} \mid \cdots \mid d_{k}$ pins down how they are grouped. Drop that chain and you could regroup the cyclic pieces differently, and uniqueness would be lost.

There are two standard ways to record the torsion, both in use. The $d_{i}$ above are the invariant factors. Alternatively split each $R/(d_{i})$ along the prime-power factors of $d_{i}$; the prime powers that appear are the elementary divisors. Neither list is more correct than the other, each is recoverable from the other, and it pays to be fluent in both directions.

Over $R = \mathbb{Z}$ a finitely generated module is just a finitely generated abelian group, so the theorem specializes to the classification of those.

Ways to work on it

• Walkthrough. How finitely generated modules over a PID decompose into cyclic pieces, in two normal forms.
• Practice. Count the abelian groups of a given order via partition numbers.
• Hardest. Convert an elementary-divisor list into invariant factors.

Not sure where to start? Take the ten-question placement test.