PID modules

Overview

The theory of modules over a PID is very interesting since it is simple enough to understand while giving a feel for deeper ring theory questions.

PID’s show up everywhere and many common examples of modules are over PID’s so this is a good place to start anyways.

Throughout these notes consider a ring $R$ to be a PID. The main goal (and final theorem) is to classify all modules over $R$.

A notion of dimension

Let $M$ be a free R-module, and say that $M\cong R^{\oplus X}$ for some finite set $X$. The cardinality of $X$ is well-defined (that is if $M\cong R^{\oplus Y}$ then $|X|=|Y|$.) Thus, we write ${\dim }_{R}M$ for the cardinality of $X$.

This builds in a canonical way off the idea of dimension for a vector space. We know that dimension is invariant for any choice of basis in a vector space, and that it “plays nice” with subspaces, quotienting, etc.

Proof

todo

Finitely generated submodules

Let $M$ be a (finitely generated) free $R$-module. Then any submodule $N\subset M$ is free and ${\dim }_{R}N\le {\dim }_{R}M$

Proof

Torsion and free modules

A finitely generated torsion free $R$-module is free.

Proof

Pulling back generators

Let $f:M\to N$ be a surjective module homomorphism. If $N$ is free, then there exists a free submodule $F\subset M$ such that $f{|}_F$ is an isomorphism and

$$M\cong \ker f\oplus F$$

Proof

Quotienting out torsion

Let $M$ be a finitely generated $R$-module. Then $M/M_{\text{tor}}$ is free. Moreover, there exists a free submodule $F\subset M$ such that

$$M\cong F\oplus M_{\text{tor}}.$$

Lastly, ${\dim }_{R}F$ is well-defined.

Proof

Classifying torsion submodules

Let $M$ be a finitely generated torsion $R$-module. Then

$$M\cong \underset{p\text{prime}}{⨁}M(p).$$

Moreover, fore each prime $p$ (up to a multiplying by a unit), there is an isomorphism

$$M(p)\cong \underset{i=i}{\overset{n_p}{⨁}}R/(p^{r_i})$$

for unique integers $1\le r_1\le \cdots \le r_{n_p}$

Proof

Classification of finitely generated modules

Let $M$ be a finitely generated $R$-module. then

$$M\cong R^{\oplus b}\oplus R/(p_1^{r_1})\oplus \cdots \oplus R/(p_n^{r_n})$$

for a unique $b\in \mathbb{N}$, which is called the rank or Betti number of $M$, and unique primes $p_i$ and natural numbers $r_i\in \mathbb{N}$ (these are unique up to permutation and multiplication by units of $R$). Moreover, the isomorphism type of $M$ is unique determined by its rand and elementary divisors $p_1^{r_1},\dots ,p_n^{r_n}$.

Proof