## Finitely Generated Modules over Principal Ideal Domains

### Main Theorem

Our goal is to prove the classification theorem for finitely generated modules over PID’s, which asserts that every finitely generated module over a PID is the direct sum of a free module and a finite set of cyclic modules. Depending on how you describe the cyclic modules you get different uniqueness statements.

Theorem: Let $R$ be a principal ideal domain and let $M$ be a finitely generated $R$ module. Then there is an integer $k$ and elements $\pi_1,\ldots, \pi_m$ in $R$ such that $\pi_1\vert \pi_2\vert\cdots\vert \pi_m$ such that

$M=R^{k}\oplus R/\pi_1 R\oplus\cdots \oplus R/\pi_m R.$

Further, the integer $k$ and the ideals $\pi_i R$ are uniquely determined by $M$. The ideals $\pi_i R$ are called the invariant factors of $M$, and the integer $k$ is its rank.

Notice that if $R=\Z$ and $M$ is finite then this is the fundamental theorem of finite abelian groups with the $\pi_i$ being the invariant factors.

### Alternative formulation

Theorem: Let $R$ be a PID and let $M$ be a finitely generated $R$ module. Then there is an integer $k$ and elements $\pi_i\in R$ such that $\pi_i$ is a prime power and

$M=R^{k}\oplus R/\pi_1 R\oplus\cdots \oplus R/\pi_m R.$

Again, the rank $k$ and the prime power factors $\pi_i$ are unique (up to ordering in this case).

The prime powers $\pi_i$ are called the elementary divisors of $M$.

If $R=\Z$ this is the fundamental theorem of finite abelian groups, asserting that every such group is a finite product of cyclic groups of prime power order, and that the prime powers are unique up to ordering.

### Strategy

Our strategy is to use ideas from linear algebra and approach the problem algorithmically.

Suppose that $M$ is generated by $n$ elements $e_1,\ldots, e_n$ over the PID $R$. Then there is a surjective map $$\pi: R^{n}\to M$$ defined by $\pi((r_1,\ldots, r_n))=\sum_{i=1}^{n} r_i e_i$.

If $f=(r_1,\ldots, r_n)$ is in the kernel of $\pi$, then

$\sum_{i=1}^{n} r_{i}e_{i}=0.$

### Relations

Because of this, elements of the kernel $N$ of $\pi$ are called relations for the generators $e_{i}$, and $N$ is called the module of relations for $M$.

Since the relation module $N$ of this map is a submodule of $R^{n}$, we know from our discussion of finite generation is generated by (at most) $n$ elements $f_1,\ldots, f_n$.

Let’s assume that our relation module has $n$ generators $f_{1},\ldots, f_{n}$, some of which might be zero.

### The relation matrix

Expressing $f_j$ in terms of the $e_i$ yields an $n\times n$ matrix $A=(a_{ij})$ defined by:

$f_j = \sum a_{ji} e_i$

The columns of the matrix $A$ express the generators $f_j$ of the kernel of $\pi$ in terms of the basis $e_i$ for $R^{n}$.

$A$ is called a relation matrix for $M$.

### The kernel as column space of the relation matrix

If, as we do in linear algebra, we express elements of $R^{n}$ as column vectors with $R$ entries, we have a map

$a: R^{n}\to R^{n}$

defined by $a(v)=Av$ (matrix multiplication by $A$ on a column vector $v$ with entries in $R$).
If the entries of $v$ are $(r_1,\ldots, r_n)$ then $a(v)=\sum_{i=1}^{n} r_i f_i$ and therefore the image of the $R$-linear map $a$ is $N$.

### Standard form

We’ve reached a point where our module $M$ is isomorphic to $R^{n}/N$ where $N$ is generated by the columns of our matrix $A$.

We will show the following:

• $N$ is free of rank $m$ where $m\le n$.
• $M$ has a basis $y_1,\ldots, y_m$ with the property that there are elements $b_1,\ldots, b_m\in R$ such that $b_1\vert b_2\vert\cdots \vert b_m$ and $b_1 y_1, b_2 y_2,\ldots, b_m y_m$ are a basis for $N$.

In terms of the relation matrix, we are saying that if we choose our basis $e_1,\ldots, e_n$ and $f_1,\ldots, f_n$ properly, then the corresponding matrix $A$ is diagonal with entries $b_1, b_2, \ldots, b_m, 0,0\ldots 0$ and $b_1\vert b_2\vert\cdots\vert b_m$.

We will do this by modifying the set of generators $f_j$ and $e_i$ so that, at each stage, they continue to be sets of generators, but eventually they have the desired relation.

### The result from standard form

If we achieve the standard form, then we have the picture $$R^{n}\to M$$ where

$(r_1,\ldots, r_n)\mapsto \sum r_{i}y_{i}$

and the kernel of this map is

$N=b_1 y_1 \oplus b_2 y_2 \oplus \cdots\oplus b_m y_m.$

Therefore $R^{n}/N=R/b_1 R \oplus\cdots R/b_m R\oplus R^{n-m}$ which is the structure we are trying to establish.

Alternatively, we can think of $M$ as having generators $e_{1},\ldots, e_{n}$ and relations $b_i e_i=0$

## Reduction Operations

We begin with our chosen generators $e_1,\ldots, e_n$ for $M$ and the corresponding generators for the relation module $f_1,\ldots, f_n$, which are related by the matrix $A=(a_{ij})\in M_{n}(R)$ where $$f_{j} = \sum_{i=1}^{n} a_{ji} e_{i}.$$

Suppose that $x,y,z,w$ are elements of $R$ such that $xw-yz=1$.

### Modifying the generators of $M$

Lemma: Suppose $1\le t,s\le n$ with $i\not=j$. If we let elements $e_{i}^{\ast}=e_{i}$ for $i\not=t,s$, and also \begin{aligned} e_{t}^{\ast} &= xe_{t}+ye_{s}\\ e_{s}^{\ast} &= ze_{t}+we_{s}\\ \end{aligned} Then $e_{1}^{\ast},\ldots, e_{n}^{\ast}$ are also generators of $M$.

Proof: Write

Since $e_{i}=e_{i}^{\ast}$ for $i\not=t,s$ and \begin{aligned} e_{t} &= we_{t}^{\ast}-ye_{s}^{\ast} \\ e_{s} &= -ze_{t}^{\ast}+xe_{s}^{\ast}. \end{aligned} wee see that all of the $e_{i}$ are in the submodule of $M$ generated by the $e_{i}^{\ast}$, and vice versa, so the $e_{i}^{\ast}$ are again a set of generators of $M$.

### Row operations

Let’s examine the effect of this change on the relation matrix $A$. If $$m=r_1 e_1+\cdots+r_n e_n.$$ then $$m=\sum_{i\not=t,s} r_{i} e_{i}^{\ast} + (r_{t}w-r_{s}z)e_{r}^{\ast} + (-yr_{t}+xr_{s})e_{s}^{\ast}.$$

This means that if we construct the relation matrix $A^{\ast}$ by writing $$f_{j}=\sum a_{ji}^{\ast} e_{i}^{\ast}$$ we see that $A^{\ast}$ is obtained from $A$ by modifying rows $t$ and $s$. If we use subscripts to describe rows of matrices then \begin{aligned} A^{\ast}_{t} &= wA_{t}-zA_{s} \\ A^{\ast}_{s} &= -yA_{t}+xA_{s} \end{aligned}

### Column Operations

More generally, we see that, given any relation matrix $A$, and $x,y,z,w$ such that $xw-yz=1$, modifying $A$ by changing rows $t$ and $s$ according to these formulas yields a new relation matrix giving rise to an isomorphic module $M$.

A similar line of argument shows that if we make the same type of modification to the generators $f_{j}$ for the relations, then we modify the relation matrix $A$ by column operations of the same type.

## Outline of proof of standard form

### Initial remarks

Now suppose we are given an $n\times n$ matrix $A$ with entries in a PID $R$. There is a sequence of row and column operations that reduces it to standard form, so that the reduced matrix is diagonal, the first $k$ diagonal elements are nonzero and the remaining $n-k$ are zero, and the nonzero diagonal elements satisfy

$a_{11}\vert a_{22}\vert\cdots\vert a_{kk}$

### Main Steps

1. If $A=0$, we’re done, otherwise swap rows and columns so $a_{11}$ is not zero.

### Clear out the first row

1. If all $a_{1i}$ for $i>1$ are divisible by $a_{11}$, replace each column $A^{j}$ where $a_{1j}$ is not zero by $A^{j}-a_{11}/a_{1j}A^{1}$. Otherwise, for each column $j=2,\ldots, n$ where $a_{1j}$ is not zero, use the fact that $R$ is a PID to find a generator $d$ for the ideal $(a_{11},a_{1j})$ for each column and write $a_{11}x-a_{1j}y=d$. Then make a column operation using this $x$ and $y$ with $w=a_{11}/d$ and $z=a_{1j}/d$ to obtain a matrix with $a_{11}=d$ and $a_{1j}=0$. At the end of this step, the only nonzero entry in the first row is $a_{11}$.

### Clear out the first column

1. If all $a_{i1}$ for $i>1$ are divisible by $a_{11}$, replace each row $A_{j}$ with $A_{j}-a_{j1}/a_{11}A_{1}$. Now you’ve got a matrix so that the first row and column are all zero, except for $a_{11}.$ Go to step 4. Otherwise, use the fact that $R$ is a PID to find a generator $d=a_{11}x-a_{j1}y$ and make a row operation using this $x$ and $y$ with $w=a_{11}/d$ and $z=a_{j1}/d$ to obtain a matrix with $a_{11}=d$ and $a_{j1}=0$. At the end of this process, you’ve got a matrix so that $a_{11}$ is the only nonzero entry in the first column; but you may have messed up the first row. So go back to step 2.

### Check divisibility; descend to submatrix

1. At this point the first row and column of $A$ are zero except for $a_{11}$. If $a_{11}$ divides every entry in the lower right $(n-1)\times (n-1)$ submatrix, then apply this algorithm to that submatrix and continue. If $a_{11}$ does NOT divide every entry in lower submatrix, find a row $A_{j}$ containing an element not divisible by $a_{11}$ and replace the first row $A_{1}$ by $A_{1}+A_{j}$. Now go back to step 2 and continue.

### Remarks on the algorithm

There are two things to consider in this algorithm.

First, the loop through steps 2 and 3 must eventually terminate because each time you go through it, you replace $a_{11}$ by a divisor of $a_{11}$. This cannot continue indefinitely, so eventually you will reach step 4.

Second, if $a_{11}$ divides everything in the lower submatrix, then by induction, once that matrix is in standard form, the whole matrix will be in standard form. If $a_{11}$ does not divide everything in the lower submatrix, then the return to step 2 will replace $a_{11}$ by a proper divisor of $a_{11}$ and again, that can’t continue indefinitely.

### Constructive for Euclidean rings

The only non-constructive part of this “algorithm” is that we invoke the PID property of $R$ so that, given $a,b$ we can find $ax+by=d$ where $d$ is the gcd of $a$ and $b$. If $R$ is Euclidean, this can be done constructively, and so this algorithm can be carried out in practice.

## Uniqueness

### Uniqueness in DF

Proof of uniqueness is given in DF, Section 12.1 Theorem 9.