Structure Theorem for Finitely Generated Modules over a PID

Smith Normal Form.

Theorem. Let $R$ be a Principal Ideal Domain (PID) and let $A$ be an $m\times n$ matrix with entries in $R$. Then there exist invertible matrices $P\in {\mathrm{GL}}_m(R)$ and $Q\in {\mathrm{GL}}_n(R)$ such that $S=PAQ$ is a diagonal matrix with diagonal entries $d_1,d_2,\dots ,d_r,0,\dots ,0$, where $r=\min (m,n)$, and the non-zero entries satisfy the divisibility condition $d_1|d_2|\dots |d_r$. The elements $d_i$ are unique up to multiplication by units.

Proof. The proof is by induction on the size of the matrix, $m\times n$. If $A$ is the zero matrix, it is already in the required form. Otherwise, let $a_{ij}$ be a non-zero entry of $A$ for which the ideal $(a_{ij})$ is maximal in the set of all principal ideals generated by entries of $A$. Such an element exists because $R$ is Noetherian. By swapping rows and columns, move this entry to the $(1,1)$ position. Let $d_1=a_{11}$.

For each $j>1$, since $R$ is a PID, the ideal $(d_1,a_{1j})$ is principal, generated by some $g\in R$. Thus $g=x{d}_1+y{a}_{1j}$ for some $x,y\in R$. Applying column operations, we can replace column $j$ with $y(\text{column }j)+x(\text{column }1)$, which makes the new $(1,j)$ entry equal to $g$. By the choice of $d_1$, we must have $(d_1)\subseteq (g)$, which means $g|d_1$. But also $d_1$ is a component of the linear combination for $g$, so $d_1|g$. Thus $g$ and $d_1$ are associates. We can perform column operations to make $a_{1j}$ a multiple of $d_1$, and then subtract a multiple of the first column to make $a_{1j}=0$. This is repeated for all $j>1$ to make the first row zero, except for $d_1$. A similar process using row operations makes the first column zero, except for $d_1$.

If after this process, some entry $a_{ij}$ (for $i,j>1$) is not divisible by $d_1$, add row $i$ to row 1. The new entry $a_{1j}$ is not divisible by $d_1=a_{11}$. Applying the procedure from the previous paragraph to clear the first row will produce a new $(1,1)$ entry that is a proper divisor of the original $d_1$. This generates a new maximal ideal larger than $(d_1)$, a contradiction unless the process terminates. Since $R$ is Noetherian, any strictly ascending chain of ideals must be finite. Thus, this process must terminate, at which point $d_1$ divides all other entries in the matrix.

The matrix now has the form:

$$\left(\begin{array}{cccc}{d}_1& 0& \cdots & 0\\ 0& & & \\ ⋮& & A^{\prime }& \\ 0& & & \end{array}\right)$$

where all entries of the $(m-1)\times (n-1)$ submatrix $A^{\prime }$ are divisible by $d_1$. By the inductive hypothesis, $A^{\prime }$ can be diagonalized to have diagonal entries $d_2,\dots ,d_r$ with $d_2|\dots |d_r$. Since any elementary row or column operation on $A^{\prime }$ corresponds to one on the full matrix that does not affect the first row or column, we can diagonalize $A^{\prime }$ in place. The resulting matrix is diagonal with entries $d_1,d_2,\dots ,d_r$. The condition $d_1|a_{ij}^{\prime }$ for all entries of $A^{\prime }$ ensures that $d_1|d_2$, since $d_2$ is a linear combination of the entries of $A^{\prime }$. The full divisibility chain $d_1|d_2|\dots |d_r$ follows. $\square$

Structure Theorem for Finitely Generated Modules over a PID (Existence).

Theorem. Let $M$ be a finitely generated module over a PID $R$. Then $M$ is isomorphic to a direct sum of cyclic modules:

$$M\cong R^r\oplus R/(d_1)\oplus R/(d_2)\oplus \cdots \oplus R/(d_s)$$

where $r\ge 0$, the $d_i$ are non-zero non-units in $R$, and they satisfy the divisibility condition $d_1|d_2|\dots |d_s$.

Proof. Let $M$ be generated by $n$ elements. There is a surjective $R$-module homomorphism $\varphi :R^n\to M$. The kernel, $K=\ker (\varphi )$, is a submodule of the free module $R^n$. Since $R$ is a PID, $K$ is also free, of rank $m\le n$. By the First Isomorphism Theorem, $M\cong R^n/K$.

Let $\{e_1,\dots ,e_n\}$ be the standard basis for $R^n$ and $\{k_1,\dots ,k_m\}$ be a basis for $K$. Each $k_j$ can be written as a linear combination of the $e_i$: $k_j={\sum }_{i=1}^n{a}_{ij}{e}_i$. This defines an $n\times m$ matrix $A=(a_{ij})$, called the relation matrix.

By the Smith Normal Form theorem, there exist invertible matrices $P\in {\mathrm{GL}}_n(R)$ and $Q\in {\mathrm{GL}}_m(R)$ such that $S=PAQ$ is a diagonal matrix with diagonal entries $d_1,\dots ,d_s,0,\dots ,0$, where $s\le \min (n,m)$ and $d_1|d_2|\dots |d_s$. The matrix $P$ defines a change of basis for $R^n$ from $\{e_i\}$ to a new basis $\{y_1,\dots ,y_n\}$. The matrix $Q$ defines a change of basis for $K$ from $\{k_j\}$ to $\{k_j^{\prime }\}$. In these new bases, the relation matrix is $S$. This means the basis vectors for $K$ are given by $k_j^{\prime }=d_j{y}_j$ for $j=1,\dots ,s$, and $k_j^{\prime }=0$ for $j>s$. Thus, $K$ is generated by $\{d_1{y}_1,\dots ,d_s{y}_s\}$.

The isomorphism $M\cong R^n/K$ can now be expressed using the new basis for $R^n$:

$$M\cong \frac{R{y}_1\oplus \cdots \oplus R{y}_n}{d_{1}R{y}_1\oplus \cdots \oplus d_{s}R{y}_s}$$

This quotient decomposes into a direct sum:

$$M\cong \left(\frac{R{y}_1}{d_{1}R{y}_1}\right)\oplus \cdots \oplus \left(\frac{R{y}_s}{d_{s}R{y}_s}\right)\oplus R{y}_{s+1}\oplus \cdots \oplus R{y}_n$$

Since $R{y}_i\cong R$, we have $R{y}_i/d_{i}R{y}_i\cong R/(d_i)$. If any $d_i$ is a unit, $R/(d_i)$ is the zero module and can be omitted. Let’s assume the $d_i$ listed are the non-unit diagonal entries. The remaining $n-s$ summands are isomorphic to $R$. Let $r=n-s$. This gives the desired decomposition. $\square$

Invariant Factors and Elementary Divisors.

Definition. Let $M$ be a finitely generated module over a PID $R$, with decomposition $M\cong R^r\oplus {⨁}_{i=1}^{s}R/(d_i)$ where the $d_i$ are non-zero non-units satisfying $d_1|d_2|\dots |d_s$. The elements $d_1,\dots ,d_s$ (or the ideals they generate) are the invariant factors of $M$. The number $r$ is the free rank or Betti number of $M$.

Definition. Let $R$ be a PID and $M$ be a finitely generated torsion $R$-module. For each non-zero non-unit $d_i$ in the invariant factor decomposition, let its prime factorization be $d_i=u{p}_{i1}^{e_{i1}}\dots p_{i{k}_i}^{e_{i{k}_i}}$, where $u$ is a unit and the $p_{ij}$ are distinct primes. By the Chinese Remainder Theorem, $R/(d_i)\cong {⨁}_{j=1}^{k_i}R/(p_{ij}^{e_{ij}})$. The elements $p_{ij}^{e_{ij}}$ (or the ideals they generate), collected from all invariant factors $d_i$, are the elementary divisors of $M$.

Structure Theorem for Finitely Generated Modules over a PID (Uniqueness).

Theorem. Let $M$ be a finitely generated module over a PID $R$.

1. (Invariant Factor Decomposition) There is an isomorphism $M\cong R^r\oplus {⨁}_{i=1}^{s}R/(d_i)$, where $r\ge 0$, each $d_i$ is a non-zero non-unit, and $d_1|d_2|\dots |d_s$. The integer $r$ and the sequence of ideals $(d_1),(d_2),\dots ,(d_s)$ are uniquely determined by $M$.
2. (Elementary Divisor Decomposition) There is an isomorphism $M\cong R^r\oplus {⨁}_{j=1}^{t}R/(q_j)$, where each $q_j$ is a power of a prime element in $R$. The integer $r$ and the multiset of ideals $\{(q_1),\dots ,(q_t)\}$ are uniquely determined by $M$.

Proof. The existence of the decompositions was previously established. We prove uniqueness. Let $T(M)$ be the torsion submodule of $M$. Then $M/T(M)$ is a finitely generated torsion-free module, which is free. The rank of this free module is $r$. The rank is an invariant of $M$, for example, $r={\dim }_K(M{\otimes }_{R}K)$ where $K$ is the field of fractions of $R$. Thus, $r$ is unique.

We only need to show the uniqueness of the decomposition for the torsion part $T(M)\cong {⨁}_{i=1}^{s}R/(d_i)$. Let $T(M)=N$. For any prime $p\in R$, let $N[p]=\{x\in N\mid px=0\}$. This is a submodule of $N$ and a vector space over the field $R/(p)$. The dimension ${\dim }_{R/(p)}N[p]$ is an invariant of $N$. If $N\cong {⨁}_{i=1}^{s}R/(d_i)$, then $N[p]\cong {⨁}_{i=1}^s(R/(d_i))[p]$. An element in $R/(d_i)$ is annihilated by $p$ if its representative in $R$ is a multiple of $d_i/\gcd (p,d_i)$. Thus $(R/(d_i))[p]$ is non-zero if and only if $p$ divides $d_i$. If $p|d_i$, $(R/(d_i))[p]\cong R/(p)$, which has dimension 1 over $R/(p)$. Therefore, ${\dim }_{R/(p)}N[p]$ equals the number of invariant factors $d_i$ divisible by $p$.

Similarly, one can show that for any $k\ge 1$, the dimension of $p^{k-1}N[p]=(p^{k-1}N)[p]$ over $R/(p)$ is equal to the number of invariant factors divisible by $p^k$. Let ${\lambda }_k(p)={\dim }_{R/(p)}(p^{k-1}N[p])$. The sequence of integers ${\lambda }_1(p),{\lambda }_2(p),\dots$ is an invariant of $M$ for each prime $p$. This sequence determines the number of elementary divisors that are powers of $p$. Specifically, ${\lambda }_k(p)-{\lambda }_{k+1}(p)$ is the number of elementary divisors equal to $p^k$ (up to associates). Since these numbers are determined for every prime $p$, the multiset of elementary divisors is unique.

The invariant factors can be recovered from the elementary divisors. For each prime $p$, let the powers of $p$ among the elementary divisors be $p^{e_{p,1}},p^{e_{p,2}},\dots$ with $e_{p,1}\le e_{p,2}\le \dots$. The largest invariant factor $d_s$ is the product of the largest powers of all distinct primes, $d_s={\prod }_p{p}^{e_{p,\max }}$. The next invariant factor $d_{s-1}$ is the product of the next-largest powers, and so on. Since the elementary divisors are unique, the invariant factors are also unique. $\square$

Application.

Theorem. Every finitely generated torsion-free module $M$ over a PID $R$ is free.

Proof. By the structure theorem, any finitely generated module $M$ over a PID $R$ has a decomposition of the form:

$$M\cong R^r\oplus R/(d_1)\oplus \cdots \oplus R/(d_s)$$

The torsion submodule of $M$, denoted $T(M)$, consists of all elements $m\in M$ for which there exists a non-zero $a\in R$ such that $am=0$. In the decomposed form, the torsion submodule is precisely the direct sum of the cyclic torsion components:

$$T(M)\cong R/(d_1)\oplus \cdots \oplus R/(d_s)$$

The module $M$ is torsion-free by hypothesis, which means $T(M)=\{0\}$. This implies that the direct sum ${⨁}_{i=1}^{s}R/(d_i)$ must be the zero module. This occurs only if there are no such summands, i.e., $s=0$. Therefore, the decomposition of $M$ simplifies to $M\cong R^r$. A module isomorphic to $R^r$ for some $r\ge 0$ is, by definition, a free module of rank $r$. $\square$