Modules over a PID

Introduction

A principal ideal domain is the most general ring in which the module theory of a field survives in a recognisable form. Every submodule of a free module is free, every finitely generated torsion-free module is free, and every finitely generated module is a direct sum of a free module and cyclic torsion modules whose invariants are unique. The price of leaving the field case is that the classification acquires a second invariant, the torsion, which a field does not have; the reward is that the same theorem classifies finitely generated abelian groups, rational canonical forms and Jordan forms, and computes indices of lattices.

Throughout, $R$ is a principal ideal domain: a commutative integral domain with $1 \neq 0$ in which every ideal is principal, so that every ideal is of the form $(a)=aR$. The fraction field is $K=\operatorname{Frac}(R)$. Every module is a left $R$-module, which over a commutative ring is the same as a right module. The two examples that matter most are $R=\mathbb{Z}$, with $K=\mathbb{Q}$, and $R=K[x]$ over a field $K$, with fraction field $K(x)$; the first is worked out in the applications article on finitely generated abelian groups and the second in the article on modules over $k[x]$ and the Jordan form.

The theorem proved here is the structure theorem for finitely generated modules over a principal ideal domain. Its existence half is a matrix reduction, the Smith normal form, and its uniqueness half is an analysis prime by prime. Both are constructive, and the algorithm is recorded in a form that can be run on a presentation matrix.

Torsion and Annihilators

Torsion Elements and the Torsion Submodule

Definition. Let $R$ be an integral domain and $M$ an $R$-module. An element $m \in M$ is a torsion element if $rm=0$ for some $0 \neq r \in R$. Write

$$ M_{\mathrm{tor}}=\{m \in M : rm=0 \text{ for some } 0 \neq r \in R\}. $$

Proposition. $M_{\mathrm{tor}}$ is a submodule of $M$, and $M/M_{\mathrm{tor}}$ is torsion-free.

Proof. If $rm=0$ and $sn=0$ with $r,s \neq 0$, then $rs(m+n)=s(rm)+r(sn)=0$ and $rs \neq 0$ because $R$ is a domain, so $m+n \in M_{\mathrm{tor}}$; and $r(am)=a(rm)=0$ for $a \in R$. For the quotient, if the class of $m$ is torsion then $rm \in M_{\mathrm{tor}}$, so $s(rm)=0$ for some $s \neq 0$, whence $(sr)m=0$ and $m \in M_{\mathrm{tor}}$.

The domain hypothesis is exactly what fails over a general commutative ring: in $\mathbb{Z}/6\mathbb{Z}$ as a module over itself, the elements $2$ and $3$ are torsion but their sum $5$ is a unit and is not, as noted. The module $M$ is torsion-free if $M_{\mathrm{tor}}=0$ and torsion if $M=M_{\mathrm{tor}}$; the quotient $M/M_{\mathrm{tor}}$ is always torsion-free.

Annihilators and Cyclic Modules

Definition. For $m \in M$ the annihilator is the ideal $\operatorname{Ann}(m)=\{r \in R : rm=0\}$; for a module, $\operatorname{Ann}(M)=\{r \in R : rm=0 \text{ for all } m \in M\}=\bigcap_{m}\operatorname{Ann}(m)$.

Both are ideals, and $\operatorname{Ann}(M)$ is the kernel of the ring homomorphism $R \to \operatorname{End}_R(M)$ that gives the module structure, so $M$ is a faithful module over $R/\operatorname{Ann}(M)$.

Proposition. Over a principal ideal domain the following hold. (i) Every ideal is principal, so $\operatorname{Ann}(m)=(a)$ and $Rm \cong R/(a)$ via $r \mapsto rm$. (ii) $Rm$ is free of rank $1$ if and only if $\operatorname{Ann}(m)=0$, and otherwise $Rm$ is torsion. (iii) The annihilator of a finitely generated module contains the product of the annihilators of a generating set, hence is nonzero for a torsion module.

Proof. (i) The first isomorphism theorem applied to $R \to Rm$, $r \mapsto rm$, whose kernel is $\operatorname{Ann}(m)$. (ii) $R \cong Rm$ exactly when the kernel is zero. (iii) If $M=Rm_1+\cdots+Rm_k$ and $a_i m_i=0$ with $a_i \neq 0$, then $a_1\cdots a_k$ annihilates every generator, hence $M$.

So cyclic modules over a principal ideal domain are exactly the modules $R/(a)$, with $a$ well defined up to multiplication by a unit, and the torsion cyclic modules are those with $a \neq 0$. The unit ambiguity is unavoidable and is the reason the invariant factors below are unique only up to units.

Primary Components

Definition. Let $p \in R$ be prime and let $n \ge 1$. The $p$-primary component of $M$ is

$$ M_p=\{m \in M : p^n m=0 \text{ for some } n \ge 1\}, $$

equivalently the set of elements annihilated by some power of $p$.

Proposition. Each $M_p$ is a submodule; if $p \neq q$ are distinct primes then $M_p \cap M_q=0$; and if $M$ is a finitely generated torsion module then only finitely many $M_p$ are nonzero and $M=\bigoplus_p M_p$.

Proof. That $M_p$ is a submodule and the intersection statement are immediate from unique factorisation: an element of $M_p\cap M_q$ is annihilated by coprime powers of $p$ and $q$, hence by a unit, hence is zero. For the sum, if $M$ is torsion and finitely generated, its annihilator $(a)$ is nonzero; the primes with $M_p \neq 0$ divide $a$, so there are finitely many. That every torsion element lies in the sum uses Bézout: if $rm=0$ and $r=\prod p_i^{n_i}$, the Chinese remainder theorem gives $m=\sum_i m_i$ with $p_i^{n_i}m_i=0$, so $m_i \in M_{p_i}$.

The decomposition into primary components is the first step of the structure theorem: it separates the primes, and within one prime the classification is that of a finite abelian $p$-group.

Free Modules over a PID

Submodules of Free Modules are Free

Theorem. Let $R$ be a principal ideal domain. Every submodule of a free $R$-module is free, of rank at most the rank of the ambient module.

Proof. Let $F$ be free with basis $u_1,\dots,u_n$ and let $N \subseteq F$. Induct on $n$. For $n=0$ there is nothing to prove. For $n \ge 1$ let $F_{n-1}=\langle u_1,\dots,u_{n-1}\rangle$ and let $\pi:F \to Ru_n \cong R$ be the projection onto the last coordinate. The image $\pi(N)$ is an ideal of $R$, hence $\pi(N)=(a)$ for some $a \in R$. If $a=0$ then $N=N \cap F_{n-1}$ and induction applies. Otherwise choose $x \in N$ with $\pi(x)=a$ and consider $N \cap F_{n-1}$, which is free of rank at most $n-1$ by induction, with basis $v_1,\dots,v_k$. Every $y \in N$ has $\pi(y)=ra$ for some $r \in R$, and $y-rx \in N \cap F_{n-1}$, so $y \in (N \cap F_{n-1})+Rx$; the sum is direct because $Rx \cap F_{n-1} \subseteq \ker \pi$ and $Rx \cap \ker\pi=0$ by $\pi(x)=a \neq 0$. Hence $v_1,\dots,v_k,x$ is a basis of $N$, and $N$ is free.

The theorem is false without the principal ideal hypothesis, as the ideal $(x,y) \subseteq k[x,y]$ shows, and it is the reason presentations over a principal ideal domain are finite matrices with a completely computable normal form.

Ranks of Submodules

Corollary. Over a principal ideal domain, if $N \subseteq F$ are free of finite ranks $m$ and $n$ then $m \le n$. Moreover $N$ is a direct summand of $F$ if and only if $F/N$ is torsion-free, and $F/N$ is torsion if and only if $m=n$. The two conditions are opposite: equal ranks do not make $N$ a summand, and a summand need not have full rank.

Proof. The structure theorem writes $F/N \cong R^{n-m} \oplus T$ with $T$ torsion, since $F/N$ is finitely generated and $N$ is free of rank $m$; tensoring $0 \to N \to F \to F/N \to 0$ with the flat module $K$ gives $m \le n$. From the display, $F/N$ is torsion precisely when $n-m=0$, that is when $m=n$, and it is torsion-free precisely when $T=0$, that is when $F/N \cong R^{n-m}$ is free. If $F=N \oplus P$ then $P \cong F/N$ is free, being a submodule of a free module; conversely if $F/N$ is torsion-free then it is free, so the surjection $F \to F/N$ splits and $N$ is a summand. Equal rank alone gives no splitting: for $F=R$ and $N=2R$ one has $m=n=1$ and $F/N=R/2R$ is torsion, while $N=R \oplus 0 \subseteq R^2$ is a summand with $m=1

Corollary (index). Let $L' \subseteq L$ be free of the same finite rank $n$ over a principal ideal domain, and let $A$ be the matrix of the inclusion with respect to bases of $L'$ and $L$. Then $L/L'$ is torsion if and only if $\det A \neq 0$, and then

$$ L/L' \cong R/(d_1)\oplus\cdots\oplus R/(d_n), $$

where $d_1 \mid \cdots \mid d_n$ are the invariant factors of $A$; in particular the order ideal of $L/L'$, the ideal generated by its order, is $(\det A)$, which is the product of the invariant factors and not in general their least common multiple. Over $R=\mathbb{Z}$ this gives $|L/L'|=|\det A|$, the product of the invariant factors, equivalently the gcd of the $n \times n$ minors of $A$.

This is the statement used for lattices: a sublattice of $\mathbb{Z}^n$ of full rank has finite index equal to the absolute value of the determinant of its basis matrix. The proof is an application of the Smith normal form below; the determinant and index are computed simultaneously.

The Structure Theorem

Statement

Theorem (structure theorem for finitely generated modules over a PID). Let $R$ be a principal ideal domain and let $M$ be a finitely generated $R$-module. Then

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

for a unique $r \ge 0$ and a unique finite sequence of nonzero non-units $d_1,\dots,d_k \in R$ with

$$ d_1 \mid d_2 \mid \cdots \mid d_k . $$

The sequence of $d_i$ is unique up to multiplication of each $d_i$ by a unit of $R$. The integer $r$ is the free rank, or Betti number, of $M$, and the $d_i$ are its invariant factors. In particular $M$ is the direct sum of a free module and a torsion module, and the torsion part is a finite direct sum of cyclic torsion modules.

The uniqueness statement has two parts: the number $r$ of free summands and the multiset of principal ideals $(d_i)$. Both are recovered from invariants of $M$ without reference to a presentation, and that is the content of the uniqueness proof.

Existence via the Smith Normal Form

Choose a finite generating set $m_1,\dots,m_n$ of $M$, which exists by finite generation, and let $\epsilon:R^n \to M$ be the surjection sending the standard basis $e_i$ to $m_i$. Its kernel $K=\ker \epsilon$ is a submodule of the free module $R^n$, hence free of rank $m \le n$ by the theorem above. Choose bases of $K$ and $R^n$ and let $A$ be the $n \times m$ matrix of the inclusion $K \hookrightarrow R^n$ in those bases. The Smith normal form of $A$ (constructed in the next section) gives invertible matrices $P,Q$ over $R$ with

$$ P A Q=\operatorname{diag}(d_1,\dots,d_m), \qquad d_1 \mid d_2 \mid \cdots \mid d_m . $$

Changing the basis of $R^n$ by $P$ and of $K$ by $Q$ does not change the quotient, so

$$ M \cong R^n/K \cong R^n / \langle d_1 e_1,\dots,d_m e_m\rangle \cong R/(d_1)\oplus\cdots\oplus R/(d_m)\oplus R^{n-m}. $$

Discarding the $d_i$ that are units, for which $R/(d_i)=0$, and taking $r=n-m$ gives the decomposition of the theorem, and the divisibility of the surviving diagonal entries is inherited from the Smith form. This proves existence.

Uniqueness

Suppose

$$ M \cong R^r \oplus T, \qquad T=\bigoplus_{i=1}^{k} R/(d_i), \qquad d_1 \mid \cdots \mid d_k \neq 0 . $$

The free rank. Tensor the decomposition with the fraction field $K$, which is a flat $R$-module, so the decomposition is preserved. Since $R \otimes_R K \cong K$ and $(R/(d_i)) \otimes_R K=0$ for $d_i \neq 0$ because $d_i$ becomes invertible in $K$, we get $M \otimes_R K \cong K^r$. Hence

$$ r=\dim_K (M \otimes_R K), $$

which depends only on $M$. Consequently $T=M_{\mathrm{tor}}$ is determined by $M$, and it remains to determine the $d_i$ from $T$.

The primary components. The torsion module $T$ is the direct sum of its primary components $T_p$, and the decomposition $T=\bigoplus_p T_p$ is determined by $T$. Fix a prime $p$ and write $T_p \cong \bigoplus_{j=1}^{s} R/(p^{e_j})$ with $e_1 \le \cdots \le e_s$, which is legitimate because a cyclic summand $R/(d)$ with $d=\prod p^{n_p}$ decomposes by the Chinese remainder theorem into $\bigoplus_p R/(p^{n_p})$. For $n \ge 1$ let $k(p)=R/(p)$ be the residue field, a field because $p$ is prime, and consider

$$ p^{n-1}T_p / p^n T_p, $$

an $R/(p)$-vector space. A cyclic summand $R/(p^{e})$ contributes a one-dimensional space for $n \le e$ and zero for $n>e$, because $p^{n-1}R/(p^e)$ is generated by the class of $p^{n-1}$, which is nonzero exactly when $n-1

$$ \dim_{k(p)}\bigl(p^{n-1}T_p/p^nT_p\bigr)=\#\{j : e_j \ge n\}, $$

which depends only on $T_p$, hence only on $M$. Subtracting the dimensions for consecutive $n$ recovers the multiset $\{e_j\}$, hence $T_p$ up to isomorphism. Since this holds for every prime and only finitely many are nonzero, $T$, and therefore $M$, is determined.

The same argument applied with $n-1$ replaced by $0$ gives the useful formula for the number of cyclic summands of $T_p$, namely $\dim_{k(p)}(T_p/pT_p)$.

Elementary Divisors

The Chinese remainder theorem rewrites each invariant factor through its prime factorisation. If $d=\prod_p p^{n_p(d)}$ with $n_p(d) \ge 0$, then

$$ R/(d) \cong \bigoplus_{p} R/(p^{n_p(d)}), $$

the sum being finite. Applying this to every invariant factor expresses $M$ in the elementary divisor form

$$ M \cong R^r \oplus \bigoplus_{p}\ \bigoplus_{j} R/(p^{e_{p,j}}), \qquad e_{p,1} \le e_{p,2} \le \cdots, $$

where the prime powers $p^{e_{p,j}}$ are the elementary divisors of $M$. They are unique up to units and reordering. The invariant factors are recovered from them prime by prime: for each $p$ arrange the exponents $e_{p,j}$ in nondecreasing order and pad the list on the left with zeros to length $k$, where $k$ is the largest number of primary summands for any single prime; the $i$-th invariant factor is

$$ d_i=\prod_p p^{\,e_{p,i}}, $$

with the convention $p^0=1$. Padding on the left is what keeps each list nondecreasing and makes the divisibility $d_i \mid d_{i+1}$ hold; appending the zeros at the end would reverse the alignment. Thus the two descriptions carry the same information, and the elementary divisor form is the one in which a computation naturally ends.

The Smith Normal Form

Elementary Operations

Let $A$ be an $n \times m$ matrix over $R$. The elementary row operations are: interchange two rows; add a multiple of one row to another; multiply a row by a unit of $R$. The elementary column operations are the same for columns. Row operations are exactly left multiplication by elementary matrices and column operations are exactly right multiplication by them; these matrices are invertible, so equivalence by elementary operations is the same relation as equivalence by left and right multiplication by invertible matrices. Over a general principal ideal domain the Smith form still exists, by an argument that uses Bézout's identity in place of Euclidean division.

Definition. A matrix is in Smith normal form if it is diagonal, $D=\operatorname{diag}(d_1,\dots,d_q)$, and $d_1 \mid d_2 \mid \cdots \mid d_q$.

Theorem. Every matrix over a principal ideal domain is equivalent by elementary row and column operations to a Smith normal form, and the form is unique up to multiplication of the $d_i$ by units.

Existence over a Euclidean Domain

When $R$ carries a Euclidean function $\nu$, the proof is an algorithm. If some entry of $A$ is nonzero, choose a nonzero entry of least Euclidean value and move it to position $(1,1)$ by row and column interchanges. Use it as a pivot: for each other entry $a_{i1}$ in the first column, divide $a_{i1}=q a_{11}+r$ with $r=0$ or $\nu(r)<\nu(a_{11})$, subtract $q$ times the first row from row $i$, and if $r \neq 0$ interchange so that the smaller entry becomes the pivot and repeat. The Euclidean value of the pivot strictly decreases, so after finitely many steps the first column is zero except for the pivot; the same is done to the first row. If some remaining entry $a_{ij}$ is not divisible by the pivot $a_{11}$, add row $i$ to row $1$ and repeat the reduction, which either makes the pivot smaller or makes all entries divisible. When the pivot divides every remaining entry, subtract suitable multiples of row $1$ and column $1$ to clear its row and column, and continue on the smaller submatrix. The divisibility of each pivot by the previous one is ensured by the clearing step, so the process terminates in Smith form.

Uniqueness and Fitting Invariants

For $1 \le k \le \min(n,m)$ let $D_k(A)$ be the ideal generated by all $k \times k$ minors of $A$, with $D_0=R$. These determinantal ideals are unchanged by elementary row and column operations, because each operation multiplies the set of $k \times k$ minors by a unit or replaces it by a set with the same generated ideal. If $A=\operatorname{diag}(d_1,\dots,d_q)$ in Smith form then the $k \times k$ minors are the products of $k$ distinct diagonal entries: every such product is divisible by $d_1d_2\cdots d_k$, and the product of the first $k$ diagonal entries is itself a minor, so

$$ D_k(A)=(d_1 d_2 \cdots d_k) $$

with the convention that $D_k=0$ if $k>q$. Hence the products $d_1\cdots d_k$ are determined by $A$, and therefore so are the quotients $d_k=D_k/D_{k-1}$, up to units. This proves uniqueness of the Smith form and gives the invariant factors of the cokernel directly from the determinantal ideals of a presentation matrix.

Worked Examples

Over $\mathbb{Z}$. Let

$$ A=\begin{pmatrix} 2 & 4 \\ 6 & 8 \end{pmatrix}, \qquad M=\mathbb{Z}^2/A\mathbb{Z}^2 . $$

The determinantal ideals are $D_1=(2,4,6,8)=(2)$ and $D_2=(\det A)=(2\cdot 8-4\cdot 6)=(-8)=(8)$, so the Smith form is $\operatorname{diag}(2,4)$ and

$$ M \cong \mathbb{Z}/2\mathbb{Z}\oplus \mathbb{Z}/4\mathbb{Z}. $$

The same computation with $A=\operatorname{diag}(6,10)$ gives $D_1=(2)$, $D_2=(60)$, hence $\operatorname{diag}(2,30)$ and $M \cong \mathbb{Z}/2\mathbb{Z}\oplus\mathbb{Z}/30\mathbb{Z}$; with $A=\operatorname{diag}(2,3)$ it gives $D_1=(1)$ and $D_2=(6)$, hence $\operatorname{diag}(1,6)$ and $M \cong \mathbb{Z}/6\mathbb{Z}$. The diagonal matrix with non-coprime entries produces two invariant factors, the coprime one a single cyclic summand; this is the Chinese remainder theorem seen through the Smith form.

Elementary divisors and invariant factors. Take

$$ T=\mathbb{Z}/2\mathbb{Z}\oplus\mathbb{Z}/4\mathbb{Z}\oplus\mathbb{Z}/3\mathbb{Z}. $$

The elementary divisors are $2,4,3$. Grouping by prime, the exponents are $\{1,2\}$ for $p=2$ and $\{1\}$ for $p=3$; padding the shorter list with a zero gives $\{1,2\}$ and $\{0,1\}$, so the invariant factors are $d_1=2^1\cdot 3^0=2$ and $d_2=2^2\cdot 3^1=12$:

$$ T \cong \mathbb{Z}/2\mathbb{Z}\oplus\mathbb{Z}/12\mathbb{Z}. $$

Conversely, splitting each invariant factor into prime powers recovers the elementary divisors. The two forms describe one group.

Over $K[x]$. Let $R=K[x]$ and let

$$ A=\begin{pmatrix} x-1 & 0 \\ 0 & (x-1)^2 \end{pmatrix}. $$

Both diagonal entries are powers of the same irreducible $x-1$, so the Smith form is already diagonal with the required divisibility. The determinantal ideals are $D_1=(x-1)$ and $D_2=((x-1)^3)$, so $\operatorname{diag}(x-1,(x-1)^2)$ is the Smith form and

$$ K[x]^2/AK[x]^2 \cong K[x]/(x-1)\oplus K[x]/((x-1)^2). $$

The first summand is one-dimensional over $K$, the second two-dimensional, and this pair is the invariant-factor data of an operator whose eigenvalue-$1$ part consists of one Jordan block of size $1$ and one of size $2$, as the applications article on the Jordan form records.

Consequences

Torsion-Free Modules are Free

Corollary. Over a principal ideal domain, a finitely generated module is free if and only if it is torsion-free.

Proof. A free module is torsion-free. Conversely, if $M$ is finitely generated and torsion-free then $k=0$ in the structure theorem, so $M \cong R^r$.

This is a genuinely principal-ideal statement: over $k[x,y]$ the ideal $(x,y)$ is finitely generated and torsion-free but not free.

Submodules and Quotients

Corollary. Over a principal ideal domain: (i) every submodule of a finitely generated module is finitely generated, so $R$ is Noetherian; (ii) a submodule of a free module of rank $n$ has rank at most $n$; (iii) if $M$ is generated by $n$ elements then every submodule of $M$ can be generated by $n$ elements; (iv) the rank function is additive on direct sums.

Proof. (i) A submodule of a quotient of $R^n$ is the image of a submodule of $R^n$, which is free, hence finitely generated. (ii) is the rank corollary above. (iii) A submodule of $M=R^n/K$ is $N/K$ for a submodule $N \subseteq R^n$ with $K \subseteq N$; the free module $N$ has rank at most $n$, so $N/K$ is generated by at most $n$ elements. (iv) Immediate on ranks.

The Classification of Finitely Generated Abelian Groups

For $R=\mathbb{Z}$ the theorem reads: every finitely generated abelian group is

$$ \mathbb{Z}^r \oplus \mathbb{Z}/(d_1)\oplus\cdots\oplus \mathbb{Z}/(d_k), \qquad d_1 \mid \cdots \mid d_k,\ d_i \ge 2, $$

with $r$ and the $d_i$ unique. The elementary divisor form is $\mathbb{Z}^r \oplus \bigoplus_i \mathbb{Z}/(p_i^{e_i})$ with $p_i$ prime. The applications article of this category computes with this classification, counts the groups of a given order, and reads off subgroups and quotients.

The Rational Canonical Form and the Jordan Form

For $R=K[x]$ the finitely generated torsion modules are the finite-dimensional $K$-vector spaces with a linear operator, and the invariant factors are polynomials. The applications article of this category turns the theorem into the rational canonical form and, when the characteristic polynomial splits, the Jordan form.

Non-Finitely-Generated Modules

Finite generation is essential. The $\mathbb{Z}$-module $\mathbb{Q}$ is torsion-free but not free, and $\mathbb{Q}/\mathbb{Z}$ is torsion but is not a direct sum of finitely many cyclic modules; neither is covered by the theorem. What survives is the theory of divisible and injective modules: over a principal ideal domain a module is injective precisely when it is divisible, and every module has an injective envelope, which is the dual statement to the existence of free presentations. The classification of all modules over a principal ideal domain is not finite in content — already over $\mathbb{Z}$ it contains the classification of all abelian groups, which is not a tame problem — and the structure theorem is best understood as the finitely generated case of a theory that has no finitely generated analogue.

Summary

Over a principal ideal domain $R$, every submodule of a free module is free, with rank at most that of the ambient module. The torsion elements of a module form a submodule, the quotient by it is torsion-free, and a finitely generated torsion module is the direct sum of its finitely many $p$-primary components, one for each prime dividing its annihilator.

The structure theorem states that a finitely generated module is

$$ M \cong R^r \oplus R/(d_1)\oplus\cdots\oplus R/(d_k), \qquad d_1 \mid \cdots \mid d_k, $$

with $r$ and the invariant factors $d_i$ unique up to units. Existence is proved by presenting $M=R^n/K$, taking the Smith normal form of the matrix of the inclusion $K \hookrightarrow R^n$ and reading the quotient off the diagonal. Uniqueness is proved by tensoring with the fraction field to recover $r$, and by counting $\dim_{k(p)}(p^{n-1}T_p/p^nT_p)$ to recover the exponents of each prime. The elementary divisor form replaces the invariant factors by the prime powers occurring in them, and the two descriptions are equivalent.

The Smith normal form is the algorithmic content: elementary row and column operations put any matrix into a diagonal form with $d_1 \mid \cdots \mid d_q$, the form is unique, and its diagonal entries are read from the determinantal ideals $D_k(A)=(d_1\cdots d_k)$. Consequences include: a finitely generated torsion-free module is free; a submodule of a module generated by $n$ elements is generated by $n$ elements; the index of a full-rank sublattice equals the absolute value of the determinant of its basis matrix. Specialising to $\mathbb{Z}$ recovers the classification of finitely generated abelian groups and specialising to $K[x]$ the rational and Jordan canonical forms.

Summary of Notation

Symbol	Meaning
$R$	a principal ideal domain, commutative with $1 \neq 0$
$K=\operatorname{Frac}(R)$	fraction field of $R$
$M$, $N$, $T$	$R$-modules, left (hence right, $R$ commutative)
$M_{\mathrm{tor}}$	torsion submodule of $M$
$M_p$	$p$-primary component, $p$ prime
$\operatorname{Ann}(m)$, $\operatorname{Ann}(M)$	annihilators; $\operatorname{Ann}(m)=(a)$ over a PID
$R/(a)$	cyclic module, $Rm \cong R/\operatorname{Ann}(m)$
$(a)$	principal ideal generated by $a$
$d_1 \mid \cdots \mid d_k$	invariant factors of a finitely generated module
$r$	free rank; $r=\dim_K(M\otimes_R K)$
$p^{e_{p,j}}$	elementary divisors
$k(p)=R/(p)$	residue field at a prime $p$
$D_k(A)$	determinantal ideal generated by the $k \times k$ minors of $A$
$\operatorname{diag}(d_1,\dots,d_q)$	diagonal matrix with the given diagonal
$\mathbb{Z}/n\mathbb{Z}$	integers modulo $n$
$\mathbb{Q}/\mathbb{Z}$	the torsion divisible abelian group