Modules over a Principal Ideal Domain and the Canonical Forms

The zero element is torsion. If $am=0$ and $bn=0$ with $a,b\mathrm{\ne }0$, then $ab\mathrm{\ne }0$ and $ab(m+n)=b(am)+a(bn)=0$; if $c\in R$, then $a(cm)=c(am)=0$. Closure under negatives is the scalar case $c=-1$, so $\mathrm{Tor}(M)$ is a submodule.

The zero element lies in $M[p^{\mathrm{\infty }}]$. If $p^{k}m=0$ and $p^{\mathrm{\ell }}n=0$ with $k,\mathrm{\ell }\ge 1$, then $p^{\max (k,\mathrm{\ell })}(m+n)=0$, and $p^k(cm)=0$ for every $c\in R$. Thus $M[p^{\mathrm{\infty }}]$ is a submodule contained in $\mathrm{Tor}(M)$. The proof includes $k=1$, the zero module, and replacing $p$ by an associate.

A submodule of the left regular module ${}_{R}R$ is an ideal. By [F1] it is generated by one element; this includes the zero ideal $(0)$ and unit ideal $(1)$.

Thus every submodule of ${}_{R}R$ is finitely generated, so [L1] makes the regular module Noetherian and hence makes $R$ a Noetherian ring. Commutativity makes the same statement valid on the right.

Let $(p)\subseteq I\subseteq R$. By [F1], write $I=(a)$, so $p=ab$ for some $b$. Irreducibility makes $a$ or $b$ a unit. If $a$ is a unit, $I=R$; if $b$ is a unit, $(a)=(p)$. Since $p$ is a nonzero nonunit, $(p)$ is proper and therefore maximal.

By [L1], the maximal ideal $(p)$ is prime.

If $p\mid xy$, then $xy\in (p)$, so primality of the ideal gives $x\in (p)$ or $y\in (p)$, equivalently $p\mid x$ or $p\mid y$. Thus $p$ is a prime element. Associates generate the same ideal and give the same conclusion.

Suppose, for contradiction, that some nonzero nonunit is not a product of irreducibles. The principal ideals generated by such bad elements form a nonempty family; by [L1] and [L2], choose a maximal member $(a)$. This use of the maximal condition spends DC, supplied by the assumed AC. The element $a$ is not irreducible, so $a=bc$ with nonunit nonzero $b,c$. Then $(a)$ is strictly contained in both $(b)$ and $(c)$: for example, $(a)=(b)$ would give $b=at=bct$ and cancellation would make $c$ a unit. Maximality therefore makes $b$ and $c$ products of irreducibles, and their product is a factorization of $a$, a contradiction. Thus every nonzero nonunit factors into irreducibles.

For uniqueness, if $p_1\cdots p_r=q_1\cdots q_s$ are irreducible factorizations, primality of $p_1$ from [L3] makes it divide some $q_j$, hence the two factors are associates. Cancel them in the domain and repeat on the remaining finite product. This induction pairs every factor and shows $r=s$ up to order and associates; the empty product is the unit case.

Step 1.1 gives factorization existence and step 1.2 gives the uniqueness required in the UFD definition. Therefore every PID is a UFD; the contradictory assumption in step 1.1 is discharged.

Some coordinate functional has a nonzero value on $N$. Fix one such nonzero value ideal $I_0$. By [L1], a nonzero generator of $I_0$ has only finitely many divisor classes, so only finitely many principal ideals can contain $I_0$. Choose a maximal value ideal among them, write it as $\psi (N)=(a)$ with $a\mathrm{\ne }0$, and choose $v\in N$ with $\psi (v)=a$.

For any coordinate functional $\phi$, let $d$ generate $(a,\phi (v))$ and choose $r,s$ with $d=ra+s\phi (v)$. The functional $r\psi +s\phi$ takes $v$ to $d$, so its value ideal contains $(d)\supseteq (a)$; maximality in step 1.1 forces $(d)=(a)$, hence $a\mid \phi (v)$.

Divisibility of every coordinate of $v$ gives $v=a{e}_1$ for some $e_1\in M$. Since $a=\psi (v)=a\psi (e_1)$ and $a\mathrm{\ne }0$, cancellation gives $\psi (e_1)=1$, so $e_1$ is primitive.

The ideal generated by the coordinates of $e_1$ in a basis of $M$ is $\{\phi (e_1):\phi \in M^{\ast }\}$, so it does not depend on the basis, and $\psi (e_1)=1$ makes it all of $R$. Fix a basis $f_1,\dots ,f_n$ of $M$ and let $b_1,\dots ,b_n$ be the coordinates of $e_1$ in it. For $j\ge 2$ let $d$ generate $(b_1,b_j)$. If $d=0$ then $b_1=b_j=0$ and nothing is done; otherwise write $d=r{b}_1+s{b}_j$ and put $A=\left(\begin{array}{cc}r& s\\ -b_j\mathrm{/}d& b_1\mathrm{/}d\end{array}\right)$, whose entries lie in $R$ and whose determinant is $(r{b}_1+s{b}_j)\mathrm{/}d=1$, so $A^{-1}$ has entries in $R$ as well. Replacing the basis pair $(f_1,f_j)$ by the pair whose coordinates are the columns of $A^{-1}$ again gives a basis of $M$, in which the coordinates of $e_1$ at $1$ and $j$ are the entries of $A(b_1,b_j{)}^{\mathsf{T}}=(d,0{)}^{\mathsf{T}}$ and the other coordinates are unchanged. Doing this for $j=2,\dots ,n$ in turn clears the coordinates $2,\dots ,n$, so the resulting basis $g_1,\dots ,g_n$ has $e_1=c{g}_1$ with $(c)$ the coordinate ideal, which is $R$. Hence $c$ is a unit and $e_1,g_2,\dots ,g_n$ is a basis of $M$.

Put $u_j=g_j-\psi (g_j)e_1$ for $2\le j\le n$. The passage from $e_1,g_2,\dots ,g_n$ to $e_1,u_2,\dots ,u_n$ is triangular with $1$ on the diagonal, so the latter is again a basis of $M$, and each $u_j$ lies in $\ker \psi$. If $m={\lambda }_1{e}_1+{\sum }_{j\ge 2}{\lambda }_j{u}_j$ lies in $\ker \psi$, applying $\psi$ gives ${\lambda }_1=0$; hence $u_2,\dots ,u_n$ is a basis of $\ker \psi$, which is therefore free of rank $n-1$, and zero when $n=1$. Every $m\in M$ is $\psi (m)e_1+(m-\psi (m)e_1)$ with the second summand in $\ker \psi$, and $R{e}_1\cap \ker \psi =0$ because $\psi (r{e}_1)=r$, so $M=R{e}_1\oplus \ker \psi$. For $w\in N$ one has $\psi (w)\in \psi (N)=(a)$, say $\psi (w)=ra$, and then $w=rv+(w-rv)$ with $w-rv\in N\cap \ker \psi$; also $Rv\cap \ker \psi =0$, since $\psi (rv)=ra$ and $a\mathrm{\ne }0$ in the domain $R$. Hence $N=Ra{e}_1\oplus (N\cap \ker \psi )$. When $a$ is a unit, $v$ and $e_1$ generate the same submodule and the second decomposition reads $N=R{e}_1\oplus (N\cap \ker \psi )$.

If $n=0$, then $M=N=0$ and both bases are empty, giving $r=0$. If $N=0$ for any $n$, choose any basis of $M$ and the empty basis of $N$.

Assume the theorem for free ambient modules of rank less than a fixed $n\ge 1$.

For nonzero $N\le M$ of rank $n$, apply [L1] to obtain $M=R{e}_1\oplus M_1$ and $N=R{a}_1{e}_1\oplus N_1$, where $M_1=\ker \psi$ is free of rank $n-1$ and $N_1=N\cap M_1$.

Apply the induction hypothesis of step 1.2 to $N_1\le M_1$, obtaining basis vectors $e_2,\dots ,e_n$ and, when $N_1\mathrm{\ne }0$, factors $a_2\mid \cdots \mid a_r$. If $N_1=0$, then $r=1$ and the required chain consists only of $a_1$. Otherwise let $e_1^{\ast },e_2^{\ast }$ be the coordinate functionals of the combined ambient basis and put $\phi =e_1^{\ast }+e_2^{\ast }$. Since $\phi (a_1{e}_1)=a_1$, one has $(a_1)\subseteq \phi (N)$; maximality of the pivot value ideal in [L1] forces equality, and $a_2=\phi (a_2{e}_2)\in (a_1)$, so $a_1\mid a_2$. Concatenating the bases gives the required aligned bases and chain, with $r\le n$, and completes the induction.

The family in [L1] is an $R$-basis of $N$, so $N$ is free and has a basis of length $r\le n$. The zero submodule has the empty basis, and ambient rank zero forces $r=0$.

Invariant basis number makes the finite basis length unique, so $r$ is the rank of $N$. The endpoints $r=0$ and $r=n$, including $N=M$, are permitted.

Choose generators $m_1,\dots ,m_n$ of $M$ and define the surjection $\pi :R^n\to M$ by sending the standard basis to them; put $N=\ker \pi$. For the zero module one may take $n=0$.

Apply [L1] to $N\le R^n$. There is a basis $e_1,\dots ,e_n$ of $R^n$ and a basis $a_1{e}_1,\dots ,a_r{e}_r$ of $N$, with nonzero $a_1\mid \cdots \mid a_r$.

The first isomorphism theorem and coordinatewise quotient give $M\cong R^n\mathrm{/}N\cong R\mathrm{/}(a_1)\oplus \cdots \oplus R\mathrm{/}(a_r)\oplus R^{n-r}$. Any unit $a_i$ contributes the zero quotient and is removed; the remaining nonunit factors preserve the divisibility chain. This constructs the stated decomposition, including purely free, purely torsion, cyclic, and empty cases.

In the decomposition from [L1], every nonzero quotient $R\mathrm{/}(a_i)$ with $a_i\mathrm{\ne }0$ consists of torsion elements, whereas the free summand is torsion-free because $R$ is a domain. Torsion-freeness therefore forces every cyclic quotient summand to be zero.

Only the finite free summand remains, so $M$ is finite free. The zero module is the free module on the empty basis, and unit invariant factors already give zero summands.

In [L1], every cyclic quotient $R\mathrm{/}(a_i)$ is torsion, and every torsion element has zero component in the free summand because a free module over a domain is torsion-free. Thus the direct sum of the cyclic quotients is exactly $\mathrm{Tor}(M)$.

Substituting that identification into the invariant-factor decomposition gives $M\cong \mathrm{Tor}(M)\oplus R^r$. It includes pure torsion when $r=0$, pure free modules when $\mathrm{Tor}(M)=0$, and the zero module when both vanish.

The torsion submodule is canonical because it is defined by $M$ alone, while a free complement is not. Suppose $a\in R$ is a nonzero nonunit and put $M=R\oplus R\mathrm{/}(a)$. Since $R$ is a domain, an element $(x,y+(a))$ killed by some nonzero scalar has $x=0$, and $a$ kills every $(0,y+(a))$; hence $\mathrm{Tor}(M)=0\oplus R\mathrm{/}(a)$. Both $C_1=R(1,0+(a))$ and $C_2=R(1,1+(a))$ are free of rank one, because $r(1,\ast )=0$ forces $r=0$, and each meets $\mathrm{Tor}(M)$ only in $0$ while $(x,y+(a))=(0,(y-x)+(a))+x(1,1+(a))$ shows $\mathrm{Tor}(M)+C_2=M$ and likewise for $C_1$. They are distinct: $(1,1+(a))\in C_2$ lies in $C_1$ only if $a\mid 1$, contrary to $a$ being a nonunit.

Coprimality means $(a,b)=R$, so choose $r,s\in R$ with $ra+sb=1$.

The map $\mathrm{\Phi }:R\mathrm{/}(ab)\to R\mathrm{/}(a)\oplus R\mathrm{/}(b)$ given by $x+(ab)\mapsto (x+(a),x+(b))$ is well defined. If both residues vanish, $a\mid x$ and $b\mid x$; multiplying the Bezout identity appropriately gives $ab\mid x$, so $\mathrm{\Phi }$ is injective. Given residues $u+(a)$ and $v+(b)$, the element $usb+vra$ maps to them, so $\mathrm{\Phi }$ is surjective and is an $R$-module isomorphism.

Repeatedly apply step 2.1 to a finite pairwise coprime family; the product of any subfamily remains coprime to the next factor. This gives the displayed finite direct sum, with the empty and singleton conventions stated above and with unit factors contributing zero quotients.

Take the invariant-factor decomposition from [L1]. Factor each nonunit $a_i$ uniquely up to associates as a finite product of powers of pairwise nonassociate irreducibles. The free part is kept separate, and no factorization is assigned to zero or to a unit.

The prime-power factors of one $a_i$ are pairwise coprime, so [L2] splits $R\mathrm{/}(a_i)$ into the corresponding quotients $R\mathrm{/}(p^e)$. Applying this to every invariant factor gives a finite elementary-divisor direct sum.

Grouping the summands by the associate class of $p$ gives exactly $M[p^{\mathrm{\infty }}]$, because an element is killed by a power of $p$ precisely in those summands. Thus the torsion submodule is the direct sum of its primary components, each having the asserted form.

Conversely, order the powers for each prime by exponent, align the largest powers in the last column, pad missing entries by units, and multiply down columns. Each column divides the next, and repeated use of [L2] recovers the original elementary-divisor sum. This constructs the invariant factors, including empty torsion data.

Since $p$ is prime, $(p)$ is a prime ideal; in a PID it is maximal by the same divisibility argument as for irreducibles, so $R\mathrm{/}(p)$ is a field. Every element of $p^{k-1}M[p^k]$ is killed by $p$, making this set an $R\mathrm{/}(p)$-vector space.

On one summand $C=R\mathrm{/}(p^e)$, if $e\ge k$ then $C[p^k]=p^{e-k}C$ and $p^{k-1}C[p^k]=p^{e-1}C$, a one-dimensional $R\mathrm{/}(p)$-space. If $e<k$, then $p^{k-1}C=0$, so the contribution is zero. For a $q$-primary summand with $q$ not associated to $p$, unique factorisation [L2] makes $p^k$ and $q^e$ coprime, so choose $u,v$ with $u{p}^k+v{q}^e=1$; multiplication by $u$ is inverse to multiplication by $p^k$ on $R\mathrm{/}(q^e)$, and its $p^k$-torsion is zero. Free summands likewise contribute no $p$-power torsion.

Direct sums commute with $M[p^k]$ and multiplication by $p^{k-1}$, so dimensions add. By step 2.1, $d_k$ counts exactly the exponents $e\ge k$; for $k=1$ it counts every $p$-primary cyclic summand, and beyond the largest exponent it is zero.

The summands counted by $d_k$ but not by $d_{k+1}$ are exactly those with exponent $e=k$, so $d_k-d_{k+1}$ recovers their multiplicity. Empty primary data and the zero module give the zero sequence.

An isomorphism carries torsion elements to torsion elements, so it induces an isomorphism of torsion submodules and of the quotients by torsion. The latter quotients are finite free, and invariant basis number recovers their common free rank.

For every associate class of irreducible $p$ and every $k\ge 1$, an isomorphism preserves $M[p^k]$, multiplication by $p^{k-1}$, and the resulting $R\mathrm{/}(p)$-dimension. By [L1], it therefore preserves every elementary-divisor multiplicity.

Aligning the unique prime-power columns as in the primary-decomposition theorem recovers one divisibility chain of invariant factors up to associates. Hence invariant factors are unique as well. Empty torsion data, pure torsion, pure free, and the zero module are included.

Conversely, equal free ranks and equal invariant-factor or elementary-divisor data give termwise isomorphisms between the corresponding direct-sum decompositions; their direct sum constructs a module isomorphism. This proves both directions of the classification.

Write $M\cong R^r\oplus T$ by [L1], with $T={⨁}_{i=1}^{t}R\mathrm{/}(a_i)$ and each $a_i$ a nonzero nonunit. This is an invariant-factor decomposition, so $r={\mathrm{rank}}_{R}M$ by the free-rank definition in the Given. Extension of scalars and [L4] give $Q{\otimes }_{R}M\cong (Q{\otimes }_{R}T)\oplus (Q{\otimes }_R{R}^r)$.

Every $m\in T$ is killed by the nonzero product $a_1\cdots a_t\in R$; in $Q$ that scalar is invertible by [L2], so each simple tensor satisfies $q\otimes m=a^{-1}q\otimes am=0$ for that $a$. Simple tensors generate the tensor product, hence $Q{\otimes }_{R}T=0$.

By [L3] and [L4], $Q{\otimes }_R{R}^r\cong Q^r$. Therefore $Q{\otimes }_{R}M\cong Q^r$ and its $Q$-dimension is $r$, which step 1.1 identified with ${\mathrm{rank}}_{R}M$. This includes $r=0$, pure torsion, pure free, rank one, and the zero module, where $t=0$ makes $T$ the zero module.

Regard $A$ as $\phi :R^n\to R^m$ and put $I=\mathrm{im}\phi$. By [L1], both $I\le R^m$ and $\ker \phi \le R^n$ are finite free; the sequence $0\to \ker \phi \to R^n\to I\to 0$ is exact.

Apply [L1] to align $I$ in the codomain: choose a basis $e_1,\dots ,e_m$ of $R^m$ such that $d_1{e}_1,\dots ,d_r{e}_r$ is a basis of $I$ with $d_1\mid \cdots \mid d_r$.

Since $I$ is free and projective, the surjection $R^n\to I$ has a section. By [L2], $R^n=\ker \phi \oplus s(I)$. Place the lifted basis $s(d_1{e}_1),\dots ,s(d_r{e}_r)$ first and then a basis of the kernel to obtain a domain basis.

In the bases from steps 2.1 and 2.2, $\phi$ sends the lifted vector for $d_i{e}_i$ to $d_i{e}_i$ and kills the kernel basis, so its matrix is the displayed Smith diagonal. Empty matrices, the zero map, and all rank deficiencies give the appropriate empty or trailing-zero diagonal.

Every $k$-minor of $PA$ is an $R$-linear combination of $k$-minors of $A$ by determinant multilinearity, so $D_k(PA)\subseteq D_k(A)$; applying the same argument to $P^{-1}$ gives equality. Right multiplication is identical. Thus equivalent matrices have the same determinantal ideals.

For a Smith diagonal matrix with $d_1\mid \cdots \mid d_r$, every nonzero $k$-minor is a product of $k$ diagonal entries and is divisible by $d_1\cdots d_k$, while the leading $k$-minor equals that product. Hence ${\mathrm{\Delta }}_k$ is associate to $d_1\cdots d_k$ for $k\le r$ and is $0$ for $k>r$.

Step 1.1 makes the ${\mathrm{\Delta }}_k$ invariants of equivalence. Step 1.2 recovers $r$ as the last nonzero index and recovers each $d_k$ up to a unit from successive products, proving uniqueness. It includes $k=0$, zero matrices, one-by-one matrices, and all rectangular ranks.