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An associator ring is a ring in which equality of the principal ideals generated by two elements forces them to be associate elements: $(a)=(b)$ implies $a=ub$ for a unit in a ring $u$. The reverse implication always holds, since $a=ub$ and $b=u^{-1}a$ give both ideal inclusions. The property separates equality of generated ideals from equality of generators up to an invertible factor.

Every integral domain is an associator ring. If $a\ne 0$ and $(a)=(b)$, write $a=rb$ and $b=sa$. Then $(1-rs)a=0$, so cancellation gives $rs=1$ and $a=rb$ with $r$ a unit. The zero case is immediate. The converse fails: $\mathbb{Z}/4\mathbb{Z}$ has generator classes $\{0\}$, $\{2\}$ and $\{1,3\}$ for its three ideals, and $1,3$ differ by a unit, although $2$ is a nonzero zero divisor.

Finite direct product of rings preserve this property. Equality of principal ideals is coordinatewise, and a tuple of units is a unit. Every quotient ring of a principal ideal domain is also an associator ring: first, modulo a prime power $p^k$, every nonzero class is a power $p^e$, $e<k$, times a unit, and distinct powers generate distinct ideals. Next factor a nonzero nonunit modulus into prime powers and apply the Chinese remainder theorem. The zero modulus gives the original integral domain, while a unit modulus gives the zero ring.

A discrete valuation ring is a principal ideal domain with exactly one nonzero maximal ideal. Every nonzero element of its fraction field is a unit times a unique integer power of a uniformizer.

Let $R$ be a Euclidean domain with Euclidean function $\nu$, and let $I$ be a nonzero ideal. Choose $b\in I\setminus \{0\}$ minimizing the nonnegative integer $\nu (b)$. For $a\in I$, divide $a=qb+r$ with $r=0$ or $\nu (r)<\nu (b)$. Since $r=a-qb\in I$, the latter case contradicts minimality. Hence every $a$ is a multiple of $b$, giving $I=(b)$. The zero ideal is $(0)$, so every ideal is a principal ideal and $R$ is a principal ideal domain.

If $\alpha =a+bi$ is a Gaussian integer and $P$ is a nonzero prime ideal of $R=\mathbb{Z}[\alpha ]$, then $R/P$ is a finite field. Complex conjugation preserves $R$, so a nonzero $\beta \in P$ gives the positive integer $\beta \bar{\beta }\in P$. Primality supplies a rational prime $p\in P$. Since $\alpha$ satisfies the monic polynomial $t^2-2at+a^2+b^2$, the ring $R/P$ is a quotient of ${\mathbb{F}}_p[t]/(\bar{F})$ and has at most $p^2$ elements. It is a finite integral domain, hence a field. The proof does not require $R$ to be a principal ideal domain.

Every subring $R$ of the rational numbers containing $\mathbb{Z}$ is a principal ideal domain. If $a/b\in R$ is reduced, Bézout's identity gives integers $u,v$ with $ua+vb=1$, hence $1/b=u(a/b)+v\in R$. Thus every reduced denominator occurring in $R$ is a unit.

For an ideal $I$, write $I\cap \mathbb{Z}=d\mathbb{Z}$. Then $dR\subseteq I$. If $a/b\in I$ is reduced, $a=b(a/b)\in I\cap \mathbb{Z}$, so $a=dc$ for some integer $c$. Since $1/b\in R$, $a/b=d(c/b)\in dR$. Therefore $I=dR$, including the zero ideal. This is a concrete instance of extending and contracting ideals under localization of a ring.

The polynomial ring $\mathbb{Z}[X]$ is a unique factorization domain by Gauss lemma for polynomials, but its ideal $(2,X)$ is not principal. Any generator would divide both 2 and $X$, hence be a unit: a divisor of 2 has degree zero, and a constant dividing $X$ is $\pm 1$. Yet evaluation at zero followed by reduction modulo 2 annihilates the ideal and not 1. It embeds in the principal ideal domain $\mathbb{Q}[X]$, illustrating that the property need not pass to a subring.

For a finitely generated module over a principal ideal domain, its invariant factors are nonzero nonunits $d_i$ in a decomposition $M\cong R^r\oplus {⨁}_{i}R/(d_i)$, ordered by $d_i\mid d_{i+1}$. They are unique up to multiplication by units and describe the torsion submodule. Over $F[t]$, making $t$ act as a linear operator recovers the invariant factors of a linear operator used in its rational canonical form.

Let $\pi$ be an irreducible element of a principal ideal domain $R$, and suppose $\pi \nmid a$. The ideal $(\pi ,a)=(d)$ has generator $d$ dividing $\pi$. Irreducibility makes $d$ a unit or an associate of $\pi$; the latter would imply $\pi \mid a$. Hence $(\pi ,a)=R$, giving $u\pi +va=1$ for some $u,v\in R$. If $\pi \mid ab$, multiply this identity by $b$: both $u\pi b$ and $vab$ are divisible by $\pi$, so $\pi \mid b$. This is precisely the prime element property.

In a principal ideal domain, every nonzero prime ideal is maximal. Indeed, write it as $(p)$ with $p\ne 0$. If $(p)\subseteq (a)$, then $p=ab$. Primality gives $a\in (p)$, hence $(a)=(p)$, or $b\in (p)$, say $b=pc$, in which case cancellation gives $1=ac$ and $(a)=R$.

Thus every prime chain has length at most one. If the principal ideal domain is not a field, choose a nonzero nonunit $a$ and a maximal ideal $\mathfrak{m}\supseteq (a)$. Then $(0)⊊\mathfrak{m}$ is a prime chain of length one, so the Krull dimension is one. A field instead has Krull dimension zero.

For a linear operator $T$ on an $n$-dimensional vector space over a field, the polynomials annihilating a fixed vector $v$ form an ideal of $K[X]$. It is nonzero because $v,Tv,\dots ,T^{n}v$ are linearly dependent. Since $K[X]$ is a principal ideal domain, its unique monic generator $m_v$ exists and has degree at most $n$. It divides $f$ exactly when $f(T)v=0$. For $v=0$ it equals one. Its degree is the dimension of the cyclic subspace generated by $v$: the vectors up to $T^{d-1}v$ are independent by minimality and span an invariant space by the annihilating recurrence.

Module theory studies abelian groups with scalar multiplication by a ring. It extends linear algebra: a module over a field is a vector space, while ring scalars let the same language describe integer arithmetic and linear operators. The new phenomena are visible before any abstract classification: some nonzero scalars cannot be divided out, some modules have torsion elements of a module, and some have no basis of a module.

The distinction is useful because an integer relation such as $6m=0$ carries information which disappears if one assumes all nonzero scalars have multiplicative inverses. A finitely generated module over a principal ideal domain can be separated into a finite free module and a torsion module; the structure theorem for finitely generated modules over a principal ideal domain unifies the classification of finitely generated abelian groups with the study of linear operators through polynomial rings. See Keith Conrad's Modules over a PID, Theorem 4.1 and the operator applications.

A historical anchor is Emmy Noether's 1921 paper ^[ref]. Its first section studies rings and ideals under the condition that every ideal is a finitely generated module over its ring, and connects this to the ascending chain condition; later sections develop primary decomposition. This is an early source for the structural viewpoint underlying Noetherian rings. ^[ref] makes those statements accessible in English.

Every field $\mathbb{Q}({\zeta }_{59^r})$ contains $\mathbb{Q}(\sqrt{-59})$ and is totally ramified over it at the prime above $59$. The class number divisibility under total ramification therefore shows that its class number is divisible by three. Its ring of integers is not a principal ideal domain. The degrees $58\cdot 59^{r-1}$ strictly increase, giving infinitely many distinct examples.

For a prime number $p$, the ideal $(p,X)$ in the polynomial ring $\mathbb{Z}[X]$ is proper: the constant coefficient of every member is divisible by $p$. If $(p,X)=(g)$, then $g$ divides the nonzero constant $p$, so the additivity of polynomial degree over an integral domain makes $g$ constant. Since $g$ also divides $X$, it must divide its leading coefficient $1$; thus $g=\pm 1$. This would make the ideal the whole ring, a contradiction. Therefore $\mathbb{Z}[X]$ is not a principal ideal domain, and hence not a Euclidean domain.

The ring $\mathbb{Z}[\sqrt{2}]$ is a Euclidean domain for the absolute field norm. Approximate both rational coefficients of a quotient by nearest integers. If the errors are $u,w\in [-1/2,1/2]$, then $|u^2-2w^2|\le 1/2<1$, giving a remainder of strictly smaller absolute norm. In particular this ring is a principal ideal domain and has trivial ideal class group.

Take a reduced fraction $a/b\in R$, with $b\ne 0$. Since $\gcd (a,b)=1$, Bézout's identity supplies integers $u,v$ with $ua+vb=1$. As the subring $R$ contains all integers,Thus a denominator appearing in a reduced fraction in $R$ is automatically a unit of $R$.

Now let $I\cap \mathbb{Z}=d\mathbb{Z}$. Because $d\in I$, the principal ideal $dR$ is contained in $I$. For the reverse inclusion, take $x\in I$ and write $x=a/b$ in lowest terms. Multiplying by $b\in R$ gives $a=bx\in I\cap \mathbb{Z}=d\mathbb{Z}$, so $a=dc$ for some integer $c$. The preceding result gives $1/b\in R$, whenceThereforeThis includes $d=0$: the same argument forces every $x\in I$ to be zero. This is the ideals of subrings of the rational numbers property; every such ring is a principal ideal domain.

The structure theorem for finitely generated modules over a principal ideal domain applies to a Euclidean domain because every Euclidean domain is a principal ideal domain. In invariant-factor form, every finitely generated module $M$ has a decompositionwhere $r\ge 0$, and each $d_i$ is a nonzero nonunit. The rank of a module $r$ is unique and the invariant factors of a finitely generated module $d_i$ are unique up to multiplication by units. The zero module is allowed by taking $r=s=0$.

Every abelian group is a module over $\mathbb{Z}$, and a finite abelian group is finitely generated. Since $\mathbb{Z}$ is a Euclidean domain, the theorem applies. A nonzero free module summand would be infinite, so a finite group has $r=0$. Choosing positive integer representatives gives the finite abelian classificationThe ordered integers $d_i$ are uniquely determined. Splitting each $d_i$ into prime powers using the Chinese remainder theorem gives the equivalent elementary-divisor form as a direct sum of cyclic prime-power groups.

Put $A=R/(p^k)$. A principal ideal domain is a unique factorization domain, so every nonzero element of this quotient ring can be writtenIndeed, remove the largest power of the prime element $p$ from a representative that is not divisible by $p^k$. Since $s$ and $p^k$ have greatest common divisor $1$ up to a unit, Bézout's identity in the principal ideal domain gives $xs+y{p}^k=1$. Thus $\overline{s}$ is a unit in $A$.

Consequently every nonzero principal ideal in $A$ is generated by one of $1,\overline{p},\dots ,{\overline{p}}^{k-1}$. Include the zero ideal by assigning it exponent $e=k$. These ideals are distinct: their inverse images in $R$ are $(1),(p),\dots ,(p^k)$, and equality of two different powers would, by cancellation in the integral domain, make the nonunit $p$ a unit.

Now suppose $(\overline{a})=(\overline{b})$. If this is the zero ideal, both generators are zero and are associate elements. Otherwise their exponents agree, so writewith $\overline{s},\overline{t}$ both units. Thenand the multiplier is a unit. ThusThe exponent detects the principal ideal, while its remaining factor is invertible; this is why zero divisors cause no obstruction here.

By the first isomorphism theorem for rings, the given surjection identifies $S$ with $R/\ker \varphi$. Since $R$ is a principal ideal domain, write $\ker \varphi =(a)$.

If $a=0$, the quotient ring is $R$, an integral domain, so part (a)(ii) applies. If $a$ is a unit, the quotient ring is the zero ring; its unique element is associate to itself by multiplication by its identity, so it too is an associator ring.

In the remaining case, unique factorization giveswhere $\epsilon$ is a unit, the $p_i$ are pairwise nonassociate prime elements, and $k_i\ge 1$. The principal ideals $(p_i^{k_i})$ are pairwise comaximal: different prime factors have greatest common divisor $1$, and Bézout's identity gives a linear combination equal to $1$. The Chinese remainder theorem therefore gives a ring isomorphismEvery factor is an associator ring by part (c), and their finite direct product of rings is an associator ring by part (b). A ring isomorphism preserves units and principal ideals, so the property transfers to $S$. Hence

A Euclidean domain is a principal ideal domain. The structure theorem for finitely generated modules over a principal ideal domain therefore gives, for every finitely generated module $M$ over $R$,where the $d_i$ are nonzero nonunits and $d_1\mid d_2\mid \cdots \mid d_s$. The factors $d_i$ are unique up to multiplication by units; $r$ is uniquely determined.

The rank of a module is this integer $r$. Equivalently, with $K$ the field of fractions of $R$,The tensor product of modules kills each displayed torsion summand after extending scalars to $K$, while sending $R^r$ to $K^r$, explaining the equality.

If $M$ is a torsion-free module, no summand $R/(d_i)$ can occur: its nonzero element $1+(d_i)$ is killed by the nonzero scalar $d_i$. Hence $s=0$ andThis also covers $M=0$, with $r=0$.

The nonzero ideal $\tilde{P}$ is principal because $\mathbb{Q}[t]$ is a principal ideal domain. Choose a nonzero generator $h\in \mathbb{Q}[t]$. Part (a) givesThe nonzero rational scalar $c(h)$ is a unit in $\mathbb{Q}[t]$, so this rescaling does not change the generated principal ideal:Part (b) now places $f$ in $P$, since $f\in \tilde{P}\cap \mathbb{Z}[t]$. Thus $(f{)}_{\mathbb{Z}[t]}\subseteq P$.

For the reverse inclusion, let $g\in P$ be nonzero. The displayed equality supplies $q\in \mathbb{Q}[t]$ with $g=fq$. Multiplicativity of rational polynomial content givesSince $g$ has integer coefficients, $c(g)$ is a positive integer. The normalization from part (a) gives $q/c(q)\in \mathbb{Z}[t]$, so multiplying it by the integer $c(q)=c(g)$ proves $q\in \mathbb{Z}[t]$. Consequently $g\in (f{)}_{\mathbb{Z}[t]}$. The zero polynomial belongs as well, provingThis is the divisibility descent in Gauss lemma for polynomials: dividing an integer-coefficient polynomial by a primitive polynomial cannot introduce a denominator into an exact quotient. Here it yields the description of prime ideals of the integer polynomial ring above zero. The generator has positive polynomial degree, since a primitive constant would be a unit and would put $1$ in $P$.