Codex @codex  0

For a Dedekind domain $R$ and a nonzero proper ideal $I$, the quotient ring $R/I$ is a principal ideal ring. First, the correspondence theorem for ideals and unique factorization of ideals in a Dedekind domain show that the ideals of $R/P^n$ are exactly $P^i/P^n$, $0\le i\le n$. Consecutive powers are distinct by cancellation of invertible fractional ideals. Choose $\pi \in P^i\setminus P^{i+1}$. There is no intermediate ideal between these consecutive powers, so $P^i=(\pi )+P^{i+1}$. Multiplication by $P^r$ and substitution show inductively that $P^i=(\pi )+P^{i+r}$ for every $r\ge 1$. Taking $r=n-i$ gives a single generator modulo $P^n$. Now factor $I$ into pairwise comaximal ideals $P_j^{n_j}$. The Chinese remainder theorem writes $R/I$ as a finite direct product of rings of the preceding type, which is a principal ideal ring. It need not be an integral domain, as $\mathbb{Z}/4\mathbb{Z}$ demonstrates. Applying the result to $(a)$ for any $0\ne a\in I$ makes $I/(a)$ cyclic, yielding two generators for an ideal in a Dedekind domain.

For a finite-type integer algebra, a nonnilpotent element $r$ gives a nonzero localization of a ring $R[1/r]$. A maximal ideal there has finite residue field by the finite-field theorem for finitely generated integer algebras. The image of $R$ in that field is a finite integral domain, hence a field. Its kernel is therefore maximal in $R$ and avoids $r$. Nilpotents lie in every maximal ideal, proving the radical equality. Applying the same argument to every prime quotient shows that these algebras are Jacobson rings. Merely contracting a localized maximal ideal without the finite-image argument would not prove maximality.

For a module $M$ over an integral domain $R$ with field of fractions $K$, its rank is the dimension of $K{\otimes }_{R}M$. The tensor product of modules here extends scalars to a field, turning the question into a dimension calculation. For a free module $R^r$, the result is $r$.

For a finitely generated module over a principal ideal domain, the structure theorem for finitely generated modules over a principal ideal domain gives $M\cong R^r\oplus {⨁}_{i}R/(d_i)$, with $d_i$ nonzero nonunits. Each torsion element of a module vanishes after extension of scalars, because its annihilating nonzero scalar becomes invertible in $K$. Thus $K{\otimes }_{R}M\cong K^r$, proving that the rank is the number of free summands. The rank does not record the torsion submodule: for example, $R\oplus R/(d)$ and $R$ have the same rank.

Let $A$ be a Noetherian ring that is an integral domain. Ifis exact, $N$ is a finitely generated free module, and $P$ is a torsion-free module, then $M$ is a reflexive module. In particular, the dual of every finitely generated $A$-module is reflexive.

A Noetherian local ring $(R,\mathfrak{m})$ is regular when the minimal number of generators of its maximal ideal equals its Krull dimension. Every regular local ring is an integral domain and a unique factorization domain.

A property of rings is a ring isomorphism invariant when any two rings related by a ring isomorphism either both have the property or both lack it. Being an integral domain is such a property: a ring isomorphism preserves nonzero elements, products, and zero.

For an additive semigroup $D$ with identity zero, the semigroup algebra $k[D]$ has a $k$-basis of formal symbols $t^q$, $q\in D$, and multiplication $t^q{t}^{q^{\prime }}=t^{q+q^{\prime }}$. For $D=\mathbb{Z}[1/2{]}_{\ge 0}$ it is the integral domain ${\bigcup }_{h\ge 0}k[t^{1/2^h}]$. Its positive-exponent ideal is a nonzero idempotent ideal, and the strictly ascending chain $(t)⊊(t^{1/2})⊊\cdots$ shows that the algebra is not a Noetherian ring.

Let $M$ be an $m$-by-$n$ matrix over a principal ideal domain $R$. Choose Smith normal form $PMQ=D$ with exactly $r$ nonzero diagonal entries. Since $P$ and $Q$ are invertible matrices, $Mx=0$ is equivalent to $D(Q^{-1}x)=0$. The integral domain property forces the first $r$ coordinates of $Q^{-1}x$ to vanish, so $Q{e}_{r+1},\dots ,Q{e}_n$ are a basis of a module for the kernel. These vectors form part of a basis of a module of $R^n$, which is stronger than merely finding a nonzero vector in the kernel over the fraction field.

If $C$ is a commutative algebra of finite dimension over a field $K$, every prime ideal is maximal and $|\mathrm{Spec}C|\le {\dim }_{K}C$. Indeed, the quotient by a prime is a finite-dimensional integral domain. Multiplication by any nonzero element is an injective linear endomorphism, hence surjective, so the quotient is a field. Distinct maximal ideals are pairwise comaximal. For any finite list ${\mathfrak{m}}_1,\dots ,{\mathfrak{m}}_r$, construct $e_i\equiv 1(\mathrm{mod}{\mathfrak{m}}_i)$ and $e_i\equiv 0(\mathrm{mod}{\mathfrak{m}}_j)$ for $j\ne i$ by multiplying elements in ${\mathfrak{m}}_j$ congruent to one modulo ${\mathfrak{m}}_i$. These give a surjection $C\to {\prod }_{i}C/{\mathfrak{m}}_i$. Each factor has positive $K$-dimension, so $r\le {\dim }_{K}C$, excluding an infinite list.

Let $R$ be a Noetherian ring that is an integral domain, and let $I$ be a nonzero proper ideal. If $I^n=I^{n+1}$ for some $n\ge 1$, the nonzero finitely generated module $M=I^n$ satisfies $M=IM$. Choose generators $u_1,\dots ,u_r$ and write $u_i={\sum }_j{a}_{ij}{u}_j$ with $a_{ij}\in I$. The adjugate matrix identity implies that $\det ({\mathrm{Id}}_r-(a_{ij}))$ annihilates $M$. This determinant is $1-b$ for some $b\in I$, since every term except the constant term contains an entry from $I$.

The determinant trick has thus supplied $(1-b)M=0$. A nonzero element of $M$ and multiplicative cancellation in an integral domain force $b=1$, contradicting properness of $I$. Therefore all positive powers of an ideal are distinct in this setting. Since they have the same radical of an ideal, there are infinitely many ideals with radical $\sqrt{I}$.

A subset closed under addition, additive inverses and multiplication, with the induced operations. Under the unital convention used here it also contains the identity of $R$. A subring of an integral domain is again an integral domain. The ring of invariants is one example: fixed elements remain fixed under all ring operations.

For a module $M$ over an integral domain $R$, a torsion element is an $m\in M$ with $rm=0$ for some $0\ne r\in R$. This includes zero. Equivalently, its annihilator ${\mathrm{Ann}}_R(m)=\{r\in R:rm=0\}$ is a nonzero ideal. The torsion elements of a module form the torsion submodule.

For $M=\mathbb{Z}/6\mathbb{Z}$, computed annihilators areFor these three elements, the additive order of a group element is respectively $6$, $3$, and $2$: these are the smallest positive integers killing them. Torsion over $\mathbb{Z}$ agrees exactly with finite additive order of a group element. Over a general integral domain, torsion concerns the chosen scalars instead; the polynomial module from a linear operator gives a torsion example whose nonzero elements have infinite additive order of a group element.

A module $M$ over an integral domain $R$ is torsion-free when $rm=0$ with $0\ne r\in R$ implies $m=0$, equivalently when its torsion submodule is zero. Every free module over an integral domain is torsion-free, as one can check in each coefficient. Every vector space is torsion-free over its scalar field.

The converse requires hypotheses. A finitely generated module over a principal ideal domain is free if it is torsion-free, but the module $\mathbb{Q}$ over $\mathbb{Z}$ is torsion-free and not free. Any two nonzero rational numbers have a nontrivial integer relation, so a possible basis of a module could have at most one element; no single $q$ generates all rational numbers, since $q/2\notin \mathbb{Z}q$ for $q\ne 0$. The finite-generation hypothesis cannot be dropped. See Keith Conrad's Modules over a PID, Corollary 2.6.

On an integral scheme, a quasi-coherent sheaf is torsion-free when its module on every affine chart is a torsion-free module over that chart's integral domain. It embeds into its generic fibre. Therefore a torsion-free sheaf with zero generic fibre is zero; locally free sheaves and their subsheaves are torsion-free.

A torsion module over an integral domain $R$ is a module in which every element is a torsion element of a module. Each element may require a different nonzero annihilating scalar. The module $\mathbb{Z}/6\mathbb{Z}$ is torsion over $\mathbb{Z}$, and the polynomial module from a linear operator $\mathbb{Q}[t]/(t^2)$ is torsion over $\mathbb{Q}[t]$.

If the torsion module is a finitely generated module, one common nonzero scalar kills the whole module: multiply annihilating scalars for a finite generating set. Without finite generation this can fail. The quotient module $\mathbb{Q}/\mathbb{Z}$ is torsion, since the class of $a/b$ is killed by $b$, but no nonzero integer $n$ kills it all: the class of $1/(2|n|)$ survives multiplication by $n$. See Keith Conrad's Modules over a PID, exercise 6.5(c).

For a module $M$ over an integral domain $R$, its torsion submodule isThis is a submodule. If $rm=0$ and $sn=0$ with $r,s\ne 0$, then $rs(m+n)=0$ and $rs\ne 0$; also $r(am)=a(rm)=0$ for every $a\in R$. The integral domain hypothesis supplies both the nonzero product and commutativity. These are the conventions of ^[ref].

A mixed example is the direct sum $M={\mathbb{Z}}^2\oplus \mathbb{Z}/6\mathbb{Z}$. If a nonzero integer kills $(a,b,\overline{c})$, then $a=b=0$. Conversely, $6$ kills every $(0,0,\overline{c})$. ThusThe torsion submodule isolates the finite-order part while the quotient module keeps the independent integer directions. In general $M/T(M)$ is a torsion-free module: if $r(m+T(M))=0$ with $r\ne 0$, some $s\ne 0$ kills $rm$, so $sr\ne 0$ kills $m$.

The hypothesis matters. In the ring $R=\mathbb{Z}/6\mathbb{Z}$ viewed as a module over itself, the elements $\overline{2}$ and $\overline{3}$ are killed by nonzero scalars $\overline{3}$ and $\overline{2}$, but their sum $\overline{5}$ is a unit and is killed by no nonzero scalar. Thus using arbitrary nonzero scalars in a ring with zero divisors need not give a submodule. A common broader convention uses regular elements of a ring as annihilators; the integral domain convention here avoids that ambiguity.

For an integral domain $R$, the breadth of a Laurent polynomial adds under multiplication. If two Laurent polynomials multiply to one, both therefore have breadth zero: they are monomials, and their coefficients must be units of $R$. Conversely every displayed monomial is invertible. For $R=\mathbb{Z}$ the units are precisely $\pm t^k$, explaining the normalization ambiguity of an Alexander polynomial.

A valuation ring is an integral domain $R$ such that, for every nonzero element $x$ of its fraction field $K$, either $x\in R$ or $x^{-1}\in R$. Equivalently, its ideals are totally ordered by inclusion.

The ring with a single element, which is both its additive and multiplicative identity. If a unital ring has $0=1$, then every element $r$ equals $r1=r0=0$, so it is this ring. Under the usual definition it is excluded from integral domains and fields. The quotient ring $\mathbb{Z}/(1)$ is an example.