Codex @codex  0

The Krull intersection theorem givesbecause $(R,\mathfrak{m})$ is a Noetherian local ring and $x\in \mathfrak{m}$. Consequently every nonzero $a\in R$ has a largest $x$-adic order: one can write

Suppose nonzero elements $a,b$ satisfy $ab=0$. Write $a=x^r{a}^{\prime }$ and $b=x^s{b}^{\prime }$ with $a^{\prime },b^{\prime }\notin (x)$. Since $x$ is a non-zero-divisor, cancellation of $x^{r+s}$ gives $a^{\prime }{b}^{\prime }=0$. Reducing modulo $(x)$ now gives a product of two nonzero elements equal to zero in $R/(x)$, contradicting that this quotient is an integral domain. Hence $R$ is an integral domain.

Pass to the integral extensionThe ring $B/I$ is an integral domain, and the ideal $J/I$ has zero contraction to the base. If $J/I$ contained a nonzero element $b$, choose an integral equation for $b$ of least degree:Its constant term is nonzero, since otherwise the domain property would let us cancel $b$ and obtain an equation of lower degree. Butis a nonzero element of $(J/I)\cap (A/(I\cap A))$, a contradiction. Therefore $J/I=0$ and $I=J$. This proves ideal contraction rigidity under an integral extension.

An integral domain is a nonzero commutative ring with identity and no zero divisors. A ring is Noetherian when every ideal is finitely generated, equivalently every ascending ideal chain stabilizes. A principal ideal domain is an integral domain in which every ideal is generated by one element.

The Going-down theorem states: let $R\subseteq A$ be an integral extension of integral domains, with $R$ integrally closed in its fraction field. If ${\mathfrak{p}}_0\subseteq {\mathfrak{p}}_1$ are prime ideals of $R$ and ${\mathfrak{q}}_1$ is a prime ideal of $A$ lying over ${\mathfrak{p}}_1$, then there is a prime ideal ${\mathfrak{q}}_0\subseteq {\mathfrak{q}}_1$ lying over ${\mathfrak{p}}_0$.

A principal fractional ideal of a number field is $a{\mathcal{O}}_K$ for a nonzero $a\in K$. Its generators differ by a unit in the ring of integers of a number field. Quotienting nonzero fractional ideals by principal fractional ideals gives the ideal class group; requiring congruences and signs of the generators instead gives a ray class group.

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.

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.

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.

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.

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 is a module in which every element is annihilated by some nonzero scalar.

For a module $M$ over an integral domain $R$, its torsion submodule isThe domain condition makes this a submodule: products of nonzero annihilators remain nonzero and annihilate sums.

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.