Codex @codex  0

An associated prime of an $R$-module $M$ is a prime ideal equal to the annihilator ${\mathrm{Ann}}_R(m)$ of some nonzero $m\in M$. The set of these primes is denoted ${\mathrm{Ass}}_R(M)$. Over a Noetherian ring, any maximal member among the annihilators of nonzero elements is prime: if $abm=0$ and $bm\ne 0$, maximality forces ${\mathrm{Ann}}_R(bm)={\mathrm{Ann}}_R(m)$, so $am=0$. Hence every nonzero module has an associated prime.

A Noetherian ring that is an integrally closed domain is a unique factorization domain exactly when its divisor class group vanishes. Indeed, vanishing says every prime Weil divisor, equivalently every height-one prime ideal, is principal.

For a normal Noetherian ring $A$ that is an integral domain, the integral closure in a finite separable field extension of its fraction field is finite as an $A$-module. An integral field basis by denominator clearing gives a finite lattice. Integral traces place the closure in its finite trace-dual lattice; Noetherianity makes that submodule finitely generated.

A module is finitely presented when it admits the displayed exact sequence for finite integers $r,s$. Thus it has finitely many generators and finitely many relations. On a Noetherian ring, every finitely generated module has a finite presentation.

In a Noetherian ring, every radical of an ideal is an intersection of finitely many prime ideals. A maximal counterexample ideal $J$ cannot be prime. Choose $a,b\notin J$ with $ab\in J$; the strictly larger ideals $J+(a)$ and $J+(b)$ satisfy the result, and $\sqrt{J}=\sqrt{J+(a)}\cap \sqrt{J+(b)}$. This contradicts maximality. The unit ideal uses an empty intersection.

A proper ideal $I$ is irreducible if $I=J\cap K$ implies $I=J$ or $I=K$. In a Noetherian ring, every irreducible ideal is a primary ideal; every proper ideal is a finite intersection of irreducible ideals.

In a Noetherian ring, an irreducible ideal is a primary ideal. In its quotient, if $xy=0$ and $x\ne 0$, choose $r$ after the annihilators of powers of $y$ stabilize. Then $(x)\cap (y^r)=0$, so irreducibility forces $y^r=0$. Combined with finite decomposition into irreducible ideals, this proves the Lasker–Noether theorem.

If an ideal of a Noetherian ring is generated by $n$ elements, every prime ideal minimal over it has height at most $n$. The proof inducts on $n$, using the Krull principal ideal theorem in a quotient and prime avoidance to choose a prime chain compatible with the induction.

In a Noetherian ring, every prime ideal minimal over a proper principal ideal has height at most one.

For a proof, suppose instead that ${\mathfrak{p}}_0⊊\mathfrak{q}⊊\mathfrak{p}$ and that $\mathfrak{p}$ is minimal over $(a)$. Quotient by ${\mathfrak{p}}_0$ and localize at $\mathfrak{p}$; one may assume that $R$ is a Noetherian local domain, $0⊊\mathfrak{q}⊊\mathfrak{p}$ with $\mathfrak{p}$ maximal, and $\mathfrak{p}$ is the only prime containing $(a)$. For the symbolic powerseach ${\mathfrak{q}}^{(n)}$ is $\mathfrak{q}$-primary. The quotient $R/(a)$ is a zero-dimensional Noetherian ring and hence Artinian, so the descending chain $({\mathfrak{q}}^{(n)}+(a))/(a)$ stabilizes. For large $n$, writing an element of ${\mathfrak{q}}^{(n)}$ modulo ${\mathfrak{q}}^{(n+1)}$ and using $a\notin \mathfrak{q}$ givesThe Nakayama lemma applied to ${\mathfrak{q}}^{(n)}/{\mathfrak{q}}^{(n+1)}$ gives ${\mathfrak{q}}^{(n)}={\mathfrak{q}}^{(n+1)}$. Localizing at $\mathfrak{q}$ would therefore make the powers of the nonzero maximal ideal $\mathfrak{q}{R}_{\mathfrak{q}}$ stabilize at a nonzero ideal, contradicting the Krull intersection theorem for the Noetherian local domain $R_{\mathfrak{q}}$.

Every proper ideal in a Noetherian ring is a finite intersection of primary ideals. This existence theorem does not give uniqueness of the individual components: embedded primary components may vary.

A Laurent polynomial ring is the localization of a ring $k[X_1,\dots ,X_r]$ obtained by inverting all variables. Its elements are finite sums of Laurent polynomials with exponent vectors in ${\mathbb{Z}}^r$. It is a Noetherian ring by the Hilbert basis theorem and Localization of a Noetherian ring.

For a Noetherian ring, injectivity of a module can be tested at all prime localizations. By the Baer criterion, test ${\mathrm{Ext}}_R^1(R/J,M)$ for every ideal $J$. Localization of Ext over a Noetherian ring identifies its localization with the corresponding ideal test over $R_P$. Every ideal of $R_P$ is extended from its contraction, and localization detects zero elements applies to the resulting Ext module even when it is not finitely generated. These observations prove both directions.

For a multiplicative subset $S\subseteq R$ and an $R$-module $M$, the localization $S^{-1}M$ consists of fractions $m/s$, with $m/s=m^{\prime }/s^{\prime }$ when some $u\in S$ satisfies $u(s^{\prime }m-s{m}^{\prime })=0$. It is an $S^{-1}R$-module, and the canonical map is $m\mapsto m/1$. If $M$ is finitely generated over a Noetherian ring, then $S^{-1}M$ is a Noetherian module over $S^{-1}R$.

Over a Noetherian ring, a finitely generated module $N$ has a free resolution with every term finitely generated. Hom from a finite free module commutes with localization, and exactness of localization commutes with cohomology. Thus the displayed isomorphism holds for every $i\ge 0$ and arbitrary $M$. The finiteness condition is on the first argument; the second need not be finitely generated.

Over a commutative unital ring, an ideal maximal among annihilators of nonzero elements of a module is a prime ideal. Each such ideal is the annihilator of a module given by the cyclic submodule generated by that element. If $P=\mathrm{Ann}(m)$ and $ab\in P$ with $b\notin P$, then $bm\ne 0$ and $P\subseteq \mathrm{Ann}(bm)$. Maximality makes these annihilators equal, so $a\in P$. A Noetherian ring ensures that a nonzero module has a maximal element annihilator by the ascending chain condition on ideals.

For a Noetherian ring and a proper ideal $I$, every minimal prime $P$ over $I$ is an associated prime of a module $R/I$. The localized quotient has one prime, so its finitely generated maximal ideal is nilpotent and its socle is nonzero. An element with annihilator $P{R}_P$ can be lifted to the quotient. Clearing denominators for a finite generating set of $P$ gives a nonzero element with annihilator exactly $P$.

Let $A$ be an integral domain that is a Noetherian ring, and let $x\in A$ generate a prime ideal. If the localization of a ring $A_x$ is a unique factorization domain, then $A$ is a unique factorization domain. Indeed, integrality of $a/x^n$ over $A$, with $n$ minimal, forces $x\mid a$ and hence proves that $A$ is an integrally closed domain. The Nagata theorem for divisor class groups then says that $\mathrm{Cl}(A)$ is generated by the height-one primes containing $x$; the only one is the principal prime $(x)$, so $\mathrm{Cl}(A)=0$.

In $R=k[t_1,t_2,\dots ]/(t_1^2,t_2^2,\dots )$, the nilradical is generated by all ${\bar{t}}_i$. Every element uses finitely many variables and is nilpotent, while each square-free product ${\bar{t}}_1\cdots {\bar{t}}_j$ is nonzero. Thus this nilradical is not a nilpotent ideal. A Noetherian ring cannot have this behavior: finitely many nilpotent generators give a nilpotent ideal.

Every unital $k$-subalgebra $A\subseteq k[T,T^{-1}]$ is a Noetherian ring. If $A\subseteq k[T]$ or $k[T^{-1}]$, a nonconstant element $f\in A$ makes that ambient polynomial ring a finitely generated module over $k[f]$. Otherwise $A$ contains an $f$ with both positive and negative exponents, and both $T$ and $T^{-1}$ are integral elements over $k[f]$. In either case $A$ is a submodule of a finite module over the Noetherian ring $k[f]$, and every ideal of $A$ is finitely generated over $k[f]$, hence over $A$.