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An affine algebraic curve is an affine variety of Krull dimension one. An irreducible affine curve has a finitely generated one-dimensional domain as its coordinate ring. Quotienting this ring by a nonzero element gives a zero-dimensional Noetherian ring, hence an Artinian ring. Consequently a nonzero regular function has finitely many zeros. No nonsingularity assumption is part of the definition.

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.

For a finite minimal primary decomposition $I={\bigcap }_i{Q}_i$, these are the prime ideals $\sqrt{Q_i}$. This terminology applies without a Noetherian hypothesis when the decomposition exists. Its invariance follows from the intrinsic description as those prime ideals of the form $\sqrt{(I:x)}$, where $x\notin I$ and $(I:x)$ is a colon ideal. Indeed, $(I:x)={\bigcap }_i(Q_i:x)$, and each factor is either the whole ring or a primary ideal with radical of an ideal $\sqrt{Q_i}$. A finite intersection of prime ideals can be prime only if it equals one of them: otherwise multiplying one element outside the intersection from each factor contradicts the prime ideal condition. Conversely, irredundancy supplies $x\in {\bigcap }_{j\ne i}{Q}_j\setminus Q_i$, giving $\sqrt{(I:x)}=\sqrt{Q_i}$. In a Noetherian ring these agree with the associated primes of a module $R/I$. For arbitrary rings, the annihilator definition of an associated prime of a module must not be substituted without a further hypothesis.

For an ideal $I$ in $R[x]$, let $J_n$ contain the coefficients of $x^n$ in its elements of degree at most $n$. These are ideals of $R$ and form an ascending chain, since multiplying by $x$ embeds one into the next. If $R$ is a Noetherian ring, the chain stabilizes at $N$, and each $J_n$ for $0\le n\le N$ has finitely many generators. Choose corresponding polynomials in $I$. Their union generates $I$: for an element of degree $d$, subtract a combination of the chosen polynomials for $J_d$ if $d<N$, or the polynomials for $J_N$ multiplied by $x^{d-N}$ otherwise. This cancels its leading coefficient and lowers its degree. Induction completes the proof without assuming that $R$ is an integral domain.

For an ideal $K\subseteq R[t]$, let $A_n(K)$ consist of the coefficients at $t^n$ of elements of $K$ of polynomial degree at most $n$, including the zero polynomial. Each $A_n(K)$ is an ideal of $R$, and multiplication of representing polynomials by $t$ shows that $A_n(K)\subseteq A_{n+1}(K)$.

If $R$ is a Noetherian ring, this chain stabilizes at some $N$. Choose finite generating sets of an ideal for $A_0(K),\dots ,A_N(K)$, and for each nonzero generator choose a representing polynomial $f_{nj}\in K$ with that leading coefficient and polynomial degree $n$. These finitely many polynomials generate $K$: for any nonzero $f\in K$ of polynomial degree $d$, use the chosen generators at level $\min (d,N)$, multiplied by suitable powers of $t$, to cancel its leading coefficient. Induction on polynomial degree finishes the argument. This proves the Hilbert basis theorem without assuming that $R$ is an integral domain.

For a Noetherian ring $A$, a finitely generated module $M$ has a finite presentation of a module, and kernels of maps from finite free modules to $M$ are finitely generated. Its affine module sheaf is therefore a coherent sheaf. Conversely, if $\tilde{M}$ is coherent, choose a finite principal open subset cover on which its section modules are finitely generated. Lift their finitely many generators to numerators in $M$. The submodule generated by these numerators becomes all of $M$ on every chart, and hence is all of $M$: its quotient has zero localization at every point. This proves finite generation without assuming that global sections preserve arbitrary sheaf surjections.

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.

LetThis ring consists of finite sums of rational multiples of nonnegative dyadic rational powers of $t$, interpreted as functions on $t>0$. Every pair of elements lies in one polynomial ring $\mathbb{Q}[u_n]$, so $R$ is an integral domain. The constant term defines a surjective ring homomorphism $R\to \mathbb{Q}$. Its kernel of a ring homomorphism $\mathfrak{m}$ is nonzero and is a maximal ideal, by the maximal ideal quotient criterion.

Every $f\in \mathfrak{m}$ has the form $u_{n}g(u_n)$ for some $n$ and some polynomial $g$. The identityexpresses it as a product of two elements of $\mathfrak{m}$. Therefore $\mathfrak{m}={\mathfrak{m}}^2$, and ${\mathfrak{m}}^n=\mathfrak{m}$ for all $n\ge 1$. This shows that strict descent of powers of an ideal can fail in an integral domain, even for a nonzero maximal ideal. By the determinant trick, such an ideal cannot be finitely generated, so this ring is not a Noetherian ring.

For a prime ideal $P$, suppose the kernel of a ring homomorphism $K=\ker (R\to R_P)$ is a finitely generated ideal, say $(a_1,\dots ,a_t)$. Choose $s_i\notin P$ killing $a_i$ and let $f={\prod }_i{s}_i$. Then $f\notin P$ and $fK=0$. If a fraction $a/f^n$ maps to zero in $R_P$, then $a\in K$, so $fa=0$ and the fraction was already zero in $R_f$. Thus the canonical ring homomorphism $R_f\to R_P$ is injective. The empty product $f=1$ handles $K=0$. Every Noetherian ring satisfies the finite generation hypothesis.

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.

Let $S_0$ be a commutative Noetherian ring, let a graded $S$ be generated over it by finitely many positive-degree homogeneous elements, and let $M$ be a finitely generated graded module. For an integer-valued function $\lambda$ additive on exact sequences of finitely generated $S_0$-modules, its graded-piece series has the displayed form. The exact sequence for multiplication by the last variable gives $(1-t^d)P_M=P_{M/xM}-t^d{P}_{(0{:}_{M}x)}$. Those two modules are finite over the ring with one fewer variable, and induction proves rationality. Finite graded pieces and the finite-generation assumptions cannot simply be omitted.

Let $\varphi :R\to S$ be a surjective ring homomorphism and let $R$ be a Noetherian ring. For every ideal $J$ of $S$, its inverse image is a finitely generated ideal of $R$. If ${\varphi }^{-1}(J)=(a_1,\dots ,a_r)$, then $J=(\varphi (a_1),\dots ,\varphi (a_r))$. Thus $S$ is a Noetherian ring. In particular, every quotient ring of a Noetherian ring is a Noetherian ring. For a ring homomorphism that is not surjective, apply the same argument with its image as codomain.

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.