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The Hilbert basis theorem makes $k[x_1,\dots ,x_n]$ a Noetherian ring. Reversing the inclusion of ideals shows that its Zariski-closed subsets satisfy the descending-chain condition, so every affine variety is a Noetherian topological space.

Suppose some nonempty closed subset had no finite decomposition into irreducible closed subsets. It would be reducible, so it could be writtenwith $V_1,V_2$ proper closed subsets. At least one of these would again admit no finite irreducible decomposition. Repeating that choice would produce an infinite strictly descending chain of closed subsets, contradicting Noetherianity. ThereforeRemoving any member contained in another leaves exactly the irreducible components.

Assume (iii). Because $R$ is a Noetherian ring, every ideal $I\subseteq R$ is finitely generated. Regard $I$ as an $R$-submodule of $F$. By (iii), $I=Rx$ for some $x\in F$; since $x\in I\subseteq R$ unless $I=0$, this is a principal ideal of $R$. Thus every ideal is principal and (i) holds.

Together with the previous two implications, this proves (i), (ii), and (iii) equivalent.

False. The integers $\mathbb{Z}$ form a Noetherian ring, since every ideal is principal, but this ring is not an Artinian ring: the strictly descending chainnever stabilizes.

Let $A$ be a principal ideal domain that is not a field. A PID is Noetherian and is a unique factorization domain. Every nonzero prime ideal is generated by a prime element. Ifthen $a\mid p$, so primality of $p$ makes $a$ a unit or an associate of $p$. The proper alternative is $(a)=(p)$, and every nonzero prime is therefore maximal. Since $A$ has a nonzero prime ideal, its Krull dimension is exactly one.

The cases $n=0$ and $n=d$ can indeed be finite: a Noetherian ring has finitely many minimal primes, and a semilocal ring may have finitely many maximal ideals. We prove that every intermediate height occurs infinitely often.

Because $d$ is a finite integer equal to the supremum of prime-chain lengths, there is a chainEach ${\mathfrak{p}}_i$ has height exactly $i$: its displayed lower chain gives height at least $i$, while any longer lower chain could be extended by the remaining displayed primes and would contradict $\mathrm{ht}{\mathfrak{p}}_d=d$.

Suppose $0<n<d$. The three primesfall under prime ideals between a three-prime chain, so infinitely many primes $\mathfrak{q}$ satisfyEvery such $\mathfrak{q}$ has height exactly $n$: the lower inclusion gives height at least $n$, and height at least $n+1$ would, after appending ${\mathfrak{p}}_{n+1}$, contradict its height $n+1$. Therefore there are infinitely many height-$n$ primes. The assumed finiteness forcesThis is infinitude of intermediate-height prime ideals.

For a prime ideal $\mathfrak{p}$, its height is the supremum of the lengths of strict chains of prime ideals ending at $\mathfrak{p}$. For a proper ideal $I$,This is the height of an ideal.

The Krull height theorem states that if $R$ is Noetherian, $I=(a_1,\dots ,a_m)$, and $\mathfrak{p}$ is minimal over $I$, then

We prove it by induction on $m$. The case $m=1$ is the Krull principal ideal theorem. For the induction step, let ${\mathfrak{q}}_1,\dots ,{\mathfrak{q}}_s$ be the finitely many minimal primes over $(a_2,\dots ,a_m)$ that lie below $\mathfrak{p}$. By induction each has height at most $m-1$; if one equals $\mathfrak{p}$, we are done.

Otherwise, suppose $\mathfrak{p}$ has finite height $d$ and choose a chainwhose first nonminimal term ${\mathfrak{p}}_1$ is contained in none of the ${\mathfrak{q}}_i$. Such a chain is obtained by prime avoidance and the principal ideal theorem: a three-term segment can be replaced by a prime minimal over ${\mathfrak{p}}_0+(b)$ for an element $b$ avoiding the finitely many unwanted primes. Choose

The prime $\mathfrak{p}$ is minimal over $(b,a_2,\dots ,a_m)$. Otherwise a prime strictly between some ${\mathfrak{q}}_i$ and $\mathfrak{p}$ would show that $\mathfrak{p}/(a_2,\dots ,a_m)$ has height at least two, although it is minimal over the principal ideal generated by $a_1$; this contradicts the principal ideal theorem. In $R/(b)$, the prime $\mathfrak{p}/(b)$ is therefore minimal over an ideal generated by $m-1$ elements, so induction bounds its height by $m-1$. The $d-1$ strict inclusions from ${\mathfrak{p}}_1/(b)$ to $\mathfrak{p}/(b)$ giveand hence $d\le m$. If the height were infinite, the same argument applied to arbitrarily long finite chains would give the same fixed bound, which is impossible. This completes the proof.

Let $J$ be an ideal of $S^{-1}A$ and contract it to an ideal $I$ of $A$. Every element $a/s\in J$ has $a/1=(s/1)(a/s)\in J$, so $a\in I$ andBecause $A$ is a Noetherian ring, write $I=(a_1,\dots ,a_r)$. Thenso every ideal of $S^{-1}A$ is finitely generated. Hence every localization of a Noetherian ring is Noetherian, proving the Localization of a Noetherian ring theorem.

The valuation ring and its maximal ideal areIf ${\mathcal{O}}_K$ is Noetherian, its maximal ideal is finitely generated. Ideals in a valuation ring are totally ordered, so every finitely generated ideal is generated by one of its generators; write $\mathfrak{m}=(\pi )$. Then $v(\pi )$ is the smallest positive element of the value group. Subtracting integral multiples of this value shows that every value is an integral multiple of $v(\pi )$, so $v$ is discrete.

Now assume $K$ is complete and discretely valued. Parts i and ii, applied to $v$ and to the other discrete valuation $v^{\prime }$, giveIf $v(a)=0$ and $\pi$ is a uniformizer for $v$, then $\pi a^r\in {\mathfrak{m}}_v$ for every $r\in \mathbb{Z}$. Hencefor every integer $r$, forcing $v^{\prime }(a)=0$ and in particular $v^{\prime }(a)\ge 0$. Every nonzero $x$ has the form $x={\pi }^{m}u$ with $v(u)=0$, soThe proportionality constant is positive because $v^{\prime }$ is nontrivial. Thus the valuations, and their associated absolute values, are equivalent.

Noetherian means every ideal is finitely generated, equivalently every ascending ideal chain stabilises. Quotients preserve this; for example $\mathbb{Z}[\sqrt{-5}]$ is Noetherian as a quotient of $\mathbb{Z}[X]$ but not a UFD. A UFD need not be Noetherian (a polynomial ring in infinitely many variables is a counterexample). Stabilisation of kernels proves every surjective endomorphism of a Noetherian ring injective. The shift $k[x_1,x_2,\dots ]\to k[x_1,x_2,\dots ]$, $x_1\mapsto 0$, $x_{i+1}\mapsto x_i$, is surjective noninjective; $x\mapsto x^2$ on $k[x]$ shows injective need not mean surjective. Finally $f_n(X)=\left(\genfrac{}{}{0ex}{}{X}{n}\right)$ is integer-valued. Finite differences prove uniquely that every integer-valued polynomial is an integral linear combination of the $f_n$, so they are a $\mathbb{Z}$-basis. Int$(\mathbb{Z})$ is not Noetherian: the denominators in $f_p$ yield an ideal chain requiring new prime denominators.

The Second uniqueness theorem for primary decomposition says that in a minimal primary decomposition of an ideal in a Noetherian ring, every primary component belonging to an isolated prime is unique. Here an isolated prime is a minimal member of the set $\{\sqrt{Q_i}\}$.

Let $\mathfrak{p}=\sqrt{Q_i}$ be isolated and apply localization at a prime ideal. If $j\ne i$, minimality of $\mathfrak{p}$ gives $\sqrt{Q_j}⊈\mathfrak{p}$, so some element of $Q_j$ becomes a unit in $R_{\mathfrak{p}}$. ConsequentlyBecause $Q_i$ is $\mathfrak{p}$-primary, multiplication by any $s\notin \mathfrak{p}$ cannot carry an element outside $Q_i$ into $Q_i$. ThereforeThe right side depends only on $I$ and $\mathfrak{p}$, proving uniqueness.

No. Take the Noetherian ring $R=k[x]$, the finitely generated $R$-moduleand $N=0$. Its annihilator of a module iswhich is a prime ideal and hence a primary ideal. But with $r=x$ and $m=(0,\overline{1})$ we have $m\ne 0$ and $rm=0$, while $x^{k}M\ne 0$ for every $k$ because the free summand survives. Thus $N$ is not a primary submodule.

A prime filtration of a module over a commutative ring is a finite increasing chain from zero to the module with each nonzero successive quotient module isomorphic to $R/P$ for a prime ideal $P$. Every finitely generated module over a Noetherian ring has one. In each nonzero quotient, a maximal annihilator of a module element is prime supplies a cyclic submodule $R/P$; lift it and repeat. The ascending chain condition for a Noetherian module makes the process terminate. This is not necessarily a composition series: the prime ideals need not be maximal.

In a Noetherian ring, an ideal $I$ disjoint from a nonempty multiplicative subset $S$ extends to an ideal $P$ maximal among those disjoint from $S$, by the ascending chain condition. It is proper. If $ab\in P$ but neither factor belongs to $P$, both $P+(a)$ and $P+(b)$ meet $S$. Multiplying such representatives puts an element of $S$ in $P$, a contradiction. Thus $P$ is a prime ideal. For a general commutative ring the same argument follows after applying the Zorn lemma to disjoint ideals; the union of a chain is disjoint from $S$.

The Rees module of an $R$-module $M$ with respect to an ideal $I$ is ${\mathcal{R}}_I(M)={⨁}_{j\ge 0}{I}^{j}M{t}^j$, a graded module over the Rees ring $\mathcal{R}(I)$. Generators of $M$ in degree zero generate this module. For $N\subseteq M$, the graded submodule ${⨁}_j(I^{j}M\cap N)t^j$ is finitely generated when $R$ is a Noetherian ring; a bound on its generators' degrees yields the Artin-Rees lemma.

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 ring homomorphism from a ring to itself. Surjectivity need not imply injectivity in general, but a surjective ring endomorphism of a Noetherian ring is a ring automorphism.

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.

For a surjective ring endomorphism $f$ of a Noetherian ring, the ascending kernel chain $\ker f\subseteq \ker f^2\subseteq \cdots$ stabilizes. If $\ker f^m=\ker f^{m+1}$ and $f(a)=0$, choose $b$ with $f^m(b)=a$. Then $b\in \ker f^{m+1}=\ker f^m$, so $a=0$. Thus the endomorphism is a ring automorphism.