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A ring is Artinian when its ideals satisfy the descending chain condition: every chain $I_1\supseteq I_2\supseteq \cdots$ eventually becomes constant. It is Noetherian when its ideals satisfy the ascending chain condition, equivalently when every ideal is finitely generated. We first prove finite length of a commutative Artinian ring; this gives the stronger structural reason for its Noetherian property.

If $\mathfrak{p}$ is a prime ideal of an Artinian ring, the quotient $R/\mathfrak{p}$ is an Artinian integral domain. For a nonzero element $x$ of that domain, the chain $(x)\supseteq (x^2)\supseteq \cdots$ stabilizes. Thus $x^n=b{x}^{n+1}$ for some $b$, and cancellation gives $bx=1$. The quotient is a field, so every prime ideal is a maximal ideal.

There are only finitely many maximal ideals. Otherwise, choose distinct ones ${\mathfrak{m}}_1,{\mathfrak{m}}_2,\dots$. The intersections ${\mathfrak{m}}_1\cap \cdots \cap {\mathfrak{m}}_n$ form a strictly descending chain. To see strictness, for each $i\le n$ choose $a_i\in {\mathfrak{m}}_i\setminus {\mathfrak{m}}_{n+1}$; their product belongs to the first $n$ ideals but not to the next one, since ${\mathfrak{m}}_{n+1}$ is prime. This contradicts the descending chain condition.

Put $J={\bigcap }_i{\mathfrak{m}}_i$. This Jacobson radical is also the nilradical, since all primes are maximal. We need the stronger conclusion that $J$ is nilpotent, without assuming Noetherianity. Its powers stabilize, say $K=J^n=J^{n+1}=JK$. Suppose $K\ne 0$. By the descending chain condition, choose an ideal $L$ minimal subject to $KL\ne 0$. Some $x\in L$ has $Kx\ne 0$, so minimality gives $L=Rx$. Moreover $K(JL)=KL\ne 0$, and minimality gives $JL=L$. Hence $x=jx$ for some $j\in J$. But $1-j$ is a unit: it cannot lie in any maximal ideal, because $j$ lies in all of them. Thus $x=0$, a contradiction. Therefore $J^n=0$.

The Chinese remainder theorem gives $R/J\cong {\prod }_{i}R/{\mathfrak{m}}_i$, a finite product of fields. Each quotient $J^i/J^{i+1}$ is an Artinian module over this product, and each field component must be a finite-dimensional vector space; an infinite-dimensional vector space admits a strictly descending chain of subspaces. Consequently every layer has finite composition length. The finite filtrationshows that $R$ itself has finite composition length. A strict inclusion of submodules strictly increases length, so an ascending chain cannot continue indefinitely. Every commutative Artinian ring is therefore Noetherian. This is the Artinian rings are Noetherian result.

For the formal power series ring, let $B=R[[X]]$ with $R$ Noetherian, and let $H$ be any ideal of $B$. For $n\ge 0$ define a coefficient idealThese coefficient ideals of a formal power series ideal satisfy $A_n\subseteq A_{n+1}$, by multiplication by $X$. The ascending chain condition gives $A_n=A_N$ for all $n\ge N$. For each $0\le n\le N$, choose finitely many series $f_{n,j}\in H\cap X^{n}B$ whose coefficients at $X^n$ generate $A_n$.

We claim that these finitely many series generate $H$ as an ordinary ideal. Given $f\in H$, cancel its coefficient at $X^k$ successively. After coefficients below $k$ have vanished, its coefficient at $X^k$ lies in $A_k$. If $k\le N$, use an $R$-linear combination of the $f_{k,j}$. If $k>N$, use a combination of $X^{k-N}{f}_{N,j}$, since $A_k=A_N$. The remainder then belongs to $H\cap X^{k+1}B$.

Collect all the cancellations against each fixed generator. For $n<N$ its multiplier is a polynomial, while the multipliers of the $f_{N,j}$ are well-defined formal power series: at any fixed degree, only finitely many cancellation steps contribute. The remainder has every coefficient zero. ThusThis is a finite sum of ideal generators, rather than merely a topological closure assertion. Since $H$ was arbitrary,This coefficient-cancellation argument proves Noetherianity of a formal power series ring.

The corresponding Artinian assertion is false. For any nonzero ring $R$, the idealsin $R[[X]]$ are strictly decreasing, since $X^n$ has a nonzero coefficient in degree $n$ and no multiple of $X^{n+1}$ does. In particular, a field $k$ is Artinian, but $k[[X]]$ is not. The zero ring is the harmless exception.

A Noetherian ring is a ring whose ideals satisfy the ascending chain condition; equivalently every ideal is finitely generated. For the surjective ring endomorphism, put $K_j=\ker f^j$. The ascending chain of ideals $K_1\subseteq K_2\subseteq \cdots$ stabilizes, say $K_m=K_{m+1}$. If $a\in \ker f$, surjectivity of $f^m$ gives $b$ with $f^m(b)=a$. Then $f^{m+1}(b)=0$, so $b\in K_{m+1}=K_m$ and $a=0$. The endomorphism is therefore injective, hence a ring automorphism.

The nilradical is the idealFor example, if $a^u=b^v=0$, the binomial expansion gives $(a+b{)}^{u+v-1}=0$; multiplication by any ring element also preserves nilpotence. Every prime ideal contains each nilpotent element, since $a^j\in P$ implies $a\in P$. Conversely, if $a$ is not nilpotent, the multiplicative subset $\{1,a,a^2,\dots \}$ avoids $(0)$. By Zorn's lemma, choose an ideal maximal among those disjoint from this subset. It is a prime ideal: if $xy\in J$ but neither factor is in $J$, then $J+(x)$ and $J+(y)$ both meet the subset, and multiplying their two witnesses puts a power of $a$ in $J$, a contradiction. This prime ideal excludes $a$. HenceFor the zero ring the empty intersection is the whole ring, as required.

Here is a finite prime-intersection representation of a radical argument that does not assume the result. In a Noetherian ring, suppose some ideal $J$ has ideal radical not expressible as a finite intersection of prime ideals, and choose such a $J$ maximal by the ascending chain condition. It is proper and cannot be prime. Choose $a,b\notin J$ with $ab\in J$. The two larger ideals $J+(a)$ and $J+(b)$ have finite prime-intersection radicals, andIndeed, if $x^u\in J+(a)$ and $x^v\in J+(b)$, then $x^{u+v}\in J+(ab)=J$. This contradicts the choice of $J$. Applying the result to $(0)$ proves that $N(R)$ is the intersection of finitely many prime ideals. Empty intersections handle the unit ideal.

For reducedness of a formal power series ring, the inclusion of constants gives one implication: a nonzero nilpotent element of $R$ remains nonzero and nilpotent in $R[[X]]$. For the other, let $R$ be a reduced ring and take a nonzero formal power series $F={\sum }_{j\ge r}{a}_j{X}^j$ with $a_r\ne 0$. The coefficient of $X^{mr}$ in $F^m$ is $a_r^m\ne 0$, because $R$ is reduced. Thus $F$ is not nilpotent. ThereforeNo Noetherian ring hypothesis is needed for this argument.

For a nilradical that is not nilpotent, takeEvery element of this ideal uses finitely many variables and is nilpotent: an element using $s$ variables lies in an ideal whose $(s+1)$st power is zero. The quotient by this ideal is the field $k$, so it is exactly the nilradical. Nevertheless ${\bar{t}}_1\cdots {\bar{t}}_m\ne 0$ for every $m$, because the square-free monomials form a vector space basis. Thus $N(R{)}^m\ne 0$ for all $m$.

If $M=0$, take the empty filtration. Otherwise a nonzero quotient $Q=M/M_i$ is a finitely generated module. Because $R$ has the ascending chain condition on ideals, the nonempty collection of annihilators of nonzero elements of $Q$ has a maximal element. Part (b), applied to $Q$, supplies $q\ne 0$ with a prime ideal $P=\mathrm{Ann}(q)$.

The map $R\to Rq$, $r\mapsto rq$, is surjective with kernel $P$. The first isomorphism theorem gives $Rq\cong R/P$. Let $M_{i+1}$ be the inverse image of $Rq$ under $M\to Q$. ThenRepeat whenever the quotient remains nonzero. An infinite repetition would give a strictly ascending chain of submodules of the Noetherian module $M$, contradicting part (a). Thus the process stops after finitely many steps, necessarily at $M$.

This constructs the required prime filtration $0=M_0⊊\cdots ⊊M_l=M$, with each successive quotient $R/P_i$ for a prime $P_i$.

Consider all ideals containing $I$ and disjoint from the nonempty multiplicative subset $S$. This collection is nonempty because it contains $I$. The ascending chain condition for the Noetherian ring gives a maximal member $P$: otherwise repeatedly choosing a strictly larger member would give an infinite strictly ascending chain of ideals. Since $S$ is nonempty and $P$ misses it, $P\ne R$.

Suppose $ab\in P$ but $a,b\notin P$. Maximality implies that $P+(a)$ and $P+(b)$ both meet $S$, so write $s=p+ra\in S$ and $t=q+ub\in S$ with $p,q\in P$. Thenwhile closure under products gives $st\in S$, a contradiction. HenceThis proves the prime ideal avoiding a multiplicative subset result without needing $1\in S$. The assumptions themselves rule out $0\in S$, since every ideal contains zero.

A right Noetherian ring satisfies the ascending chain condition on right ideals; equivalently, every right ideal is a finitely generated module.

Here is a noncommutative Hilbert basis theorem proof adapted to the stated hypothesis. Set $F_{-1}=0$ andThe equality follows inductively from $A+xA=A+Ax$, by moving one coefficient past one $x$ at a time. These spaces form an exhaustive filtered algebra structure on $B$, with $F_n{F}_m\subseteq F_{n+m}$. This does not assert uniqueness of the displayed expressions or the existence of a coefficient-moving automorphism.

For a right ideal $I\subseteq B$, defineEach $L_n$ is a right ideal of $A$. In fact, if $a\in L_n$ and $c\in A$, write $c{x}^n=x^{n}b+u$ with $b\in A$, $u\in F_{n-1}$; then $ac{x}^n=(a{x}^n)b+au\in I+F_{n-1}$. Also $L_n\subseteq L_{n+1}$ by right multiplication by $x$. Since $A$ is a right Noetherian ring, this ascending chain stabilizes at some $L_N$.

Choose finite generators $a_{nj}$ for each $L_n$, $0\le n\le N$, and choose $f_{nj}\in I$ with $f_{nj}-a_{nj}{x}^n\in F_{n-1}$. These finitely many elements generate $I$ as a right ideal. To see this, induct on $m$ for $f\in I\cap F_m$. Write $f=a{x}^m+u$ with $u\in F_{m-1}$; then $a\in L_m$. Put $d=\min (m,N)$ and express $a={\sum }_j{a}_{dj}{c}_j$. Write $c_j{x}^d=x^d{b}_j+u_j$ with $u_j\in F_{d-1}$. The differencelies in $I\cap F_{m-1}$, so the induction applies. At $m=0$ the remainder is zero. Hence $B$ is right Noetherian.

For the quantum torus, take $q\in k^{\times }$ and the convention $YX=qXY$. Begin with the polynomial ring $k[X]$, which is Noetherian by the Hilbert basis theorem. Adjoining $X^{-1}$ preserves the hypothesis because it commutes with $k[X]$, and gives $k[X^{\pm 1}]$. Adjoin $Y$ next. The relation $Yf(X)=f(qX)Y$ and its inverse coefficient-moving relation give $A+YA=A+AY$ for $A=k[X^{\pm 1}]$. Finally adjoin $Y^{-1}$ to $k[X^{\pm 1}][Y;\sigma ]$, where $\sigma (X)=qX$. On a monomial, $Y^{-1}{X}^i{Y}^j=q^{-i}{X}^i{Y}^{j-1}$; when $j=0$ this is in $A{Y}^{-1}$, and when $j>0$ it is in $A$. The reverse inclusion follows by the same relation. The preceding argument applies at each step, proving the quantum torus is right Noetherian. Nonzero $q$ is required for this notation.

For a noncommutative ring, a prime ideal of a noncommutative ring means a proper two-sided ideal $P$ such that $UV\subseteq P$ for two-sided ideals $U,V$ implies $U\subseteq P$ or $V\subseteq P$. Equivalently, $aAb\subseteq P$ implies $a\in P$ or $b\in P$. This definition does not require $A/P$ to be a noncommutative domain.

Retain the right Noetherian ring hypothesis for the last assertion. More generally, the ascending chain condition on two-sided ideals suffices. We claim that every proper two-sided ideal $I$ contains a product of finitely many prime ideals of a noncommutative ring, each containing $I$. If not, choose a maximal counterexample $I$. It cannot be a prime ideal of a noncommutative ring. Thus there are two-sided ideals $U,V$ strictly containing $I$ with $UV\subseteq I$: add $I$ to the two witnesses for failure of the defining condition for a prime ideal of a noncommutative ring. By maximality, both $U$ and $V$ contain products of finitely many prime ideals of a noncommutative ring containing them. Concatenating these products gives a product inside $UV\subseteq I$, a contradiction.

Apply the claim to $I=0$ in a nonzero $A$, obtaining $P_1\cdots P_r=0$. Every prime ideal of a noncommutative ring $Q$ contains one of the $P_i$, by repeated application of the definition of a prime ideal of a noncommutative ring. For the prime radical of a noncommutative ring $N$, it follows thatIndeed, $N\subseteq P_i$ gives $N^r\subseteq P_1\cdots P_r=0$, and ${\bigcap }_i{P}_i\subseteq Q$ for every $Q$ gives equality of the intersections. If $A=0$, the empty intersection is the whole zero ring and the conclusion is immediate.

The final assertion is false for arbitrary algebras without the preceding chain condition. For example, in the commutative ring $k[z_1,z_2,\dots ]/(z_i^2:i\ge 1)$ the nilradical is $(z_1,z_2,\dots )$, its only prime ideal, but the product of any number of distinct $z_i$ is nonzero. Thus this nilradical is not a nilpotent ideal.

Use $q\in k^{\times }$ and $YX=qXY$ for the quantum plane. Its monomials $X^i{Y}^j$, $i,j\ge 0$, form a basis, with multiplicationOne way to verify the basis assertion without assuming it is to define this multiplication on the vector space with the displayed formal basis. This defines an associative algebra: for a third monomial $X^w{Y}^z$, the two products of three monomials have the same exponent $ju+jw+vw$ of $q$, and its generators satisfy the required relation. Conversely, the relation puts every word into this form, establishing the presentation. Order exponent pairs by the lexicographic order. The largest monomials of two nonzero finite sums give the uniquely largest monomial of their product, with coefficient $cd{q}^{ju}\ne 0$. Therefore the quantum plane is a domain. If one allows $q=0$, the assertion fails because $YX=0$ with both factors nonzero.

A uniform module is a nonzero module in which any two nonzero submodules have nonzero intersection. Suppose the right regular module of a right Noetherian domain $A$ were not a uniform module. Choose nonzero $a,b$ from two right ideals with zero intersection. Then $aA\cap bA=0$. The right idealsform a direct sum. For if ${\sum }_{i=0}^m{b}^{i}a{c}_i=0$, then $a{c}_0\in aA\cap bA$ is zero, so $c_0=0$ by the noncommutative domain property. Cancel the nonzero factor $b$ on the left and repeat to obtain every $c_i=0$. Each summand is nonzero, so their finite partial sums form a strictly ascending chain of right ideals. This contradicts the ascending chain condition of a right Noetherian ring. Hence $A$ is a uniform right module.

It follows that $aA\cap sA\ne 0$ whenever $a,s\ne 0$: there are nonzero $u,v$ with $au=sv$. This is the right Ore condition for the multiplicative set $S=A\setminus \{0\}$; zero numerators cause no difficulty. The Ore localization theorem therefore constructs the ring of right fractionsThe map is injective because an element mapping to zero is annihilated on the right by some nonzero denominator, impossible in a noncommutative domain. To make the denominator convention concrete, if $su=tv\ne 0$ thenFor multiplication, choose $bu=sc$ with $u\ne 0$; thenCommon right multiples make these operations independent of the chosen representatives. Every nonzero $a{s}^{-1}$ has inverse $s{a}^{-1}$, so $Q$ is a division ring. The original field $k$ is central in $A$ and therefore in the inverses as well, making $Q$ a division algebra over $k$. No commutative fraction field construction is being assumed.

First, every nonzero nonunit in a principal ideal domain factors into irreducible elements. Otherwise repeatedly choose a proper nonunit factor $a_{j+1}$ of $a_j$; then the principal idealsform a strictly increasing chain, contradicting the ascending chain condition, since every principal ideal domain is a Noetherian ring.

Next every irreducible $p$ is a prime element. If $p\mid ab$ but $p\nmid a$, then $\gcd (p,a)=1$. The Bezout identity gives $up+va=1$, so multiplying by $b$ shows $p\mid b$. Thus irreducibles are prime.

Existence of an irreducible factorization and primality give uniqueness: an irreducible on one side divides the product on the other, hence is associate to one factor; cancel and continue. Therefore every principal ideal domain is a unique factorization domain.

True. An irreducible ideal $I$ is a proper ideal such that $I=J\cap K$ forces $I=J$ or $I=K$. Passing to the quotient ring $A=R/I$ turns this into the assertion that zero cannot be the intersection of two nonzero ideals. Also $A$ is Noetherian: ascending chains of ideals lift to ascending chains of ideals containing $I$ in $R$.

Suppose $ab=0$ in $A$, and suppose $b$ is not a nilpotent element. The ascending chain condition makes the chain of annihilatorsstabilize. Choose $n\ge 1$ with $\mathrm{Ann}(b^n)=\mathrm{Ann}(b^{n+1})$. We claimIf $z=ua=v{b}^n$ lies in this intersection, then $bz=uab=0$, so $v{b}^{n+1}=0$. The equality of annihilators implies $v{b}^n=0$, hence $z=0$. Since $b$ is not nilpotent, $(b^n)\ne 0$. Irreducibility of zero therefore forces $(a)=0$, or $a=0$.

We have proved that $ab=0$ and $a\ne 0$ force $b$ to be a nilpotent element. This says precisely that zero is a primary ideal in $A$, and therefore that $I$ is a primary ideal in $R$.

If its radical of an ideal is to be denoted by a prime ideal, this requires no extra hypothesis: for any primary ideal $I$, $uv\in \sqrt{I}$ and $u\notin \sqrt{I}$ imply $u^n{v}^n\in I$ with $u^n\notin I$, so some power of $v$ lies in $I$. Thus $v\in \sqrt{I}$.

False: existence holds, but uniqueness fails. First, the Lasker–Noether theorem supplies existence; here is an argument from the preceding part. If some proper ideal in a Noetherian ring were not a finite intersection of irreducible ideals, the ascending chain condition would let us choose such an ideal $I$ maximal under inclusion. It is not irreducible, so $I=J\cap K$ with both $J$ and $K$ strictly larger than $I$. By maximality, both are finite intersections of irreducible ideals, and therefore so is $I$, a contradiction. Part (d) makes each factor a primary ideal. The whole ring is represented by the empty intersection if that convention is needed.

For a counterexample to uniqueness, work in $R=k[x,y]$ over a field $k$ and put $I=(x^2,xy)$. The Hilbert basis theorem makes $R$ a Noetherian ring. There are two distinct minimal primary decompositions:The ideal $(x)$ is a prime ideal, hence a primary ideal. Each of the other factors has quotient ring isomorphic to $k[x]/(x^2)$, by substituting respectively $y=0$ and $y=x$. As in part (b), each is $(x,y)$-primary.

To check both intersections, let $Q_c=(x^2,y-cx)$, for $c=0,1$. Substitution identifies $R/Q_c$ with $k[x]/(x^2)$. The class of $xg(x,y)$ is zero exactly when $g(0,0)=0$, equivalently when $g\in (x,y)$. Therefore $(Q_c:x)=(x,y)$ andThe two $Q_c$ are different, since $y$ belongs to $Q_0$ but has nonzero class $x$ modulo $Q_1$. Both decompositions are irredundant: $x\notin Q_c$, while $y-cx\notin (x)$. Applying the radical of an ideal operation to the two components in either decomposition gives $(x)$ and $(x,y)$, respectively.

The second uniqueness theorem for primary decomposition explains the distinction: the component at an isolated prime of a primary decomposition is unique, but embedded primary components need not be. The counterexample varies an embedded component.

A ring is a Noetherian ring if it satisfies the ascending chain condition on ideals: every chaineventually becomes constant. Equivalently, every ideal has a finite generating set of an ideal. To see this equivalence, if an ideal were not finitely generated, one could repeatedly choose an element outside the ideal generated by the preceding choices, producing a strictly ascending chain. Conversely, the union of an ascending chain is an ideal; if it is generated by finitely many elements, all of them lie in one term of the chain, and every later term equals that term.

A ring is an Artinian ring if it satisfies the descending chain condition on ideals: every chaineventually becomes constant. These are stabilization conditions on all ideals, not merely on principal ideals.

Suppose first that every ideal of $R$ is finitely generated. For an ascending chainthe union $I={\bigcup }_n{I}_n$ is an ideal. Its finite generating set lies in one $I_N$, so $I_n=I_N$ for all $n\ge N$. Thus $R$ satisfies the ascending chain condition and is a Noetherian ring.

Conversely, if an ideal $I$ is not finitely generated, choose $a_1\in I$, and after choosing $a_1,\dots ,a_n$, chooseThis creates a strictly ascending chain of ideals, contradicting Noetherianity. Hence every ideal is finitely generated.

If $\phi :R\to S$ is surjective and $J⊲S$, then ${\phi }^{-1}(J)$ is an ideal of $R$. If it is generated by $a_1,\dots ,a_m$, then $J$ is generated by $\phi (a_1),\dots ,\phi (a_m)$. Thus every ideal of $S$ is finitely generated, so $S$ is Noetherian.

The Hilbert basis theorem states that if $R$ is a commutative Noetherian ring, then $R[X]$ is Noetherian. To prove it, let $I⊲R[X]$. The leading coefficients of polynomials in $I$ generate an ideal of $R$; choose generators that occur as leading coefficients of $f_1,\dots ,f_s\in I$, and let $N={\max }_i\mathrm{deg}{f}_i$. For each degree below $N$, the leading coefficients of members of $I$ of at most that degree likewise form a finitely generated ideal; choose finitely many corresponding polynomials. Any $f\in I$ of degree at least $N$ can have its leading term cancelled by a linear combination of monomial multiples of the $f_i$. Repeating lowers its degree below $N$, where the second finite list completes the reduction. These finitely many selected polynomials generate $I$, proving the theorem.

For every $\alpha \in \mathbb{C}$, the ring $\mathbb{Z}[\alpha ]$ is Noetherian. If $\alpha$ is transcendental, evaluation identifies it with $\mathbb{Z}[X]$. If $\alpha$ is algebraic, it is a quotient of $\mathbb{Z}[X]$. The integers are Noetherian, the Hilbert basis theorem handles $\mathbb{Z}[X]$, and quotients preserve Noetherianity.

For a unique factorization domain that is not Noetherian, takeEvery polynomial and every factorization uses only finitely many variables, so existence and uniqueness of factorization reduce to a finite-variable polynomial ring, which is a UFD. Butis a strictly ascending chain, so the ring is not Noetherian.

Finally, the ring $C(\mathbb{R},\mathbb{R})$ is not Noetherian. For $n\ge 1$, letThese are ideals and $I_n⊊I_{n+1}$: the distance function to $[-1/(n+1),1/(n+1)]$ belongs to $I_{n+1}$ but not to $I_n$. This strict ascending chain proves the claim.

Suppose every submodule of $M$ is finitely generated. For an ascending chain $N_1\subseteq N_2\subseteq \cdots$, its union $N$ is a submodule. Finitely many generators of $N$ all lie in one $N_r$, so $N=N_r$ and the chain stabilizes. Thus the ascending chain condition holds.

If the ascending chain condition holds, any nonempty collection of submodules has a maximal member: otherwise, starting from one member and repeatedly choosing a strictly larger one constructs a nonstationary ascending chain.

Finally, assume the maximal condition. Among the finitely generated submodules of a given $N\subseteq M$, choose a maximal one $F$. If $F\ne N$, then $F+Rx$ is a larger finitely generated submodule for any $x\in N\setminus F$, a contradiction. Hence every $N$ is finitely generated. These are the three equivalent characterizations of a Noetherian module.

True. This is the Artinian commutative ring is Noetherian theorem. One proof uses the nilpotent nilradical $N$ of an Artinian ring $A$. The quotient $A/N$ is a finite product of fields. Each quotient $N^j/N^{j+1}$ is an Artinian module over the semisimple ring $A/N$, hence has finite length and is Noetherian. The finite filtrationthen makes $A$ a Noetherian module over itself, which is exactly the ascending chain condition on its ideals.

Suppose first that $A$ is a unique factorization domain, and let $\mathfrak{p}$ be a minimal nonzero prime ideal. Choose $0\ne a\in \mathfrak{p}$ and factor it into irreducibles. In a UFD each irreducible is a prime element, so one factor $\pi$ belongs to $\mathfrak{p}$. The nonzero prime ideal $(\pi )\subseteq \mathfrak{p}$ must equal $\mathfrak{p}$ by minimality.

Conversely, the ascending chain condition in the Noetherian domain implies that every nonzero nonunit factors into irreducibles. Let $\pi$ be irreducible and choose a prime $\mathfrak{p}$ minimal over $(\pi )$. The Krull principal ideal theorem gives $\mathrm{ht}\mathfrak{p}\le 1$. Since $A$ is a domain and $\mathfrak{p}\ne (0)$, it is a minimal nonzero prime and hence is principal, say $\mathfrak{p}=(q)$. The divisibility $q\mid \pi$ and irreducibility of $\pi$ force $q$ to be associate to $\pi$, so $(\pi )=\mathfrak{p}$ is prime. Thus every irreducible is prime, proving that $A$ is a UFD. This is the Minimal-prime criterion for a Noetherian unique factorization domain.

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 prime radical is the intersection of all prime ideals of a noncommutative ring. For a nonzero ring satisfying the ascending chain condition on two-sided ideals, there exist prime ideals of a noncommutative ring $P_1,\dots ,P_r$ with $P_1\cdots P_r=0$. Indeed, a maximal counterexample among two-sided ideals could not itself be a prime ideal of a noncommutative ring, and two larger witness ideals would contradict its maximality. Every prime ideal of a noncommutative ring contains one $P_i$, so the intersection equals ${\bigcap }_i{P}_i$, and its $r$th power is zero. This nilpotence assertion fails without a chain condition.