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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.

All rings below are nonzero and commutative with identity; the zero ring is trivially Artinian. A primary ideal is proper. An ideal $Q$ is primary precisely when every zero divisor of $R/Q$ is a nilpotent element. Its radical of an ideal $\mathfrak{p}=\sqrt{Q}$ is prime: if $ab\in \sqrt{Q}$ and $a\notin \sqrt{Q}$, apply the primary property to $(ab{)}^r\in Q$ to obtain a power of $b$ in $Q$. In a Noetherian ring, finite generation of $\mathfrak{p}$ gives ${\mathfrak{p}}^s\subseteq Q$ for some $s$. For generators $u_i$ with $u_i^{s_i}\in Q$, one can take $s=1+{\sum }_i(s_i-1)$.

We first prove existence of primary decomposition. The ascending chain condition implies that every proper ideal is a finite intersection of irreducible ideals. Otherwise choose a maximal counterexample $I$ under inclusion. It is not irreducible, so $I=J\cap K$ with both $J,K$ strictly larger; their finite irreducible decompositions give one for $I$, a contradiction.

An irreducible ideal in a Noetherian ring is primary. Pass to the quotient and suppose zero is irreducible. If $xy=0$ with $x\ne 0$, the ascending chain of annihilators of $y^j$ stabilizes, say at $j=r$. Any element of $(x)\cap (y^r)$ can be written $z=ax=b{y}^r$. Then $yz=0$, so $b\in \mathrm{Ann}(y^{r+1})=\mathrm{Ann}(y^r)$ and $z=0$. Irreducibility of zero implies $(y^r)=0$, since $(x)\ne 0$. Thus $y$ is nilpotent. This proves irreducible ideals are primary in Noetherian rings, and hence the Lasker–Noether theorem.

An intersection of finitely many primary ideals with the same radical $\mathfrak{p}$ is again primary: if $ab$ belongs to all of them and $b\notin \mathfrak{p}$, the primary property forces $a$ into every component. Combining such components and removing redundant ones yields a minimal primary decompositionThe radicals are uniquely determined, but the components need not all be unique. Here is a proof identifying the invariant radicals as the associated primes of a module $R/I$.

Every nonzero module over a Noetherian ring has a nonzero element with a prime annihilator: maximize the annihilator of a nonzero element using the ascending chain condition. If $abm=0$ and $bm\ne 0$, maximality gives $\mathrm{Ann}(bm)=\mathrm{Ann}(m)$ and hence $am=0$. This is the maximal annihilator of a module element is prime argument. For a $\mathfrak{p}$-primary quotient $R/Q$, the radical of the annihilator of every nonzero element is $\mathfrak{p}$: it is contained in $\mathfrak{p}$ by the primary property and contains a power of $\mathfrak{p}$ because ${\mathfrak{p}}^s\subseteq Q$. Thus the only possible associated prime is $\mathfrak{p}$, and it does occur.

The diagonal injection $R/I\hookrightarrow {⨁}_{i}R/Q_i$ shows that every associated prime of $R/I$ is one of the ${\mathfrak{p}}_i$. Indeed, for an element whose annihilator $P$ is prime, that annihilator is the intersection of the finitely many component annihilators, all containing $P$. Their product is contained in $P$, so primality forces one of them to equal $P$; its component element is nonzero and has associated prime ${\mathfrak{p}}_i$. Conversely, irredundancy gives $b\in {\bigcap }_{j\ne i}{Q}_j\setminus Q_i$. The nonzero cyclic submodule generated by $b+I$ embeds in $R/Q_i$. It has an associated prime, necessarily ${\mathfrak{p}}_i$, which is then an associated prime of $R/I$. We have proved the first uniqueness theorem for primary decomposition:Also the zero divisors on $R/I$ are exactly ${\bigcup }_i{\mathfrak{p}}_i$. For a scalar killing a nonzero element, extend its element annihilator to a maximal element annihilator containing it; the preceding argument gives an associated prime containing that scalar. The converse is immediate from the definition of an associated prime.

The minimal members of the set $\{{\mathfrak{p}}_i\}$ are the isolated primes of a primary decomposition. They are exactly the primes minimal over $I$: if a prime contains ${\bigcap }_i{\mathfrak{p}}_i$, it contains one ${\mathfrak{p}}_i$ by the product argument. A component belonging to an isolated prime is unique. Localize at ${\mathfrak{p}}_i$. Every other component becomes the whole ring, since its radical contains an element outside ${\mathfrak{p}}_i$ whose power lies in that component. A ${\mathfrak{p}}_i$-primary ideal contracts unchanged from this localization, because $sa\in Q_i$ with $s\notin {\mathfrak{p}}_i$ implies $a\in Q_i$. Hence the second uniqueness theorem for primary decomposition givesAn embedded primary component can vary. For example, in $k[x,y]$,Both second components are $(x,y)$-primary, since their quotients are dual-number algebras. For the second equality, reducing modulo $(x^2,y+x)$ makes $y=-x$; an element $xg$ vanishes exactly when $g$ has zero constant term, giving $(x^2,xy)$. The isolated component is $(x)$ and the embedded associated prime is $(x,y)$.

The relevance to Artinian rings is particularly sharp in Krull dimension zero. Suppose $R$ is Noetherian and all its primes are maximal. Apply a minimal primary decomposition to zero. Its radicals are distinct maximal ideals, so the components are pairwise comaximal ideals: each contains a power of its radical, and expanding $1=(a+b{)}^N$ for $a+b=1$ shows that powers of comaximal ideals remain comaximal. The Chinese remainder theorem gives the Artinian decomposition into local factorsEach factor has one maximal ideal, and that ideal is nilpotent. Its finite radical filtration has successive layers finitely generated over its residue field, hence of finite vector-space dimension. It follows that the factor, and therefore $R$, has a finite composition series of a module, so $R$ is Artinian. This proves the Noetherian dimension-zero criterion for an Artinian ring in this direction. Applied to $R/I$, it says that the quotient is Artinian exactly when all primes over $I$ are maximal. In that case no primary component is embedded, so all components are unique.

For completeness, the converse does not require initially assuming Noetherianity. In an Artinian ring, an Artinian domain is a field: stabilization of $(a^j)$ and cancellation gives an inverse to every nonzero $a$. Thus every prime is maximal. There are finitely many maximal ideals, since infinitely many distinct ones would give strictly descending finite intersections; comaximality guarantees strictness by the Chinese remainder theorem. Its nilradical $N$ is nilpotent. Indeed, its powers stabilize at $J=N^r$, with $J^2=J$. If $J\ne 0$, choose an ideal $C$ minimal among those satisfying $JC\ne 0$. Then $JC=C$, since $J(JC)=JC\ne 0$. Choose $x\in C$ with $Jx\ne 0$; minimality also gives $Rx=C$. Therefore $x=ax$ for some $a\in J\subseteq N$. But $1-a$ is a unit, contradicting $x\ne 0$. Hence $J=0$.

Now $R/N$ is a finite product of residue fields. Each $N^j/N^{j+1}$ is an Artinian module over that product, and therefore a finite direct sum of finite-dimensional vector spaces: an infinite-dimensional vector space would have a strictly descending chain of subspaces. The finite filtration by powers of $N$ makes $R$ a module of finite composition length, in particular Noetherian. Combining both directions yieldsThus primary decomposition separates the local pieces of a zero-dimensional ring, while their nilpotent maximal ideals record the multiplicities that the reduced set of primes alone does not detect.

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.