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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.

A component in a minimal primary decomposition is embedded if its radical of an ideal strictly contains another radical occurring in that decomposition. Such components need not be unique. For example,in $k[x,y]$ gives different $(x,y)$-primary ideals as embedded components, with the same isolated component $(x)$.

For a commutative Noetherian ring, the distinct radicals in a minimal primary decomposition are exactly the associated primes of a module $R/I$, and hence are invariant. The injection into the sum of primary quotients proves one inclusion; an element supported on one component, supplied by irredundancy, proves the other. This does not imply uniqueness of embedded primary components.

In the terminology of a minimal primary decomposition, an associated prime of a primary decomposition of $I$ is a prime ideal occurring as $\sqrt{{\mathfrak{q}}_j}$ in such a decomposition. The first uniqueness theorem for primary decomposition says that this collection is independent of the chosen minimal primary decomposition. Among these prime ideals, an isolated prime of a primary decomposition is minimal under inclusion, equivalently a minimal prime over an ideal $I$; an embedded prime ideal strictly contains another member of the collection. ThusWhen $R$ is a Noetherian ring, these are precisely the associated primes of a module $R/I$. In that case the equivalent definition iswhere $(I:x)$ is a colon ideal. The decomposition terminology applies when a minimal primary decomposition exists; a general ring need not provide one for every ideal.

A minimal primary decomposition is a finite expressionin proper primary ideals such that the prime ideals ${\mathfrak{p}}_j=\sqrt{{\mathfrak{q}}_j}$ are pairwise distinct and no primary ideal can be omitted. More explicitly, for every $j$,Thus minimality means both irredundancy and distinct radicals of ideals; it does not assert that the primary ideals themselves are uniquely determined. For $I=R$, one may use the empty intersection convention.

Put $A=\mathbb{C}[X,Y]$, $P=(X)$ and $Q=(X^2,Y)$. The quotient ring $A/P\cong \mathbb{C}[Y]$ is an integral domain, so $P$ is a prime ideal and hence a primary ideal. Put $\mathfrak{m}=(X,Y)$. ThenThe polynomial ring $A$ is a Noetherian ring by the Hilbert basis theorem, and $\mathfrak{m}$ is a maximal ideal since $A/\mathfrak{m}\cong \mathbb{C}$. Therefore the maximal-primary ideal criterion from c makes $Q$ $\mathfrak{m}$-primary. In particular its radical of an ideal is $\mathfrak{m}$, distinct from $P$.

Neither primary ideal can be omitted: $X\in P\setminus Q$, while $Y\in Q\setminus P$. Thus the given intersection is already a minimal primary decomposition. To express the ideal by generators, write an element of $P\cap Q$ as $Xh=X^{2}u+Yv$. Reducing modulo the prime ideal $(X)$ gives $Y\bar{v}=0$ in the integral domain $\mathbb{C}[Y]$, so $v=Xw$. Cancelling $X$ in the integral domain $A$ gives $h=Xu+Yw$. HenceBy the first uniqueness theorem for primary decomposition, the associated primes of a module $A/J$ are exactly the two radicals of ideals. Their strict inclusion $P⊊\mathfrak{m}$ givesOne can also see both associated primes of a module directly through annihilators:For the first colon ideal, reducing $hY\in J$ modulo $(X)$ forces $h\in (X)$, and the reverse inclusion follows from $XY\in J$. For the second colon ideal, cancel $X$ in $hX\in X(X,Y)$. Both represented elements of the quotient module $A/J$ are nonzero.

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.

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 minimal primary decomposition is an expressionin which every $Q_i$ is a primary ideal, the prime ideals $\sqrt{Q_i}$ are pairwise distinct, and the decomposition is irredundant: deleting any $Q_i$ changes the intersection.