= Solution
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 $x\ne0$, the ascending chain of <annihilators> of $y^j$ stabilizes, say at $j=r$. Any element of $(x)\cap(y^r)$ can be written $z=ax=by^r$. Then $yz=0$, so $b\in\operatorname{Ann}(y^{r+1})=\operatorname{Ann}(y^r)$ and $z=0$. Irreducibility of zero implies $(y^r)=0$, since $(x)\ne0$. 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 decomposition>
$$
I=Q_1\cap\cdots\cap Q_r,\qquad\mathfrak p_i=\sqrt{Q_i}\text{ pairwise distinct}.
$$
The 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 $bm\ne0$, maximality gives $\operatorname{Ann}(bm)=\operatorname{Ann}(m)$ 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 $R/I\hookrightarrow\bigoplus_iR/Q_i$ 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>:
$$
\boxed{\operatorname{Ass}_R(R/I)=\{\mathfrak p_1,\ldots,\mathfrak p_r\}.}
$$
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> gives
$$
Q_i=IR_{\mathfrak p_i}\cap R\qquad(\mathfrak p_i\text{ isolated}).
$$
An <embedded primary component> can vary. For example, in $k[x,y]$,
$$
(x^2,xy)=(x)\cap(x^2,y)=(x)\cap(x^2,y+x).
$$
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 factors>
$$
\boxed{R\cong\prod_iR/Q_i.}
$$
Each 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 $J\ne0$, choose an ideal $C$ minimal among those satisfying $JC\ne0$. Then $JC=C$, since $J(JC)=JC\ne0$. Choose $x\in C$ with $Jx\ne0$; minimality also gives $Rx=C$. Therefore $x=ax$ for some $a\in J\subseteq N$. But $1-a$ is a unit, contradicting $x\ne0$. 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 yields
$$
\boxed{R\text{ nonzero commutative Artinian}\iff R\text{ Noetherian and }\dim R=0.}
$$
Thus <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.
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