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Module theory studies abelian groups with scalar multiplication by a ring. It extends linear algebra: a module over a field is a vector space, while ring scalars let the same language describe integer arithmetic and linear operators. The new phenomena are visible before any abstract classification: some nonzero scalars cannot be divided out, some modules have torsion elements of a module, and some have no basis of a module.

The distinction is useful because an integer relation such as $6m=0$ carries information which disappears if one assumes all nonzero scalars have multiplicative inverses. A finitely generated module over a principal ideal domain can be separated into a finite free module and a torsion module; the structure theorem for finitely generated modules over a principal ideal domain unifies the classification of finitely generated abelian groups with the study of linear operators through polynomial rings. See Keith Conrad's Modules over a PID, Theorem 4.1 and the operator applications.

A historical anchor is Emmy Noether's 1921 paper ^[ref]. Its first section studies rings and ideals under the condition that every ideal is a finitely generated module over its ring, and connects this to the ascending chain condition; later sections develop primary decomposition. This is an early source for the structural viewpoint underlying Noetherian rings. ^[ref] makes those statements accessible in English.

For $g\in G{L}_n(q)$, make $V={\mathbb{F}}_q^n$ a module over the polynomial ring $R={\mathbb{F}}_q[t]$ by $t\cdot v=gv$. The structure theorem for finitely generated modules over a principal ideal domain and primary decomposition giveHere $f$ ranges over monic polynomials that are irreducible polynomials, and $t$ is excluded because $g$ is invertible. Associate the integer partition ${\lambda }_f$ having $m_j(f)$ parts equal to $j$. It has finite support and satisfies ${\sum }_f\mathrm{deg}(f)|{\lambda }_f|=n$. Two matrices are conjugate exactly when their R-modules are isomorphic, so this gives a bijective parameterization of conjugacy classes.

The centralizer is the R-module automorphism group. Homomorphisms between distinct primary components vanish: coprimeness of their annihilating polynomials makes each such map zero. Fix one $f$ of degree $d$, put $Q=q^d$, and write $m_j=m_j(f)$. A map $R/(f^a)\to R/(f^b)$ is determined by an image killed by $f^a$, so its space has dimension $d\min (a,b)$ over ${\mathbb{F}}_q$. Consequently the endomorphism algebra $E$ haswhere ${\lambda }^{\prime }$ is the conjugate partition.

Its semisimple quotient is ${\prod }_j{M}_{m_j}({\mathbb{F}}_Q)$: reduce the maps between the equal-length summands modulo $f$ and discard the maps between unequal lengths. Composition through a different length acquires a factor of $f$ when it returns to the original length, so these reductions define an algebra homomorphism. It is onto, and its kernel is nilpotent. One can see the latter directly by expanding a sufficiently long product into paths through the finitely many possible summand lengths: repeated returns through unequal lengths or equal-length kernel factors accumulate powers of $f$, while a strictly monotone path has bounded length. Eventually every path vanishes in every summand.

An element is invertible exactly when its image in that quotient is invertible, since a nilpotent error is inverted by a finite geometric series. The fraction of units is therefore ${\prod }_j|G{L}_{m_j}(Q)|/Q^{m_j^2}={\prod }_j{\prod }_{r=1}^{m_j}(1-Q^{-r})$. Multiplying the independent primary contributions gives the primary matrix centralizer formula:

The annihilating polynomial excludes any free $\mathbb{C}[X]$-summand. In a cyclic primary decomposition summand $\mathbb{C}[X]/(P^a)$, the same condition requires $P^a$ to divide $(X-2{)}^4$. Thus $P=X-2$, up to a scalar unit, and $1\le a\le 4$. The complex dimension of this summand is $a$.

Write $A_j=\mathbb{C}[X]/((X-2{)}^j)$. The exponents must form a partition of an integer $4$, giving exactlyThey are distinct module isomorphism types by uniqueness in the structure theorem for finitely generated modules over a principal ideal domain; equivalently they are the five possible Jordan block patterns for eigenvalue $2$ in dimension four.

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.