Artinian ring

An Artinian ring satisfies the descending chain condition on ideals. A commutative ring is Artinian exactly when it is Noetherian and has Krull dimension zero.

If $C$ is a commutative algebra of finite dimension over a field $K$, every prime ideal is maximal and $|\mathrm{Spec}C|\le {\dim }_{K}C$. Indeed, the quotient by a prime is a finite-dimensional integral domain. Multiplication by any nonzero element is an injective linear endomorphism, hence surjective, so the quotient is a field. Distinct maximal ideals are pairwise comaximal. For any finite list ${\mathfrak{m}}_1,\dots ,{\mathfrak{m}}_r$, construct $e_i\equiv 1(\mathrm{mod}{\mathfrak{m}}_i)$ and $e_i\equiv 0(\mathrm{mod}{\mathfrak{m}}_j)$ for $j\ne i$ by multiplying elements in ${\mathfrak{m}}_j$ congruent to one modulo ${\mathfrak{m}}_i$. These give a surjection $C\to {\prod }_{i}C/{\mathfrak{m}}_i$. Each factor has positive $K$-dimension, so $r\le {\dim }_{K}C$, excluding an infinite list.

Every nonzero commutative Artinian ring is a finite product of Artinian local rings. The primary components of zero have distinct maximal radicals and are pairwise comaximal ideals, so the Chinese remainder theorem gives the product. Each local factor has nilpotent maximal ideal and finite composition length. These factors and the components of zero are unique.

A family of ideals satisfies the descending chain condition when every descending sequence of ideals eventually becomes constant. A ring is Artinian precisely when all its ideals satisfy this condition.

A commutative Artinian ring has finitely many prime ideals, and every one is a maximal ideal. If $\mathfrak{p}$ is prime, the Artinian domain $R/\mathfrak{p}$ is a field: stabilization of $(x)\supseteq (x^2)\supseteq \cdots$ gives $x^n=x^{n+1}a$, and cancellation gives $xa=1$. Infinitely many distinct maximal ideals ${\mathfrak{m}}_i$ would give a strictly descending chainbecause distinct maximal ideals are comaximal.

The nilradical $N$ of a commutative Artinian ring is a nilpotent ideal. Its powers stabilize, say $N^r=N^{r+1}=I$. If $I\ne 0$, choose an ideal $J$ minimal subject to $IJ\ne 0$, and then $x\in J$ with $Ix\ne 0$. Minimality and $I^2=I$ give $Rx=Ix$, so $x=ax$ for some $a\in I$. Since $a$ is nilpotent, $1-a$ is a unit, contradicting $(1-a)x=0$ and $x\ne 0$.

Every commutative Artinian ring is Noetherian. Its nilradical $N$ is nilpotent by the nilradical of a commutative Artinian ring, and the Chinese remainder theorem identifies the reduced quotient $R/N$ with the finite product of the residue fields at the finitely many maximal ideals. Each $N^i/N^{i+1}$ is an Artinian module over that product of fields and hence a finite-dimensional vector space. The finite filtration by these layers makes every ideal finitely generated.

A commutative Artinian ring has finite composition length as a module over itself. It has finitely many maximal ideals, and its Jacobson radical $J$ is nilpotent. The Chinese remainder theorem makes $R/J$ a finite product of fields; each Artinian module $J^i/J^{i+1}$ has finite-dimensional components over those fields. Adding their lengths along the finite radical filtration proves the assertion, and in particular the ascending chain condition.

An Artinian local ring is an Artinian ring with a unique maximal ideal. Its maximal ideal is nilpotent.

Let $(R,\mathfrak{m})$ be an Artinian local ring. If ${\dim }_{R/\mathfrak{m}}\mathfrak{m}/{\mathfrak{m}}^2=1$, then every ideal of $R$ is principal. The Nakayama lemma gives $\mathfrak{m}=(x)$, and nilpotence gives a finite chainFor a nonzero ideal $J$, choose the largest $r$ for which $J\subseteq (x^r)$ and take $y\in J\setminus (x^{r+1})$. Writing $y=x^{r}u$, one has $u\notin \mathfrak{m}$, so $u$ is a unit and $J=(x^r)=(y)$.

An Artinian local ring is a finite set exactly when its residue field is finite. If ${\mathfrak{m}}^r=0$, each layer ${\mathfrak{m}}^i/{\mathfrak{m}}^{i+1}$ is a finite-dimensional vector space over that residue field.

A commutative Noetherian ring is Artinian if and only if every prime ideal is maximal. Equivalently, it is Artinian exactly when its Krull dimension is zero.