Artinian ring

• An integral domain is Artinian if and only if it is a field.
• A ring with finitely many, say left, ideals is left Artinian. In particular, a finite ring (e.g., ${\displaystyle \mathbb {Z} /n\mathbb {Z} }$) is left and right Artinian.
• Let k be a field. Then ${\displaystyle k[t]/(t^{n})}$ is Artinian for every positive integer n.
• Similarly, ${\displaystyle k[x,y]/(x^{2},y^{3},xy^{2})=k\oplus k\cdot x\oplus k\cdot y\oplus k\cdot xy\oplus k\cdot y^{2}}$ is an Artinian ring with maximal ideal ${\displaystyle (x,y)}$.
• Let ${\displaystyle x}$ be an endomorphism between a finite-dimensional vector space V. Then the subalgebra ${\displaystyle A\subset \operatorname {End} (V)}$ generated by ${\displaystyle x}$ is a commutative Artinian ring.
• If I is a nonzero ideal of a Dedekind domain A, then ${\displaystyle A/I}$ is a principal Artinian ring.[2]
• For each ${\displaystyle n\geq 1}$, the full matrix ring ${\displaystyle M_{n}(R)}$ over a left Artinian (resp. left Noetherian) ring R is left Artinian (resp. left Noetherian).[3]

The following two are examples of non-Artinian rings.

• If R is any ring, then the polynomial ring R[x] is not Artinian, since the ideal generated by ${\displaystyle x^{n+1}}$ is (properly) contained in the ideal generated by ${\displaystyle x^{n}}$ for all natural numbers n. In contrast, if R is Noetherian so is R[x] by the Hilbert basis theorem.
• The ring of integers ${\displaystyle \mathbb {Z} }$ is a Noetherian ring but is not Artinian.

Let M be a left module over a left Artinian ring. Then the following are equivalent (Hopkins' theorem): (i) M is finitely generated, (ii) M has finite length (i.e., has composition series), (iii) M is Noetherian, (iv) M is Artinian.[4]

Let A be a commutative Noetherian ring with unity. Then the following are equivalent.

• A is Artinian.
• A is a finite product of commutative Artinian local rings.[5]
• A / nil(A) is a semisimple ring, where nil(A) is the nilradical of A.
• Every finitely generated module over A has finite length. (see above)
• A has Krull dimension zero.[6] (In particular, the nilradical is the Jacobson radical since prime ideals are maximal.)
• ${\displaystyle \operatorname {Spec} A}$ is finite and discrete.
• ${\displaystyle \operatorname {Spec} A}$ is discrete.[7]

Let k be a field and A finitely generated k-algebra. Then A is Artinian if and only if A is finitely generated as k-module.

An Artinian local ring is complete. A quotient and localization of an Artinian ring is Artinian.

One version of the Wedderburn–Artin theorem states that a simple Artinian ring A is a matrix ring over a division ring. Indeed,[8] let I be a minimal (nonzero) right ideal of A, which exists since A is Artinian (and the rest of the proof does not use the fact that A is Artinian). Then, since ${\displaystyle AI}$ is a two-sided ideal, ${\displaystyle AI=A}$ since A is simple. Thus, we can choose ${\displaystyle a_{i}\in A}$ so that ${\displaystyle 1\in a_{1}I+\cdots +a_{k}I}$. Assume k is minimal with respect that property. Consider the map of right A-modules:

${\displaystyle {\begin{cases}I^{\oplus k}\to A,\\(y_{1},\dots ,y_{k})\mapsto a_{1}y_{1}+\cdots +a_{k}y_{k}\end{cases}}}$

It is surjective. If it is not injective, then, say, ${\displaystyle a_{1}y_{1}=a_{2}y_{2}+\cdots +a_{k}y_{k}}$ with nonzero ${\displaystyle y_{1}}$. Then, by the minimality of I, we have: ${\displaystyle y_{1}A=I}$. It follows:

${\displaystyle a_{1}I=a_{1}y_{1}A\subset a_{2}I+\cdots +a_{k}I}$,

which contradicts the minimality of k. Hence, ${\displaystyle I^{\oplus k}\simeq A}$ and thus ${\displaystyle A\simeq \operatorname {End} _{A}(A)\simeq M_{k}(\operatorname {End} _{A}(I))}$.

• Artin algebra
• Artinian ideal
• Serial module
• Semiperfect ring
• Gorenstein ring
• Noetherian ring