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., \mathbb{Z}/n \mathbb{Z}) is left and right Artinian. • Let k be a field. Then k[t]/(t^n) is Artinian for every positive integer n. • Similarly, 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 (x,y). • Let x be an endomorphism between a finite-dimensional vector space V. Then the subalgebra A \subset \operatorname{End}(V) generated by x is a commutative Artinian ring. • If I is a nonzero ideal of a Dedekind domain A, then A/I is a principal Artinian ring. • For each n \ge 1, the full matrix ring M_n(R) over a left Artinian (resp. left Noetherian) ring R is left Artinian (resp. left Noetherian). 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 x^{n+1} is (properly) contained in the ideal generated by 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 \mathbb{Z} is a Noetherian ring but is not Artinian. == Modules over Artinian rings ==

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. == Commutative Artinian rings ==

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. • 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. (In particular, the nilradical is the Jacobson radical since prime ideals are maximal.) • \operatorname{Spec}A is finite and discrete. • \operatorname{Spec}A is discrete. Let k be a field and A a finitely generated k-algebra. Then A is Artinian if and only if A is finitely generated as a k-module. An Artinian local ring is complete. A quotient and localization of an Artinian ring is Artinian. == Simple Artinian ring ==

One version of the Wedderburn–Artin theorem states that a simple Artinian ring A is a matrix ring over a division ring. Indeed, 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 AI is a two-sided ideal, AI = A since A is simple. Thus, we can choose a_i \in A so that 1 \in a_1 I + \cdots + a_k I. Assume k is minimal with respect to that property. Now consider the map of right A-modules: : \begin{cases} I^{\oplus k} \to A, \\ (y_1, \dots, y_k) \mapsto a_1y_1 + \cdots + a_k y_k \end{cases} This map is surjective, since the image is a right ideal and contains 1. If it is not injective, then, say, a_1y_1 = a_2y_2 + \cdots + a_k y_k with nonzero y_1. Then, by the minimality of I, we have y_1 A = I. It follows: : a_1 I = a_1 y_1 A \subset a_2 I + \cdots + a_k I, which contradicts the minimality of k. Hence, I^{\oplus k} \simeq A and thus A \simeq \operatorname{End}_A(A) \simeq M_k(\operatorname{End}_A(I)). == See also ==