In abstract algebra, an Artinian ring is a ring that satisfies the descending chain condition on ideals; that is, there is no infinite descending sequence of ideals. Artinian rings are named after Emil Artin, who first discovered that the descending chain condition for ideals simultaneously generalizes finite rings and rings that are finite-dimensional vector spaces over fields. The definition of Artinian rings may be restated by interchanging the descending chain condition with an equivalent notion: the minimum condition. A ring is left Artinian if it satisfies the descending chain condition on left ideals, right Artinian if it satisfies the descending chain condition on right ideals, and Artinian or two-sided Artinian if it is both left and right Artinian. For commutative rings the left and right definitions coincide, but in general they are distinct from each other. The Artin–Wedderburn theorem characterizes all simple Artinian rings as the ring of matrices over a division ring. This implies that a simple ring is left Artinian if and only if it is right Artinian. The same definition and terminology can be applied to modules, with ideals replaced by submodules. Although the descending chain condition appears dual to the ascending chain condition, in rings it is in fact the stronger condition. Specifically, a consequence of the Akizuki–Hopkins–Levitzki theorem is that a left Artinian ring is automatically a left Noetherian ring. This is not true for general modules; that is, an Artinian module need not be a Noetherian module.
Examples
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 is left and right Artinian.
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.
Simple Artinian ring
A simple Artinian ring A is a matrix ring over a division ring. Indeed, let I be a minimal right ideal of A. Then, since is a two-sided ideal, since A is simple. Thus, we can choose so that. Assume k is minimal with respect that property. Consider the map of right A-modules: It is surjective. If it is not injective, then, say, with nonzero. Then, by the minimality of I, we have:. It follows: which contradicts the minimality of k. Hence, and thus.
