In
mathematics, specifically
abstract algebra, an Artinian ring (sometimes Artin ring) is a
ring that satisfies the
descending chain condition on (one-sided)
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-dimensionalvector spaces over
fields. The definition of Artinian rings may be restated by interchanging the descending chain condition with an equivalent notion: the
minimum condition.
Precisely, 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.[1] For
commutative rings the left and right definitions coincide, but in general they are distinct from each other.
For each , the full
matrix ring 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 ringRx] is not Artinian, since the ideal generated by is (properly) contained in the ideal generated by for all
natural numbersn. In contrast, if R is Noetherian so is Rx] by the
Hilbert basis theorem.
The ring of integers is a Noetherian ring but is not 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
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 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 .