Artinian ring

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In abstract algebra, an Artinian ring is a ring that satisfies the descending chain condition on ideals. They are also called Artin rings and 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 matrix rings over a division ring. This implies that a simple ring is left Artinian if and only if it is right Artinian.

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 (right) Artinian ring is automatically a left (right) Noetherian ring. This is not true for general modules, that is, an Artinian module need not be a Noetherian module.

Examples[edit]

  • 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.
  • If I is a nonzero ideal of a Dedekind domain A, then A/I is a principal Artinian ring.[1]
  • 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).[2]

The ring of integers \mathbb{Z} is a Noetherian ring but is not Artinian.

Modules over Artinian rings[edit]

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.[3]

Commutative Artinian rings[edit]

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.[4]
  • A / nil(A) is a semisimple ring, where nil(A) is the nilradical of A.[citation needed][5]
  • 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.)
  • \operatorname{Spec}A is finite and discrete.
  • \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.

Simple Artinian ring[edit]

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. 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 that property. Consider the map of right A-modules:

I^{\oplus k} \to A, \, (y_1, \dots, y_k) \mapsto a_1y_1 + \cdots + a_k y_k.

It is surjective. 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[edit]

Notes[edit]

  1. ^ Theorem 459 of http://math.uga.edu/~pete/integral.pdf
  2. ^ Cohn 2003, 5.2 Exercise 11
  3. ^ Bourbaki, VIII, pg 7
  4. ^ Atiyah & Macdonald 1969, Theorems 8.7
  5. ^ Sketch: In commutative rings, nil(A) is contained in the Jacobson radical of A. Since A/nil(A) is semisimple, nil(A) is actually equal to the Jacobson radical of A. By Levitzky's theorem, nil(A) is a nilpotent ideal. These last two facts show that A is a semiprimary ring, and by the Hopkins–Levitzki theorem A is Artinian.
  6. ^ Atiyah & Macdonald 1969, Theorems 8.5
  7. ^ Atiyah & Macdonald 1969, Ch. 8, Exercise 2.
  8. ^ Milnor, John Willard (1971), Introduction to algebraic K-theory, Annals of Mathematics Studies 72, Princeton, NJ: Princeton University Press, p. 144, MR 0349811, Zbl 0237.18005 

References[edit]