Local Rings, The Structure of an Artinian Ring

A Finite Product of Maximal Ideals

We know that artinian implies noetherian, but in some cases we can infer the converse. Throughout this page, rings are assumed to be commutative.

A module over a field is free, with a fixed rank. Such a module is acc iff it has finite rank iff it is dcc.

Let R be a noetherian ring where the product of finitely many maximal ideals equals 0. Some of these maximal ideals could be duplicated in the product. We will prove R is artinian.

Let M1 be the first maximal ideal, with Q1 = R/M1. Note that Q1, the first quotient, is a field.

Naturally n1 is a noetherian R module. The product ideal M1M2 is a noetherian submodule of M1. Similarly, M1/(M1M2) is a noetherian R module. However, the action of R depends only on the coset of M2 in R. In other words, M1/(M1M2) is essentially a Q2 = R/M2 module. Since Q2 is a field, and M1/(M1M2) is acc, it is also dcc.

Start with M1M2, which is acc. Multiply by M3 and find another acc module. The quotient, (M1M2)/(M1M2M3) is also acc. Yet the action depends only on the coset of M3, hence it is a Q3 module, and a vector space. It therefore becomes dcc.

Repeat this process until the product of M1 through Mn becomes 0. This is obviously dcc, and the quotient is an acc vector space that becomes dcc, hence the product of ideals from M1 through Mn-1 is dcc. Combine this with the prior quotient to show the product of M1 through Mn-2 is dcc. This moves all the way up to M1, which is dcc. Combine this with Q1, and R is artinian.

Noetherian and Dimension 0

Let R be a noetherian ring with dimension 0. This means a prime cannot properly contain another prime, and every prime ideal is maximal.

Since R is laskerian, write a primary decomposition Q1 through Qn for 0. Their intersection is 0, hence their product, which lies in the intersection, is 0.

The radical of each Qi is some prime Pi, which is also maximal. Let Q be one of these primary ideals with radical P. Since P is finitely generated, some exponent e satisfies xe ∈ Q for all the generators x of P. Multiply e by the number of generators and use this as your new exponent. By the multinomial theorem, anything in P, raised to this exponent, lies in Q. Thus each Pi, raised to a sufficiently high power, lies in Qi. Since the product over Qi is 0, a finite product of maximal ideals is 0. By the above, R is artinian.

Conversely, an artinian ring is noetherian (we already knew that), and it has dimension 0. Mod out by a prime ideal P and find an artinian integral domain. This is a field, hence P is maximal, and R has dimension 0. A commutative ring is artinian iff it is noetherian and has dimension 0.


Consider the collection of finite intersections of maximal ideals in an artinian ring R. Let H be a minimal ideal in this collection. Suppose a maximal ideal M is not part of the finite intersection that produces H. bring in M, and H does not get any smaller. In other words, M contains H. Since M contains the product of maximal ideals it contains one of them. Thus M is part of the intersection after all. Since H is a finite intersection, R has finitely many maximal ideals. In other words, every artinian ring is semilocal.

Note that H is both the nil radical and the jacobson radical. Since R is artinian, H is nilpotent.

Spec R is finite and discrete.


We are about to analyze the structure of an artinian ring, but only after it has been reduced. The nil radical is not part of this characterization. Consider the following example.

Let R be an artinian ring with 0 nil radical. adjoin x and y to create polynomials in two indeterminants. Let x3 = 0, and let y7 = 0. Every prime ideal contains x and y, and all polynomials generated by x and y. Call this ideal H. If c is a nonzero constant from R, find a prime ideal P that does not contain c. Extend this to the prime ideal comprising polynomials with coefficients in P. This still misses c, hence c is not part of the nil radical. Similarly, a polynomial with c as constant cannot be in the nil radical, since we could subtract away everything except c. Therefore H is the nil radical. Reduce the ring and get R back again.

One can adjoin an infinite number of indeterminants and relations to build H. The only constraint is that H must be a nilpotent ideal. The result is another artinian ring with a wild nil radical. Let's set these radicals aside and look at reduced artinian rings.

The Structure of an Artinian Ring

Let R be a reduced artinian ring. As mentioned above, spec R is finite, with the discrete topology. Therefore R splits into a direct product of rings.

Each component ring has but one point in spec R, which is one prime/maximal ideal. Therefore R is a finite direct product of local rings.

Each component is a submodule of R, and inherits acc and dcc from R. By the krull schmidt theorem, the decomposition is unique. In other words, R is uniquely (up to order and isomorphism) a direct product of local artinian rings.

Let S be one of these local artinian rings. Its maximal ideal M is its jacobson radical, which is nilpotent. The powers of M reach 0 in finitely many steps.

Let H be an ideal of S. Since rad(H) = M, H is M primary. But don't assume H is automatically a power of M. Start with a field F and adjoin x and y, such that x3 = y7 = 0. Let M be the ideal generated by x and y, which is maximal. Every prime containing 0 pulls in x and y, hence M is the only prime ideal. The ring is a finite dimensional F vector space, hence an artinian ring. We have an artinian local ring, yet the ideal generated by x is not a power of M.