# Artinian Rings

The main result we wish to prove is the following.

Theorem.A ring A is artinian if and only if it is noetherian and , where denotes the Krull dimension.

**Note**

Recall that means all prime ideals of *A* are maximal.

Since the proof is long we will break it up into steps.

# Artinian ⟹ Dim = 0

**Step 1: An artinian integral domain A is a field.**

Let . From the sequence we have for some *n* > 0. Thus for some . Since *A* is a domain and we have so *x* is a unit.

**Step 2: An artinian ring A has finitely many maximal ideals.**

Let be the collection of ideals of *A* of the form where are maximal ideals. Since *A* is artinian has a minimal element . By minimality for any maximal ideal we have and so . We claim that for some which would complete our claim.

Since and are all maximal it suffices to show for some *i*. Indeed if not we can pick for . Then

a contradiction.

**Step 3: An artinian ring A has finitely many prime ideals, all maximal.**

For any prime ideal , is an artinian integral domain, hence a field by step 1. Thus is maximal so all prime ideals of *A* are maximal. By step 2, there are only finitely many of them.

Thus, we have shown that an artinian ring has Krull dimension 0. Since the connected components of Spec *A* are all singleton sets, we have proven:

Intermediate Result.An artinian ring A is a finite product , where each is a local artinian ring with a unique prime ideal.

Our next target is noetherianness.

# Artinian ⟹ Noetherian

By the intermediate result, we may assume *A* is a local artinian ring, with a unique prime ideal *which is the nilradical of A. *[ Recall (proposition 5 here) that in any ring, the nilradical is the intersection of all its prime ideals. ]

**Step 4: is nilpotent.**

We claim for some *N* > 0. Note that this is not a trivial result: since is the nilradical, for each we can find *N* such that but we have to find an *N* which works for all *x*.

For that take . Since *A* is artinian for some *N*>0. Write for this ideal so . It remains to show .

If not, among all ideals such that , pick a minimal . Pick any such that . Since minimality of forces . Also

by minimality of again. Thus for some . Since *y* is nilpotent we have for some *N*; hence , a contradiction.

**Step 5: An artinian ring A is noetherian.**

Continuing step 4, we now have

Each is an artinian *A*-module. Furthermore is a vector space over ; by exercise B.2 here this is artinian and hence finite-dimensional. Thus is a noetherian module over and hence over *A* (again by exercise B.2 here). This shows that *A* is noetherian.

# Noetherian + Dim 0 ⟹ Artinian

**Step 6: A noetherian ring A of Krull dimension 0 is artinian.**

Since the connected components of Spec *A* are all singleton sets, *A* is a direct product of noetherian rings , where each has a unique prime ideal. It suffices to prove each is artinian so we assume *A* is noetherian and has a unique prime ideal .

Since *A* is noetherian, has a finite generating set . There is an *M* > 0 such that for each *i*. It follows that . [ Indeed this ideal is generated by all over all , . Since for some *i*, we have . ]

Let *N* = *nM* so that . As above, form the sequence

Each is now a noetherian vector space over so it is finite-dimensional. Thus it is also artinian over and hence over *A*. This shows that *A* is artinian. ♦

This completes our proof. Together with the previous article, we have:

Corollary 1.Every artinian ring has a composition series and thus a well-defined length and set of composition factors.

Although all artinian rings are noetherian, there are artinian modules which are not noetherian, as we saw in the previous article.

**Optional Note**

For non-commutative rings, it is also true that a (left) artinian ring is (left) noetherian, but its proof is much more involved.

# Special Case

Now suppose *A* is a *reduced *artinian ring. We factor

as above, where each is an artinian ring with a unique prime ideal . Since each is also reduced, its nilradical is zero so is a field. Hence we have shown:

Corollary 2.The ring A is reduced and artinian if and only if it is isomorphic to a finite product of fields.

We also have the following special case.

Corollary 3.Let A be an algebra over a field k such that as a vector space. Then A is noetherian, and

where each has a unique prime ideal . Furthermore,

- is the number of prime ideals of A;
- if A is reduced then each is a finite field extension of k;
- if A is reduced and k is algebraically closed, then each so .

Note that in the context of algebraic geometry, if for an affine *k*-scheme *V* (*k* algebraically closed), then each corresponds to a point .