# 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 $\dim A = 0$, where $\dim$ denotes the Krull dimension.

Note

Recall that $\dim A = 0$ 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 $x \in A - \{0\}$. From the sequence $(x) \supseteq (x^2) \supseteq (x^3) \supseteq \ldots$ we have $(x^n) = (x^{n+1})$ for some n > 0. Thus $x^n = x^{n+1}y$ for some $y\in A$. Since A is a domain and $x\ne 0$ we have $xy = 1$ so x is a unit.

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

Let $\Sigma$ be the collection of ideals of A of the form $\mathfrak m_1 \cap \ldots \cap \mathfrak m_k$ where $\mathfrak m_i \subset A$ are maximal ideals. Since A is artinian $\Sigma$ has a minimal element $\mathfrak a = \mathfrak m_1 \cap \ldots \cap \mathfrak m_k$. By minimality for any maximal ideal $\mathfrak m$ we have $\mathfrak a \cap \mathfrak m = \mathfrak a$ and so $\mathfrak a \subseteq \mathfrak m$. We claim that $\mathfrak m = \mathfrak m_i$ for some $1\le i \le k$ which would complete our claim.

Since $\mathfrak m$ and $\mathfrak m_i$ are all maximal it suffices to show $\mathfrak m \supseteq \mathfrak m_i$ for some i. Indeed if not we can pick $x_i \in \mathfrak m_i - \mathfrak m$ for $i=1,\ldots, k$. Then

$x_1 \ldots x_k \in \mathfrak m_1 \ldots \mathfrak m_k \subseteq \mathfrak m_1 \cap \ldots \cap \mathfrak m_k = \mathfrak a \subseteq \mathfrak m$

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

For any prime ideal $\mathfrak p \subset A$, $A/\mathfrak p$ is an artinian integral domain, hence a field by step 1. Thus $\mathfrak p$ 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 $A_1 \times \ldots \times A_k$, where each $A_i$ 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 $\mathfrak p$ 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: $\mathfrak p$ is nilpotent.

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

For that take $\mathfrak p \supseteq \mathfrak p^2 \supseteq \ldots$. Since A is artinian $\mathfrak p^N = \mathfrak p^{N+1} = \ldots$ for some N>0. Write $\mathfrak a$ for this ideal so $\mathfrak a = \mathfrak a^2 = \ldots$. It remains to show $\mathfrak a = 0$.

If not, among all ideals $\mathfrak b$ such that $\mathfrak {ab} \ne 0$, pick a minimal $\mathfrak b$. Pick any $x\in \mathfrak b$ such that $x\mathfrak a \ne 0$. Since $(x)\mathfrak a \ne 0$ minimality of $\mathfrak b$ forces $\mathfrak b = (x)$. Also

$(x\mathfrak a)\mathfrak a = x\mathfrak a^2 = x\mathfrak a \ne 0 \implies x\mathfrak a = \mathfrak b = (x)$

by minimality of $\mathfrak b$ again. Thus $x = xy$ for some $y\in \mathfrak a$. Since y is nilpotent we have $y^N = 0$ for some N; hence $x=0$, a contradiction.

Step 5: An artinian ring A is noetherian.

Continuing step 4, we now have

$0 = \mathfrak p^N \subseteq \mathfrak p^{N-1} \subseteq \ldots \subseteq \mathfrak p \subseteq A.$

Each $\mathfrak p^i/\mathfrak p^{i+1}$ is an artinian A-module. Furthermore $\mathfrak p^i / \mathfrak p^{i+1}$ is a vector space over $A/\mathfrak p$; by exercise B.2 here this is artinian and hence finite-dimensional. Thus $\mathfrak p^i / \mathfrak p^{i+1}$ is a noetherian module over $A/\mathfrak p$ 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 $A_1, \ldots, A_k$, where each $A_i$ has a unique prime ideal. It suffices to prove each $A_i$ is artinian so we assume A is noetherian and has a unique prime ideal $\mathfrak p$.

Since A is noetherian, $\mathfrak p$ has a finite generating set $x_1, \ldots, x_n$. There is an M > 0 such that $x_i^M = 0$ for each i. It follows that $\mathfrak p^{nM} = 0$. [ Indeed this ideal is generated by all $t = \prod x_i^{m_i}$ over all $m_1 + \ldots + m_n = nM$, $m_i \ge 0$. Since $m_i \ge M$ for some i, we have $t=0$. ]

Let N = nM so that $\mathfrak p^N = 0$. As above, form the sequence

$0 = \mathfrak p^N \subseteq \mathfrak p^{N-1} \subseteq \ldots \subseteq \mathfrak p \subseteq A.$

Each $\mathfrak p^i/\mathfrak p^{i+1}$ is now a noetherian vector space over $A/\mathfrak p$ so it is finite-dimensional. Thus it is also artinian over $A/\mathfrak p$ 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

$A = A_1 \times \ldots \times A_k$

as above, where each $A_i$ is an artinian ring with a unique prime ideal $\mathfrak p_i$. Since each $A_i$ is also reduced, its nilradical $\mathfrak p_i$ is zero so $A_i$ 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 $\dim_k A < \infty$ as a vector space. Then A is noetherian, $\dim A = 0$ and

$A = A_1 \times \ldots \times A_m$

where each $A_i$ has a unique prime ideal $\mathfrak p_i$. Furthermore,

• $m \le \dim_k A$ is the number of prime ideals of A;
• if A is reduced then each $A_i$ is a finite field extension of k;
• if A is reduced and k is algebraically closed, then each $A_i \cong k$ so $m = \dim_k A$.

Note that in the context of algebraic geometry, if $A = k[V]$ for an affine k-scheme V (k algebraically closed), then each $A_i$ corresponds to a point $P\in V$.

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