Noetherian and Artinian Rings and Modules

We saw the case of the semisimple ring R, which is a (direct) sum of its simple left ideals. Such a ring turned out to be nothing more than a finite product of matrix algebras. One asks if there is a more intrinsic property of R which makes it semisimple. Heuristically, it turns out that the ring must be “suitably small” and must not have any “obstructing elements”. We will consider the first property in this article, i.e. what does it mean for a ring to be “small”? [ The other property will be discussed in the next article. ]

Again, R denotes a ring, possibly non-commutative. All modules are left modules. We begin with a:

Theorem. For an R-module M, the following are equivalent:

any non-empty collection Σ of submodules of M has a maximal element N (i.e. N ∈ Σ, and whenever M’ ∈ Σ we have M’ ⊆ N);

for any increasing sequence $M_0 \subseteq M_1 \subseteq M_2 \subseteq \ldots$ of submodules of M, there is an n such that $M_n = M_{n+1} = M_{n+2} = \ldots.$ We say that the sequence is eventually constant.

Proof

⇒: assume the first property; given $M_0 \subseteq M_1\subseteq \ldots$ , let Σ be the collection of all M_n. This has a maximal element, say M_n∈Σ. Being maximal, all subsequent terms $M_{n+1}, M_{n+2}, \ldots$ must be equal to M_n.

⇐ : suppose Σ is non-empty and has no maximal element; pick M₀∈Σ; this is not maximal, so we can pick M₁∈Σ which properly contains M₀; again this is not maximal, so pick M₂∈Σ properly containing M₁; repeat. ♦

Definition. A module M which satisfies the two properties in the above theorem is said to be (left) noetherian. A ring is (left) noetherian if it is noetherian as a module over itself.

The following result is a basic property of noetherian modules.

Theorem.

If M is noetherian, so is any submodule and quotient module of M.

Conversely, if N ⊆ M is such that N and M/N are noetherian, then so is M.

Proof

First statement: let N ⊆ M. Any increasing sequence of submodules of N is also an increasing sequence of submodules of M, so it must terminate. Similarly, any increasing sequence of submodules of M/N corresponds to a sequence of submodules of M containing N, so it must terminate.

Second statement: let (M_n) be an increasing sequence of submodules of M. Then (N ∩ M_n) is an increasing sequence of submodules of N so it is eventually constant. Also, ((N+M_n)/N) is an increasing sequence of submodules of M/N so it is eventually constant. So for large n, we have:

$M_n \subseteq M_{n+1},\quad N\cap M_n = N\cap M_{n+1}, \quad N + M_n = N + M_{n+1}.$

This implies M_n = M_n+1. [ Proof : if x∈M_n+1, then by third equality x = y+z for y∈N and z∈M_n. So y = x–z is in N ∩ M_n+1 = N ∩ M_n, and x–z ∈ M_n means x∈M_n. ] So (M_n) is eventually constant. ♦

Corollary.

If M, N are noetherian, so is their direct sum M ⊕ N.

If M, N are noetherian submodules of P, so is M+N.

If M is a finitely generated module over a noetherian ring, then M is noetherian.

Proof

Indeed, M ⊆ M⊕N is a submodule whose quotient is isomorphic to N. Since M and N are noetherian, so is M⊕N. The second statement follows from that M+N is a quotient of M⊕N.

For the third statement, let M be generated by $x_1, \ldots, x_n$ . Then M is a sum of Rx_i, as submodules of M. Each Rx_i is a quotient of the form R/I for some left ideal I ⊂ R; since R is noetherian, so is R/I, and M. ♦

Examples

1. A simple module is noetherian since it has only two submodules. Thus a finitely generated semisimple module is noetherian. [#] In particular, a semisimple ring is noetherian.

[#] Subtle point: show that a finitely generated semisimple module M must be a direct sum of finitely many simple submodules. Warning: even if M is generated by k elements, it is not true that M is a direct sum of k or less simple submodules. E.g. as Z-module, Z/6 is generated by 1 element but Z/6 = Z/2 ⊕ Z/3.

2. The Z-module Z is noetherian, i.e. Z is a noetherian ring. Thus, a finitely generated abelian group is a noetherian Z-module.

3. The Z-module Q is not noetherian, for we have an infinite increasing sequence Z ⊂ (1/2)Z ⊂ (1/4)Z ⊂ … . This example also shows that $M := \{\frac a {2^m} : a, m \in\mathbf{Z}\}$ is not noetherian. Since Z is noetherian, it implies M/Z is non-noetherian.

4. The Q-module Q is obviously noetherian though. More generally, all division rings are noetherian.

5. Z[√2] is a finitely generated Z-module, so it is noetherian as a Z-module. This implies it is a noetherian ring, since every (left) ideal of Z[√2] is also a Z-module.

6. The infinite polynomial ring $\mathbf{R}[x_1, x_2, \ldots] := \cup_{n\ge 1} \mathbf{R}[x_1, \ldots, x_n]$ is a non-noetherian ring since the sequence of ideals $(x_1) \subset (x_1, x_2) \subseteq \ldots$ never terminates.

Artinian Modules and Rings

Reversing the direction of inclusion in the definition of noetherian rings, we get a similar concept. We will merely state the results since the proofs are identical to the above.

Theorem. For an R-module M, the following are equivalent:

any non-empty collection Σ of submodules of M has a minimal element N (i.e. N ∈ Σ, and whenever M’ ∈ Σ we have M’ ⊇ N);

for any decreasing sequence $M_0 \supseteq M_1 \supseteq M_2 \supseteq \ldots$ of submodules of M, there is an n such that $M_n = M_{n+1} = M_{n+2} = \ldots.$

Definition. A module which satisfies the above two properties is said to be (left) artinian. A ring is (left) artinian if it is artinian as a module over itself.

Again we have the following basic property.

Theorem.

If M is artinian, so is any submodule and quotient module of M.

Conversely, if N ⊆ M is such that N and M/N are artinian, then so is M.

Corollary.

If M, N are artinian, so is their direct sum M ⊕ N.

If M, N are artinian submodules of P, so is M+N.

A finitely generated module over an artinian ring is also artinian.

Examples

1. A simple module is artinian since it has only two submodules. Thus, a finitely generated semisimple module is artinian. In particular a semisimple ring is noetherian and artinian!

2. The Z-module Z is not artinian since it contains an infinite decreasing sequence of left ideals Z ⊃ 2Z ⊃ 4Z ⊃ … .

3. The module $M := \{\frac a {2^m} : a, m \in\mathbf{Z}\}$ is not artinian since it contains Z; however, M/Z is artinian! The proof is left as an exercise.

Easy Exercises

Prove that if R is a noetherian (resp. artinian) ring, then for any two-sided ideal I, R/I is also noetherian (resp. artinian).

Prove that if R and S are noetherian (resp. artinian) rings, so is R × S.

Summary. Noetherian and artinian modules are both concepts of “finite” modules. Finite sums, submodules and quotients of noetherian modules are noetherian. A finitely generated module over a noetherian ring is noetherian. All the above holds when we replace “noetherian” with “artinian”.

In case you missed the above examples, let us reiterate that semisimple rings are noetherian and artinian.

Finally, to further emphasize the fact that noetherian / artinian modules are the correct analogy for “finite”, we have the following important lemma. Recall that for a finite set X, a function f : X → X is bijective ⇔ f is injective ⇔ f is surjective. Likewise:

Lemma. Let M be a noetherian and artinian module. The following are equivalent for a module map f : M → M.

f is bijective;

f is injective;

f is surjective.

Proof

Suppose f is injective. We get

$\text{im} f\supseteq \text{im} f^2 \supseteq \text{im} f^3 \supseteq \ldots.$

Since M is artinian, eventually $\text{im} f^k = \text{im} f^{k+1}.$ To prove that f is surjective, let x∈M. Then $f^k(x) \in \text{im} f^k = \text{im} f^{k+1}$ so $f^k(x) = f^{k+1}(y)$ for some y∈M. Now $f^k(x - f(y)) = 0$ and since f is injective we have x = f(y) ∈ im f.

The case where f is surjective ⇒f is injective, is left as an exercise for the reader. Hint: replace im with ker and you get an increasing sequence. ♦

Subtleties on Noetherian and Artinian

In the above examples, we saw that a noetherian module may not be artinian, and vice versa. But when it comes to rings, an artinian ring must be noetherian! [ Hopefully we will eventually get around to proving this. ] The apparent asymmetry is rather surprising at first glance, but it may be partially explained by the following heuristics.

Suppose R is a commutative ring which is artinian (thus, all left ideals are two-sided). If we let ∑ be any collection of ideals of R, then the collection of products $I_1 I_2 \ldots I_n$ of ideals from ∑ has a lower bound, so eventually $I_1 I_2 \ldots I_n = I_1 I_2\ldots I_{n+1} = I_1 I_2 \ldots I_{n+2} = \ldots.$ This suggests that R has only finitely many ideals, so the artinian condition is a rather strong one.

Let us mention a result for noetherian modules which has no parallel for artinian ones.

Theorem. An R-module M is noetherian ⇔ all its submodules are finitely generated.

Proof

⇒: since a submodule of M is noetherian, it suffices to show a noetherian module is finitely generated. Now, if M is noetherian, let ∑ be the collection of all finitely generated submodules of M. This has a maximal N∈∑, which is finitely generated. If N≠M, pick x in M outside N; then N + Rx is a finitely generated submodule of M which is strictly bigger than N, contradicting its maximality. Hence N=M, so M is finitely generated.

⇐: take any increase sequence $M_0 \subseteq M_1 \subseteq \ldots$ of submodules of M. Let $N := \cup_{n\ge 0} M_n$ , which is a submodule of M, so it is finitely generated by, say, $x_1, \ldots, x_k.$ Since there are only finitely many x_i, some M_n must contain all of them, but this means M_n = N so $M_n = M_{n+1} = \ldots.$ ♦

Left and Right Modules

Finally, note that we’ve been talking about left modules throughout, but we can also define the concept of noetherian and artinian for right modules. [ Or just note that a right R-module is the same as a left R^op-module. ] You may be surprised to learn that a left noetherian ring is not necessarily right noetherian. In fact, here we have a ring which is left noetherian and left artinian, but neither right noetherian nor right artinian!

$R = \begin{pmatrix} \mathbf{Q} &0 \\ \mathbf{R} &\mathbf{R}\end{pmatrix} = \left\{\begin{pmatrix} a & 0 \\ c & d\end{pmatrix} : a\in\mathbf{Q}, c,d\in\mathbf{R}\right\}.$

Proof

It’s not right artinian or right noetherian because it has right ideals of the form $\begin{pmatrix} 0 & 0 \\ A & 0\end{pmatrix}$ where A is a subspace of R as a Q-vector space. It is easy to see that the collection of such subspaces has no maximal or minimal element.

On the other hand, R is a direct sum of left ideals

$R = \overbrace{\begin{pmatrix} \mathbf{Q} & 0 \\ \mathbf{R} & 0\end{pmatrix} }^I \oplus \overbrace{\begin{pmatrix} 0 & 0 \\ 0 & \mathbf{R}\end{pmatrix}}^J.$

Clearly J is simple. And I has a simple submodule $I' := \begin{pmatrix} \mathbf{Q} & 0\\ 0 & 0\end{pmatrix},$ and the resulting quotient is I/I’ is isomorphic to J. Since I’, I/I’ and J are simple, they’re noetherian and artinian. Thus R is left noetherian and left artinian. ♦