# Artinian Modules

Instead of the ascending chain condition, we can take its reverse.

Definition.

Let M be an A-module. Consider the set $\Sigma$ of submodules of M, ordered by inclusion, i.e. $N \le N'$ if and only if $N\subseteq N'$. We say M is artinian if $\Sigma$ is noetherian.

The ring A is said to be artinian if it is artinian as a module over itself.

Again M is artinian if either of the following equivalent conditions holds.

• Every non-empty collection of submodules of M has a minimal element.
• If $N_0 \supseteq N_1 \supseteq N_2 \supseteq \ldots$ is a sequence of submodules of M, then $N_k = N_{k+1} = \ldots$ for some $k\ge 0$.

### Examples

Let $A = \mathbb Z$. This is a noetherian ring as we saw; it is not artinian because we have a decreasing sequence of ideals $(2) \supset (4) \supset (8) \supset \ldots$.

Let $M = B / \mathbb Z$ as a $\mathbb Z$-module where $B = \{\frac a {2^k} \in \mathbb Q : a, k \in \mathbb Z\}$. Then M is artinian but not noetherian (exercise). We thus see that an artinian module is not finitely generated in general.

Exercise A

1. Prove that in an exact sequence of A-modules: $0 \longrightarrow N \longrightarrow M \longrightarrow P \longrightarrow 0$,

M is artinian if and only if N and P are. In particular, the direct sum of two artinian modules is artinian.

2. Prove that if M is a noetherian A-module, any surjective linear $f:M\to M$ is also injective. State and prove the dual statement for an artinian module.

Decide if each of the following statements is true.

3. The product of any two artinian rings is artinian.

4. Any quotient of an artinian ring is artinian.

5. Any localization of an artinian ring is artinian. # Simple Modules

Let A be a fixed ring throughout this article and M be an A-module.

Definition.

M is said to be simple if it is non-zero and has no submodules except 0 and itself.

We immediately have the following.

Lemma 1.

M is simple if and only if $M \cong A/\mathfrak m$ for some maximal ideal $\mathfrak m \subset M$.

Proof

(⇒) Pick a non-zero $m\in M$ and consider the A-linear map $f:A\to M$, $a \mapsto am$. Its image is a non-zero submodule of M so it is the whole M; hence f is surjective. We have $A / \mathrm{ker} f \cong M$ and submodules of M correspond to ideals of A containing ker f. Thus ker f is maximal.

(⇐) Submodules of $A/\mathfrak m$ correspond to ideals of A containing $\mathfrak m$. Thus M is simple ⟹ $\mathfrak m$ is maximal. ♦

Example

Thus we see that simple modules over a coordinate ring k[V] (k algebraically closed) correspond to points on V. Even though we have $A/\mathfrak m_P \cong k$ as k-algebras, different points correspond to different A-modules!

Easy Exercise

Prove that for any ideals $\mathfrak a, \mathfrak b$ of A, $A / \mathfrak a \cong A/\mathfrak b$ as A-modules if and only if $\mathfrak a= \mathfrak b$. # Composition Series

Next we would like to “factor” a module into its constituents comprising of simple modules.

Definition.

composition series for M is a sequence of submodules $0 = M_0 \subsetneq M_1 \subsetneq M_2 \subsetneq \ldots \subsetneq M_n = M$

such that $M_{i+1}/M_i$ is simple for all $i=0, \ldots, n-1$. The length of the composition series is n; the composition factors of the series are the isomorphism classes of $M_{i+1}/M_i$, treated as a multi-set.

Example

Let $V = \mathbb A^2(\mathbb C)$ and $W_1, W_2 \subset V$ be varieties cut out by the equations $f = Y - X^3 + 3X$ and $g = Y - 2$ respectively. Recall that their scheme intersection (exercise D here) has coordinate ring \begin{aligned} A := k[W_1] \otimes_{k[V]} k[W_2] &\cong k[X, Y]/(Y - X^3 + 3X, Y - 2)\\ &\cong k[X]/(X^3 - 3X - 2)\\ &\cong k[X] / (X+1)^2 \times k[X] / (X-2) \end{aligned}

by Chinese Remainder Theorem. In fact, the above are all isomorphisms as algebras over $k[V] = k[X, Y]$ where we have identified $k[X]$ with $k[X, Y]/(Y)$. Thus we have the following composition series $0 \subset (X+1)^2 A \subset (X+1) A \subset A.$

This has length 3; the composition factors are two copies of $A/(X+1)$ and one copy of $A/(X-2)$.

Proposition 1.

M has a composition series if and only if it is a noetherian and artinian module.

Proof

(⇒) Suppose $0 = M_0 \subsetneq M_1 \subsetneq \ldots \subsetneq M_n = M$ is a composition series. Each $M_{i+1}/M_i$ is simple hence both artinian and noetherian. By induction one shows that all $M_i$ are noetherian and artinian.

(⇐) Conversely suppose M is both noetherian and artinian; we may assume $M\ne 0$. If M has no simple submodule, we can find an infinitely decreasing sequence of submodules $M_1 \supsetneq M_2 \supsetneq \ldots$, contradicting the fact that M is artinian. Hence we pick any simple submodule $M_1 \subseteq M$. If $M = M_1$ we are done; otherwise, we repeat the process with $M/M_1$ and obtain a simple submodule $M_2/M_1 \subseteq M/M_1$. This gives the next term for a composition series $0\subsetneq M_1 \subsetneq M_2 \subseteq M$, etc. Since M is noetherian, the process must eventually terminate. ♦

Corollary 1.

If M has a composition series, so do any submodule and quotient module of M. # Uniqueness of Composition Series

Theorem 1.

Let M be a noetherian and artinian A-module. The composition factors of any composition series are unique up to isomorphism and reordering.

Note

For a composition series $M_0 \subset M_1 \subset \ldots \subset M_n$ we write $\mathrm{CF}(M_0, \ldots, M_n)$ for the multi-set of its composition factors.

Proof

The proof is by contradiction; say a module is bad if it has two composition series with different sets of composition factors.

Step 1: Assume M is minimal.

Take the collection Σ of all bad submodules of M. Since M is artinian, we can replace M by a minimal element of Σ. Thus M has two composition series $0 = M_0 \subset M_1 \subset \ldots \subset M_m = M, \quad 0 = N_0 \subset N_1 \subset \ldots \subset N_n = M$

such that $\mathrm{CF}(M_0, \ldots, M_m) \ne \mathrm{CF}(N_0, \ldots, N_n)$; and no other submodule of M has this property.

Step 2: Consider the easy case.

Clearly $M\ne 0$ so $m,n \ge 1$. If $M_{m-1} = N_{n-1}$ then by minimality this is a good module so $M_0 \subset \ldots \subset M_m$ and $N_0 \subset \ldots \subset N_n$ have the same composition factors, a contradiction.

Step 3: Do the “diamond argument”.

Otherwise we have $M_{m-1} + N_{n-1} = M$ which gives us $M / M_{m-1} = (M_{m-1} + N_{n-1})/M_{m-1} \cong N_{n-1} / (M_{m-1} \cap N_{n-1})$

and similarly $M / N_{n-1} \cong M_{m-1} / (M_{m-1} \cap N_{n-1})$.

Step 4: Apply induction.

Pick any composition series $0 = P_0 \subset \ldots \subset P_k = M_{m-1} \cap N_{n-1}$ for $M_{m-1} \cap N_{n-1}$. Using (P) to denote the isomorphism class of a module P we have: \begin{aligned} \mathrm{CF}(M_0, \ldots, M_m) &= (M/M_{m-1}) + \mathrm {CF}(M_0, \ldots, M_{m-1}) \\ &= (M/M_{m-1}) + (M_{m-1} / (M_{m-1} \cap N_{n-1})) + \mathrm{CF}(P_0, \ldots, P_k) \\ &= (M/N_{n-1}) + (N_{n-1} / (M_{m-1} \cap N_{n-1})) + \mathrm{CF}(P_0, \ldots, P_k)\\ &= (M/N_{n-1}) + \mathrm{CF}(N_0, \ldots, N_{n-1}) \\ &= \mathrm{CF}(N_0, \ldots, N_n).\end{aligned}

The second and fourth equalities follow from the fact that $M_{m-1}$ and $N_{n-1}$ are not bad, since M is a minimal bad module. The third equality follows from step 3. In diagram form we have the following. This completes our proof. ♦

Corollary 2.

If M is a noetherian and artinian module, write $l(M)$ (length of M) for the length of any composition series of M. This is a well-defined value.

Similarly, we can define the composition factors of M and denote it by $\mathrm{CF}(M)$.

Note that if $0 \to N \to M \to P \to 0$ is an exact sequence of A-modules and M has finite length, then $l(M) = l(N) + l(P), \quad \mathrm{CF}(M) = \mathrm{CF}(N) + \mathrm{CF}(P).$

This gives another application of short exact sequences.

Exercise B

1. Compute the length and composition series of $B = \mathbb R[X]/ (X^6 + X^2)$

as an B-module.

2. Prove that for a ring quotient $B = A/\mathfrak a$, a B-module M is noetherian (resp. artinian) as a B-module if and only if it is so as an A-module.

In the next article, we will look at artinian rings. It turns out this is a small subclass of the collection of noetherian rings.

Optional Note

All the results in this section can be generalized to left modules over a non-commutative ring. This has applications in representation theory, since a simple module over the group algebra $k[G]$ (k = any field) corresponds to an irreducible k-representation of G. This entry was posted in Advanced Algebra and tagged , , , , , . Bookmark the permalink.