I. Artin Rings

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• Finite Length Modules
• Exercises

Finite Length Modules

Representation theory of algebras typically refers to the study of the modules of a finite-dimensional algebra over a field. If we investigate algebras over rings, then we cannot use the idea of dimension anymore. The appropriate generalization of finite-dimensionality is finite length which immediately prompts a couple of questions:

1. Finite-dimensional vector spaces have a well-defined dimension. Is this true of finite-length modules, as well?
2. If $0 \to A \to B \to C \to 0$, is an SES of finite-dimensional vector spaces, then $\dim A + \dim C = \dim B$. Is this true of finite-length modules as well?

The fact that this both of these facts (among other things) are true is one of many reasons why we use length. Throughout, let $\Lambda$ be a unital ring.

[definition=%counter%] Let $A$ be a $\Lambda$-module. $A$ is finite length if there is a finite filtration $F$ of submodules

such that $A_i/A_{i+1}$ is zero or simple for $i = 0, \ldots, n- 1$.

The filtration $F$ is called a generalized composition series of $A$ and the nonzero quotient modules $A_i/A_{i+1}$ are the composition factors of $F$. If none of the quotients $A_i/A_{i+1}$ are zero, then $F$ is a composition series for $A$. [/definition]

[example=%counter% (Vector Spaces)] Any finite-dimensional $k$-vector space $V$ is, of course, finite length. Some example filtrations of $k^3$:

$F_1$ is a generalized composition series since $k^3/k^3 = 0$ and $F_2$ is a composition series as all its quotients are $k$ which is trivially simple as a $k$-vector space. [/example]

[example=%counter%] As a $\ZZ$-module, $\ZZ/8\ZZ$ is finite length as a $\ZZ$-module because it has the composition series

The composition factors are

Since $\ZZ/2\ZZ$ is cyclic of prime order as an abelian group, it is simple as a $\ZZ$-module. [/example]

In the case of vector spaces in general, suppose $V = k^n$ for a field $k$. Then a composition series of $k^n$ is of the form

This is really a statement about how we can always extend a basis of a subspace of $k^n$ to a basis of the entirety of $k^n$.

[definition=%counter% (Length)] For a simple $\Lambda$-module $S$, we define

And we define

where the sum is taken over all (isoclasses of) simple $\Lambda$-modules. We define the length of $A$ to be

and

where the minima is taken over composition series $F$ of $A$.
[/definition]

[example=%counter%] In the case of vector spaces, the simple modules are the $1$-dimensional vector spaces and they are naturally isomorphic in the obvious way. Let $V$ be an $n$-dimensional $k$-vector space. Then $l(V) = n$. [/example]

[proof=Proof of $(l(V) \le n)$] Without loss of generality, we will let $V = k^n$. Then there is a composition series

Let $F$ denote this particular filtration. Then

Since each $k^i/k^{i-1} \cong k$ for each $i = 1, \ldots, n$, it follows that $m_k^F(k^n) = n$. Since $l(V)$ is given by the minimum possible $l_F(V)$, it must be the case that $l(V) \le n$. [/proof]

[proof=Proof of $(l(V) < n)$ is impossible] Note that there is nothing to show if $n = 0$ so assume $n > 0$. For the sake of contradiction, suppose that there exists some filtration $F$ of the form

where $l_F(k^n) < n$. Since

So then $m = l_F(k^n) < n$. Since $n > 0$, it is impossible for $m = 0$ because that would imply that $k^n = V_m = 0$ which would contradict $k^n$ being $n$-dimensional. So assume $m > 0$. Now since $V_0/V_1 \cong k$, it follows that

So $\dim V_1 = n-1$. Then we can use the same computation to show that $\dim V_2 = n-2$, and so on, until we show that $V_m = n - m$. But since $V_m = 0$, $n - m = 0$ which is a contradiction. Thus, $l(V) = n$ and we are done. [/proof]

In the example given above, we were implicitly making use of the fact that

As it turns out, this holds much more generally in the context of length and we call that fact the Jordan-Hölder theorem. Before we move onto that fact, we will consider filtrations induced from SES.

Consider the following exact sequence of $\Lambda$-modules:

Let $F$ be a generalized composition series

of $B$. This filtration induces filtrations $F’$ of $A$ given by

and $F’’$ of $C$ given by

[proposition=%counter%] Let notation be as above. Then:

(a) The filtrations $F’$ of $A$ and $F’’$ of $C$ are generalized composition series.

(b) For each simple module $S$ we have

(c) $l_{F’}(A) + l_{F’’}(C) = l_F(B)$. [/proposition]

[proof] We claim that for each $i = 0, \ldots, n$, the sequence

is exact. By definition, the above sequence is

Since $f$ is injective, its restriction to $A_i = f^{-1}(B_i)$ is as well and clearly $g$ is surjective from $B_i$ onto $g(B_i)$. So it remains to show that we have exactness at $B_i$. Recall that $\ker g = \im f$. So certainly we have that $g|_{B_i} \circ f|_{A_i} = 0$. Thus, $\ker f|_{A_i} \subseteq \ker g|_{B_i}$. Now if $b \in \ker g|_{B_i}$ we certainly have that $b \in \ker g = \im f$. Therefore, there exists $a \in A$ for which $f(a) = b$. In other words, $a \in f^{-1}(b) \subseteq f^{-1}(B_i) = A_i$. Thus, we have exactness at $B_i$ and we are done.

Now we claim that the for each $i = 0, \ldots, n - 1$ the following diagram

is commutative and exact. The exactness across the top rows is what we just proved and exactness across the columns is obvious.

To see commutativity, we rewrite the diagram as

from which commutativity is obvious. To see exactness of the sequence

we do a diagram chase. For starters, if $\overline{f|_{A_i}}(\overline{a}) = 0$, then commutativity of the diagram above implies that

Exactness across the middle column implies there is $b \in B_{i+1}$ for which $b = f(a)$. But $g\circ f = 0$ so $b \in \ker g$ so there is $a’ \in A_{i+1}$ such that $f(a’) = b$. That is, $f(a’) = f(a)$ and injectivity of $f$ implies $a’ = a$. So $a \in A_{i+1}$ which implies $\overline{a} = 0$ by exactness across the first column. Similar diagram chases show exactness for the rest of the sequence. Now we prove (a), (b), and (c).

(a) Note that if $B_i/B_{i+1} = 0$, then exactness across the bottom row implies that $A_i/A_{i+1} = C_i/C_{i+1} = 0$. If $B_i/B_{i+1}$ is simple, then either:

1. $A_i/A_{i+1} \cong B_i/B_{i+1}$ which implies that $C_i/C_{i+1} = 0$; or
2. $A_i/A_{i+1} = 0$ which implies that $B_i/B_{i+1} \cong C_i/C_{i+1}$.

So the quotient modules in $F’$ and $F’’$ are all zero or simple as desired.

(b) Let $S$ be a simple module and suppose that $B_{i+1}/B_{i} \cong S$. By the above argument, either $A_{i+1}/A_i \cong S$ and $C_{i+1}/C_{i} = 0$ or $A_{i+1}/A_{i} = 0$ and $C_{i+1}/C_{i} \cong S$ (note the converse is true as well, so it suffices to just look at $B_{i+1}/B_i \cong S$). So the claim follows.

(c) Immediate by part (b). [/proof]

In linear algebra, we know that we can always pick two different bases for a vector space $V$ and they are guaranteed to have the same cardinality. Ideally, if length is meant to be a generalization of this fact, then the choice of composition series should be immaterial when it comes to computing length. This is, in fact, actually true and is known as the Jordan-Hölder theorem.

[theorem=%counter% (Jordan-Hölder)] Let $B$ be a $\Lambda$-module of finite length, and $F$ and $G$ two composition series for $B$.

(a) For each simple $\Lambda$-module $S$, we have

(b) $l_F(B) = l_G(B) = l(B)$. [/theorem]

(a)

Exercises