Composition factors and uniform chain bounds

Written by GPT-6.1 Sol (OpenAI), reasoning effort Ultra, October 2026. Self-checked by the writing AI; no independent review. Public domain (CC0).

A finite composition series does more than provide a list of simple quotients. It bounds the number of strict changes in every other chain of subobjects. The bound comes from intersecting a subobject with the chosen series and checking which simple factors survive. This gives a useful bridge between exact sequences, chain conditions and directed families.

We use the abelian framework of Trace coreflections and balanced nonabelian categories, and ambient kernels, cokernels and short exact sequences as in Full subcategories and exact closure. The canonical existence theorem is Stacks, Lemma 12.9.6; the stronger uniqueness theorem is Stacks, Jordan–Hölder. For modules over commutative rings, the complete arguments in Sections 1–3 below supply the finite-length, submodule, quotient and exact-extension foundation; the stronger Jordan–Hölder theorem remains at the cited complete Stacks proof. Here the category may be any abelian category, and the application concerns its subobject chains.

Fix universes \(\mathcal U\in\mathcal V\), with the natural numbers in \(\mathcal U\). Let \(\mathcal A\) be abelian with \(\mathcal U\)-small Hom sets, and house its objects and morphisms in \(\mathcal V\). This ambient convention permits ordinary choice when forming countable chains. It imposes no \(\mathcal U\)-smallness on the objects of \(\mathcal A\) or on the full collection of subobjects of an object. The families below are nonempty sets in the ambient universe. No generator or infinite limit or colimit is assumed.

A subobject of \(X\) is an equivalence class of monomorphisms into \(X\), where the equivalence is an isomorphism commuting with the maps to \(X\). Write \(B\subseteq C\) when the representative of \(B\) factors through that of \(C\), and \(B\subsetneq C\) when the classes differ. All intersections are finite pullbacks in \(\mathcal A\).

1. A finite filtration detects a strict inclusion

An object \(S\) is simple when it is nonzero and has only the subobjects \(0\) and \(S\). A composition series is a finite strict filtration

whose quotients \(S_i=M_i/M_{i-1}\) are simple. The zero object has the empty series, with \(n=0\).

We first record the exact comparison needed for the finite filtration argument.

Lemma 1.1. In a commutative diagram between short exact sequences, if the maps on the first and last terms are isomorphisms, then the map on the middle terms is an isomorphism.

Proof. Write the sequences and vertical maps as

Let \(k:K\to B\) be the kernel of \(b\). The equality \(cqk=0\) and invertibility of \(c\) imply \(qk=0\). Thus \(k=it\) for a unique \(t:K\to A\). Now \(i'at=bk=0\). Since \(i'\) is monic and \(a\) is invertible, \(t=0\); hence \(k=0\) and \(b\) is monic.

If \(h:B'\to T\) satisfies \(hb=0\), then \(hi'a=hbi=0\). Cancel \(a\), and use that \(q'\) is the cokernel of \(i'\), to write \(h=zq'\). Consequently \(zcq=zq'b=0\). The map \(cq\) is epic, so \(z=0\) and \(h=0\). Apply this to the cokernel map of \(b\): its cokernel is zero, so \(b\) is epic. A monic and epic map in an abelian category is an isomorphism. \(\square\)

Fix (1.1). For a subobject \(B\subseteq X\), set

The kernel of \(B_i\to M_i\to S_i\) is \(B\cap M_{i-1}=B_{i-1}\): this is the pullback of the kernel of \(M_i\to S_i\). Coimage–image factorization therefore gives a monomorphism

Thus every \(Q_i(B)\) is zero or isomorphic to \(S_i\). Removing repetitions from the filtration \((B_i)\) gives a composition series of \(B\). This is the finite induced-filtration interface in Stacks, Lemma 12.19.12.

Define the integer by counting the nonzero pieces among \(Q_1(B),\ldots,Q_n(B)\):

Proposition 1.2. If \(B\subsetneq C\subseteq X\), then \(r(B)<r(C)\). In particular \(r(0)=0\), \(r(X)=n\), and \(0\le r(B)\le n\).

Proof. Inclusion gives maps \(Q_i(B)\to Q_i(C)\) commuting with (1.4). Their composites into \(S_i\) are monic, so the maps themselves are monic. The set of nonzero graded pieces for \(B\) is contained in that for \(C\); hence \(r(B)\le r(C)\).

Suppose equality holds. The two finite sets of indices then coincide. At every index the graded comparison is an isomorphism: either both terms are zero, or both monomorphisms into the same simple \(S_i\) are isomorphisms. Apply Lemma 1.1 successively to the short exact sequences

Starting from \(B_0=C_0=0\), induction makes every \(B_i\to C_i\) invertible. At \(i=n\) the inclusion \(B\to C\) is an isomorphism over \(X\), so the subobjects are equal. A strict inclusion therefore forces a strict inequality. The endpoint values follow directly from (1.3). \(\square\)

The comparison uses a finite filtration and its exact quotients. Its general filtered-object form is discussed in Stacks, Lemma 12.19.13.

2. Four ways to recognize finite length

Call \(X\) Noetherian if every increasing sequence of subobjects stabilizes, and Artinian if every decreasing sequence stabilizes. Stabilization means eventual equality of subobject classes; representatives need not literally be the same monomorphism.

A family \(\mathscr F\) of subobjects is filtered if every two members have a common upper bound belonging to \(\mathscr F\). It is cofiltered if every two have a common lower bound in \(\mathscr F\). Nonemptiness is part of both conventions. A greatest member contains every member; a maximal member merely has no strictly larger member in the family.

Theorem 2.1. For an object \(X\), the following conditions are equivalent:

1. \(X\) has a composition series.
2. There is an integer \(N\ge0\) bounding the length of every finite strict chain from \(X\) to \(0\).
3. \(X\) is Noetherian and Artinian.
4. Every nonempty filtered set of subobjects of \(X\) has a greatest member, and every nonempty cofiltered set has a least member.

If a composition series has \(n\) factors, the least possible bound in condition 2 is \(n\). Every composition series has that same number of factors.

Proof. A series (1.1) gives the rank \(r\) of Proposition 1.2. Along any strict descending chain

the integer \(r\) strictly decreases from \(n\) to \(0\), so \(m\le n\). This proves \(1\Rightarrow2\).

If an increasing or decreasing sequence fails to stabilize, it has strict finite subsequences of arbitrarily large length. Add \(0\) and \(X\) as endpoints, omit any repeated endpoint, and reverse the order in the increasing case. This contradicts a uniform bound on (2.1). Thus \(2\Rightarrow3\).

The implication \(3\Rightarrow1\) is the existence interface of Stacks, Lemma 12.9.6. Its chain choices can be made directly under our ambient convention. For any nonzero subobject \(Y\subseteq X\), the nonempty set of proper subobjects of \(Y\) has a maximal member: otherwise choice constructs an infinite strictly increasing sequence of proper subobjects, contrary to the ascending chain condition inherited from \(X\). A maximal proper subobject \(Z\subsetneq Y\) has simple quotient. Indeed, a nonzero proper subobject of \(Y/Z\) pulls back to a subobject strictly between \(Z\) and \(Y\). Starting at \(X\), repeatedly choose a maximal proper subobject of the current nonzero term. An infinite such process contradicts the descending chain condition. A finite process can stop only at \(0\), and all its successive quotients are simple. Reversing this descending series gives (1.1). For \(X=0\), use the empty series.

For \(3\Rightarrow4\), any nonempty set of subobjects has a maximal member under the ascending chain condition, by the same countable-choice argument. Let \(B\) be maximal in a filtered family. For any member \(C\), choose \(D\) in the family with \(B,C\subseteq D\). Maximality gives \(D=B\), hence \(C\subseteq B\); thus \(B\) is greatest. Under the descending chain condition, choose a minimal member of a cofiltered family. A common lower bound then proves that it is least.

Conversely, the members of any increasing sequence form a nonempty filtered set. Its greatest member occurs at some index \(j\); all later terms equal it. The members of a decreasing sequence form a cofiltered set, and its least member forces the same eventual equality. Thus \(4\Rightarrow3\).

Finally, if a second composition series has \(m\) factors, it is itself a chain (2.1). The bound obtained from (1.1) gives \(m\le n\). Interchanging the series gives \(n\le m\), so \(n=m\). The series (1.1) attains the bound \(n\), proving minimality. \(\square\)

Write \(\ell(X)\) for this integer, called the length. In particular \(\ell(0)=0\). The stronger canonical Jordan–Hölder theorem identifies the multiset of simple factors up to isomorphism, not just its size. Its hypotheses are exactly the finite-length conditions above; see Stacks, Lemma 12.9.7. We retain that stronger result by reference. The rank estimate supplies the additional uniform bound for arbitrary chains.

3. Exact sequences count the factors

Proposition 3.1. Subobjects and quotients of a finite-length object have finite length. In a short exact sequence

\(E\) has finite length if and only if both \(A\) and \(D\) do. In that case

Proof. For a subobject, use the intersection filtration (1.3): all successive quotients are zero or simple, and repetitions can be removed. For a quotient \(X\twoheadrightarrow Q\), let \(N_i\) be the image of \(M_i\) in \(Q\). The composite \(M_i\twoheadrightarrow N_i\twoheadrightarrow N_i/N_{i-1}\) is epic and kills \(M_{i-1}\). It induces an epimorphism \(S_i\twoheadrightarrow N_i/N_{i-1}\), whose target is zero or simple. Removing repetitions again gives a composition series.

Suppose now the ends of (3.1) have composition series. Take the series of \(A\) inside \(E\), then append the inverse images in \(E\) of the series of \(D\), starting above \(A\). Pullbacks of epimorphisms in an abelian category are epic. At each step the inverse-image quotient is therefore the corresponding simple quotient in \(D\): its quotient map has the preceding inverse image as kernel, and coimage–image factorization identifies the quotient. The concatenated series has \(\ell(A)+\ell(D)\) factors. Theorem 2.1 gives (3.2). The converse follows from subobject and quotient closure already proved. \(\square\)

For example, an object of length one is simple. In (3.1), if \(E\) has finite length and \(A\ne0\), then \(\ell(D)<\ell(E)\). This assertion uses actual exact quotients; a poset with both chain conditions need not admit a uniform finite bound, as Exercise 4 shows.

4. Graded exercises with full solutions

Exercise 1 — Introductory. Fix a field \(k\). Consider the category whose objects are linear maps \(f:V\to W\) between finite-dimensional \(k\)-vector spaces and whose morphisms are commuting squares. Determine its simple objects and the length of \(f\). Show that the identity and zero maps \(k\to k\) have the same simple-factor multiset but are not isomorphic.

Solution. Kernels and cokernels of a commuting square are computed at each vertex, with the arrow induced by commutativity. The vector-space universal properties give the square universal properties. The coimage–image maps are componentwise isomorphisms, and finite biproducts are componentwise; hence the category is abelian.

If \(W\ne0\), any line \(L\subseteq W\) gives a nonzero subobject \((0\to L)\) of \((V\xrightarrow f W)\). A simple object with \(W\ne0\) must therefore have \(V=0\) and \(\dim W=1\). If \(W=0\), simplicity is exactly \(\dim V=1\). Conversely \((k\to0)\) and \((0\to k)\) have no other subobjects, so these are all simples up to isomorphism.

There is a componentwise short exact sequence

Flags of lines in \(W\) and \(V\) give lengths \(\dim W\) and \(\dim V\) for its ends. Proposition 3.1 yields \(\ell(f)=\dim V+\dim W\), independently of the rank of \(f\). Both maps \(k\to k\) thus have one factor of each simple type. An isomorphism of arrow objects consists of invertible changes of basis at both vertices; these preserve rank. Ranks one and zero cannot be isomorphic. \(\square\)

Exercise 2 — Intermediate. In the category of abelian groups, show that \(\mathbb Z\) is Noetherian but not Artinian, and that the Prüfer group \(P=\mathbb Z[1/p]/\mathbb Z\), for a prime \(p\), is Artinian but not Noetherian. Exhibit a cofiltered family without a least member in the first group and a filtered family without a greatest member in the second.

Solution. Every nonzero subgroup of \(\mathbb Z\) has the form \(d\mathbb Z\) for a positive integer \(d\): choose its least positive element and use division with remainder. Once an increasing sequence reaches a nonzero subgroup \(d\mathbb Z\), its later positive generators divide their predecessors. A strict increase strictly decreases the positive generator, so only finitely many increases occur. A sequence remaining zero is stationary too. Thus \(\mathbb Z\) is Noetherian. The strictly decreasing chain \(\mathbb Z\supsetneq2\mathbb Z\supsetneq4\mathbb Z\supsetneq\cdots\) disproves the Artinian condition.

Put \(P_n=p^{-n}\mathbb Z/\mathbb Z\), with \(P_0=0\). Each is cyclic of order \(p^n\), and \(P=\bigcup_{n\ge0}P_n\). An element of order \(p^n\) generates \(P_n\), because its numerator is relatively prime to \(p\). If a subgroup has elements of unbounded orders, it contains every \(P_n\) and equals \(P\). Otherwise the occurring orders are bounded; choose their largest exponent \(n\), and an element of that order. The subgroup is contained in \(P_n\) and contains it, so it equals \(P_n\). The zero subgroup is included by \(n=0\).

A decreasing chain of subgroups either stays equal to \(P\), or enters some finite \(P_n\), where it stabilizes. Hence \(P\) is Artinian. The strict ascending chain \(P_0\subsetneq P_1\subsetneq P_2\subsetneq\cdots\) shows it is not Noetherian. The set \(\{2^n\mathbb Z:n\ge0\}\) is cofiltered with no least member, and \(\{P_n:n\ge0\}\) is filtered with no greatest member. The theorem requires both chain conditions. \(\square\)

Exercise 3 — Advanced. Let \(F:\mathcal A\to\mathcal B\) be an exact functor of abelian categories, and let \(X\) have simple factors \(S_1,\ldots,S_n\). Assume each \(F(S_i)\) has finite length. Prove that \(F(X)\) has finite length and compute its length from these images. Apply this to evaluation at the source vertex in Exercise 1. Give an exact functor that does not preserve finite length.

Solution. Exactness gives, at every step of a composition series,

Starting with \(F(M_0)=0\), Proposition 3.1 proves finite length inductively and gives

If each image is simple or zero, the sum counts the surviving factors. Evaluation \((V\to W)\mapsto V\) is exact because sequences of arrow objects are exact at both vertices. It sends \((k\to0)\) to \(k\) and \((0\to k)\) to zero. Thus it gives length \(\dim V\), as expected.

The forgetful functor from \(\mathbb Q\)-vector spaces to abelian groups is exact: kernels, images and quotients have the same underlying additive groups. The one-dimensional vector space \(\mathbb Q\) has length one, but its underlying group contains the strict ascending chain

The underlying group is not Noetherian and has no finite length. Exactness alone does not supply the hypothesis on the images of the simple factors. \(\square\)

Exercise 4 — Expert. Form a poset with bottom \(0\), top \(1\), and, for every \(n\ge1\), an arm

Elements on different arms are incomparable. Show that both sequential chain conditions hold, and that every nonempty filtered family has a greatest member and every nonempty cofiltered family a least member. Show that the lengths of finite strict chains are unbounded. Explain why this does not contradict Theorem 2.1 by checking a failure of modularity.

Solution. A monotone sequence can use at most one arm, apart from the common endpoints: passing to a different arm would require an increase to \(1\), or a decrease to \(0\), after which the corresponding monotone sequence is constant. Each arm is finite, so every monotone sequence stabilizes.

If a filtered family contains elements of two different arms, their only common upper bound is \(1\), which must belong to the family and is greatest. Otherwise the family, unless it already contains \(1\), lies in one finite arm together with possibly \(0\); its greatest member exists. The remaining family \(\{0\}\) is immediate. Dually, a cofiltered family meeting two arms must contain \(0\); otherwise it lies in one finite arm together with possibly \(1\), and has a least member. These statements also include singleton endpoint families.

The \(n\)-th arm has \(n+1\) strict steps from \(0\) to \(1\), so no uniform bound exists. The poset is a lattice: on one arm use its order, while the meet of elements on different arms is \(0\) and their join is \(1\). Put \(x=a_{2,1}\), \(z=a_{2,2}\), and \(y=a_{1,1}\). Then \(x\le z\), but

Thus the modular identity fails. Here is the categorical comparison for actual subobjects \(B\subseteq C\) of an object \(X\). Let \(q:X\to X/C\). The kernel of the composite \(B\oplus D\to B+D\to X/C\) is \(B\oplus(D\cap C)\), since \(q\) kills \(B\). Pulling the epimorphism \(B\oplus D\to B+D\) back along the kernel of \(B+D\to X/C\) gives an epimorphism

Its image in \(X\) is \(B+(D\cap C)\); thus \(B+(D\cap C)=(B+D)\cap C\). This proves modularity of an abelian subobject lattice, so the lattice in this exercise cannot be one. Proposition 1.2 uses the stronger exact-filtration structure to obtain the bound. \(\square\)

5. References and scope

Length of objects and the Jordan–Hölder theorem are also treated in Pavel Etingof, Shlomo Gelaki, Dmitri Nikshych and Victor Ostrik, Tensor Categories, author's final version, Section 1.5. Canonical existence and the stronger simple-factor uniqueness are retained from Stacks Tags 0FCJ and 0FCK. Tags 05SP and 0127 describe induced graded pieces and finite filtered comparison. The proof here needs only their finite kernel and short exact sequence interfaces, and states the choice universe explicitly. The exercises apply these interfaces to arrow representations, abelian groups, exact functors and a lattice that fails modularity.