TheoremStatement: AI-adaptedProof: AI-generatedprecheck passaudited 2026-08-28

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Length is additive along a subobject

Statement

Let $B \leq A$ be a subobject in an abelian category. If any two of the objects $A$ , $B$ , and $A / B$ have finite length, then so does the third, and whenever all three do one has

$ℓ (A) = ℓ (B) + ℓ (A / B) .$

Facts & Assumptions

Given: A subobject $B \leq A$ .

[L1]

Finite length means admitting a composition series, and length is the common number of factors in such a series (Object of finite length).

[L2]

Jordan-Hölder makes the factor count independent of the chosen composition series (Jordan-Holder theorem in an abelian category).

[L3]

The second isomorphism theorem identifies the factors that arise from pulling a chain across a quotient or intersecting with a subobject (Second isomorphism theorem in an abelian category).

Proof

technique · direct

1.1L1L3chooseconstruct

Assume $B$ and $A / B$ have finite length. Choose composition series $0 = B_{0} < \dots < B_{r} = B, 0 = C_{0} < \dots < C_{s} = A / B .$ Let $q : A \to A / B$ be the quotient map and put $D_{j} : = q^{- 1} (C_{j})$ . Then $B = D_{0} < D_{1} < \dots < D_{s} = A,$ and [L3] identifies each quotient $D_{j} / D_{j - 1}$ with $C_{j} / C_{j - 1}$ . So $0 = B_{0} < \dots < B_{r} = D_{0} < D_{1} < \dots < D_{s} = A$ is a composition series of $A$ . Therefore $A$ has finite length and $ℓ (A) = r + s = ℓ (B) + ℓ (A / B)$ .

1.2L1L3construct

Assume now that $A$ has finite length, with composition series $0 = A_{0} < A_{1} < \dots < A_{n} = A .$ Intersecting with $B$ gives an increasing chain $0 = A_{0} \land B \leq A_{1} \land B \leq \dots \leq A_{n} \land B = B,$ and quotienting by $B$ gives an increasing chain in $A / B$ . By [L3], each successive factor in either chain is a subquotient of a simple factor $A_{i} / A_{i - 1}$ , hence is either $0$ or simple. Deleting repeated adjacent terms therefore yields composition series of $B$ and of $A / B$ .

2.1L2step 1.1step 1.2∎

Step 1.1 proves the extension direction and the displayed additive formula. Step 1.2 proves that finite length passes to subobjects and quotients. The number in the formula is independent of the chosen composition series by [L2].

Depends on

Used by

Dependency tree · two levels

9 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

Pavel Etingof, Shlomo Gelaki, Dmitri Nikshych, and Victor Ostrik, Tensor Categories, Section 1.5 (standard reference, not scraped)