TheoremStatement: AI-adaptedProof: AI-generatedprecheck passaudited 2026-08-30

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Euler-Poincare formula for finite free complexes

Statement

Let $C_{∙}$ be a bounded chain complex of finite-rank free abelian groups. Assume that every homology group $H_{n} (C)$ is free of finite rank. Then $χ (C) = \sum_{n \in Z} (- 1)^{n} rank (H_{n} (C)) .$

Facts & Assumptions

Given: A bounded chain complex $C_{∙}$ of finite-rank free abelian groups whose homology groups are free of finite rank.

[L1]

The Euler characteristic is the finite alternating sum $\sum_{n} (- 1)^{n} rank (C_{n})$ (Euler characteristic of a finite complex of finite-rank free abelian groups).

[L2]

Homology is the quotient $Z_{n} (C) / B_{n} (C)$ (Homology object of a chain complex).

[L3]

Abelian groups and $Z$ -modules are the same objects and morphisms (Abelian groups and $Z$ -modules have the same objects and morphisms).

[L4]

The ring $Z$ is a principal ideal domain: it is a commutative ring (The integers form a commutative ring), it has no zero divisors (The integers have no zero divisors; multiplicative cancellation), and every ideal is generated by one integer because every subgroup of $(Z, +)$ is cyclic (Every subgroup of $(Z, +)$ is $⟨ n ⟩ = n Z$ for exactly one natural number $n$ , Principal ideal domain).

[L5]

A submodule of a finite-rank free module over a PID is again finite free (A submodule of a free module of finite rank over a PID is free of no larger rank).

[L6]

A finite free module over a commutative ring is projective (Under the stated choice boundary, free modules are projective and hence flat).

[L7]

A short exact sequence ending in a projective module splits (Equivalent characterizations of projective modules).

[L8]

A free module of rank $r$ has a basis of $r$ elements (The free module on a set and its standard basis, Invariant basis number and the rank of a free module).

Proof

technique · direct

1.1L3L4L5given

By [L3] and [L4], each $C_{n}$ is a finite-rank free $Z$ -module over a PID. Therefore [L5] makes every cycle group $Z_{n} (C) \leq C_{n}$ and every boundary group $B_{n} (C) \leq C_{n}$ finite free. Since the homology groups are finite free by the hypothesis, all ranks appearing below are defined.

2.1L2L6L7L8step 1.1algebra

By [L2], the short exact sequence $0 \to Z_{n} (C) \to C_{n} \to B_{n - 1} (C) \to 0$ has projective quotient $B_{n - 1} (C)$ by steps 1.1 and [L6], so [L7] splits it. Likewise $0 \to B_{n} (C) \to Z_{n} (C) \to H_{n} (C) \to 0$ splits because $H_{n} (C)$ is finite free by hypothesis and hence projective by [L6]. Using [L8], split exactness gives $rank (C_{n}) = rank (Z_{n} (C)) + rank (B_{n - 1} (C))$ and $rank (Z_{n} (C)) = rank (B_{n} (C)) + rank (H_{n} (C)) .$

3.1L1step 2.1algebra

Substitute the second identity of step 2.1 into the first: $rank (C_{n}) = rank (B_{n} (C)) + rank (H_{n} (C)) + rank (B_{n - 1} (C)) .$ Multiply by $(- 1)^{n}$ and sum over all $n$ . The boundedness in [L1] makes this a finite sum, and the two boundary sums cancel after the index shift $m = n - 1$ : $\sum_{n} (- 1)^{n} rank (B_{n - 1} (C)) = - \sum_{n} (- 1)^{n} rank (B_{n} (C)) .$ Hence only the homology contribution remains.

4.1L1step 3.1∎

The resulting identity is $\sum_{n} (- 1)^{n} rank (C_{n}) = \sum_{n} (- 1)^{n} rank (H_{n} (C)),$ which is exactly the formula in the statement by [L1].

Depends on

Used by

Dependency tree · two levels

70 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

P. Hekmati, Homological Algebra, section 3.1 (standard reference, not scraped)
K. Conrad, Modules over a PID, Theorem 2.2 and Corollary 2.6 (standard reference, not scraped)