# nLab free resolution

### Context

#### Homological algebra

homological algebra

Introduction

diagram chasing

# Contents

## Idea

The homological resolution of a chain complex by a complex of free modules.

## Properties

### Existence

###### Proposition

For $R$ a ring, every $R$-module admits a free resolution.

## Examples

###### Proposition

Assuming the axiom of choice, then every abelian group $A$ admits a free resolution of just length 2, hence with trivial syzygies, hence there exists a short exact sequence of abelian groups of the form

$0 \to F_2 \longrightarrow F_1 \longrightarrow A \to 0$

where both $F_1$ and $F_2$ are free abelian groups.

###### Proof

Let $F_1 = \mathbb{Z}[A]$ be the free abelian group on the underlying set of $A$, and let $F_1 \to A$ be the canonical morphism that sends a generator to itself (the adjunction counit of the free-forgetful adjunction).

Then let $F_2 \to F_1$ be the kernel of that map, hence $F_2$ a subgroup of $F_1$. Assuming the axiom of choice then every subgroup of a free abelian group is itself free abelian, hence $F_2$ is free abelian (prop.).

###### Remark

Prop. implies that all Ext-groups between abelian groups are concentrated in degrees 0 and 1.

(By the discussion at derived functor in homological algebra.)

###### Example

(Koszul complex is free resolution of quotient ring)

For $R$ a commutative ring and $(x_1, \cdots, x_d)$ a regular sequence of elements, then the Koszul complex $K(x_1, \cdots, x_d)$ is a free resolutions of the quotient ring $R/(x_1, \cdots, x_d)$ by free $R$-modules.

## References

Lecture notes include for instance

around page 5 of

Last revised on September 29, 2017 at 05:31:41. See the history of this page for a list of all contributions to it.