nLab free resolution

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Idea

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

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.

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.).

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

(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.

Lecture notes include for instance

around page 5 of

Last revised on September 29, 2017 at 09:31:41. See the history of this page for a list of all contributions to it.