syzygy

syzygy

Def:
R: commutative ring
A: R-algebra
M: A-module

R-derivation of A into M is a linear map

d: A → M s.t. d(ab) = bd(a) + ad(b)（∀a, b∈A）.

We denote the set of R-derivation of A into M by Der_R(A, M).

Remark:
For all r∈R, we have dr = 0.

∵
dr = d(1r) = rd(1) + dr (∵Leibniz rule)
∴ rd(1) = 0

dr = rd(1) (∵d is a R-linear map)

∴ dr = 0.

>>2
> R-derivation of A into M is a linear map
→ R-derivation of A into M is a R-linear map

Def:
R: commutative ring
A: R-algebra

The module of relative differential forms of A over R is a A-module Ω_A/R endowed with an R-derivation

d: A → Ω_A/R

satisfying the following universal properties:

∀M: A-module,
∀d': R-derivation of A into M,
∃φ: Ω_A/B → M: A-homomorphism s.t.

φ○d = d'.

Def:
R: commutative ring
A: R-algebra

The module of relative differential forms of A over R is a A-module Ω_A/R endowed with an R-derivation

d: A → Ω_A/R

satisfying the following universal property:

∀M: A-module,
∀d': R-derivation of A into M,
∃φ: Ω_A/R → M: A-homomorphism s.t.

φ○d = d'.

けっきょく可換環論なの？

Prop:
R: commutative ring
A: R-algebra

the module of relative differential forms (Ω_A/R, d) exists and is unique up to isomorphisms.

>>8
Proof:
The uniqueness follows from the universal property.

Let us show the existence.
Let F be a free module generated by elements of the form da (a∈A).
Let E be a submodule of F generated by elements of the form:

* dr
* d(a + b) - da - db
* d(ab) - bda - adb

(r∈R, a, b∈A). Set Ω_A/R = F/E and d: A → Ω_A/R sends a∈A to the image of da∈Ω_A/R. (Ω_A/R, d) satisfies the universal property.