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§ 10。 推导
在本段中，以及另外提到的除非，代数不被认为是结合的，也不一定具有单位元； K 表示交换环。
1. 交换因子
当在本段中我们谈论毕业但没有指定它们时，我们总是指型Δ \Delta Δ Δ \Delta Δ Δ \Delta Δ Z \mathbf{Z} Z Δ \Delta Δ Δ \Delta Δ Δ × Δ \Delta \times \Delta Δ × Δ ε : ( α , β ) ↦ ε α β = ε ( α , β ) \varepsilon : (\alpha, \beta) \mapsto \varepsilon_{\alpha \beta} = \varepsilon(\alpha, \beta) ε : ( α , β ) ↦ ε α β = ε ( α , β ) { − 1 , 1 } \{-1, 1\} { − 1 , 1 } Δ \Delta Δ α , α ′ , β , β ′ \alpha, \alpha', \beta, \beta' α , α ′ , β , β ′
{ ε ( α + α ′ , β ) = ε ( α , β ) ε ( α ′ , β ) ε ( α , β + β ′ ) = ε ( α , β ) ε ( α , β ′ ) ε ( β , α ) = ε ( α , β ) .
\begin{cases}
\varepsilon(\alpha + \alpha', \beta) = \varepsilon(\alpha, \beta)\varepsilon(\alpha', \beta) \\
\varepsilon(\alpha, \beta + \beta') = \varepsilon(\alpha, \beta)\varepsilon(\alpha, \beta') \\
\varepsilon(\beta, \alpha) = \varepsilon(\alpha, \beta).
\end{cases}
⎩ ⎨ ⎧ ε ( α + α ′ , β ) = ε ( α , β ) ε ( α ′ , β ) ε ( α , β + β ′ ) = ε ( α , β ) ε ( α , β ′ ) ε ( β , α ) = ε ( α , β ) . (1)
由此可见ε ( 2 α , β ) = ε ( α , 2 β ) = 1 \varepsilon(2\alpha, \beta) = \varepsilon(\alpha, 2\beta) = 1 ε ( 2 α , β ) = ε ( α , 2 β ) = 1
当Δ = Z \Delta = \mathbf{Z} Δ = Z ε \varepsilon ε ε ( 1 , 1 ) \varepsilon(1, 1) ε ( 1 , 1 )
(2)
ε ( p , q ) = 1 for p , q in Z
\varepsilon(p, q) = 1 \quad \text{for } p, q \text{ in } \mathbf{Z}
ε ( p , q ) = 1 for p , q in Z ε ( p , q ) = ( − 1 ) p q for p , q in Z .
\varepsilon(p, q) = (-1)^{pq} \quad \text{for } p, q \text{ in } \mathbf{Z}.
ε ( p , q ) = ( − 1 ) pq for p , q in Z . 2. 定义推导将军
考虑交换环K K K Δ : A , A ′ , A ′ ′ , B , B ′ , B ′ ′ \Delta : A, A', A'', B, B', B'' Δ : A , A ′ , A ′′ , B , B ′ , B ′′ K K K K K K
μ : A × A ′ → A ′ ′ , λ 1 : B × A ′ → B ′ ′ , λ 2 : A × B ′ → B ′ ′
\mu : A \times A' \to A'', \qquad \lambda_1 : B \times A' \to B'', \qquad \lambda_2 : A \times B' \to B''
μ : A × A ′ → A ′′ , λ 1 : B × A ′ → B ′′ , λ 2 : A × B ′ → B ′′ K K K A ⊗ K A ′ → A ′ ′ , B ⊗ K A ′ → B ′ ′ , A ⊗ K B ′ → B ′ ′
A \otimes_K A' \to A'', \qquad B \otimes_K A' \to B'', \qquad A \otimes_K B' \to B''
A ⊗ K A ′ → A ′′ , B ⊗ K A ′ → B ′′ , A ⊗ K B ′ → B ′′ 分级为次数 0。a ∈ A , a ′ ∈ A ′ a \in A, a' \in A' a ∈ A , a ′ ∈ A ′ μ ( a , a ′ ) \mu(a, a') μ ( a , a ′ ) a . a ′ a.a' a . a ′ a a ′ aa' a a ′ a . a ′ a.a' a . a ′ a a a a ′ a' a ′
鉴于上述和Δ × Δ \Delta \times \Delta Δ × Δ ε \varepsilon ε ( A , A ′ , A ′ ′ ) (A, A', A'') ( A , A ′ , A ′′ ) δ ∈ Δ \delta \in \Delta δ ∈ Δ ( B , B ′ , B ′ ′ ) (B, B', B'') ( B , B ′ , B ′′ ) ε \varepsilon ε ( K , ε ) (K, \varepsilon) ( K , ε ) δ \delta δ K K K
d : A → B , d ′ : A ′ → B ′ , d ′ ′ : A ′ ′ → B ′ ′
d : A \to B, \qquad d' : A' \to B', \qquad d'' : A'' \to B''
d : A → B , d ′ : A ′ → B ′ , d ′′ : A ′′ → B ′′ 这样，对于每个同质元素a ∈ A a \in A a ∈ A a ′ ∈ A ′ a' \in A' a ′ ∈ A ′ (4)
d ′ ′ ( a . a ′ ) = ( d a ) . a ′ + ε δ , deg ( a ) a . ( d ′ a ′ ) .
d''(a.a') = (da).a' + \varepsilon_{\delta, \deg(a)}a.(d'a').
d ′′ ( a . a ′ ) = ( d a ) . a ′ + ε δ , d e g ( a ) a . ( d ′ a ′ ) . 被以下引用 3 被以下引用 习题 1, § 10 , 习题 2, § 10 , 例 5
当a a a a ′ a' a ′ A A A A ′ A' A ′
通常，用同一个字母 d d d d , d ′ , d ′ ′ d, d', d'' d , d ′ , d ′′ d d d δ \delta δ K K K
( a , a ′ , a ′ ′ ) ↦ ( d a , d ′ a ′ , d ′ ′ a ′ ′ )
(a, a', a'') \mapsto (da, d'a', d''a'')
( a , a ′ , a ′′ ) ↦ ( d a , d ′ a ′ , d ′′ a ′′ ) A ⊕ A ′ ⊕ A ′ ′ A \oplus A' \oplus A'' A ⊕ A ′ ⊕ A ′′ B ⊕ B ′ ⊕ B ′ ′ B \oplus B' \oplus B'' B ⊕ B ′ ⊕ B ′′
d ( a . a ′ ) = ( d a ) . a ′ + ε δ , deg ( a ) a . ( d a ′ ) .
d(a.a') = (da).a' + \varepsilon_{\delta,\deg(a)} a.(da').
d ( a . a ′ ) = ( d a ) . a ′ + ε δ , d e g ( a ) a . ( d a ′ ) . 给定 次数形式 ( A , A ′ , A ′ ′ ) (A, A', A'') ( A , A ′ , A ′′ ) ( B , B ′ , B ′ ′ ) (B, B', B'') ( B , B ′ , B ′′ ) ε \varepsilon ε Hom K ( A ⊕ A ′ ⊕ A ′ ′ , B ⊕ B ′ ⊕ B ′ ′ ) .
\operatorname{Hom}_{\mathbf{K}}(A \oplus A' \oplus A'', B \oplus B' \oplus B'').
Hom K ( A ⊕ A ′ ⊕ A ′′ , B ⊕ B ′ ⊕ B ′′ ) . ε ( α , β ) = 1 \varepsilon(\alpha, \beta) = 1 ε ( α , β ) = 1 Δ \Delta Δ α , β \alpha, \beta α , β ε \varepsilon ε Hom K ( A ⊕ A ′ ⊕ A ′ ′ , B ⊕ B ′ ⊕ B ′ ′ ) .
\operatorname{Hom}_{\mathbf{K}}(A \oplus A' \oplus A'', B \oplus B' \oplus B'').
Hom K ( A ⊕ A ′ ⊕ A ′′ , B ⊕ B ′ ⊕ B ′′ ) . Δ = Z \Delta = \mathbf{Z} Δ = Z ε ( p . q ) = ( − 1 ) p q \varepsilon(p.q) = (-1)^{pq} ε ( p . q ) = ( − 1 ) pq ε \varepsilon ε 奇数 次数的每个ε \varepsilon ε d d d d ( a . a ′ ) = ( d a ) . a ′ + ( − 1 ) deg ( a ) a . ( d a ′ )
d(a.a') = (da).a' + (-1)^{\deg(a)} a.(da')
d ( a . a ′ ) = ( d a ) . a ′ + ( − 1 ) d e g ( a ) a . ( d a ′ ) a ∈ A a \in A a ∈ A 3. 例 / 推导
设A是R \mathbf{R} R R \mathbf{R} R R \mathbf{R} R x 0 x_0 x 0 R \mathbf{R} R R \mathbf{R} R ( f , a ) ↦ f ( x 0 ) a (f, a) \mapsto f(x_0)a ( f , a ) ↦ f ( x 0 ) a f ↦ D f ( x 0 ) f \mapsto Df(x_0) f ↦ D f ( x 0 )
( D ( f g ) ) ( x 0 ) = ( D f ( x 0 ) ) g ( x 0 ) + f ( x 0 ) ( D g ( x 0 ) ) . ∗
(D(fg))(x_0) = (Df(x_0))g(x_0) + f(x_0)(Dg(x_0)).*
( D ( f g )) ( x 0 ) = ( D f ( x 0 )) g ( x 0 ) + f ( x 0 ) ( D g ( x 0 )) . ∗ 被以下引用 1 被以下引用 推论
令X为类C ∞ C^\infty C ∞ R \mathbf{R} R ω \omega ω d ω d\omega d ω
被以下引用 1 被以下引用 § 10. 。 推导
设 A 为结合律 K-代数。对于所有 a ∈ A a \in A a ∈ A x ↦ a x − x a x \mapsto ax - xa x ↦ a x − x a
回到no. 2的定义1的一般情况，令K ‾ \overline{K} K ρ : K → K ‾ \rho : K \to \overline{K} ρ : K → K A ‾ , A ‾ ′ , A ‾ ′ ′ , B ‾ , B ‾ ′ , B ‾ ′ ′ \overline{A}, \overline{A}', \overline{A}'', \overline{B}, \overline{B}', \overline{B}'' A , A ′ , A ′′ , B , B ′ , B ′′ K ‾ \overline{K} K A , A ′ , A ′ ′ , B , B ′ , B ′ ′ A, A', A'', B, B', B'' A , A ′ , A ′′ , B , B ′ , B ′′ K ‾ \overline{K} K μ , λ 1 \mu, \lambda_1 μ , λ 1 λ 2 \lambda_2 λ 2 K ‾ \overline{K} K
μ ˉ : A ‾ × A ‾ ′ → A ‾ ′ ′ , λ ˉ 1 : B ‾ × A ‾ ′ → B ‾ ′ ′ , λ ˉ 2 : A ‾ × B ‾ ′ → B ‾ ′ ′
\bar{\mu} : \overline{A} \times \overline{A}' \to \overline{A}'', \quad \bar{\lambda}_1 : \overline{B} \times \overline{A}' \to \overline{B}'', \quad \bar{\lambda}_2 : \overline{A} \times \overline{B}' \to \overline{B}''
μ ˉ : A × A ′ → A ′′ , λ ˉ 1 : B × A ′ → B ′′ , λ ˉ 2 : A × B ′ → B ′′ 通过考虑张量积乘
l K ‾ l_{\overline{K}} l K 的相应的
K K K -线性映射到
μ , λ 1 \mu, \lambda_1 μ , λ 1 和
λ 2 \lambda_2 λ 2 （II, § 5, no. 1）。那么，如果
d d d 是
( A , A ′ , A ′ ′ ) (A, A', A'') ( A , A ′ , A ′′ ) 的次数
δ \delta δ 到
( B , B ′ , B ′ ′ ) (B, B', B'') ( B , B ′ , B ′′ ) 的
ε \varepsilon ε 推导，则
A ‾ ⊕ A ‾ ′ ⊕ A ‾ ′ ′ \overline{A} \oplus \overline{A}' \oplus \overline{A}'' A ⊕ A ′ ⊕ A ′′ 的映射
d ˉ = d ⊗ l K ‾ \bar{d} = d \otimes l_{\overline{K}} d ˉ = d ⊗ l K 到
B ‾ ⊕ B ‾ ′ ⊕ B ‾ ′ ′ \overline{B} \oplus \overline{B}' \oplus \overline{B}'' B ⊕ B ′ ⊕ B ′′ 是
ε \varepsilon ε -将
( A ‾ , A ‾ ′ , A ‾ ′ ′ ) (\overline{A}, \overline{A}', \overline{A}'') ( A , A ′ , A ′′ ) 的次数
δ \delta δ 推导为
( B ‾ , B ‾ ′ , B ‾ ′ ′ ) (\overline{B}, \overline{B}', \overline{B}'') ( B , B ′ , B ′′ ) 。
引用了 3 引用了 II § 11. 。分级为模和环 , 定义 1 , II § 5. 。标量的环的扩张
令A A A Z \mathbf{Z} Z K K K d : A → A d : A \to A d : A → A K K K x n ∈ A n ( n ∈ Z ) x_n \in A_n (n \in \mathbf{Z}) x n ∈ A n ( n ∈ Z ) d ( x n ) = n x n d(x_n) = nx_n d ( x n ) = n x n A A A x p ∈ A p , x q ∈ A q x_p \in A_p, x_q \in A_q x p ∈ A p , x q ∈ A q
d ( x p x q ) = ( p + q ) x p x q = d ( x p ) x q + x p d ( x q ) .
d(x_p x_q) = (p + q)x_p x_q = d(x_p)x_q + x_p d(x_q).
d ( x p x q ) = ( p + q ) x p x q = d ( x p ) x q + x p d ( x q ) . 被以下引用 1 被以下引用 命题 6 (*《欧拉恒等式》*), IV § 1
4. 合成 / 推导
我们假设在这个没有。 no. 2的情况(I)成立，即A , A ′ , A ′ ′ A, A', A'' A , A ′ , A ′′ Δ \Delta Δ K K K K K K μ : A × A ′ → A ′ ′ \mu : A \times A' \to A'' μ : A × A ′ → A ′′ K K K A ⊗ K A ′ → A ′ ′ A \otimes_K A' \to A'' A ⊗ K A ′ → A ′′ A ⊕ A ′ ⊕ A ′ ′ A \oplus A' \oplus A'' A ⊕ A ′ ⊕ A ′′ f f f f ( A ) ⊂ A , f ( A ′ ) ⊂ A ′ f(A) \subset A, f(A') \subset A' f ( A ) ⊂ A , f ( A ′ ) ⊂ A ′ f ( A ′ ′ ) ⊂ A ′ ′ f(A'') \subset A'' f ( A ′′ ) ⊂ A ′′ E n d g r K ( A ⊕ A ′ ⊕ A ′ ′ ) \mathrm{Endgr}_K(A \oplus A' \oplus A'') Endgr K ( A ⊕ A ′ ⊕ A ′′ ) ( A , A ′ , A ′ ′ ) (A, A', A'') ( A , A ′ , A ′′ ) ε \varepsilon ε ε \varepsilon ε ε \varepsilon ε
在型 Δ \Delta Δ B B B u , v u, v u , v ε \varepsilon ε ε = 1 \varepsilon = 1 ε = 1
[ u , v ] ε = u v − ε deg u , deg v v u (denoted simply by [ u , v ] if ε = 1 ) .
[u, v]_\varepsilon = uv - \varepsilon_{\deg u, \deg v} vu \text{ (denoted simply by } [u, v] \text{ if } \varepsilon = 1).
[ u , v ] ε = uv − ε d e g u , d e g v v u (denoted simply by [ u , v ] if ε = 1 ) . B × B B \times B B × B B B B K K K ( u , v ) ↦ [ u , v ] ε (u, v) \mapsto [u, v]_\varepsilon ( u , v ) ↦ [ u , v ] ε B B B u u u v v v [ v , u ] ε = − ε deg u , deg v [ u , v ] ε .
[v, u]_\varepsilon = -\varepsilon_{\deg u, \deg v}[u, v]_\varepsilon.
[ v , u ] ε = − ε d e g u , d e g v [ u , v ] ε . E n d g r K ( A ⊕ A ′ ⊕ A ′ ′ ) \mathrm{Endgr}_K(A \oplus A' \oplus A'') Endgr K ( A ⊕ A ′ ⊕ A ′′ ) ε \varepsilon ε
令d 1 , d 2 d_1, d_2 d 1 , d 2 δ 1 , δ 2 \delta_1, \delta_2 δ 1 , δ 2 ( A , A ′ , A ′ ′ ) (A, A', A'') ( A , A ′ , A ′′ ) ε \varepsilon ε ε \varepsilon ε
[ d 1 , d 2 ] ε = d 1 ∘ d 2 − ε δ 1 , δ 2 d 2 ∘ d 1
[d_1, d_2]_\varepsilon = d_1 \circ d_2 - \varepsilon_{\delta_1, \delta_2} d_2 \circ d_1
[ d 1 , d 2 ] ε = d 1 ∘ d 2 − ε δ 1 , δ 2 d 2 ∘ d 1 *是次数
δ 1 + δ 2 \delta_1 + \delta_2 δ 1 + δ 2 的
ε \varepsilon ε 派生。此外，如果
d d d 是次数
δ \delta δ 的
( A , A ′ , A ′ ′ ) (A, A', A'') ( A , A ′ , A ′′ ) 的
ε \varepsilon ε 衍生，并且如果
ε δ , δ = − 1 \varepsilon_{\delta, \delta} = -1 ε δ , δ = − 1 ，则
d 2 = d ∘ d d^2 = d \circ d d 2 = d ∘ d 是推导。
被以下引用 1 被以下引用 命题 14
假设x ∈ A x \in A x ∈ A ξ \xi ξ y ∈ A ′ y \in A' y ∈ A ′
d 1 ( d 2 ( x y ) ) = ( ( d 1 d 2 ) ( x ) ) y + ε δ 1 , δ 2 + ξ ( d 2 x ) ( d 1 y ) + ε δ 2 , ξ ( d 1 x ) ( d 2 y ) + ε δ 1 + δ 2 , ξ x ( ( d 1 d 2 ) ( y ) )
d_1(d_2(xy)) = ((d_1 d_2)(x))y + \varepsilon_{\delta_1, \delta_2 + \xi}(d_2 x)(d_1 y)
+ \varepsilon_{\delta_2, \xi}(d_1 x)(d_2 y) + \varepsilon_{\delta_1 + \delta_2, \xi} x((d_1 d_2)(y))
d 1 ( d 2 ( x y )) = (( d 1 d 2 ) ( x )) y + ε δ 1 , δ 2 + ξ ( d 2 x ) ( d 1 y ) + ε δ 2 , ξ ( d 1 x ) ( d 2 y ) + ε δ 1 + δ 2 , ξ x (( d 1 d 2 ) ( y )) 考虑no. 1的公式(1)。交换
d 1 d_1 d 1 和
d 2 d_2 d 2 的角色，我们再次使用（1）（no. 1）得到之后简化，
( d 1 d 2 ) ( x y ) − ε δ 1 , δ 2 ( d 2 d 1 ) ( x y ) = ( ( d 1 d 2 ) ( x ) ) y − ε δ 1 , δ 2 ( ( d 2 d 1 ) ( x ) ) y + ε δ 1 + δ 2 , ξ x ( ( d 1 d 2 ) ( y ) ) − ε δ 1 , δ 2 ε δ 1 + δ 2 , ξ x ( ( d 2 d 1 ) ( y ) )
(d_1 d_2)(xy) - \varepsilon_{\delta_1, \delta_2}(d_2 d_1)(xy) = ((d_1 d_2)(x))y - \varepsilon_{\delta_1, \delta_2}((d_2 d_1)(x))y
+ \varepsilon_{\delta_1 + \delta_2, \xi} x((d_1 d_2)(y))
- \varepsilon_{\delta_1, \delta_2} \varepsilon_{\delta_1 + \delta_2, \xi} x((d_2 d_1)(y))
( d 1 d 2 ) ( x y ) − ε δ 1 , δ 2 ( d 2 d 1 ) ( x y ) = (( d 1 d 2 ) ( x )) y − ε δ 1 , δ 2 (( d 2 d 1 ) ( x )) y + ε δ 1 + δ 2 , ξ x (( d 1 d 2 ) ( y )) − ε δ 1 , δ 2 ε δ 1 + δ 2 , ξ x (( d 2 d 1 ) ( y )) 即写
d = [ d 1 , d 2 ] ε d = [d_1, d_2]_\varepsilon d = [ d 1 , d 2 ] ε 和
δ = δ 1 + δ 2 \delta = \delta_1 + \delta_2 δ = δ 1 + δ 2 ，
d ( x y ) = ( d x ) y + ε δ , ξ x ( d y )
d(xy) = (dx)y + \varepsilon_{\delta, \xi} x(dy)
d ( x y ) = ( d x ) y + ε δ , ξ x ( d y ) 这证明了
d d d 是
ε \varepsilon ε 的导数。
另一方面，如果在 (11) 中，我们令 d 1 = d 2 = d d_1 = d_2 = d d 1 = d 2 = d δ 1 = δ 2 = δ \delta_1 = \delta_2 = \delta δ 1 = δ 2 = δ ε δ , δ = − 1 \varepsilon_{\delta, \delta} = -1 ε δ , δ = − 1 ε δ , δ + ξ = − ε δ , ξ \varepsilon_{\delta, \delta + \xi} = -\varepsilon_{\delta, \xi} ε δ , δ + ξ = − ε δ , ξ
d 2 ( x y ) = ( d 2 x ) y + ε 2 δ , ξ x ( d 2 y )
d^2(xy) = (d^2 x)y + \varepsilon_{2\delta, \xi} x(d^2 y)
d 2 ( x y ) = ( d 2 x ) y + ε 2 δ , ξ x ( d 2 y ) 而作为
ε 2 δ , ξ = 1 \varepsilon_{2\delta, \xi} = 1 ε 2 δ , ξ = 1 ，可以看出
d 2 d^2 d 2 是推导。
假设Δ = Z \Delta = \mathbf{Z} Δ = Z 反推导的平方是推导。 (ii) 两个推导的括号是推导。 (iii) 反推导的括号和偶数次数的推导是反推导。 (iv) *如果d 1 d_1 d 1 d 2 d_2 d 2 d 1 d 2 + d 2 d 1 d_1 d_2 + d_2 d_1 d 1 d 2 + d 2 d 1
引用了 1 引用了 III § 2. 。 例 / 代数
在这个数开头的假设下，现在考虑( A , A ′ , A ′ ′ ) (A, A', A'') ( A , A ′ , A ′′ ) D = ( d i ) 1 ≤ i ≤ n D = (d_i)_{1 \leq i \leq n} D = ( d i ) 1 ≤ i ≤ n K [ X 1 , … , X n ] K[X_1, \ldots, X_n] K [ X 1 , … , X n ] P ( X 1 , … , X n ) P(X_1, \ldots, X_n) P ( X 1 , … , X n ) E n d g r K ( A ⊕ A ′ ⊕ A ′ ′ ) \mathrm{Endgr}_K(A \oplus A' \oplus A'') Endgr K ( A ⊕ A ′ ⊕ A ′′ ) P ( d 1 , … , d n ) P(d_1, \ldots, d_n) P ( d 1 , … , d n ) P ( D ) P(D) P ( D )
根据上述假设和记号，考虑 2n 个不定式 T 1 , … , T n , T 1 ′ , … , T n ′ T_1, \ldots, T_n, T'_1, \ldots, T'_n T 1 , … , T n , T 1 ′ , … , T n ′ F ∈ K [ X 1 , … , X n ] F \in K[X_1, \ldots, X_n] F ∈ K [ X 1 , … , X n ] F ( T ) = F ( T 1 , … , T n ) , F ( T ′ ) = F ( T 1 ′ , … , T n ′ ) F(T) = F(T_1, \ldots, T_n), F(T') = F(T'_1, \ldots, T'_n) F ( T ) = F ( T 1 , … , T n ) , F ( T ′ ) = F ( T 1 ′ , … , T n ′ )
F ( T + T ′ ) = F ( T 1 + T 1 ′ , … , T n + T n ′ ) .
F(T + T') = F(T_1 + T'_1, \ldots, T_n + T'_n).
F ( T + T ′ ) = F ( T 1 + T 1 ′ , … , T n + T n ′ ) . 假设 P ( T + T ′ ) = ∑ i Q i ( T ) R i ( T ′ )
P(T + T') = \sum_i Q_i(T) R_i(T')
P ( T + T ′ ) = i ∑ Q i ( T ) R i ( T ′ ) 其中 Q i Q_i Q i R i R_i R i K [ X 1 , … , X n ] K[X_1, \ldots, X_n] K [ X 1 , … , X n ] x ∈ A x \in A x ∈ A y ∈ A ′ y \in A' y ∈ A ′ P ( D ) ( x y ) = ∑ i ( Q i ( D ) x ) ( R i ( D ) y ) .
P(D)(xy) = \sum_i (Q_i(D)x)(R_i(D)y).
P ( D ) ( x y ) = i ∑ ( Q i ( D ) x ) ( R i ( D ) y ) . 被以下引用 2 被以下引用 推论（莱布尼兹公式） , 命题 14
我们引入n n n T 1 ′ ′ , … , T n ′ ′ T_1'', \ldots, T_n'' T 1 ′′ , … , T n ′′ K [ T 1 , … , T n , T 1 ′ , … , T n ′ , T 1 ′ ′ , … , T n ′ ′ ] = B K[T_1, \ldots, T_n, T_1', \ldots, T_n', T_1'', \ldots, T_n''] = B K [ T 1 , … , T n , T 1 ′ , … , T n ′ , T 1 ′′ , … , T n ′′ ] = B A × A ′ A \times A' A × A ′ K K K M M M A ′ ′ A'' A ′′ K K K f ∈ M f \in M f ∈ M 1 ≤ i ≤ n 1 \leq i \leq n 1 ≤ i ≤ n M M M B B B
{ ( T i f ) ( a , a ′ ) = f ( d i a , a ′ ) ( T i ′ f ) ( a , a ′ ) = f ( a , d i a ′ ) ( T i ′ ′ f ) ( a , a ′ ) = d i ( f ( a , a ′ ) )
\begin{cases}
(T_i f)(a, a') = f(d_i a, a') \\
(T_i' f)(a, a') = f(a, d_i a') \\
(T_i'' f)(a, a') = d_i(f(a, a'))
\end{cases}
⎩ ⎨ ⎧ ( T i f ) ( a , a ′ ) = f ( d i a , a ′ ) ( T i ′ f ) ( a , a ′ ) = f ( a , d i a ′ ) ( T i ′′ f ) ( a , a ′ ) = d i ( f ( a , a ′ )) 由于 d i d_i d i F ∈ K [ X 1 , … , X n ] F \in K[X_1, \ldots, X_n] F ∈ K [ X 1 , … , X n ] ( F ( T ) f ) ( a , a ′ ) = f ( F ( D ) a , a ′ ) (F(T)f)(a, a') = f(F(D)a, a') ( F ( T ) f ) ( a , a ′ ) = f ( F ( D ) a , a ′ )
( F ( T ′ ) f ) ( a , a ′ ) = f ( a , F ( D ) a ′ )
(F(T')f)(a, a') = f(a, F(D)a')
( F ( T ′ ) f ) ( a , a ′ ) = f ( a , F ( D ) a ′ ) 和( F ( T ′ ′ ) f ) = F ( D ) ( f ( a , a ′ ) ) (F(T'')f) = F(D)(f(a, a')) ( F ( T ′′ ) f ) = F ( D ) ( f ( a , a ′ ))
( P ( T ′ ′ ) − ∑ i Q i ( T ) R i ( T ′ ) ) . μ = 0
(P(T'') - \sum_i Q_i(T)R_i(T')).\mu = 0
( P ( T ′′ ) − i ∑ Q i ( T ) R i ( T ′ )) . μ = 0 或B B B M M M ( P ( T ′ ′ ) − P ( T + T ′ ) ) . μ = 0 (P(T'') - P(T + T')).\mu = 0 ( P ( T ′′ ) − P ( T + T ′ )) . μ = 0 d i d_i d i 1 ≤ i ≤ n 1 \leq i \leq n 1 ≤ i ≤ n
( T i ′ ′ − T i − T i ′ ) . μ = 0
(T_i'' - T_i - T_i').\mu = 0
( T i ′′ − T i − T i ′ ) . μ = 0 在B B B M M M
P ( T 1 ′ ′ , T 2 ′ ′ , … , T n ′ ′ ) − P ( T 1 + T 1 ′ , T 2 ′ ′ , … , T n ′ ′ ) P(T_1'', T_2'', \ldots, T_n'') - P(T_1 + T_1', T_2'', \ldots, T_n'') P ( T 1 ′′ , T 2 ′′ , … , T n ′′ ) − P ( T 1 + T 1 ′ , T 2 ′′ , … , T n ′′ ) P ( T 1 + T 1 ′ , T 2 ′ ′ , … , T n ′ ′ ) − P ( T 1 + T 1 ′ , T 2 + T 2 ′ , … , T n ′ ′ ) P(T_1 + T_1', T_2'', \ldots, T_n'') - P(T_1 + T_1', T_2 + T_2', \ldots, T_n'') P ( T 1 + T 1 ′ , T 2 ′′ , … , T n ′′ ) − P ( T 1 + T 1 ′ , T 2 + T 2 ′ , … , T n ′′ ) … \ldots … P ( T 1 + T 1 ′ , … , T n − 1 + T n − 1 ′ , T n ′ ′ ) − P ( T 1 + T 1 ′ , … , T n − 1 + T n − 1 ′ , T n + T n ′ ) P(T_1 + T_1', \ldots, T_{n-1} + T_{n-1}', T_n'') - P(T_1 + T_1', \ldots, T_{n-1} + T_{n-1}', T_n + T_n') P ( T 1 + T 1 ′ , … , T n − 1 + T n − 1 ′ , T n ′′ ) − P ( T 1 + T 1 ′ , … , T n − 1 + T n − 1 ′ , T n + T n ′ ) 可见差P ( T ′ ′ ) − P ( T + T ′ ) P(T'') - P(T + T') P ( T ′′ ) − P ( T + T ′ )
∑ i = 1 n ( T i ′ ′ − T i − T i ′ ) G i ( T , T ′ , T ′ ′ )
\sum_{i=1}^n (T_i'' - T_i - T_i') G_i(T, T', T'')
i = 1 ∑ n ( T i ′′ − T i − T i ′ ) G i ( T , T ′ , T ′′ ) 其中 G i G_i G i B B B
推论（莱布尼兹公式） 066I
设 d i d_i d i 1 ≤ i ≤ n 1 \leq i \leq n 1 ≤ i ≤ n ( A , A ′ , A ′ ′ ) (A, A', A'') ( A , A ′ , A ′′ ) n n n α = ( α 1 , … , α n ) ∈ N n \alpha = (\alpha_1, \ldots, \alpha_n) \in \mathbf{N}^n α = ( α 1 , … , α n ) ∈ N n
引用了 2 引用了 I § 8. 。 环 , 命题 2
d α = d 1 α 1 d 2 α 2 … d n α n .
d^\alpha = d_1^{\alpha_1} d_2^{\alpha_2} \ldots d_n^{\alpha_n}.
d α = d 1 α 1 d 2 α 2 … d n α n . 然后，对于x ∈ A x \in \mathbf{A} x ∈ A y ∈ A ′ y \in \mathbf{A}' y ∈ A ′
d α ( x y ) = ∑ β + γ = α ( ( β , γ ) ) d β ( x ) d γ ( y )
d^\alpha(xy) = \sum_{\beta + \gamma = \alpha} ((\beta, \gamma)) d^\beta(x)d^\gamma(y)
d α ( x y ) = β + γ = α ∑ (( β , γ )) d β ( x ) d γ ( y ) 我们所写的地方（在记号在章开头引入）
( ( β , γ ) ) = ( β + γ ) ! / ( β ! γ ! ) .
((\beta, \gamma)) = (\beta + \gamma)!/(\beta! \gamma!).
(( β , γ )) = ( β + γ )! / ( β ! γ !) . 这立即由多项式公式 (I, § 8, no. 2) 得出
( T + T ′ ) α = ∑ β + γ = α ( ( β , γ ) ) T β T ′ γ
(T + T')^\alpha = \sum_{\beta + \gamma = \alpha} ((\beta, \gamma)) T^\beta {T'}^\gamma
( T + T ′ ) α = β + γ = α ∑ (( β , γ )) T β T ′ γ 和命题2。
5. 推导代数 A 变成 A-模
我们假设在这个没有。 no. 2 的情况 (II) 成立。然后有一个分级的 K-代数 A 和一个分级的 K-模 E，还有两个 K-线性映射次数 0
E ⊗ K A → E , A ⊗ K E → E
E \otimes_K A \to E, \quad A \otimes_K E \to E
E ⊗ K A → E , A ⊗ K E → E 表示为
x ⊗ a ↦ x . a and a ⊗ x ↦ a . x for a ∈ A and x ∈ E .
x \otimes a \mapsto x.a \quad \text{and} \quad a \otimes x \mapsto a.x \quad \text{for } a \in A \text{ and } x \in E.
x ⊗ a ↦ x . a and a ⊗ x ↦ a . x for a ∈ A and x ∈ E .
显然K e r ( d ) \mathrm{Ker}(d) Ker ( d ) x x x y y y K e r ( d ) \mathrm{Ker}(d) Ker ( d ) d ( x y ) = 0 d(xy) = 0 d ( x y ) = 0 x y ∈ K e r ( d ) xy \in \mathrm{Ker}(d) x y ∈ Ker ( d ) x x x y y y d ( 1 ) = d ( 1 ) + d ( 1 ) d(1) = d(1) + d(1) d ( 1 ) = d ( 1 ) + d ( 1 ) d ( 1 ) = 0 d(1) = 0 d ( 1 ) = 0
令d 1 d_1 d 1 d 2 d_2 d 2 δ \delta δ ε \varepsilon ε d 1 d_1 d 1 d 2 d_2 d 2 d 1 = d 2 d_1 = d_2 d 1 = d 2
被以下引用 1 被以下引用 推论
d 1 − d 2 d_1 - d_2 d 1 − d 2 δ \delta δ ε \varepsilon ε K e r ( d 1 − d 2 ) \mathrm{Ker}(d_1 - d_2) Ker ( d 1 − d 2 )
令 d : A → E d : A \to E d : A → E δ \delta δ ε \varepsilon ε x x x x − 1 x^{-1} x − 1
d ( x − 1 ) = − ε δ , deg ( x ) x − 1 ( ( d x ) x − 1 ) = − ε δ , deg ( x ) ( x − 1 ( d x ) ) x − 1 .
d(x^{-1}) = -\varepsilon_{\delta, \deg(x)} x^{-1}((dx)x^{-1}) = -\varepsilon_{\delta, \deg(x)} (x^{-1}(dx))x^{-1}.
d ( x − 1 ) = − ε δ , d e g ( x ) x − 1 (( d x ) x − 1 ) = − ε δ , d e g ( x ) ( x − 1 ( d x )) x − 1 . 引用了 1 引用了 命题 3
我们有 d ( x x − 1 ) = d ( 1 ) = 0 d(xx^{-1}) = d(1) = 0 d ( x x − 1 ) = d ( 1 ) = 0
( d x ) x − 1 + ε δ , deg ( x ) x ( d ( x − 1 ) ) = 0
(dx)x^{-1} + \varepsilon_{\delta, \deg(x)} x(d(x^{-1})) = 0
( d x ) x − 1 + ε δ , d e g ( x ) x ( d ( x − 1 )) = 0 这就证明了(19)的第一个公式。另一方面，
x − 1 x^{-1} x − 1 是次数
− deg ( x ) -\deg(x) − deg ( x ) 和
ε δ , deg ( x ) = ε δ , − deg ( x ) \varepsilon_{\delta, \deg(x)} = \varepsilon_{\delta, -\deg(x)} ε δ , d e g ( x ) = ε δ , − d e g ( x ) 齐次于no. 1的公式(1)；代入
d ( x − 1 x ) = 0 d(x^{-1}x) = 0 d ( x − 1 x ) = 0 ，类似地得到(19)的第二个公式。
假设A \mathbf{A} A L \mathbf{L} L A \mathbf{A} A E \mathbf{E} E L \mathbf{L} L A \mathbf{A} A L \mathbf{L} L E \mathbf{E} E
被以下引用 2 被以下引用 命题 23 , IV § 3. 。 有理分式
令d d d A \mathbf{A} A E \mathbf{E} E L \mathbf{L} L d ˉ \bar{d} d ˉ E \mathbf{E} E d d d u ∈ A , v ∈ A , v ≠ 0 u \in \mathbf{A}, v \in \mathbf{A}, v \neq 0 u ∈ A , v ∈ A , v = 0
d ˉ ( u / v ) = v − 1 d u − u v − 2 d v
\bar{d}(u/v) = v^{-1}du - uv^{-2}dv
d ˉ ( u / v ) = v − 1 d u − u v − 2 d v 这就证明了
d ˉ \bar{d} d ˉ 的唯一性。相反，我们证明
d ˉ \bar{d} d ˉ 可以由公式(20)定义；首先必须验证当
u u u 被
u ′ u' u ′ 替代，
v v v 被
v ′ v' v ′ 替代时，(20)右边的
u / v = u ′ / v ′ u/v = u'/v' u / v = u ′ / v ′ 值是否不是变换。现在，
u v ′ = v u ′ uv' = vu' u v ′ = v u ′ ，因此
v ′ ( d u ) + u ( d v ′ ) = v ( d u ′ ) + u ′ ( d v ) v'(du) + u(dv') = v(du') + u'(dv) v ′ ( d u ) + u ( d v ′ ) = v ( d u ′ ) + u ′ ( d v ) ，因此
v ′ ( d u − u v − 1 d v ) = v ( d u ′ − u ′ v ′ − 1 d v ′ ) v'(du - uv^{-1}dv) = v(du' - u'{v'}^{-1}dv') v ′ ( d u − u v − 1 d v ) = v ( d u ′ − u ′ v ′ − 1 d v ′ ) ，因为
u v ′ v − 1 = u ′ uv'v^{-1} = u' u v ′ v − 1 = u ′ 和
u ′ v ′ − 1 v = u u'{v'}^{-1}v = u u ′ v ′ − 1 v = u 。因此映射
d ˉ : L → E \bar{d}: \mathbf{L} \to \mathbf{E} d ˉ : L → E 被定义为扩展
d d d ；立即验证它是
K \mathbf{K} K -线性和推导。
假设 A \mathbf{A} A K \mathbf{K} K E \mathbf{E} E ( A , A ) (\mathbf{A}, \mathbf{A}) ( A , A ) d : A → E d : \mathbf{A} \to \mathbf{E} d : A → E δ \delta δ ε \varepsilon ε A \mathbf{A} A ( x i ) 1 ≤ i ≤ n (x_i)_{1 \leq i \leq n} ( x i ) 1 ≤ i ≤ n ξ i \xi_i ξ i 1 ≤ i ≤ n 1 \leq i \leq n 1 ≤ i ≤ n
d ( x 1 x 2 … x n ) = ∑ i = 1 n ε δ , ξ 1 + ⋯ + ξ i − 1 x 1 ⋯ x i − 1 ( d x i ) x i + 1 ⋯ x n .
d(x_1 x_2 \ldots x_n) = \sum_{i=1}^n \varepsilon_{\delta, \xi_1 + \cdots + \xi_{i-1}} x_1 \cdots x_{i-1} (dx_i) x_{i+1} \cdots x_n.
d ( x 1 x 2 … x n ) = i = 1 ∑ n ε δ , ξ 1 + ⋯ + ξ i − 1 x 1 ⋯ x i − 1 ( d x i ) x i + 1 ⋯ x n . 公式 (21) 对于
n = 0 n = 0 n = 0 来说是微不足道的，并由归纳在
n n n 上证明，考虑到 (4) (no. 2)。
被以下引用 1 被以下引用 命题 4, IV § 1
假设A \mathbf{A} A E \mathbf{E} E A \mathbf{A} A d : A → E d : \mathbf{A} \to \mathbf{E} d : A → E n ≥ 0 n \geq 0 n ≥ 0
d ( x n ) = n x n − 1 ( d x ) for all x ∈ A .
d(x^n) = nx^{n-1}(dx) \quad \text{for all } x \in \mathbf{A}.
d ( x n ) = n x n − 1 ( d x ) for all x ∈ A . 给予
A \mathbf{A} A 简单的毕业并应用 (21) 就足够了，所有
x i x_i x i 等于
x x x 。
我们回到次数δ \delta δ ε \varepsilon ε d : A → E d : \mathbf{A} \to \mathbf{E} d : A → E Z δ Z_\delta Z δ a ∈ A a \in \mathbf{A} a ∈ A α \alpha α a a a a α a_\alpha a α E \mathbf{E} E x x x
x a α = ε α , deg ( x ) a α x .
xa_\alpha = \varepsilon_{\alpha, \deg(x)} a_\alpha x.
x a α = ε α , d e g ( x ) a α x . 如果 A A A E E E ( A , A ) (A, A) ( A , A ) Z ε Z_\varepsilon Z ε A A A
假设 A A A E E E ( A , A ) (A, A) ( A , A ) d : A → E d : A \to E d : A → E δ \delta δ ε \varepsilon ε a a a α \alpha α Z ε Z_\varepsilon Z ε x ↦ a ( d x ) x \mapsto a(dx) x ↦ a ( d x ) δ + α \delta + \alpha δ + α ε \varepsilon ε
被以下引用 1 被以下引用 推论
我们写d ′ ( x ) = a ( d x ) d'(x) = a(dx) d ′ ( x ) = a ( d x ) A A A y ∈ A y \in A y ∈ A ξ \xi ξ x x x
d ′ ( x y ) = a ( ( d x ) y ) + ε δ , ξ a ( x ( d y ) ) = ( a ( d x ) ) y + ε δ + α , ξ ( x a ) ( d y ) = ( d ′ x ) y + ε δ + α , ξ x ( d ′ y ) .
d'(xy) = a((dx)y) + \varepsilon_{\delta, \xi} a(x(dy)) = (a(dx))y + \varepsilon_{\delta+\alpha, \xi}(xa)(dy)
= (d'x)y + \varepsilon_{\delta+\alpha, \xi} x(d'y).
d ′ ( x y ) = a (( d x ) y ) + ε δ , ξ a ( x ( d y )) = ( a ( d x )) y + ε δ + α , ξ ( x a ) ( d y ) = ( d ′ x ) y + ε δ + α , ξ x ( d ′ y ) . 命题 7 表示ε \varepsilon ε A A A E E E Δ \Delta Δ Z ε Z_\varepsilon Z ε
6. 推导代数
设A A A a ∈ A a \in A a ∈ A a d ε ( a ) \mathrm{ad}_\varepsilon(a) ad ε ( a ) a d ( a ) \mathrm{ad}(a) ad ( a ) A A A A A A
x ↦ [ a , x ] ε
x \mapsto [a, x]_\varepsilon
x ↦ [ a , x ] ε deg a \deg a deg a
(i) 对于ε \varepsilon ε -导数d : A → A d : A \to A d : A → A A A A a a a
[ d , a d ε ( a ) ] ε = a d ε ( d a ) .
[d, \mathrm{ad}_\varepsilon(a)]_\varepsilon = \mathrm{ad}_\varepsilon(da).
[ d , ad ε ( a ) ] ε = ad ε ( d a ) . (ii) 如果代数 A A A a d ε ( a ) \mathrm{ad}_\varepsilon(a) ad ε ( a ) deg ( a ) \deg(a) deg ( a ) A A A ε \varepsilon ε 导数。
(i) 假设d d d δ \delta δ α = deg a \alpha = \deg a α = deg a f = [ d , a d ε ( a ) ] ε f = [d, \mathrm{ad}_\varepsilon(a)]_\varepsilon f = [ d , ad ε ( a ) ] ε ξ \xi ξ x ∈ A x \in A x ∈ A
f ( x ) = d ( a x − ε ( α , ξ ) x a ) − ε δ , α ( a ( d x ) ) − ε α , δ + ξ ( d x ) a = ( d a ) x + ε δ , α a ( d x ) − ε α , ξ ( d x ) a − ε δ + α , ξ x ( d a ) − ε δ , α a ( d x ) + ε α , ξ ( d x ) a = ( d a ) x − ε δ + α , ξ x ( d a ) = [ d a , x ] ε .
f(x) = d(ax - \varepsilon(\alpha, \xi) xa) - \varepsilon_{\delta, \alpha}(a(dx)) - \varepsilon_{\alpha, \delta+\xi}(dx)a \\
= (da)x + \varepsilon_{\delta, \alpha} a(dx) - \varepsilon_{\alpha, \xi}(dx)a - \varepsilon_{\delta+\alpha, \xi} x(da) \\
- \varepsilon_{\delta, \alpha} a(dx) + \varepsilon_{\alpha, \xi}(dx)a \\
= (da)x - \varepsilon_{\delta+\alpha, \xi} x(da) = [da, x]_\varepsilon.
f ( x ) = d ( a x − ε ( α , ξ ) x a ) − ε δ , α ( a ( d x )) − ε α , δ + ξ ( d x ) a = ( d a ) x + ε δ , α a ( d x ) − ε α , ξ ( d x ) a − ε δ + α , ξ x ( d a ) − ε δ , α a ( d x ) + ε α , ξ ( d x ) a = ( d a ) x − ε δ + α , ξ x ( d a ) = [ d a , x ] ε . (ii) 对于A A A ξ \xi ξ x x x η \eta η y y y
a d ε ( a ) ( x y ) = a ( x y ) − ε α , ξ + η ( x y ) a = ( a x − ε α , ξ x a ) y + ε α , ξ x ( a y − ε α , η y a ) = a d ε ( a ) ( x ) . y + ε α , ξ x . a d ε ( a ) ( y )
\mathrm{ad}_\varepsilon(a)(xy) = a(xy) - \varepsilon_{\alpha, \xi+\eta}(xy)a \\
= (ax - \varepsilon_{\alpha, \xi} xa)y + \varepsilon_{\alpha, \xi} x(ay - \varepsilon_{\alpha, \eta} ya) \\
= \mathrm{ad}_\varepsilon(a)(x).y + \varepsilon_{\alpha, \xi} x.\mathrm{ad}_\varepsilon(a)(y)
ad ε ( a ) ( x y ) = a ( x y ) − ε α , ξ + η ( x y ) a = ( a x − ε α , ξ x a ) y + ε α , ξ x ( a y − ε α , η y a ) = ad ε ( a ) ( x ) . y + ε α , ξ x . ad ε ( a ) ( y ) 考虑 (1) 和
A A A 的结合性。
当A A A a d ε ( a ) \mathrm{ad}_\varepsilon(a) ad ε ( a ) a a a A A A ε \varepsilon ε
令 A A A a , b a, b a , b A A A
[ a d ε ( a ) , a d ε ( b ) ] ε = a d ε ( [ a , b ] ε ) .
[\mathrm{ad}_\varepsilon(a), \mathrm{ad}_\varepsilon(b)]_\varepsilon = \mathrm{ad}_\varepsilon([a, b]_\varepsilon).
[ ad ε ( a ) , ad ε ( b ) ] ε = ad ε ([ a , b ] ε ) . (25)
在(24)中，足以满足替换d d d a d ε ( a ) \mathrm{ad}_\varepsilon(a) ad ε ( a ) a d ε ( a ) \mathrm{ad}_\varepsilon(a) ad ε ( a ) a d ε ( b ) \mathrm{ad}_\varepsilon(b) ad ε ( b )
如果deg a = α , deg b = β \deg a = \alpha, \deg b = \beta deg a = α , deg b = β γ \gamma γ c ∈ A c \in A c ∈ A
ε α , γ [ a , [ b , c ] ε ] ε + ε β , α [ b , [ c , a ] ε ] ε + ε γ , β [ c , [ a , b ] ε ] ε = 0
\varepsilon_{\alpha, \gamma}[a, [b, c]_\varepsilon]_\varepsilon + \varepsilon_{\beta, \alpha}[b, [c, a]_\varepsilon]_\varepsilon + \varepsilon_{\gamma, \beta}[c, [a, b]_\varepsilon]_\varepsilon = 0
ε α , γ [ a , [ b , c ] ε ] ε + ε β , α [ b , [ c , a ] ε ] ε + ε γ , β [ c , [ a , b ] ε ] ε = 0 称为*雅可比恒等式*。
7. 函数性质
在这个数中，所有代数都被假定为结合且单位的，并且每个代数同态被假定为单位的。
令A , B A, B A , B K K K E E E ( A , A ) (A, A) ( A , A ) F F F ( B , B ) (B, B) ( B , B ) ρ : A → B \rho : A \to B ρ : A → B θ : E → F \theta : E \to F θ : E → F A A A A A A ρ \rho ρ
被以下引用 6 被以下引用 推论 , 命题 10 , 命题 11 , 命题 17 , 命题 19 , 命题 22
(i) 对于每个 ε \varepsilon ε d ′ : B → F d' : B \to F d ′ : B → F d ′ ∘ ρ : A → ρ ∗ ( F ) d' \circ \rho : A \to \rho^*(F) d ′ ∘ ρ : A → ρ ∗ ( F ) ε \varepsilon ε (ii) 对于每个 ε \varepsilon ε d : A → E d : A \to E d : A → E θ ∘ d : A → ρ ∗ ( F ) \theta \circ d : A \to \rho^*(F) θ ∘ d : A → ρ ∗ ( F ) ε \varepsilon ε
两个断言紧随关系而来
d ′ ( ρ ( x y ) ) = d ′ ( ρ ( x ) ρ ( y ) ) = d ′ ( ρ ( x ) ) ρ ( y ) + ε δ ′ , ξ ρ ( x ) d ′ ( ρ ( y ) )
d'(\rho(xy)) = d'(\rho(x)\rho(y)) = d'(\rho(x))\rho(y) + \varepsilon_{\delta', \xi}\rho(x)d'(\rho(y))
d ′ ( ρ ( x y )) = d ′ ( ρ ( x ) ρ ( y )) = d ′ ( ρ ( x )) ρ ( y ) + ε δ ′ , ξ ρ ( x ) d ′ ( ρ ( y )) θ ( d ( x y ) ) = θ ( ( d x ) y + ε δ , ξ x ( d y ) ) = θ ( d x ) ρ ( y ) + ε δ , ξ ρ ( x ) θ ( d y )
\theta(d(xy)) = \theta((dx)y + \varepsilon_{\delta, \xi}x(dy)) = \theta(dx)\rho(y) + \varepsilon_{\delta, \xi}\rho(x)\theta(dy)
θ ( d ( x y )) = θ (( d x ) y + ε δ , ξ x ( d y )) = θ ( d x ) ρ ( y ) + ε δ , ξ ρ ( x ) θ ( d y ) 对于次数、
ξ \xi ξ 和
y ∈ A y \in A y ∈ A 的
x ∈ A x \in A x ∈ A 齐次，
δ \delta δ 和
δ ′ \delta' δ ′ 分别表示
d d d 和
d ′ d' d ′ 的次数。
设S S S A A A d ′ ∘ ρ = θ ∘ d d' \circ \rho = \theta \circ d d ′ ∘ ρ = θ ∘ d d ′ ( ρ ( x ) ) = θ ( d ( x ) ) d'(\rho(x)) = \theta(d(x)) d ′ ( ρ ( x )) = θ ( d ( x )) x ∈ S x \in S x ∈ S
引用了 4 · 被以下引用 1 引用了 推论 , 命题 9 , II § 1. 。 模 , 例 1
被以下引用 命题 11, III § 11
这是命题 9 和 no. 5 的直接的推论，命题 3 的推论。
在命题 9 的条件下，我们知道B B B ρ \rho ρ ( A , A ) (A, A) ( A , A )
在命题 9 的条件下，对于B B B ρ ∗ ( F ) \rho^*(F) ρ ∗ ( F ) A A A ε \varepsilon ε d ′ : B → F d' : B \to F d ′ : B → F A A A d ′ d' d ′ B B B ρ ( A ) \rho(A) ρ ( A )
引用了 2 · 被以下引用 1 引用了 命题 9 , 命题 3
被以下引用 推论
我们为左A-模结构表演证明。对于a ∈ A , b ∈ B a \in A, b \in B a ∈ A , b ∈ B
d ′ ( ρ ( a ) b ) = d ′ ( ρ ( a ) ) b + ρ ( a ) d ′ b
d'(\rho(a)b) = d'(\rho(a))b + \rho(a)d'b
d ′ ( ρ ( a ) b ) = d ′ ( ρ ( a )) b + ρ ( a ) d ′ b 因此，如果
d ′ ∘ ρ = 0 d' \circ \rho = 0 d ′ ∘ ρ = 0 ，
d ′ d' d ′ 对于 B 和
ρ ∗ ( F ) \rho^*(F) ρ ∗ ( F ) 上的左A-模结构是线性的。相反，如果是这样的话，特别是
d ′ ( ρ ( a ) ) = d ′ ( ρ ( a ) .1 ) = ρ ( a ) d ′ ( 1 ) = 0
d'(\rho(a)) = d'(\rho(a).1) = \rho(a)d'(1) = 0
d ′ ( ρ ( a )) = d ′ ( ρ ( a ) .1 ) = ρ ( a ) d ′ ( 1 ) = 0 （no. 5，命题3）。
特别地，让D K ( B , F ) D_K(B, F) D K ( B , F ) ρ ( A ) \rho(A) ρ ( A ) D K ( B , F ) D_K(B, F) D K ( B , F ) D A , ρ ( B , F ) D_{A,\rho}(B, F) D A , ρ ( B , F ) D A ( B , F ) D_A(B, F) D A ( B , F ) D K ( B , F ) = D K , ϕ ( B , F ) D_K(B, F) = D_{K,\phi}(B, F) D K ( B , F ) = D K , ϕ ( B , F ) ϕ : K → B \phi : K \to B ϕ : K → B
现在让A，B，C为三级K-代数，ρ : A → B , σ : B → C \rho : A \to B, \sigma : B \to C ρ : A → B , σ : B → C D A ( B , G ) D_A(B, G) D A ( B , G ) D B ( C , G ) D_B(C, G) D B ( C , G ) D A ( C , G ) D_A(C, G) D A ( C , G ) D A , ρ ( B , σ ∗ ( G ) ) D_{A,\rho}(B, \sigma_*(G)) D A , ρ ( B , σ ∗ ( G )) D B , σ ( C , G ) D_{B,\sigma}(C, G) D B , σ ( C , G ) D A , σ ∘ ρ ( C , G ) D_{A,\sigma \circ \rho}(C, G) D A , σ ∘ ρ ( C , G ) D B ( C , G ) D_B(C, G) D B ( C , G ) D A ( C , G ) D_A(C, G) D A ( C , G ) σ ( ρ ( A ) ) ⊂ σ ( B ) \sigma(\rho(A)) \subset \sigma(B) σ ( ρ ( A )) ⊂ σ ( B )
在上面的条件下，有正合序列个K-同态*
0 → D B ( C , G ) → u D A ( C , G ) → v D A ( B , G )
0 \to D_B(C, G) \xrightarrow{u} D_A(C, G) \xrightarrow{v} D_A(B, G)
0 → D B ( C , G ) u D A ( C , G ) v D A ( B , G ) 其中 u 为典范嵌入，v 同态 d ↦ d ∘ σ d \mapsto d \circ \sigma d ↦ d ∘ σ (命题 9)。
引用了 1 · 被以下引用 2 引用了 命题 9
被以下引用 推论 , 命题 21
v 的核是推导 d : C → G d : C \to G d : C → G b ∈ B b \in B b ∈ B d ( σ ( b ) ) = 0 d(\sigma(b)) = 0 d ( σ ( b )) = 0
8. 关系推导和代数同态之间
我们再次假设这个没有。 no. 2 的 情况 (II) 成立，且分级 K-代数 A 不假定为结合律。给定元素 δ ∈ Δ \delta \in \Delta δ ∈ Δ δ \delta δ
( E ( δ ) ) μ = E μ + δ
(E(\delta))_\mu = E_{\mu+\delta}
( E ( δ ) ) μ = E μ + δ μ ∈ Δ \mu \in \Delta μ ∈ Δ A ⊕ E ( δ ) A \oplus E(\delta) A ⊕ E ( δ ) E ( δ ) E(\delta) E ( δ ) a ∈ A a \in A a ∈ A a ′ ∈ A , x , x ′ a' \in A, x, x' a ′ ∈ A , x , x ′ ( a , x ) ( a ′ , x ′ ) = ( a a ′ , x . a ′ + ε δ , deg ( a ) a . x ′ ) ;
(a, x)(a', x') = (aa', x.a' + \varepsilon_{\delta, \deg(a)} a.x');
( a , x ) ( a ′ , x ′ ) = ( a a ′ , x . a ′ + ε δ , d e g ( a ) a . x ′ ) ; 投影p : ( a , x ) ↦ a p : (a, x) \mapsto a p : ( a , x ) ↦ a A ⊕ E ( δ ) A \oplus E(\delta) A ⊕ E ( δ ) g : A → A ⊕ E ( δ ) g : A \to A \oplus E(\delta) g : A → A ⊕ E ( δ )
A → g A ⊕ E ( δ ) → p A
A \xrightarrow{g} A \oplus E(\delta) \xrightarrow{p} A
A g A ⊕ E ( δ ) p A 是身份1 A 1_A 1 A x ↦ ( x , f ( x ) ) x \mapsto (x, f(x)) x ↦ ( x , f ( x )) f : A → E f : A \to E f : A → E δ \delta δ
对于次数δ \delta δ f : A → E f : A \to E f : A → E ε \varepsilon ε A ⊕ E ( δ ) A \oplus E(\delta) A ⊕ E ( δ ) x ↦ ( x , f ( x ) ) x \mapsto (x, f(x)) x ↦ ( x , f ( x ))
被以下引用 1 被以下引用 命题 14
利用以下事实：对于 A A A y ∈ A y \in A y ∈ A x x x
( x y , f ( x y ) ) = ( x , f ( x ) ) . ( y , f ( y ) ) ,
(xy, f(xy)) = (x, f(x)).(y, f(y)),
( x y , f ( x y )) = ( x , f ( x )) . ( y , f ( y )) , 考虑到 (28)，我们得到等价的关系
f ( x y ) = f ( x ) . y + ε δ , deg ( x ) x . f ( y ) ,
f(xy) = f(x).y + \varepsilon_{\delta, \deg(x)} x.f(y),
f ( x y ) = f ( x ) . y + ε δ , d e g ( x ) x . f ( y ) , 从而命题。
为了使代数 A ⊕ E ( δ ) A \oplus E(\delta) A ⊕ E ( δ ) A A A E E E ( A , A ) (A, A) ( A , A ) ( a , x ) ↦ a . x (a, x) \mapsto a.x ( a , x ) ↦ a . x ( a , x ) ↦ x . a (a, x) \mapsto x.a ( a , x ) ↦ x . a A ⊕ E ( δ ) A \oplus E(\delta) A ⊕ E ( δ ) ( 1 , 0 ) (1, 0) ( 1 , 0 )
被以下引用 1 被以下引用 命题 14
如果把元素( u , m ) ∈ A ⊕ E ( δ ) (u, m) \in A \oplus E(\delta) ( u , m ) ∈ A ⊕ E ( δ ) u u u A A A ( u , m ) . ( 0 , x ) = ( 0 , x ) . ( u , m ) = ( 0 , x ) (u, m).(0, x) = (0, x).(u, m) = (0, x) ( u , m ) . ( 0 , x ) = ( 0 , x ) . ( u , m ) = ( 0 , x ) x ∈ E x \in E x ∈ E u . x = x . u = x u.x = x.u = x u . x = x . u = x ( u , m ) . ( u , 0 ) = ( u , 0 ) . ( u , m ) = ( u , 0 ) (u, m).(u, 0) = (u, 0).(u, m) = (u, 0) ( u , m ) . ( u , 0 ) = ( u , 0 ) . ( u , m ) = ( u , 0 ) m = 0 m = 0 m = 0 A ⊕ E ( δ ) A \oplus E(\delta) A ⊕ E ( δ ) A A A ( x . a ′ ) . a ′ ′ = x . ( a ′ . a ′ ′ ) (x.a').a'' = x.(a'.a'') ( x . a ′ ) . a ′′ = x . ( a ′ . a ′′ ) ( ( 0 , x ) ( a ′ , 0 ) ) ( a ′ ′ , 0 ) = ( 0 , x ) ( ( a ′ , 0 ) ( a ′ ′ , 0 ) ) ((0, x)(a', 0))(a'', 0) = (0, x)((a', 0)(a'', 0)) (( 0 , x ) ( a ′ , 0 )) ( a ′′ , 0 ) = ( 0 , x ) (( a ′ , 0 ) ( a ′′ , 0 )) a . ( a ′ . x ) = ( a . a ′ ) . x a.(a'.x) = (a.a').x a . ( a ′ . x ) = ( a . a ′ ) . x
( a , 0 ) ( ( a ′ , 0 ) ( 0 , x ) ) = ( ( a , 0 ) ( a ′ , 0 ) ) ( 0 , x ) ;
(a, 0)((a', 0)(0, x)) = ((a, 0)(a', 0))(0, x);
( a , 0 ) (( a ′ , 0 ) ( 0 , x )) = (( a , 0 ) ( a ′ , 0 )) ( 0 , x ) ; 最后条件 a . ( x . a ′ ) = ( a . x ) . a ′ a.(x.a') = (a.x).a' a . ( x . a ′ ) = ( a . x ) . a ′
( a , 0 ) ( ( 0 , x ) ( a ′ , 0 ) ) = ( ( a , 0 ) ( 0 , x ) ) ( a ′ , 0 ) .
(a, 0)((0, x)(a', 0)) = ((a, 0)(0, x))(a', 0).
( a , 0 ) (( 0 , x ) ( a ′ , 0 )) = (( a , 0 ) ( 0 , x )) ( a ′ , 0 ) .
9. 扩张推导
令A A A M M M A A A B B B A A A T ( M ) T(M) T ( M ) S ( M ) S(M) S ( M ) Λ ( M ) \Lambda(M) Λ ( M ) E E E ( B , B ) (B, B) ( B , B ) d 0 : A → E d_0 : A \to E d 0 : A → E A \mathbf{A} A A \mathbf{A} A E \mathbf{E} E d 1 : M → E d_1 : M \to \mathbf{E} d 1 : M → E a ∈ A a \in \mathbf{A} a ∈ A x ∈ M x \in M x ∈ M
引用了 4 · 被以下引用 2 引用了 命题 1 , 命题 12 , 命题 13 , 命题 2
被以下引用 例 , 推论
( 29 ) d 1 ( a x ) = a d 1 ( x ) + d 0 ( a ) ⋅ x ,
(29) \quad d_1(ax) = ad_1(x) + d_0(a) \cdot x,
( 29 ) d 1 ( a x ) = a d 1 ( x ) + d 0 ( a ) ⋅ x , 进一步，当B = S ( M ) B = S(M) B = S ( M )
( 30 ) x ⋅ d 1 ( y ) + d 1 ( x ) ⋅ y = y ⋅ d 1 ( x ) + d 1 ( y ) ⋅ x
(30) \quad x \cdot d_1(y) + d_1(x) \cdot y = y \cdot d_1(x) + d_1(y) \cdot x
( 30 ) x ⋅ d 1 ( y ) + d 1 ( x ) ⋅ y = y ⋅ d 1 ( x ) + d 1 ( y ) ⋅ x 对于M M M x , y x, y x , y B = ⋀ ( M ) B = \bigwedge(M) B = ⋀ ( M )
( 31 ) x ⋅ d 1 ( x ) + d 1 ( x ) ⋅ x = 0
(31) \quad x \cdot d_1(x) + d_1(x) \cdot x = 0
( 31 ) x ⋅ d 1 ( x ) + d 1 ( x ) ⋅ x = 0 对于所有x ∈ M x \in M x ∈ M ( B , B ) (B, B) ( B , B ) E \mathbf{E} E B B B Z \mathbf{Z} Z d d d d ∣ A = d 0 d|_{\mathbf{A}} = d_0 d ∣ A = d 0 d ∣ M = d 1 d|_{M} = d_1 d ∣ M = d 1
我们采用 Z \mathbf{Z} Z B ⊕ E B \oplus \mathbf{E} B ⊕ E Z \mathbf{Z} Z
( b , t ) ( b ′ , t ′ ) = ( b b ′ , b t ′ + t b ′ )
(b, t)(b', t') = (bb', bt' + tb')
( b , t ) ( b ′ , t ′ ) = ( b b ′ , b t ′ + t b ′ ) 其中 ( 1 , 0 ) (1, 0) ( 1 , 0 ) t ↦ ( 0 , t ) t \mapsto (0, t) t ↦ ( 0 , t ) E \mathbf{E} E B ⊕ E B \oplus \mathbf{E} B ⊕ E E 2 = { 0 } \mathbf{E}^2 = \{0\} E 2 = { 0 } h 0 ( a ) = ( a , d 0 ( a ) ) h_0(a) = (a, d_0(a)) h 0 ( a ) = ( a , d 0 ( a )) h 0 : B ⊕ E h_0 : B \oplus \mathbf{E} h 0 : B ⊕ E B ⊕ E B \oplus \mathbf{E} B ⊕ E A \mathbf{A} A x ∈ M x \in M x ∈ M h 1 ( x ) = ( x , d 1 ( x ) ) h_1(x) = (x, d_1(x)) h 1 ( x ) = ( x , d 1 ( x )) h 0 h_0 h 0 h 1 ( a x ) = h 0 ( a ) h 1 ( x ) h_1(ax) = h_0(a)h_1(x) h 1 ( a x ) = h 0 ( a ) h 1 ( x ) h 1 h_1 h 1 M M M B ⊕ E B \oplus \mathbf{E} B ⊕ E A \mathbf{A} A A \mathbf{A} A h : B → B ⊕ E h : B \to B \oplus \mathbf{E} h : B → B ⊕ E h ∣ M = h 1 h|_M = h_1 h ∣ M = h 1 h ∣ A = h 0 h|_{\mathbf{A}} = h_0 h ∣ A = h 0 B = T ( M ) B = T(M) B = T ( M ) § 5 § 5 §5 B = S ( M ) B = S(M) B = S ( M ) h ( x ) h ( y ) = h ( y ) h ( x ) h(x)h(y) = h(y)h(x) h ( x ) h ( y ) = h ( y ) h ( x ) M M M x , y x, y x , y § 6 § 6 §6 B = ⋀ ( M ) B = \bigwedge(M) B = ⋀ ( M ) x ∧ x = 0 x \wedge x = 0 x ∧ x = 0 x ∈ M x \in M x ∈ M ( h ( x ) ) 2 = 0 (h(x))^2 = 0 ( h ( x ) ) 2 = 0 § 7 § 7 §7 h h h p ∘ h : B → B p \circ h : B \to B p ∘ h : B → B ρ : B ⊕ E → B \rho : B \oplus \mathbf{E} \to B ρ : B ⊕ E → B l B l_B l B p ∘ h p \circ h p ∘ h l B l_B l B A \mathbf{A} A M M M B B B h ( b ) = ( b , d ( b ) ) h(b) = (b, d(b)) h ( b ) = ( b , d ( b )) b ∈ B b \in B b ∈ B b ↦ d ( b ) b \mapsto d(b) b ↦ d ( b ) B B B E \mathbf{E} E
令M M M Δ \Delta Δ K K K K K K T ( M ) T(M) T ( M ) S ( M ) S(M) S ( M ) ⋀ ( M ) \bigwedge(M) ⋀ ( M ) Δ ′ = Δ × Z \Delta' = \Delta \times \mathbf{Z} Δ ′ = Δ × Z § 5 § 5 §5 § 6 § 6 §6 § 7 § 7 §7 M M M Δ ′ \Delta' Δ ′ α ∈ Δ \alpha \in \Delta α ∈ Δ M α , 1 = M α M_{\alpha, 1} = M_{\alpha} M α , 1 = M α α ∈ Δ \alpha \in \Delta α ∈ Δ n ≠ 1 n \neq 1 n = 1 M α , n = { 0 } M_{\alpha, n} = \{0\} M α , n = { 0 } ε ′ \varepsilon' ε ′ Δ ′ \Delta' Δ ′
引用了 4 · 被以下引用 1 引用了 命题 10 , 命题 11 , 命题 7 , 命题 14
被以下引用 命题 15
(i) 令 E 为型 Δ ′ \Delta' Δ ′ T ( M ) \mathbf{T}(M) T ( M ) δ ∈ Δ \delta \in \Delta δ ∈ Δ n ∈ Z n \in \mathbf{Z} n ∈ Z δ 1 ′ = ( δ , n ) \delta_1' = (\delta, n) δ 1 ′ = ( δ , n ) f : M → E f : M \to E f : M → E ε ′ \varepsilon' ε ′ d : T ( M ) → E d : \mathbf{T}(M) \to E d : T ( M ) → E δ ′ \delta' δ ′
(ii) 令 E 为型 Δ ′ \Delta' Δ ′ S ( M ) S(M) S ( M ) δ ′ \delta' δ ′ f : M → E f : M \to E f : M → E δ ′ \delta' δ ′ ε ′ \varepsilon' ε ′ d : S ( M ) → E d : S(M) \to E d : S ( M ) → E ( x , y ) (x, y) ( x , y )
x . f ( y ) + ε δ ′ , ( deg ( y ) , 1 ) y . f ( x ) = y . f ( x ) + ε δ ′ , ( deg ( x ) , 1 ) x . f ( y ) .
x . f(y) + \varepsilon_{\delta', (\deg(y), 1)} y . f(x) = y . f(x) + \varepsilon_{\delta', (\deg(x), 1)} x . f(y).
x . f ( y ) + ε δ ′ , ( d e g ( y ) , 1 ) y . f ( x ) = y . f ( x ) + ε δ ′ , ( d e g ( x ) , 1 ) x . f ( y ) . 那么ε ′ \varepsilon' ε ′
(iii) 令 E 为型 Δ ′ \Delta' Δ ′ Λ ( M ) \Lambda(M) Λ ( M ) δ ′ \delta' δ ′ f : M → E f : M \to E f : M → E ε ′ \varepsilon' ε ′
d : Λ ( M ) → E
d : \Lambda(M) \to E
d : Λ ( M ) → E 次数 δ ′ \delta' δ ′
x . f ( x ) + ε δ ′ , ( deg ( x ) , 1 ) f ( x ) . x = 0.
x . f(x) + \varepsilon_{\delta', (\deg(x), 1)} f(x) . x = 0.
x . f ( x ) + ε δ ′ , ( d e g ( x ) , 1 ) f ( x ) . x = 0. 那么ε ′ \varepsilon' ε ′
no. 2 的注 2 应用于 E 上的外部B-模律之一（B 等于T ( M ) \mathbf{T}(M) T ( M ) S ( M ) S(M) S ( M ) Λ ( M ) \Lambda(M) Λ ( M ) A = K A = K A = K d 0 = 0 d_0 = 0 d 0 = 0
例 (1)。在命题 14 注的应用中，如果 d 0 = 0 d_0 = 0 d 0 = 0 d 1 d_1 d 1 E = B E = B E = B d 1 d_1 d 1 M → B M \to B M → B S ( M ) S(M) S ( M ) s ( x ) s(x) s ( x ) Λ ( M ) \Lambda(M) Λ ( M ) T ( M ) \mathbf{T}(M) T ( M ) D s D_s D s S ( M ) S(M) S ( M ) Λ ( M ) \Lambda(M) Λ ( M )
[ D s , D t ] = D [ s , t ]
[D_s, D_t] = D_{[s, t]}
[ D s , D t ] = D [ s , t ] 两边都是 T ( M ) \mathbf{T}(M) T ( M ) S ( M ) S(M) S ( M ) Λ ( M ) \Lambda(M) Λ ( M ) [ s , t ] [s, t] [ s , t ]
D s D_s D s M M M x 1 , x 2 , … , x n x_1, x_2, \ldots, x_n x 1 , x 2 , … , x n
{ D s ( x 1 ⊗ x 2 ⊗ ⋯ ⊗ x n ) D s ( x 1 ⊗ x 2 ⊗ ⋯ ⊗ x n ) = ∑ i = 1 n x 1 ⊗ ⋯ ⊗ x i − 1 ⊗ s ( x i ) ⊗ x i + 1 ⊗ ⋯ ⊗ x n D s ( x 1 x 2 … x n ) = ∑ i = 1 n x 1 … x i − 1 s ( x i ) x i + 1 … x n D s ( x 1 ∧ x 2 ∧ ⋯ ∧ x n ) D s ( x 1 ∧ x 2 ∧ ⋯ ∧ x n ) = ∑ i = 1 n x 1 ∧ ⋯ ∧ x i − 1 ∧ s ( x i ) ∧ x i + 1 ∧ ⋯ ∧ x n .
\begin{cases}
D_s(x_1 \otimes x_2 \otimes \cdots \otimes x_n) \\
\phantom{D_s(x_1 \otimes x_2 \otimes \cdots \otimes x_n)} = \sum_{i=1}^n x_1 \otimes \cdots \otimes x_{i-1} \otimes s(x_i) \otimes x_{i+1} \otimes \cdots \otimes x_n \\
D_s(x_1 x_2 \ldots x_n) = \sum_{i=1}^n x_1 \ldots x_{i-1} s(x_i) x_{i+1} \ldots x_n \\
D_s(x_1 \wedge x_2 \wedge \cdots \wedge x_n) \\
\phantom{D_s(x_1 \wedge x_2 \wedge \cdots \wedge x_n)} = \sum_{i=1}^n x_1 \wedge \cdots \wedge x_{i-1} \wedge s(x_i) \wedge x_{i+1} \wedge \cdots \wedge x_n.
\end{cases}
⎩ ⎨ ⎧ D s ( x 1 ⊗ x 2 ⊗ ⋯ ⊗ x n ) D s ( x 1 ⊗ x 2 ⊗ ⋯ ⊗ x n ) = ∑ i = 1 n x 1 ⊗ ⋯ ⊗ x i − 1 ⊗ s ( x i ) ⊗ x i + 1 ⊗ ⋯ ⊗ x n D s ( x 1 x 2 … x n ) = ∑ i = 1 n x 1 … x i − 1 s ( x i ) x i + 1 … x n D s ( x 1 ∧ x 2 ∧ ⋯ ∧ x n ) D s ( x 1 ∧ x 2 ∧ ⋯ ∧ x n ) = ∑ i = 1 n x 1 ∧ ⋯ ∧ x i − 1 ∧ s ( x i ) ∧ x i + 1 ∧ ⋯ ∧ x n . 以代数Λ ( M ) \Lambda(M) Λ ( M ) D s D_s D s
令( e j ) 1 ≤ j ≤ n (e_j)_{1 \leq j \leq n} ( e j ) 1 ≤ j ≤ n M M M e = e 1 ∧ e 2 ∧ ⋯ ∧ e n e = e_1 \wedge e_2 \wedge \cdots \wedge e_n e = e 1 ∧ e 2 ∧ ⋯ ∧ e n
s ( e j ) = ∑ k = 1 n α j k e k ,
s(e_j) = \sum_{k=1}^n \alpha_{jk} e_k,
s ( e j ) = k = 1 ∑ n α j k e k , (36) 中的第三个公式给出
D s ( e ) = ∑ i = 1 n e 1 ∧ ⋯ ∧ e i − 1 ∧ s ( e i ) ∧ e i + 1 ∧ ⋯ ∧ e n = ( ∑ j = 1 n α j j ) e .
D_s(e) = \sum_{i=1}^n e_1 \wedge \cdots \wedge e_{i-1} \wedge s(e_i) \wedge e_{i+1} \wedge \cdots \wedge e_n = \left( \sum_{j=1}^n \alpha_{jj} \right) e.
D s ( e ) = i = 1 ∑ n e 1 ∧ ⋯ ∧ e i − 1 ∧ s ( e i ) ∧ e i + 1 ∧ ⋯ ∧ e n = ( j = 1 ∑ n α j j ) e . 例 (2). 在命题 14、部分 (iii) 的推论中，设Δ = { 0 } \Delta = \{0\} Δ = { 0 } Λ ( M ) \Lambda(M) Λ ( M ) Z \mathbf{Z} Z ε ( p , q ) = ( − 1 ) p q \varepsilon(p, q) = (-1)^{pq} ε ( p , q ) = ( − 1 ) pq M M M x ∗ ∈ M ∗ x^* \in M^* x ∗ ∈ M ∗ x ↦ ⟨ x , x ∗ ⟩ x \mapsto \langle x, x^* \rangle x ↦ ⟨ x , x ∗ ⟩ M M M − 1 -1 − 1 Λ ( M ) \Lambda(M) Λ ( M ) K K K Λ ( M ) \Lambda(M) Λ ( M ) − 1 -1 − 1 i ( x ∗ ) i(x^*) i ( x ∗ )
i ( x ∗ ) ( x 1 ∧ ⋯ ∧ x n ) = ∑ i = 1 n ( − 1 ) i − 1 ⟨ x i , x ∗ ⟩ x 1 ∧ ⋯ ∧ x i − 1 ∧ x i + 1 ∧ ⋯ ∧ x n
i(x^*)(x_1 \wedge \cdots \wedge x_n)
= \sum_{i=1}^n (-1)^{i-1} \langle x_i, x^* \rangle x_1 \wedge \cdots \wedge x_{i-1} \wedge x_{i+1} \wedge \cdots \wedge x_n
i ( x ∗ ) ( x 1 ∧ ⋯ ∧ x n ) = i = 1 ∑ n ( − 1 ) i − 1 ⟨ x i , x ∗ ⟩ x 1 ∧ ⋯ ∧ x i − 1 ∧ x i + 1 ∧ ⋯ ∧ x n 这是内部乘积的特殊情况，在§ 11, no. 9，公式（68）中定义。
令A A A K K K M i M_i M i 1 ≤ i ≤ n 1 \leq i \leq n 1 ≤ i ≤ n P P P A A A H H H A A A A A A M 1 × M 2 × ⋯ × M n M_1 \times M_2 \times \cdots \times M_n M 1 × M 2 × ⋯ × M n d 0 : A → A d_0 : A \to A d 0 : A → A d i : M i → M i d_i : M_i \to M_i d i : M i → M i D : P → P D : P \to P D : P → P 1 ≤ i ≤ n 1 \leq i \leq n 1 ≤ i ≤ n ( d 0 , d i , d i ) (d_0, d_i, d_i) ( d 0 , d i , d i ) ( A , M i , M i ) (A, M_i, M_i) ( A , M i , M i ) ( d 0 , D , D ) (d_0, D, D) ( d 0 , D , D ) ( A , P , P ) (A, P, P) ( A , P , P ) D ′ : H → H D' : H \to H D ′ : H → H ( d 0 , D ′ , D ′ ) (d_0, D', D') ( d 0 , D ′ , D ′ ) ( A , H , H ) (A, H, H) ( A , H , H )
被以下引用 1 被以下引用 例
(37) D ( f ( x 1 , … , x n ) ) D(f(x_1, \ldots, x_n)) D ( f ( x 1 , … , x n ))
= ( D ′ f ) ( x 1 , … , x n ) + ∑ i = 1 n f ( x 1 , … , x i − 1 , d i x i , x i + 1 , … , x n )
= (D'f)(x_1, \ldots, x_n) + \sum_{i=1}^n f(x_1, \ldots, x_{i-1}, d_ix_i, x_{i+1}, \ldots, x_n)
= ( D ′ f ) ( x 1 , … , x n ) + i = 1 ∑ n f ( x 1 , … , x i − 1 , d i x i , x i + 1 , … , x n ) 对于所有x i ∈ M i x_i \in M_i x i ∈ M i 1 ≤ i ≤ n 1 \leq i \leq n 1 ≤ i ≤ n f ∈ H f \in H f ∈ H
我们证明，对于 f ∈ H f \in H f ∈ H M 1 × M 2 × ⋯ × M n M_1 \times M_2 \times \cdots \times M_n M 1 × M 2 × ⋯ × M n D ′ f D'f D ′ f a ∈ A a \in A a ∈ A
( D ′ f ) ( a x 1 , x 2 , … , x n ) = D ( a f ( x 1 , … , x n ) ) − f ( d 1 ( a x 1 ) , x 2 , … , x n )
(D'f)(ax_1, x_2, \ldots, x_n) = D(af(x_1, \ldots, x_n)) - f(d_1(ax_1), x_2, \ldots, x_n)
( D ′ f ) ( a x 1 , x 2 , … , x n ) = D ( a f ( x 1 , … , x n )) − f ( d 1 ( a x 1 ) , x 2 , … , x n ) − a ∑ i = 2 n f ( x 1 , … , x i − 1 , d i x i , x i + 1 , … , x n )
-a \sum_{i=2}^n f(x_1, \ldots, x_{i-1}, d_ix_i, x_{i+1}, \ldots, x_n)
− a i = 2 ∑ n f ( x 1 , … , x i − 1 , d i x i , x i + 1 , … , x n ) 到假设
D ( a f ( x 1 , … , x n ) ) = ( d 0 a ) f ( x 1 , … , x n ) + a D ( f ( x 1 , … , x n ) )
D(af(x_1, \ldots, x_n)) = (d_0a)f(x_1, \ldots, x_n) + aD(f(x_1, \ldots, x_n))
D ( a f ( x 1 , … , x n )) = ( d 0 a ) f ( x 1 , … , x n ) + a D ( f ( x 1 , … , x n )) 和 d 1 ( a x 1 ) = ( d 0 a ) x 1 + a . d 1 x 1 d_1(ax_1) = (d_0a)x_1 + a.d_1x_1 d 1 ( a x 1 ) = ( d 0 a ) x 1 + a . d 1 x 1
( D ′ f ) ( a x 1 , x 2 , … , x n ) = a . ( D ′ f ) ( x 1 , … , x n )
(D'f)(ax_1, x_2, \ldots, x_n) = a.(D'f)(x_1, \ldots, x_n)
( D ′ f ) ( a x 1 , x 2 , … , x n ) = a . ( D ′ f ) ( x 1 , … , x n ) 并且类似地证明了每个x i x_i x i
( D ′ ( a f ) ) ( x 1 , … , x n ) = D ( a f ( x 1 , … , x n ) )
(D'(af))(x_1, \ldots, x_n) = D(af(x_1, \ldots, x_n))
( D ′ ( a f )) ( x 1 , … , x n ) = D ( a f ( x 1 , … , x n )) − ∑ i = 1 n a f ( x 1 , … , x i − 1 , d i x i , x i + 1 , … , x n )
- \sum_{i=1}^n af(x_1, \ldots, x_{i-1}, d_ix_i, x_{i+1}, \ldots, x_n)
− i = 1 ∑ n a f ( x 1 , … , x i − 1 , d i x i , x i + 1 , … , x n ) = ( d 0 a ) f ( x 1 , … , x n ) ) + a D ( f ( x 1 , … , x n ) )
= (d_0a)f(x_1, \ldots, x_n)) + aD(f(x_1, \ldots, x_n))
= ( d 0 a ) f ( x 1 , … , x n )) + a D ( f ( x 1 , … , x n )) − ∑ i = 1 n a f ( x 1 , … , x i − 1 , d i x i , x i + 1 , … , x n )
- \sum_{i=1}^n af(x_1, \ldots, x_{i-1}, d_ix_i, x_{i+1}, \ldots, x_n)
− i = 1 ∑ n a f ( x 1 , … , x i − 1 , d i x i , x i + 1 , … , x n ) = ( d 0 a ) f ( x 1 , … , x n ) + a ( D ′ f ) ( x 1 , … , x n )
= (d_0a)f(x_1, \ldots, x_n) + a(D'f)(x_1, \ldots, x_n)
= ( d 0 a ) f ( x 1 , … , x n ) + a ( D ′ f ) ( x 1 , … , x n ) 其他字
D ′ ( a f ) = ( d 0 a ) f + a ( D ′ f )
D'(af) = (d_0a)f + a(D'f)
D ′ ( a f ) = ( d 0 a ) f + a ( D ′ f ) 从而建立命题。
(3) 应用命题 16 到情况n = 1 n = 1 n = 1 M 1 = M M_1 = M M 1 = M P = A P = A P = A H = M ∗ H = M^* H = M ∗ ( A , M , M ) (A, M, M) ( A , M , M ) ( d 0 , d , d ′ ) (d_0, d, d') ( d 0 , d , d ′ ) ( d 0 , d ∗ , d ∗ ) (d_0, d^*, d^*) ( d 0 , d ∗ , d ∗ ) ( A , M ∗ , M ∗ ) (A, M^*, M^*) ( A , M ∗ , M ∗ )
(38) d 0 ⟨ m , m ∗ ⟩ = ⟨ d m , m ∗ ⟩ + ⟨ m , d ∗ m ∗ ⟩ d_0\langle m, m^*\rangle = \langle dm, m^*\rangle + \langle m, d^*m^*\rangle d 0 ⟨ m , m ∗ ⟩ = ⟨ d m , m ∗ ⟩ + ⟨ m , d ∗ m ∗ ⟩
对于 m ∈ M m \in M m ∈ M m ∗ ∈ M ∗ m^* \in M^* m ∗ ∈ M ∗ M ⊕ M ∗ M \oplus M^* M ⊕ M ∗ M M M d d d M ∗ M^* M ∗ d ∗ d^* d ∗ A A A K K K D D D T ( M ⊕ M ∗ ) T(M \oplus M^*) T ( M ⊕ M ∗ ) A A A d 0 d_0 d 0 M M M d d d M ∗ M^* M ∗ d ∗ d^* d ∗ D D D d J I d_J^I d J I T ( M ⊕ M ∗ ) T(M \oplus M^*) T ( M ⊕ M ∗ ) A A A T J I ( M ) T_J^I(M) T J I ( M ) § 5 § 5 §5 T J I ( M ) T_J^I(M) T J I ( M ) K K K ( d 0 , d J I , d J I ) (d_0, d_J^I, d_J^I) ( d 0 , d J I , d J I ) ( A , T J I ( M ) , T J I ( M ) ) (A, T_J^I(M), T_J^I(M)) ( A , T J I ( M ) , T J I ( M )) K K K i ∈ I , j ∈ J i \in I, j \in J i ∈ I , j ∈ J I ′ = I − { i } , J ′ = J − { j } I' = I - \{i\}, J' = J - \{j\} I ′ = I − { i } , J ′ = J − { j } c j i c_j^i c j i § 5 § 5 §5
c j i ( d J I ( z ) ) = d J ′ I ′ ( c j i ( z ) ) for all z ∈ T J I ( M ) .
c_j^i(d_J^I(z)) = d_{J'}^{I'}(c_j^i(z)) \quad \text{for all } z \in T_J^I(M).
c j i ( d J I ( z )) = d J ′ I ′ ( c j i ( z )) for all z ∈ T J I ( M ) . (4) 设M i M_i M i 1 ≤ i ≤ 3 1 \leq i \leq 3 1 ≤ i ≤ 3 A A A i i i ( d 0 , d i , d i ) (d_0, d_i, d_i) ( d 0 , d i , d i ) ( A , M i , M i ) (A, M_i, M_i) ( A , M i , M i ) n = 1 n = 1 n = 1 ( A , H i j , H i j ) (A, H_{ij}, H_{ij}) ( A , H ij , H ij ) ( d 0 , d i j , d i j ) (d_0, d_{ij}, d_{ij}) ( d 0 , d ij , d ij ) H i j = Hom A ( M i , M j ) H_{ij} = \operatorname{Hom}_A(M_i, M_j) H ij = Hom A ( M i , M j ) ( i , j ) (i, j) ( i , j ) u ∈ Hom A ( M 1 , M 2 ) u \in \operatorname{Hom}_A(M_1, M_2) u ∈ Hom A ( M 1 , M 2 ) v ∈ Hom A ( M 2 , M 3 ) v \in \operatorname{Hom}_A(M_2, M_3) v ∈ Hom A ( M 2 , M 3 )
d 13 ( v ∘ u ) = ( d 23 v ) ∘ u + v ∘ ( d 12 u )
d_{13}(v \circ u) = (d_{23}v) \circ u + v \circ (d_{12}u)
d 13 ( v ∘ u ) = ( d 23 v ) ∘ u + v ∘ ( d 12 u ) 从定义立即得到验证。
引用了 1 · 被以下引用 1 引用了 命题 16
被以下引用 LIE III App. 0. 附录
10. 通用问题适用于推导；非交换情况
在 § 10 § 10 §10
令 A A A K K K A ⊗ K A A \otimes_K A A ⊗ K A ( A , A ) (A, A) ( A , A )
x . ( u ⊗ v ) . y = ( x u ) ⊗ ( v y )
x . (u \otimes v) . y = (xu) \otimes (vy)
x . ( u ⊗ v ) . y = ( xu ) ⊗ ( v y ) A A A x , y , u , v x, y, u, v x , y , u , v § 4 § 4 §4 A A A K K K m : A ⊗ K A → A m : A \otimes_K A \to A m : A ⊗ K A → A m ( x ⊗ y ) = x y m(x \otimes y) = xy m ( x ⊗ y ) = x y ( A , A ) (A, A) ( A , A ) I I I A ⊗ K A A \otimes_K A A ⊗ K A
映射δ A : x ↦ x ⊗ 1 − 1 ⊗ x \delta_A : x \mapsto x \otimes 1 - 1 \otimes x δ A : x ↦ x ⊗ 1 − 1 ⊗ x A A A I I I K K K I I I δ A \delta_A δ A A A A
被以下引用 2 被以下引用 § 10. 。 推导 , 命题 17
第一个断言源自以下事实：
( x y ) ⊗ 1 − 1 ⊗ ( x y ) = ( x ⊗ 1 − 1 ⊗ x ) . y + x . ( y ⊗ 1 − 1 ⊗ y )
(xy) \otimes 1 - 1 \otimes (xy) = (x \otimes 1 - 1 \otimes x) . y + x . (y \otimes 1 - 1 \otimes y)
( x y ) ⊗ 1 − 1 ⊗ ( x y ) = ( x ⊗ 1 − 1 ⊗ x ) . y + x . ( y ⊗ 1 − 1 ⊗ y ) 由（40）。另一方面，如果元素
∑ i x i ⊗ y i \sum_i x_i \otimes y_i ∑ i x i ⊗ y i （
A A A 中的
x i , y i x_i, y_i x i , y i ）属于到
I I I ，则定义
∑ i x i y i = 0 \sum_i x_i y_i = 0 ∑ i x i y i = 0 ，因此
∑ i ( x i ⊗ y i ) = ∑ i x i ( 1 ⊗ y i − y i ⊗ 1 )
\sum_i (x_i \otimes y_i) = \sum_i x_i (1 \otimes y_i - y_i \otimes 1)
i ∑ ( x i ⊗ y i ) = i ∑ x i ( 1 ⊗ y i − y i ⊗ 1 ) 通过 (40)，完成引理的证明。
推导* δ A \delta_A δ A 具有以下泛性质 ：对于每个( A , A ) (A, A) ( A , A ) E E E K K K d : A → E d : A \to E d : A → E ( A , A ) (A, A) ( A , A ) f : I → E f : I \to E f : I → E d = f ∘ δ A d = f \circ \delta_A d = f ∘ δ A
引用了 2 · 被以下引用 1 引用了 命题 9 , 引理 1
被以下引用 § 10. 。 推导
注首先，对于每个( A , A ) (A, A) ( A , A ) f : I → E f : I \to E f : I → E f ∘ δ A f \circ \delta_A f ∘ δ A d : A → E d : A \to E d : A → E ( A , A ) (A, A) ( A , A ) f : I → E f : I \to E f : I → E d = f ∘ δ A d = f \circ \delta_A d = f ∘ δ A f f f δ A \delta_A δ A
f ( x ⊗ 1 − 1 ⊗ x ) = d x
f(x \otimes 1 - 1 \otimes x) = dx
f ( x ⊗ 1 − 1 ⊗ x ) = d x 而我们的断言是从δ A \delta_A δ A
f ( ∑ i x i ⊗ y i ) = ∑ i x i ⋅ f ( 1 ⊗ y i − y i ⊗ 1 ) = − ∑ i x i ⋅ d y i .
f\left( \sum_i x_i \otimes y_i \right) = \sum_i x_i \cdot f(1 \otimes y_i - y_i \otimes 1) = -\sum_i x_i \cdot dy_i.
f ( i ∑ x i ⊗ y i ) = i ∑ x i ⋅ f ( 1 ⊗ y i − y i ⊗ 1 ) = − i ∑ x i ⋅ d y i . 相反，由于A × A A \times A A × A E E E ( x , y ) ↦ − x ⋅ d y (x, y) \mapsto -x \cdot dy ( x , y ) ↦ − x ⋅ d y K K K K K K g : A ⊗ K A → E g : A \otimes_K A \to E g : A ⊗ K A → E g ( x ⊗ y ) = − x ⋅ d y g(x \otimes y) = -x \cdot dy g ( x ⊗ y ) = − x ⋅ d y g g g I I I f f f ( x x ′ ) ⋅ d y = x ⋅ ( x ′ ⋅ d y ) (xx') \cdot dy = x \cdot (x' \cdot dy) ( x x ′ ) ⋅ d y = x ⋅ ( x ′ ⋅ d y ) ∑ i x i ⊗ y i ∈ I \sum_i x_i \otimes y_i \in I ∑ i x i ⊗ y i ∈ I x ∈ A x \in A x ∈ A
∑ i x i ⋅ d ( y i x ) = ∑ i x i ⋅ d y i ⋅ x + ∑ i ( x i y i ) ⋅ d x
\sum_i x_i \cdot d(y_i x) = \sum_i x_i \cdot dy_i \cdot x + \sum_i (x_i y_i) \cdot dx
i ∑ x i ⋅ d ( y i x ) = i ∑ x i ⋅ d y i ⋅ x + i ∑ ( x i y i ) ⋅ d x 但从∑ i x i y i = 0 \sum_i x_i y_i = 0 ∑ i x i y i = 0 I I I
因此，我们定义了一个 规范 K K K 同构 f ↦ f ∘ δ A f \mapsto f \circ \delta_A f ↦ f ∘ δ A
Hom ( A , A ) ( I , E ) → D K ( A , E )
\operatorname{Hom}_{(A, A)}(I, E) \to D_K(A, E)
Hom ( A , A ) ( I , E ) → D K ( A , E ) 左侧是( A , A ) (A, A) ( A , A ) K K K E E E A A A
11. 通用问题适用于推导；交换案例
现在假设 A A A K K K E E E A A A E E E ( A , A ) (A, A) ( A , A ) A A A A ⊗ K A A \otimes_K A A ⊗ K A ( A , A ) (A, A) ( A , A ) A ⊗ K A A \otimes_K A A ⊗ K A ( A ⊗ K A ) (A \otimes_K A) ( A ⊗ K A ) A A A x , y , u , v x, y, u, v x , y , u , v
x ⋅ ( u ⊗ v ) ⋅ y = ( x u ) ⊗ ( v y ) = ( x u ) ⊗ ( y v ) = ( x ⊗ y ) ( u ⊗ v ) .
x \cdot (u \otimes v) \cdot y = (xu) \otimes (vy) = (xu) \otimes (yv) = (x \otimes y)(u \otimes v).
x ⋅ ( u ⊗ v ) ⋅ y = ( xu ) ⊗ ( v y ) = ( xu ) ⊗ ( y v ) = ( x ⊗ y ) ( u ⊗ v ) . 因此，在这种情况下，m m m J \mathfrak{J} J A ⊗ K A A \otimes_K A A ⊗ K A m : A ⊗ K A → A m : A \otimes_K A \to A m : A ⊗ K A → A ( A ⊗ K A ) / J (A \otimes_K A)/\mathfrak{J} ( A ⊗ K A ) / J A A A E E E m m m ( A ⊗ K A ) (A \otimes_K A) ( A ⊗ K A ) ( A ⊗ K A ) (A \otimes_K A) ( A ⊗ K A ) m ∗ ( E ) m_*(E) m ∗ ( E ) ( A , A ) (A, A) ( A , A ) J → E \mathfrak{J} \to E J → E ( A ⊗ K A ) (A \otimes_K A) ( A ⊗ K A ) J → E \mathfrak{J} \to E J → E K K K
Hom ( A , A ) ( J , E ) → Hom A ⊗ K A ( J , E ) .
\operatorname{Hom}_{(A, A)}(\mathfrak{J}, E) \to \operatorname{Hom}_{A \otimes_K A}(\mathfrak{J}, E).
Hom ( A , A ) ( J , E ) → Hom A ⊗ K A ( J , E ) . 另一方面，J E = { 0 } \mathfrak{J}E = \{0\} J E = { 0 } 1 ⊗ x − x ⊗ 1 1 \otimes x - x \otimes 1 1 ⊗ x − x ⊗ 1 J \mathfrak{J} J ( A ⊗ K A ) (A \otimes_K A) ( A ⊗ K A ) z ∈ E z \in E z ∈ E
( 1 ⊗ x − x ⊗ 1 ) z = 0
(1 \otimes x - x \otimes 1)z = 0
( 1 ⊗ x − x ⊗ 1 ) z = 0 凭借E E E ( A ⊗ K A ) (A \otimes_K A) ( A ⊗ K A ) J \mathfrak{J} J ( A ⊗ K A ) (A \otimes_K A) ( A ⊗ K A ) E E E E E E ( ( A ⊗ K A ) / J ) ((A \otimes_K A)/\mathfrak{J}) (( A ⊗ K A ) / J ) E E E A A A
J ⊗ K ( ( A ⊗ K A ) / J )
\mathfrak{J} \otimes_K ((A \otimes_K A)/\mathfrak{J})
J ⊗ K (( A ⊗ K A ) / J ) 到J / J 2 \mathfrak{J}/\mathfrak{J}^2 J / J 2 K K K
Hom A ⊗ K A ( J , E ) → Hom A ( J / J 2 , E ) .
\operatorname{Hom}_{A \otimes_K A}(\mathfrak{J}, E) \to \operatorname{Hom}_A(\mathfrak{J}/\mathfrak{J}^2, E).
Hom A ⊗ K A ( J , E ) → Hom A ( J / J 2 , E ) . 考虑到no. 10的命题 17，可以看出我们已经证明了以下命题：
令 A A A K K K J \mathfrak{J} J m : A ⊗ K A → A m : A \otimes_K A \to A m : A ⊗ K A → A A A A ( A ⊗ K A ) / J (A \otimes_K A)/\mathfrak{J} ( A ⊗ K A ) / J J / J 2 \mathfrak{J}/\mathfrak{J}^2 J / J 2 A A A d A / K : A → I / I 2 d_{A/K} : A \to I/I^2 d A / K : A → I / I 2 K K K x ⊗ 1 − 1 ⊗ x x \otimes 1 - 1 \otimes x x ⊗ 1 − 1 ⊗ x J 2 \mathfrak{J}^2 J 2 x ∈ A x \in A x ∈ A d A / K d_{A/K} d A / K K K K A A A E E E K K K D : A → E D : A \to E D : A → E A A A g : J / J 2 → E g : \mathfrak{J}/\mathfrak{J}^2 \to E g : J / J 2 → E D = g ∘ d A / K D = g \circ d_{A/K} D = g ∘ d A / K
被以下引用 2 被以下引用 命题 19 , 命题 20
设M为K-模；从 no. 9 的命题 14 可以得出，对于每个 S(M)-模 E，映射 D ↦ D | M 定义 D_K(S(M), E) 的 S(M) 模同构到 Hom_K(M, E)；另一方面，由于 E 是 S(M) 模，因此 Hom_K(M, E) 规范地同构于
Hom S ( M ) ( M ⊗ K S ( M ) , E ) ,
\operatorname{Hom}_{S(M)}(M \otimes_K S(M), E),
Hom S ( M ) ( M ⊗ K S ( M ) , E ) , M 到 E 的每个 K 同态都可以在形式 x ↦ h ( x ⊗ 1 ) x \mapsto h(x \otimes 1) x ↦ h ( x ⊗ 1 )
h ∈ Hom S ( M ) ( M ⊗ K S ( M ) , E )
h \in \operatorname{Hom}_{S(M)}(M \otimes_K S(M), E)
h ∈ Hom S ( M ) ( M ⊗ K S ( M ) , E ) （II, § 5, no. 1）。令 D_0 为 K 导数 S ( M ) → M ⊗ K S ( M ) S(M) \to M \otimes_K S(M) S ( M ) → M ⊗ K S ( M ) x ↦ x ⊗ 1 x \mapsto x \otimes 1 x ↦ x ⊗ 1 h ∘ D 0 h \circ D_0 h ∘ D 0
h ∈ Hom S ( M ) ( M ⊗ K S ( M ) , E ) .
h \in \operatorname{Hom}_{S(M)}(M \otimes_K S(M), E).
h ∈ Hom S ( M ) ( M ⊗ K S ( M ) , E ) . 由普遍映射问题的解唯一性可知，存在唯一的S(M)模同构
ω : M ⊗ K S ( M ) → Ω K ( S ( M ) )
\omega : M \otimes_K S(M) \to \Omega_K(S(M))
ω : M ⊗ K S ( M ) → Ω K ( S ( M )) 使得 D 0 ∘ ω = d S ( M ) / K D_0 \circ \omega = d_{S(M)/K} D 0 ∘ ω = d S ( M ) / K x ∈ M x \in M x ∈ M ω ( x ⊗ 1 ) = d x \omega(x \otimes 1) = dx ω ( x ⊗ 1 ) = d x
特别是，*如果 M 是自由K-模且基* ( e λ ) λ ∈ L (e_\lambda)_{\lambda \in L} ( e λ ) λ ∈ L Ω K ( S ( M ) ) \Omega_K(S(M)) Ω K ( S ( M )) 是一个自由 S(M) 模，具有基集合微分 d e λ de_\lambda d e λ L = { 1 , n } L = \{1, n\} L = { 1 , n } K [ X 1 , … , X n ] K[X_1, \ldots, X_n] K [ X 1 , … , X n ] P ∈ K [ X 1 , … , X n ] P \in K[X_1, \ldots, X_n] P ∈ K [ X 1 , … , X n ]
d P = ∑ i = 1 n D i P . d X i
dP = \sum_{i=1}^n D_i P . dX_i
d P = i = 1 ∑ n D i P . d X i 与D i P ∈ K [ X 1 , … , X n ] D_i P \in K[X_1, \ldots, X_n] D i P ∈ K [ X 1 , … , X n ] P ↦ D i P P \mapsto D_i P P ↦ D i P K [ X 1 , … , X n ] K[X_1, \ldots, X_n] K [ X 1 , … , X n ] Ω K ( S ( M ) ) \Omega_K(S(M)) Ω K ( S ( M )) ( d X i ) (dX_i) ( d X i ) ∂ P ∂ X i \frac{\partial P}{\partial X_i} ∂ X i ∂ P D i P D_i P D i P X i X_i X i
引用了 3 引用了 命题 14 , II § 5. 。标量的环的扩张 , III § 6. 。对称代数
12. K-差速器的功能特性
K → ρ K ′ ↓ η ↓ η ′ A → u A ′
\begin{array}{ccc}
K & \xrightarrow{\rho} & K' \\
\downarrow \eta & & \downarrow \eta' \\
A & \xrightarrow{u} & A'
\end{array}
K ↓ η A ρ u K ′ ↓ η ′ A ′ 是交换环同态 、η \eta η η ′ \eta' η ′
A（分别为A ′ A' A ′ K K K K ′ K' K ′ A A A v : Ω K ( A ) → Ω K ′ ( A ′ ) v : \Omega_K(A) \to \Omega_{K'}(A') v : Ω K ( A ) → Ω K ′ ( A ′ )
A → u A ′ d A / K ↓ d A ′ / K ′ ↓ Ω K ( A ) → v Ω K ′ ( A ′ )
\begin{array}{ccc}
A & \xrightarrow{u} & A' \\
d_{A/K} \downarrow & & d_{A'/K'} \downarrow \\
\Omega_K(A) & \xrightarrow{v} & \Omega_{K'}(A')
\end{array}
A d A / K ↓ Ω K ( A ) u v A ′ d A ′ / K ′ ↓ Ω K ′ ( A ′ ) d A ′ / K ′ ∘ u d_{A'/K'} \circ u d A ′ / K ′ ∘ u A A A Ω K ′ ( A ′ ) \Omega_{K'}(A') Ω K ′ ( A ′ ) A A A K K K v v v
命题 19 中的映射 v v v Ω ( u ) \Omega(u) Ω ( u )
K → σ K ′ → σ ′ K ′ ′ η ↓ η ′ ↓ η ′ ′ ↓ A → u A ′ → u ′ A ′ ′
\begin{array}{cccccc}
K & \xrightarrow{\sigma} & K' & \xrightarrow{\sigma'} & K'' \\
\eta \downarrow & & \eta' \downarrow & & \eta'' \downarrow \\
A & \xrightarrow{u} & A' & \xrightarrow{u'} & A''
\end{array}
K η ↓ A σ u K ′ η ′ ↓ A ′ σ ′ u ′ K ′′ η ′′ ↓ A ′′ 从命题 19 的唯一性性质立即得出
Ω ( u ′ ∘ u ) = Ω ( u ′ ) ∘ Ω ( u ) .
\Omega(u' \circ u) = \Omega(u') \circ \Omega(u).
Ω ( u ′ ∘ u ) = Ω ( u ′ ) ∘ Ω ( u ) . 由于 Ω K ′ ( A ′ ) \Omega_{K'}(A') Ω K ′ ( A ′ ) A ′ A' A ′ Ω ( u ) \Omega(u) Ω ( u ) A ′ A' A ′
Ω 0 ( u ) : Ω K ( A ) ⊗ A A ′ → Ω K ′ ( A ′ )
\Omega_0(u) : \Omega_K(A) \otimes_A A' \to \Omega_{K'}(A')
Ω 0 ( u ) : Ω K ( A ) ⊗ A A ′ → Ω K ′ ( A ′ ) 这样Ω ( u ) \Omega(u) Ω ( u ) Ω 0 ( u ) \Omega_0(u) Ω 0 ( u ) i A : Ω K ( A ) → Ω K ( A ) ⊗ A A ′ i_A : \Omega_K(A) \to \Omega_K(A) \otimes_A A' i A : Ω K ( A ) → Ω K ( A ) ⊗ A A ′ A ′ A' A ′ E ′ E' E ′
Hom A ′ ( Ω K ′ ( A ′ ) , E ′ ) → Hom ( Ω 0 ( u ) , 1 E ′ ) Hom A ′ ( Ω K ( A ) ⊗ A A ′ , E ′ ) ϕ A ′ ↓ ϕ A ∘ r A ↓ D K ′ ( A ′ , E ′ ) → C ( u ) D K ( A , E )
\begin{array}{ccc}
\operatorname{Hom}_{A'}(\Omega_{K'}(A'), E') & \xrightarrow{\operatorname{Hom}(\Omega_0(u), 1_{E'})} & \operatorname{Hom}_{A'}(\Omega_K(A) \otimes_A A', E') \\
\phi_{A'} \downarrow & & \phi_A \circ r_A \downarrow \\
D_{K'}(A', E') & \xrightarrow{C(u)} & D_K(A, E)
\end{array}
Hom A ′ ( Ω K ′ ( A ′ ) , E ′ ) ϕ A ′ ↓ D K ′ ( A ′ , E ′ ) Hom ( Ω 0 ( u ) , 1 E ′ ) C ( u ) Hom A ′ ( Ω K ( A ) ⊗ A A ′ , E ′ ) ϕ A ∘ r A ↓ D K ( A , E ) 其中 C ( u ) C(u) C ( u ) D ↦ D ∘ u D \mapsto D \circ u D ↦ D ∘ u r A r_A r A
Hom ( i A , 1 E ′ ) : Hom A ′ ( Ω K ( A ) ⊗ A A ′ , E ′ ) → Hom A ( Ω K ( A ) , E ′ ) ;
\operatorname{Hom}(i_A, 1_{E'}) : \operatorname{Hom}_{A'}(\Omega_K(A) \otimes_A A', E') \to \operatorname{Hom}_A(\Omega_K(A), E');
Hom ( i A , 1 E ′ ) : Hom A ′ ( Ω K ( A ) ⊗ A A ′ , E ′ ) → Hom A ( Ω K ( A ) , E ′ ) ; 这紧接着命题 19 和定义的同构 ϕ A \phi_A ϕ A ϕ A ′ \phi_{A'} ϕ A ′
Ω 0 ( u ) : Ω K ( A ) ⊗ A A ′ → Ω K ′ ( A ′ )
\Omega_0(u): \Omega_K(A) \otimes_A A' \to \Omega_{K'}(A')
Ω 0 ( u ) : Ω K ( A ) ⊗ A A ′ → Ω K ′ ( A ′ ) 是同构。
由于图表 (42) 中的垂直箭头是双射的，约化到证明，对于每个 A ′ A' A ′ E ′ E' E ′ C ( u ) : D ↦ D ∘ u C(u): D \mapsto D \circ u C ( u ) : D ↦ D ∘ u H o m ( u , 1 E ′ ) : H o m K ′ ( A ⊗ K K ′ , E ′ ) → H o m K ( A , E ′ ) \mathrm{Hom}(u, 1_{E'}): \mathrm{Hom}_{K'}(A \otimes_K K', E') \to \mathrm{Hom}_K(A, E') Hom ( u , 1 E ′ ) : Hom K ′ ( A ⊗ K K ′ , E ′ ) → Hom K ( A , E ′ ) C ( u ) C(u) C ( u ) D K ′ ( A ′ , E ′ ) D_{K'}(A', E') D K ′ ( A ′ , E ′ ) f : A ′ → E ′ f: A' \to E' f : A ′ → E ′ K ′ K' K ′
f ∘ u : A → E ′
f \circ u: A \to E'
f ∘ u : A → E ′ 是 K K K f f f K ′ K' K ′ f ( ( x ⊗ 1 ) ( y ⊗ 1 ) ) = ( y ⊗ 1 ) f ( x ⊗ 1 ) + ( x ⊗ 1 ) f ( y ⊗ 1 ) f((x \otimes 1)(y \otimes 1)) = (y \otimes 1)f(x \otimes 1) + (x \otimes 1)f(y \otimes 1) f (( x ⊗ 1 ) ( y ⊗ 1 )) = ( y ⊗ 1 ) f ( x ⊗ 1 ) + ( x ⊗ 1 ) f ( y ⊗ 1 ) A A A x , y x, y x , y f f f K ′ K' K ′ x ⊗ 1 x \otimes 1 x ⊗ 1 x ∈ A x \in A x ∈ A K ′ K' K ′ A ′ A' A ′ C ( u ) C(u) C ( u )
从现在开始，我们将注意力集中在ρ : K → K ′ \rho: K \to K' ρ : K → K ′ K K K K K K u : A → B u: A \to B u : A → B B B B
Ω 0 ( u ) : Ω K ( A ) ⊗ K B → Ω K ( B ) .
\Omega_0(u): \Omega_K(A) \otimes_K B \to \Omega_K(B).
Ω 0 ( u ) : Ω K ( A ) ⊗ K B → Ω K ( B ) . 另一方面，我们可以考虑 A A A Ω A ( B ) \Omega_A(B) Ω A ( B ) B B B B B B u u u A A A d B / A : B → Ω A ( B ) d_{B/A}: B \to \Omega_A(B) d B / A : B → Ω A ( B ) 更不必说 K K K
B → d B / K Ω K ( B ) → Ω u Ω A ( B )
B \xrightarrow{d_{B/K}} \Omega_K(B) \xrightarrow{\Omega_u} \Omega_A(B)
B d B / K Ω K ( B ) Ω u Ω A ( B ) 其中Ω u \Omega_u Ω u B B B B B B E E E
H o m B ( Ω A ( B ) , E ) → H o m ( Ω u , 1 E ) H o m B ( Ω K ( B ) , E ) ↓ ϕ A , B ↓ ϕ K , B D A ( B , E ) → j u D K ( B , E )
\begin{array}{ccc}
\mathrm{Hom}_B(\Omega_A(B), E) & \xrightarrow{\mathrm{Hom}(\Omega_u, 1_E)} & \mathrm{Hom}_B(\Omega_K(B), E) \\
\downarrow \phi_{A,B} & & \downarrow \phi_{K,B} \\
D_A(B, E) & \xrightarrow{j_u} & D_K(B, E)
\end{array}
Hom B ( Ω A ( B ) , E ) ↓ ϕ A , B D A ( B , E ) Hom ( Ω u , 1 E ) j u Hom B ( Ω K ( B ) , E ) ↓ ϕ K , B D K ( B , E ) 其中 j u j_u j u
Ω K ( A ) ⊗ A B → Ω 0 ( u ) Ω K ( B ) → Ω u Ω A ( B ) → 0
\Omega_K(A) \otimes_A B \xrightarrow{\Omega_0(u)} \Omega_K(B) \xrightarrow{\Omega_u} \Omega_A(B) \to 0
Ω K ( A ) ⊗ A B Ω 0 ( u ) Ω K ( B ) Ω u Ω A ( B ) → 0 准确。
约化验证，对于每个 B-模 E，序列
0 → Hom B ( Ω A ( B ) , E ) → Hom ( Ω u , 1 E ) Hom B ( Ω K ( B ) , E ) → Hom ( Ω 0 ( u ) , 1 E ) Hom B ( Ω K ( A ) ⊗ A B , E )
0 \to \operatorname{Hom}_B(\Omega_A(B), E) \xrightarrow{\operatorname{Hom}(\Omega_u, 1_E)} \operatorname{Hom}_B(\Omega_K(B), E) \xrightarrow{\operatorname{Hom}(\Omega_0(u), 1_E)} \operatorname{Hom}_B(\Omega_K(A) \otimes_A B, E)
0 → Hom B ( Ω A ( B ) , E ) Hom ( Ω u , 1 E ) Hom B ( Ω K ( B ) , E ) Hom ( Ω 0 ( u ) , 1 E ) Hom B ( Ω K ( A ) ⊗ A B , E ) 是精确的 (II, § 2, no. 1, 定理 1)；但由于在交换式图表 (42) 和 (44) 中垂直箭头是同构，就足以证明序列
0 ⟶ D A ( B , E ) → j u D K ( B , E ) → C ( u ) D K ( A , E )
0 \longrightarrow D_A(B, E) \xrightarrow{j_u} D_K(B, E) \xrightarrow{C(u)} D_K(A, E)
0 ⟶ D A ( B , E ) j u D K ( B , E ) C ( u ) D K ( A , E ) 是准确的，正好是 no. 7 中的命题 11。
我们现在考虑 K-代数同态 u : A → B u : A \to B u : A → B J \mathfrak{J} J A / J A/\mathfrak{J} A / J d = d A / K d = d_{A/K} d = d A / K d ∣ J : J → Ω K ( A ) d|_{\mathfrak{J}} : \mathfrak{J} \to \Omega_K(A) d ∣ J : J → Ω K ( A )
d ′ : J → d ∣ J Ω K ( A ) → i A Ω K ( A ) ⊗ A B .
d' : \mathfrak{J} \xrightarrow{d|_{\mathfrak{J}}} \Omega_K(A) \xrightarrow{i_A} \Omega_K(A) \otimes_A B.
d ′ : J d ∣ J Ω K ( A ) i A Ω K ( A ) ⊗ A B . 然后d ′ ( J 2 ) = 0 d'(\mathfrak{J}^2) = 0 d ′ ( J 2 ) = 0 J \mathfrak{J} J x , y x, y x , y
d ′ ( x y ) = d ( x y ) ⊗ 1 = ( x . d y + y . d x ) ⊗ 1 = d y ⊗ u ( x ) + d x ⊗ u ( y ) = 0
d'(xy) = d(xy) \otimes 1 = (x.dy + y.dx) \otimes 1 = dy \otimes u(x) + dx \otimes u(y) = 0
d ′ ( x y ) = d ( x y ) ⊗ 1 = ( x . d y + y . d x ) ⊗ 1 = d y ⊗ u ( x ) + d x ⊗ u ( y ) = 0 从u ( x ) = u ( y ) = 0 u(x) = u(y) = 0 u ( x ) = u ( y ) = 0 d ′ d' d ′
d ˉ : J / J 2 → Ω K ( A ) ⊗ A B
\bar{d} : \mathfrak{J}/\mathfrak{J}^2 \to \Omega_K(A) \otimes_A B
d ˉ : J / J 2 → Ω K ( A ) ⊗ A B 当J \mathfrak{J} J J / J 2 \mathfrak{J}/\mathfrak{J}^2 J / J 2 d ˉ \bar{d} d ˉ
J / J 2 → d ˉ Ω K ( A ) ⊗ A B → Ω 0 ( u ) Ω K ( B ) → 0
\mathfrak{J}/\mathfrak{J}^2 \xrightarrow{\bar{d}} \Omega_K(A) \otimes_A B \xrightarrow{\Omega_0(u)} \Omega_K(B) \to 0
J / J 2 d ˉ Ω K ( A ) ⊗ A B Ω 0 ( u ) Ω K ( B ) → 0 那么准确。
注，Ω K ( A ) ⊗ A B \Omega_K(A) \otimes_A B Ω K ( A ) ⊗ A B Ω K ( A ) / J Ω K ( A ) \Omega_K(A)/\mathfrak{J}\Omega_K(A) Ω K ( A ) / J Ω K ( A ) d ˉ \bar{d} d ˉ d ( J ) d(\mathfrak{J}) d ( J ) Ω K ( A ) ⊗ A B \Omega_K(A) \otimes_A B Ω K ( A ) ⊗ A B Im ( d ˉ ) \operatorname{Im}(\bar{d}) Im ( d ˉ ) Ω K ( A ) / N \Omega_K(A)/N Ω K ( A ) / N J Ω K ( A ) \mathfrak{J}\Omega_K(A) J Ω K ( A ) d ( J ) d(\mathfrak{J}) d ( J )
A → d A / K Ω K ( A ) ⟶ Ω K ( A ) / N
A \xrightarrow{d_{A/K}} \Omega_K(A) \longrightarrow \Omega_K(A)/N
A d A / K Ω K ( A ) ⟶ Ω K ( A ) / N 是 K 导数 (no. 7, 命题 9)，并且由于它是 J \mathfrak{J} J D 0 : B → Ω K ( A ) / N D_0 : B \to \Omega_K(A)/N D 0 : B → Ω K ( A ) / N
考虑普遍映射问题的解的唯一性，约化到证明，对于每个B-模E和每个K-导数D：B→E，存在唯一的B-线性映射g： Ω_K(A)/N → E 使得 D = g ∘ D_0。但是，复合映射 D ∘ u: A → E 是 K 导数 (no. 7, 命题 9)，因此存在且仅有一个 A-线性映射 f: Ω_K(A) → E 使得 f ∘ d_{A/K} = D ∘ u。这个关系已经表明f在d(Σ)上是零；同样 ΣE = {0}，因为 E 为 B-模，f 在 ΣΩ_K(A) 上为零；因此 f 在 N 和定义上是零，当经过为商时，B-线性映射 g: Ω_K(A)/N → E 使得 g ∘ D_0 = D； g 的唯一性由 f 的唯一性推出。
千万不要认为，即使 u: A → B 是单同态，Ω_0(u): Ω_K(A) ⊗_A B → Ω_K(B) 是单射的 (习题 5)。然而我们有以下命题：
设A是K-代数的积分，B是分数的域，u：A→B典范嵌入。则 Ω_0(u): Ω_K(A) ⊗_A B → Ω_K(B) 为同构。
引用了 1 引用了 命题 5
利用图表 (42) 中垂直箭头是双射的事实，约化到证明，对于 B 上的每个向量空间 E，映射 C(u)： D_K(B, E) → D_K(A, E) 是双射的。但这源于以下事实：A 到 E 的每个 K 推导都可以唯一地扩展到 B 到 E 的 K 推导（no. 5，命题 5）。
习题
参见习题为 § 10 。