Google 翻译生成的粗略草稿，质量较低。术语依照术语表，但句子可能有误；之后会由仔细的译文取代。
§ 2。 例 / 代数
在本段中，A 表示交换环。
1. 内同态代数
令 B 为结合律 A-代数，其中单位元用 1 表示，并令 M 为右B-模。我们知道，B 的环 E = E n d B ( M ) E = \mathrm{End}_B(M) E = End B ( M ) h : α ↦ α .1 h : \alpha \mapsto \alpha . 1 h : α ↦ α .1 h h h α ∈ A \alpha \in A α ∈ A f , g f, g f , g
α ( f ∘ g ) = f ∘ ( α g ) = ( α f ) ∘ g ;
\alpha(f \circ g) = f \circ (\alpha g) = (\alpha f) \circ g;
α ( f ∘ g ) = f ∘ ( α g ) = ( α f ) ∘ g ; 因此，E 中的乘法和 E 上的 A-模结构定义是 E 上的联想 A-代数结构； M 的恒等映射是这个代数的单位元。
2. 矩阵元素
设 B 是 B (II, § 10, no. 7) 上阶 n 的平方矩阵的酉结合 A-代数和 M n ( B ) \mathbf{M}_n(B) M n ( B ) M n ( B ) \mathbf{M}_n(B) M n ( B ) α . ( b i j ) = ( α b i j ) \alpha . (b_{ij}) = (\alpha b_{ij}) α . ( b ij ) = ( α b ij ) α ∈ A , b i j ∈ B , 1 ≤ i ≤ n , 1 ≤ j ≤ n \alpha \in A, b_{ij} \in B, 1 \leq i \leq n, 1 \leq j \leq n α ∈ A , b ij ∈ B , 1 ≤ i ≤ n , 1 ≤ j ≤ n M n ( B ) \mathbf{M}_n(B) M n ( B ) M n ( B ) \mathbf{M}_n(B) M n ( B ) E n d B ( B d n ) \mathrm{End}_B(B_d^n) End B ( B d n )
当B = A B = A B = A M n ( A ) \mathbf{M}_n(A) M n ( A ) ( E i j ) (E_{ij}) ( E ij )
E i j E h k = δ j h E i k .
E_{ij} E_{hk} = \delta_{jh} E_{ik}.
E ij E hk = δ j h E ik . 单位元 I n I_n I n ∑ i = 1 n E i i \sum_{i=1}^n E_{ii} ∑ i = 1 n E ii
3. 二次代数
令α , β \alpha, \beta α , β A 2 A^2 A 2 ( e 1 , e 2 ) (e_1, e_2) ( e 1 , e 2 ) ( α , β ) (\alpha, \beta) ( α , β ) A 2 A^2 A 2
e 1 2 = e 1 , e 1 e 2 = e 2 e 1 = e 2 , e 2 2 = α e 1 + β e 2 .
e_1^2 = e_1, \quad e_1 e_2 = e_2 e_1 = e_2, \quad e_2^2 = \alpha e_1 + \beta e_2.
e 1 2 = e 1 , e 1 e 2 = e 2 e 1 = e 2 , e 2 2 = α e 1 + β e 2 . 与二次代数同构的A-代数E也称为二次代数。这相当于说 E 承认两个元素中的基，其中一个是单位元。
可以证明，每个允许两个元素中的基的单位A-代数都是二次代数（习题1）。
如果A-代数的基( e 1 , e 2 ) (e_1, e_2) ( e 1 , e 2 ) ( α , β ) (\alpha, \beta) ( α , β ) ( α , β ) (\alpha, \beta) ( α , β ) ( α , β ) (\alpha, \beta) ( α , β )
E 是可交换的，这一事实由 (2) 中的方程 e 1 e 2 = e 2 e 1 e_1 e_2 = e_2 e_1 e 1 e 2 = e 2 e 1 x , y , z x, y, z x , y , z e 1 e_1 e 1 e 2 e_2 e 2 x ( y z ) = ( x y ) z x(yz) = (xy)z x ( y z ) = ( x y ) z x , y , z x, y, z x , y , z e 1 e_1 e 1 x = y = z = e 2 x = y = z = e_2 x = y = z = e 2
令e e e ( e , i ) (e, i) ( e , i ) ( α , β ) (\alpha, \beta) ( α , β ) e e e ( e , j ) (e, j) ( e , j ) j = γ e + δ i j = \gamma e + \delta i j = γ e + δ i ( e , j ) (e, j) ( e , j ) δ \delta δ E / A e E/Ae E / A e i ˉ \bar{i} i ˉ j ˉ = δ i ˉ \bar{j} = \delta \bar{i} j ˉ = δ i ˉ i ˉ \bar{i} i ˉ i i i E / A e E/Ae E / A e
j 2 = ( γ 2 + α δ 2 ) e + ( 2 γ δ + β δ 2 ) i = ( α δ 2 − γ 2 − β γ δ ) e + ( 2 γ + β δ ) j ;
j^2 = (\gamma^2 + \alpha \delta^2)e + (2\gamma \delta + \beta \delta^2)i = (\alpha \delta^2 - \gamma^2 - \beta \gamma \delta)e + (2\gamma + \beta \delta)j;
j 2 = ( γ 2 + α δ 2 ) e + ( 2 γ δ + β δ 2 ) i = ( α δ 2 − γ 2 − β γ δ ) e + ( 2 γ + β δ ) j ; 由此可知E为型
( α δ 2 − γ 2 − β γ δ , 2 γ + β δ )
(\alpha \delta^2 - \gamma^2 - \beta \gamma \delta, 2\gamma + \beta \delta)
( α δ 2 − γ 2 − β γ δ , 2 γ + β δ ) 对于所有可逆的δ ∈ A \delta \in A δ ∈ A γ ∈ A \gamma \in A γ ∈ A E E E ( α , 2 β ′ ) (\alpha, 2\beta') ( α , 2 β ′ ) i i i ( α + β ′ 2 , 0 ) (\alpha + {\beta'}^2, 0) ( α + β ′ 2 , 0 ) γ = − β ′ \gamma = -\beta' γ = − β ′ δ = 1 \delta = 1 δ = 1
设E E E A A A e e e u ∈ E u \in E u ∈ E T ( u ) T(u) T ( u ) A A A E E E m u : x ↦ u x m_u : x \mapsto ux m u : x ↦ ux u ∈ E u \in E u ∈ E s ( u ) = T ( u ) . e − u s(u) = T(u).e - u s ( u ) = T ( u ) . e − u s s s E E E s 2 ( u ) = u s^2(u) = u s 2 ( u ) = u
引用了 3 · 被以下引用 1 引用了 II § 4. 。 关系张量积和同态模之间 , 命题 3, II § 4 , III § 8. 。决定因素
被以下引用 例
设( e , i ) (e, i) ( e , i ) ( α , β ) (\alpha, \beta) ( α , β ) E E E T ( e ) = 2 T(e) = 2 T ( e ) = 2 s ( e ) = e s(e) = e s ( e ) = e T ( i ) = β T(i) = \beta T ( i ) = β s ( i ) = β e − i s(i) = \beta e - i s ( i ) = β e − i ( e , s ( i ) ) (e, s(i)) ( e , s ( i )) E E E γ = β \gamma = \beta γ = β δ = − 1 \delta = -1 δ = − 1 ( α , β ) (\alpha, \beta) ( α , β ) s s s E E E A A A E E E m s ( u ) = s m u s − 1 m_{s(u)} = s m_u s^{-1} m s ( u ) = s m u s − 1 m u m_u m u m s ( u ) m_{s(u)} m s ( u )
s 2 ( u ) = T ( u ) . e − s ( u ) = T ( u ) . e − ( T ( u ) . e − u ) = u
s^2(u) = T(u).e - s(u) = T(u).e - (T(u).e - u) = u
s 2 ( u ) = T ( u ) . e − s ( u ) = T ( u ) . e − ( T ( u ) . e − u ) = u 对于所有u ∈ E u \in E u ∈ E
自同构 s s s A A A E E E s ( u ) s(u) s ( u ) u u u
如果u = ξ e + η i u = \xi e + \eta i u = ξ e + η i A A A ξ , η \xi, \eta ξ , η s ( u ) = ( ξ + β η ) e − η i s(u) = (\xi + \beta \eta)e - \eta i s ( u ) = ( ξ + β η ) e − η i
(4)
T ( u ) e = u + s ( u ) = ( 2 ξ + β η ) e
T(u)e = u + s(u) = (2\xi + \beta \eta)e
T ( u ) e = u + s ( u ) = ( 2 ξ + β η ) e (5)
u . s ( u ) = ( ξ 2 + β ξ η − α η 2 ) e = N ( u ) e
u.s(u) = (\xi^2 + \beta \xi \eta - \alpha \eta^2)e = N(u)e
u . s ( u ) = ( ξ 2 + β ξ η − α η 2 ) e = N ( u ) e 我们写了N ( u ) = ξ 2 + β ξ η − α η 2 N(u) = \xi^2 + \beta \xi \eta - \alpha \eta^2 N ( u ) = ξ 2 + β ξ η − α η 2 T ( u ) T(u) T ( u ) N ( u ) N(u) N ( u ) A A A A e Ae A e T ( u ) e T(u)e T ( u ) e N ( u ) e N(u)e N ( u ) e u u u
当β = 0 \beta = 0 β = 0
(6)
s ( ξ e + η i ) = ξ e − η i , T ( ξ e + η i ) = 2 ξ , N ( ξ e + η i ) = ξ 2 − α η 2 .
s(\xi e + \eta i) = \xi e - \eta i, \quad T(\xi e + \eta i) = 2\xi, \quad N(\xi e + \eta i) = \xi^2 - \alpha \eta^2.
s ( ξ e + η i ) = ξ e − η i , T ( ξ e + η i ) = 2 ξ , N ( ξ e + η i ) = ξ 2 − α η 2 . 显然，T T T E E E N N N E E E E E E
(7)
N ( u v ) = N ( u ) N ( v ) .
N(uv) = N(u)N(v).
N ( uv ) = N ( u ) N ( v ) . 对于u u u E E E N ( u ) N(u) N ( u ) A A A N ( e ) = 1 N(e) = 1 N ( e ) = 1 v = u − 1 v = u^{-1} v = u − 1 N ( u ) N(u) N ( u ) A A A u u u
(8)
u − 1 = ( N ( u ) ) − 1 s ( u ) .
u^{-1} = (N(u))^{-1}s(u).
u − 1 = ( N ( u ) ) − 1 s ( u ) . 可以证明N ( u ) N(u) N ( u ) m u m_u m u
以下命题给出二次代数与交换域的结构：
设 E 为型* ( α , β ) (\alpha, \beta) ( α , β )
引用了 1 引用了 V § 3. 。代数扩张
(i) 如果 A 是域并且包含没有元素 ζ \zeta ζ 使得 ζ 2 = α + β ζ \zeta^2 = \alpha + \beta \zeta ζ 2 = α + β ζ
(ii) 如果环 A 包含元素 ζ \zeta ζ 使得 ζ 2 = α + β ζ \zeta^2 = \alpha + \beta \zeta ζ 2 = α + β ζ 且 β − 2 ζ \beta - 2\zeta β − 2 ζ 可逆 （分别零），*E 与* A × \times × 同构的。为型 ( 0 , 0 ) ) (0, 0)) ( 0 , 0 ))
我们证明（i）。设ξ , η \xi, \eta ξ , η u = ξ e + η i u = \xi e + \eta i u = ξ e + η i η ≠ 0 \eta \neq 0 η = 0 θ = − ξ η − 1 \theta = -\xi \eta^{-1} θ = − ξ η − 1 N ( u ) = η 2 ( θ 2 − β θ − α ) \mathrm{N}(u) = \eta^2 (\theta^2 - \beta \theta - \alpha) N ( u ) = η 2 ( θ 2 − β θ − α ) N ( u ) ≠ 0 \mathrm{N}(u) \neq 0 N ( u ) = 0 η = 0 \eta = 0 η = 0 N ( u ) = ξ 2 \mathrm{N}(u) = \xi^2 N ( u ) = ξ 2 u ≠ 0 u \neq 0 u = 0 N ( u ) ≠ 0 \mathrm{N}(u) \neq 0 N ( u ) = 0 N ( u ) \mathrm{N}(u) N ( u ) u u u
我们现在证明(ii)。 代数 A × \times × ( e 1 , e 2 ) (e_1, e_2) ( e 1 , e 2 ) ( 0 , 1 ) (0, 1) ( 0 , 1 )
( α δ 2 − γ 2 − β γ δ , 2 γ + β δ )
(\alpha \delta^2 - \gamma^2 - \beta \gamma \delta, 2\gamma + \beta \delta)
( α δ 2 − γ 2 − β γ δ , 2 γ + β δ ) 对于A中所有γ ∈ A \gamma \in A γ ∈ A δ \delta δ β − 2 ζ \beta - 2\zeta β − 2 ζ δ = ( β − 2 ζ ) − 1 \delta = (\beta - 2\zeta)^{-1} δ = ( β − 2 ζ ) − 1 γ = − ζ ( β − 2 ζ ) − 1 \gamma = -\zeta (\beta - 2\zeta)^{-1} γ = − ζ ( β − 2 ζ ) − 1 2 γ + β δ = 1 2\gamma + \beta \delta = 1 2 γ + β δ = 1
α δ 2 − γ 2 − β γ δ = δ 2 ( α − ζ 2 + β ζ ) = 0 ;
\alpha \delta^2 - \gamma^2 - \beta \gamma \delta = \delta^2 (\alpha - \zeta^2 + \beta \zeta) = 0;
α δ 2 − γ 2 − β γ δ = δ 2 ( α − ζ 2 + β ζ ) = 0 ; 因此E是型( 0 , 1 ) (0, 1) ( 0 , 1 ) × \times × β − 2 ζ = 0 \beta - 2\zeta = 0 β − 2 ζ = 0 ( α + ζ 2 , 0 ) (\alpha + \zeta^2, 0) ( α + ζ 2 , 0 ) α + ζ 2 = 2 ζ 2 − β ζ = 0 \alpha + \zeta^2 = 2\zeta^2 - \beta \zeta = 0 α + ζ 2 = 2 ζ 2 − β ζ = 0 ( 0 , 0 ) (0, 0) ( 0 , 0 )
型 ( 0 , 0 ) (0, 0) ( 0 , 0 )
4. 凯利代数
Cayley 代数 over* A 是有序对 (E, s)，*其中* E 是代数 over A 与单位元 e 和 s 是 E *的反自同构，使得
被以下引用 1 被以下引用 VIII § 19. 。 四元数代数
u + s ( u ) ∈ A e and u . s ( u ) ∈ A e
u + s(u) \in Ae \quad \text{and} \quad u.s(u) \in Ae
u + s ( u ) ∈ A e and u . s ( u ) ∈ A e 对于所有人 u ∈ E u \in E u ∈ E
s s s s ( u ) s(u) s ( u ) u u u 共轭 。 条件 u + s ( u ) ∈ A e u + s(u) \in Ae u + s ( u ) ∈ A e u u u s ( u ) s(u) s ( u )
(9)
T ( u ) = u + s ( u )
T(u) = u + s(u)
T ( u ) = u + s ( u ) N ( u ) = u . s ( u ) = s ( u ) . u
N(u) = u.s(u) = s(u).u
N ( u ) = u . s ( u ) = s ( u ) . u 子代数 A e Ae A e u u u
二次代数 E 的有序对由……组成及其共轭 s（这是反自同构，因为 E 是可交换的）(no. 3) 是凯莱代数。
令 (E, s) 为凯莱代数；如s ( e ) = e , s ( u + s ( u ) ) = u + s ( u ) s(e) = e,\ s(u + s(u)) = u + s(u) s ( e ) = e , s ( u + s ( u )) = u + s ( u ) s ( u ) + s 2 ( u ) = u + s ( u ) s(u) + s^2(u) = u + s(u) s ( u ) + s 2 ( u ) = u + s ( u )
s 2 ( u ) = u
s^2(u) = u
s 2 ( u ) = u 因此 s 2 s^2 s 2
T ( s ( u ) ) = T ( u ) , N ( s ( u ) ) = N ( u ) .
T(s(u)) = T(u), \quad N(s(u)) = N(u).
T ( s ( u )) = T ( u ) , N ( s ( u )) = N ( u ) . 最后，关系( u − u ) ( u − s ( u ) ) = 0 (u - u)(u - s(u)) = 0 ( u − u ) ( u − s ( u )) = 0
u 2 − T ( u ) . u + N ( u ) = 0
u^2 - T(u).u + N(u) = 0
u 2 − T ( u ) . u + N ( u ) = 0 对于所有u ∈ E u \in E u ∈ E
设 E 为 A-代数，并且 s 和 s' E 的反自同构，使得 (E, s) 和 (E, s') 为 Cayley 代数。如果 E 承认基包含单位元 e，则 s' = s.
显然 s ′ ( u ) = s ( u ) = u s'(u) = s(u) = u s ′ ( u ) = s ( u ) = u u ∈ A e u \in Ae u ∈ A e
( T ( u ) − T ′ ( u ) ) . u − ( N ( u ) − N ′ ( u ) ) = 0.
(T(u) - T'(u)).u - (N(u) - N'(u)) = 0.
( T ( u ) − T ′ ( u )) . u − ( N ( u ) − N ′ ( u )) = 0. 设 B 为 E 包含 e 的基，且不同于 e 的 B 的 u 元素；然后T ( u ) − T ′ ( u ) = 0 T(u) - T'(u) = 0 T ( u ) − T ′ ( u ) = 0 s ( u ) = s ′ ( u ) s(u) = s'(u) s ( u ) = s ′ ( u )
接下来，我们写u ˉ = s ( u ) \bar{u} = s(u) u ˉ = s ( u )
{ u + u ˉ = T ( u ) , u u ˉ = u ˉ u = N ( u ) , u ˉ = u u + v = u ˉ + v ˉ , α u = α u ˉ , u v = v ˉ . u ˉ
\left\{\begin{array}{lll}
u + \bar{u} = T(u), & u\bar{u} = \bar{u}u = N(u), & \bar{u} = u \\
u + v = \bar{u} + \bar{v}, & \alpha u = \alpha \bar{u}, & uv = \bar{v}.\bar{u}
\end{array}\right.
{ u + u ˉ = T ( u ) , u + v = u ˉ + v ˉ , u u ˉ = u ˉ u = N ( u ) , α u = α u ˉ , u ˉ = u uv = v ˉ . u ˉ 对于E中的u , v u, v u , v α ∈ A \alpha \in A α ∈ A
T ( e ) = 2 e , N ( e ) = e .
T(e) = 2e, \quad N(e) = e.
T ( e ) = 2 e , N ( e ) = e . 来自公式
T ( u v ) = u v + u v ‾ = u v + v ˉ . u ˉ = u v + ( T ( v ) − v ) ( T ( u ) − u ) ,
T(uv) = uv + \overline{uv} = uv + \bar{v}.\bar{u} = uv + (T(v) - v)(T(u) - u),
T ( uv ) = uv + uv = uv + v ˉ . u ˉ = uv + ( T ( v ) − v ) ( T ( u ) − u ) , 我们推断
u v + v u = T ( u ) v + T ( v ) u + ( T ( u v ) − T ( u ) T ( v ) )
uv + vu = T(u)v + T(v)u + (T(uv) - T(u)T(v))
uv + v u = T ( u ) v + T ( v ) u + ( T ( uv ) − T ( u ) T ( v )) 从而，交换u u u v v v
T ( v u ) = T ( u v ) .
T(vu) = T(uv).
T ( v u ) = T ( uv ) . 另一方面，N ( u + v ) = ( u + v ) ( u ˉ + v ˉ ) = N ( u ) + N ( v ) + T ( u v ˉ ) N(u + v) = (u + v)(\bar{u} + \bar{v}) = N(u) + N(v) + T(u\bar{v}) N ( u + v ) = ( u + v ) ( u ˉ + v ˉ ) = N ( u ) + N ( v ) + T ( u v ˉ )
(18)
T ( v u ˉ ) = T ( u v ˉ ) = N ( u + v ) − N ( u ) − N ( v ) .
T(v\bar{u}) = T(u\bar{v}) = N(u + v) - N(u) - N(v).
T ( v u ˉ ) = T ( u v ˉ ) = N ( u + v ) − N ( u ) − N ( v ) . 现在，将 (16) 应用的 u u u u ˉ \bar{u} u ˉ
T ( u v ˉ ) = T ( u ) T ( v ) + u ˉ v + v u ˉ − T ( u ) v − T ( v ) u ˉ = T ( u ) T ( v ) − u v − v ˉ . u ˉ ;
T(u\bar{v}) = T(u)T(v) + \bar{u}v + v\bar{u} - T(u)v - T(v)\bar{u} = T(u)T(v) - uv - \bar{v}.\bar{u};
T ( u v ˉ ) = T ( u ) T ( v ) + u ˉ v + v u ˉ − T ( u ) v − T ( v ) u ˉ = T ( u ) T ( v ) − uv − v ˉ . u ˉ ; 从而
(19)
T ( v u ˉ ) = T ( u v ˉ ) = N ( u + v ) − N ( u ) − N ( v ) = T ( u ) T ( v ) − T ( u v ) .
T(v\bar{u}) = T(u\bar{v}) = N(u + v) - N(u) - N(v) = T(u)T(v) - T(uv).
T ( v u ˉ ) = T ( u v ˉ ) = N ( u + v ) − N ( u ) − N ( v ) = T ( u ) T ( v ) − T ( uv ) . 最后，对于所有α ∈ A \alpha \in A α ∈ A
(20)
N ( α u ) = α 2 N ( u ) ;
N(\alpha u) = \alpha^2 N(u);
N ( α u ) = α 2 N ( u ) ; 特别是
N ( 2 u ) = 4 N ( u ) N(2u) = 4N(u) N ( 2 u ) = 4 N ( u ) ，因此公式 (19) 给出
(21)
( T ( u ) ) 2 − T ( u 2 ) = 2 N ( u ) .
(T(u))^2 - T(u^2) = 2N(u).
( T ( u ) ) 2 − T ( u 2 ) = 2 N ( u ) . 显然，T T T ( A e ) (Ae) ( A e ) E E E ( u , v ) ↦ T ( v u ˉ ) (u, v) \mapsto T(v\bar{u}) ( u , v ) ↦ T ( v u ˉ ) N N N
5. 凯利的构造代数。四元数
令( E , s ) (E, s) ( E , s ) A A A γ ∈ A \gamma \in A γ ∈ A F F F A A A E × E E \times E E × E
(22)
( x , y ) ( x ′ , y ′ ) = ( x x ′ + γ y ˉ ′ y , y x ˉ ′ + y ′ x ) ;
(x, y)(x', y') = (xx' + \gamma \bar{y}'y, y\bar{x}' + y'x);
( x , y ) ( x ′ , y ′ ) = ( x x ′ + γ y ˉ ′ y , y x ˉ ′ + y ′ x ) ; ( e , 0 ) (e, 0) ( e , 0 ) F F F E × { 0 } E \times \{0\} E × { 0 } F F F E E E E E E x ∈ E x \in E x ∈ E ( x , 0 ) (x, 0) ( x , 0 ) e e e F F F 令 t t t F F F
(23)
t ( ( x , y ) ) = ( x ˉ , − y ) ( x ∈ E , y ∈ E ) .
t((x, y)) = (\bar{x}, -y) \quad (x \in E, y \in E).
t (( x , y )) = ( x ˉ , − y ) ( x ∈ E , y ∈ E ) .
(ii) 设 j = ( 0 , e ) j = (0, e) j = ( 0 , e ) ( x , y ) = x e + y j (x, y) = xe + yj ( x , y ) = x e + y j x ∈ E , y ∈ E x \in E, y \in E x ∈ E , y ∈ E F F F T F T_F T F N F N_F N F 给出
(24)
T F ( x e + y j ) = T ( x ) , N F ( x e + y j ) = N ( x ) − γ N ( y ) .
T_F(xe + yj) = T(x), \quad N_F(xe + yj) = N(x) - \gamma N(y).
T F ( x e + y j ) = T ( x ) , N F ( x e + y j ) = N ( x ) − γ N ( y ) . (iii) 要使 F F F E E E
对于( x , y ) ∈ F (x, y) \in \mathbf{F} ( x , y ) ∈ F
( x , y ) + t ( ( x , y ) ) = ( x + x ‾ , 0 ) = T ( x ) e
(x, y) + t((x, y)) = (x + \overline{x}, 0) = T(x)e
( x , y ) + t (( x , y )) = ( x + x , 0 ) = T ( x ) e (25)
( x , y ) t ( ( x , y ) ) = ( x , y ) ( x ‾ , − y ) = ( x x − γ y ‾ y , y x ‾ − y x ) = ( N ( x ) − γ N ( y ) , 0 ) = ( N ( x ) − γ N ( y ) ) e .
(x, y)t((x, y)) = (x, y)(\overline{x}, -y) = (xx - \gamma \overline{y}y, y\overline{x} - yx)
= (\mathrm{N}(x) - \gamma \mathrm{N}(y), 0) = (\mathrm{N}(x) - \gamma \mathrm{N}(y))e.
( x , y ) t (( x , y )) = ( x , y ) ( x , − y ) = ( xx − γ y y , y x − y x ) = ( N ( x ) − γ N ( y ) , 0 ) = ( N ( x ) − γ N ( y )) e . (26)
因此，为了证明 (i) 和 (ii)，只需证明 t t t F \mathbf{F} F t t t
t ( ( x , y ) ⋅ ( x ′ , y ′ ) ) = t ( ( x x ′ + γ y ‾ ′ y , y x ‾ ′ + y ′ x ) ) = ( x ‾ ′ x ‾ + γ y ‾ y ′ , − y x ‾ − y ′ x ) = ( x ‾ ′ , − y ′ ) ( x ‾ , − y ) = t ( ( x ′ , y ′ ) ) t ( ( x , y ) )
t((x, y) \cdot (x', y')) = t((xx' + \gamma \overline{y}'y, y\overline{x}' + y'x)) = (\overline{x}'\overline{x} + \gamma \overline{y}y', -y\overline{x} - y'x)
= (\overline{x}', -y')(\overline{x}, -y) = t((x', y'))t((x, y))
t (( x , y ) ⋅ ( x ′ , y ′ )) = t (( x x ′ + γ y ′ y , y x ′ + y ′ x )) = ( x ′ x + γ y y ′ , − y x − y ′ x ) = ( x ′ , − y ′ ) ( x , − y ) = t (( x ′ , y ′ )) t (( x , y )) 因此
t t t 是一个反自同构。
仍有待证明(iii)。由于 E 与 F 的子代数相同，因此可以假设 E 是结合的。令 u = ( x , y ) u = (x, y) u = ( x , y ) u ′ = ( x ′ , y ′ ) u' = (x', y') u ′ = ( x ′ , y ′ ) u ′ ′ = ( x ′ ′ , y ′ ′ ) u'' = (x'', y'') u ′′ = ( x ′′ , y ′′ )
{ ( u u ′ ) u ′ ′ = ( ( x x ′ + γ y ‾ ′ y ) x ′ ′ + γ y ‾ ′ ′ ( y ‾ x ′ + y ′ x ) , ( u u ′ ) u ′ ′ = ( y x ‾ ′ + y ′ x ) x ‾ ′ ′ + y ′ ′ ( x x ′ + γ y ‾ ′ y ) ) u ( u ′ u ′ ′ ) = ( x ( x ′ x ′ ′ + γ y ‾ ′ ′ y ′ ) + ( γ ( x ′ ′ y ‾ ′ + x ‾ ′ y ‾ ′ ′ ) y , u ( u ′ u ′ ′ ) = y ( x ‾ ′ ′ x ‾ ′ + γ y ‾ ′ y ′ ′ ) + ( y ′ x ‾ ′ ′ + y ′ ′ x ′ ) x ) .
\begin{cases}
(uu')u'' = ((xx' + \gamma \overline{y}'y)x'' + \gamma \overline{y}''(\overline{y}x' + y'x), \\
\phantom{(uu')u'' = } (y\overline{x}' + y'x)\overline{x}'' + y''(xx' + \gamma \overline{y}'y)) \\
u(u'u'') = (x(x'x'' + \gamma \overline{y}''y') + (\gamma(x''\overline{y}' + \overline{x}'\overline{y}'')y, \\
\phantom{u(u'u'') = } y(\overline{x}''\overline{x}' + \gamma \overline{y}'y'') + (y'\overline{x}'' + y''x')x).
\end{cases}
⎩ ⎨ ⎧ ( u u ′ ) u ′′ = (( x x ′ + γ y ′ y ) x ′′ + γ y ′′ ( y x ′ + y ′ x ) , ( u u ′ ) u ′′ = ( y x ′ + y ′ x ) x ′′ + y ′′ ( x x ′ + γ y ′ y )) u ( u ′ u ′′ ) = ( x ( x ′ x ′′ + γ y ′′ y ′ ) + ( γ ( x ′′ y ′ + x ′ y ′′ ) y , u ( u ′ u ′′ ) = y ( x ′′ x ′ + γ y ′ y ′′ ) + ( y ′ x ′′ + y ′′ x ′ ) x ) . (27)
检查这些公式表明，E的交换律意味着F的结合性。相反，如果F是结合律，公式(27)与y = y ′ = 0 y = y' = 0 y = y ′ = 0 x ′ ′ = 0 x'' = 0 x ′′ = 0 y ′ ′ = e y'' = e y ′′ = e ( 0 , x ′ x ) = ( 0 , x x ′ ) (0, x'x) = (0, xx') ( 0 , x ′ x ) = ( 0 , x x ′ ) x ′ x = x x ′ x'x = xx' x ′ x = x x ′ x , x ′ x, x' x , x ′
注另外，在上面的记号中，对于E中的x , y x, y x , y
y j = j y ‾ , x ( y j ) = ( y x ) j , ( x j ) y = ( x y ‾ ) j , ( x j ) ( y j ) = y ‾ x e
yj = j\overline{y}, \quad x(yj) = (yx)j, \quad (xj)y = (x\overline{y})j, \quad (xj)(yj) = \overline{y}xe
y j = j y , x ( y j ) = ( y x ) j , ( x j ) y = ( x y ) j , ( x j ) ( y j ) = y x e (28)
(29)
Cayley 代数 ( F , t ) (\mathbf{F}, t) ( F , t ) ( E , s ) (\mathbf{E}, s) ( E , s ) 由 γ \gamma γ
(1) 如果我们取E = A \mathbf{E} = \mathbf{A} E = A s = 1 A s = 1_A s = 1 A F \mathbf{F} F ( e , j ) (e, j) ( e , j ) j 2 = γ e j^2 = \gamma e j 2 = γ e
(2) 取E为型* ( α , β ) (\alpha, \beta) ( α , β ) A 2 \mathbf{A}^2 A 2 γ ∈ A \gamma \in \mathbf{A} γ ∈ A γ \gamma γ ( E , s ) (\mathbf{E}, s) ( E , s ) ( α , β , γ ) (\alpha, \beta, \gamma) ( α , β , γ ) A 4 \mathbf{A}^4 A 4 ( e , i , j , k ) (e, i, j, k) ( e , i , j , k ) A 4 \mathbf{A}^4 A 4
{ i 2 = α e + β i , i j = k , i k = α j + β k , j i = β j − k , j 2 = γ e , j k = β γ e − γ i , k i = − α j , k j = γ i , k 2 = − α γ e .
\left\{\begin{array}{lll}
i^2 = \alpha e + \beta i, & ij = k, & ik = \alpha j + \beta k, \\
ji = \beta j - k, & j^2 = \gamma e, & jk = \beta \gamma e - \gamma i, \\
ki = -\alpha j, & kj = \gamma i, & k^2 = -\alpha \gamma e.
\end{array}\right.
⎩ ⎨ ⎧ i 2 = α e + β i , j i = β j − k , k i = − α j , ij = k , j 2 = γ e , k j = γ i , ik = α j + β k , j k = β γ e − γ i , k 2 = − α γ e . (30)
此外，对于u = ρ e + ξ i + η j + ζ k u = \rho e + \xi i + \eta j + \zeta k u = ρ e + ξ i + η j + ζ k ρ , ξ , η , ζ \rho, \xi, \eta, \zeta ρ , ξ , η , ζ A \mathbf{A} A u ˉ \bar{u} u ˉ t ( u ) t(u) t ( u ) A e \mathbf{Ae} Ae A \mathbf{A} A
{ u = ( ρ + β ξ ) e − ξ i − η j − ζ k T F ( u ) = 2 ρ + β ξ N F ( u ) = ρ 2 + β ρ ξ − α ξ 2 − γ ( η 2 + β η ζ − α ζ 2 ) .
\left\{
\begin{array}{l}
u = (\rho + \beta \xi)e - \xi i - \eta j - \zeta k \\
T_F(u) = 2\rho + \beta \xi \\
N_F(u) = \rho^2 + \beta \rho \xi - \alpha \xi^2 - \gamma (\eta^2 + \beta \eta \zeta - \alpha \zeta^2).
\end{array}
\right.
⎩ ⎨ ⎧ u = ( ρ + β ξ ) e − ξ i − η j − ζ k T F ( u ) = 2 ρ + β ξ N F ( u ) = ρ 2 + β ρ ξ − α ξ 2 − γ ( η 2 + β η ζ − α ζ 2 ) . 公式 (30) 由 (28) 和 (29) 得出，公式 (31) 由 (23) 和 (24) 得出，考虑到二次代数 E 的公式。
那么，对于 F 中的 u , v u, v u , v
N F ( u v ) = N F ( u ) N F ( v )
N_F(uv) = N_F(u)N_F(v)
N F ( uv ) = N F ( u ) N F ( v ) 对于N F ( u v ) = u v . u v ‾ = u v ( v ‾ . u ‾ ) = u ( v v ‾ ) u ‾ = ( u u ‾ ) ( v v ‾ ) N_F(uv) = uv.\overline{uv} = uv(\overline{v}.\overline{u}) = u(v\overline{v})\overline{u} = (u\overline{u})(v\overline{v}) N F ( uv ) = uv . uv = uv ( v . u ) = u ( v v ) u = ( u u ) ( v v ) N F ( u ) N_F(u) N F ( u )
与四元数代数同构的A-代数也称为*四元数代数*；如果这样的代数中的基有乘法表(30)，则称为型*( α , β , γ ) (\alpha, \beta, \gamma) ( α , β , γ ) ( α , β , γ ) (\alpha, \beta, \gamma) ( α , β , γ ) ( α , β , γ ) (\alpha, \beta, \gamma) ( α , β , γ )
当 β = 0 \beta = 0 β = 0
{ i 2 = α e , i j = k , i k = α j , j i = − k , j 2 = γ e , j k = − γ i , k i = − α i , k j = γ i , k 2 = − α γ e ,
\left\{
\begin{array}{lll}
i^2 = \alpha e, & ij = k, & ik = \alpha j, \\
ji = -k, & j^2 = \gamma e, & jk = -\gamma i, \\
ki = -\alpha i, & kj = \gamma i, & k^2 = -\alpha \gamma e,
\end{array}
\right.
⎩ ⎨ ⎧ i 2 = α e , j i = − k , k i = − α i , ij = k , j 2 = γ e , k j = γ i , ik = α j , j k = − γ i , k 2 = − α γ e , 和
{ u ˉ = ρ e − ξ i − η j − ζ k T F ( u ) = 2 ρ N F ( u ) = ρ 2 − α ξ 2 − γ η 2 + α γ ζ 2 .
\left\{
\begin{array}{l}
\bar{u} = \rho e - \xi i - \eta j - \zeta k \\
T_F(u) = 2\rho \\
N_F(u) = \rho^2 - \alpha \xi^2 - \gamma \eta^2 + \alpha \gamma \zeta^2.
\end{array}
\right.
⎩ ⎨ ⎧ u ˉ = ρ e − ξ i − η j − ζ k T F ( u ) = 2 ρ N F ( u ) = ρ 2 − α ξ 2 − γ η 2 + α γ ζ 2 . 然后，在上面的表达式中，( α , β , γ ) (\alpha, \beta, \gamma) ( α , β , γ ) ( α , γ ) (\alpha, \gamma) ( α , γ ) ( α , γ ) (\alpha, \gamma) ( α , γ ) ( γ , α ) (\gamma, \alpha) ( γ , α )
注公式 (32) 证明当 − 1 ≠ 1 -1 \neq 1 − 1 = 1 A \mathbf{A} A
取A为实数和α = γ = − 1 , β = 0 \alpha = \gamma = -1, \beta = 0 α = γ = − 1 , β = 0 R \mathbf{R} R H \mathbf{H} H u = ρ e + ξ i + η j + ζ k u = \rho e + \xi i + \eta j + \zeta k u = ρ e + ξ i + η j + ζ k R \mathbf{R} R ρ , ξ , η , ζ \rho, \xi, \eta, \zeta ρ , ξ , η , ζ H \mathbf{H} H ≠ 0 \neq 0 = 0 u u ˉ = u ˉ u = ρ 2 + ξ 2 + η 2 + ζ 2 u\bar{u} = \bar{u}u = \rho^2 + \xi^2 + \eta^2 + \zeta^2 u u ˉ = u ˉ u = ρ 2 + ξ 2 + η 2 + ζ 2 R \mathbf{R} R N ( u ) ≠ 0 N(u) \neq 0 N ( u ) = 0 u u u H \mathbf{H} H u − 1 = N ( u ) − 1 u ˉ u^{-1} = N(u)^{-1}\bar{u} u − 1 = N ( u ) − 1 u ˉ H \mathbf{H} H
(3) 如果 E 取四元数代数 (cf. 例 2 )，则由元素 δ ∈ A \delta \in \mathbf{A} δ ∈ A
（命题5）；它被称为 A 上的八元数代数 (cf. 附录, no. 3)。
引用了 3 引用了 命题 1 , 命题 2 , 命题 5
6. 代数岩浆，幺半群，群
回想一下，岩浆的大小为集合，其中律为合成 (I, § 1, no. 1)。设 S 为乘法写成的岩浆，设 E = A^{(S)} 为 S 的元素形式线性组合 A-模 (II, § 1, no. 11)；我们知道典范映射 s \mapsto e_s 被 S 定义为 A^{(S)}，这样族 (e_s)_{s \in S} 就是 A^{(S)} 的基（称为规范），A^{(S)} 的每个元素都被唯一地写在形式 ∑ s ∈ S α s e s \sum_{s \in S} \alpha_s e_s ∑ s ∈ S α s e s
e s e t = e s t .
e_s e_t = e_{st}.
e s e t = e s t . 这样定义的代数 E 称为 A 上的岩浆 S 的代数。如果 x = ∑ s ∈ S ξ s e s x = \sum_{s \in S} \xi_s e_s x = ∑ s ∈ S ξ s e s y = ∑ s ∈ S η s e s y = \sum_{s \in S} \eta_s e_s y = ∑ s ∈ S η s e s
x y = ∑ s ∈ S ( ∑ t u = s ξ t η u ) e s .
xy = \sum_{s \in S} \left( \sum_{tu = s} \xi_t \eta_u \right) e_s.
x y = s ∈ S ∑ ( t u = s ∑ ξ t η u ) e s . 当 S 是幺半群（分别群）时，E 称为幺半群（分别群）S 在 A 上的代数；那么它是一个结合词代数 (§ 1, no. 7)；类似地，当S是交换幺半群时，它的代数是结合律和交换律的。最后，如果岩浆S承认单位元素u，则e u e_u e u e u e_u e u
当 A \neq \{0\} 时，S 的有时与其在单射 s \mapsto e_s 下的像一致，因此 E 的元素写为 ∑ s ∈ S α s s \sum_{s \in S} \alpha_s s ∑ s ∈ S α s s e s e^s e s e s e_s e s
设B为另一个交换环和ρ : A → B \rho : A \to B ρ : A → B ( e s ) s ∈ S (e_s)_{s \in S} ( e s ) s ∈ S ( e s ′ ) s ∈ S (e'_s)_{s \in S} ( e s ′ ) s ∈ S j ( e s ⊗ 1 ) = e s ′ j(e_s \otimes 1) = e'_s j ( e s ⊗ 1 ) = e s ′ s ∈ S s \in S s ∈ S
S 转化为 F，仅具有乘法结构。那么存在且仅有一个 A-代数同态 f ˉ : A ( S ) → F \bar{f}: A^{(S)} \to F f ˉ : A ( S ) → F
S → f F ∣ ∣ A ( S ) → f ˉ F
\begin{array}{ccc}
S & \xrightarrow{f} & F \\
| & & | \\
A^{(S)} & \xrightarrow{\bar{f}} & F
\end{array}
S ∣ A ( S ) f f ˉ F ∣ F （其中左侧的垂直箭头是典范映射 s ↦ e s s \mapsto e_s s ↦ e s
设f ˉ : A ( S ) → F \bar{f}: A^{(S)} \to F f ˉ : A ( S ) → F f ˉ ( e s ) = f ˉ ( s ) \bar{f}(e_s) = \bar{f}(s) f ˉ ( e s ) = f ˉ ( s ) f ˉ \bar{f} f ˉ f ˉ ( e s e t ) = f ˉ ( e s ) f ˉ ( e t ) \bar{f}(e_s e_t) = \bar{f}(e_s) \bar{f}(e_t) f ˉ ( e s e t ) = f ˉ ( e s ) f ˉ ( e t ) f ( s t ) = f ( s ) f ( t ) f(st) = f(s)f(t) f ( s t ) = f ( s ) f ( t )
命题 6 表示A ( S ) A^{(S)} A ( S ) s ↦ e s s \mapsto e_s s ↦ e s Σ \Sigma Σ α \alpha α
令S、S'为两块岩浆且g : S → S ′ g: S \to S' g : S → S ′ u : A ( S ) → A ( S ′ ) u: A^{(S)} \to A^{(S')} u : A ( S ) → A ( S ′ )
S → g S ′ ∣ ∣ A ( S ) → u A ( S ′ )
\begin{array}{ccc}
S & \xrightarrow{g} & S' \\
| & & | \\
A^{(S)} & \xrightarrow{u} & A^{(S')}
\end{array}
S ∣ A ( S ) g u S ′ ∣ A ( S ′ ) 引用了 2 引用了 命题 6 , I § 1. 。 律合成； 结合性； 交换律
（其中垂直箭头是典范映射）。
应用命题 6 取f f f S → g S ′ → A ( S ′ ) S \xrightarrow{g} S' \to A^{(S')} S g S ′ → A ( S ′ )
特别是，如果 T 是岩浆 S (I, § 1, no. 4) 的稳定子集，则A ( S ) A^{(S)} A ( S ) ∑ s ∈ T α s e s \sum_{s \in T} \alpha_s e_s ∑ s ∈ T α s e s A ( S ) A^{(S)} A ( S ) A ( T ) A^{(T)} A ( T )
令 V 为A-模，S 是一个作用于左侧 V 的幺半群；这意味着(I, § 5, no. 1)将S的映射( s , x ) ↦ s . x (s, x) \mapsto s.x ( s , x ) ↦ s . x s . ( x + y ) = s . x + s . y , s . ( α x ) = α ( s . x ) s.(x + y) = s.x + s.y, s.(\alpha x) = \alpha(s.x) s . ( x + y ) = s . x + s . y , s . ( α x ) = α ( s . x ) s . ( t . x ) = ( s t ) . x s.(t.x) = (st).x s . ( t . x ) = ( s t ) . x s , t s, t s , t x , y x, y x , y α ∈ A \alpha \in A α ∈ A f ˉ : A ( S ) → E n d A ( V ) \bar{f}: A^{(S)} \to \mathrm{End}_A(V) f ˉ : A ( S ) → End A ( V ) A ( S ) A^{(S)} A ( S )
这使我们能够通过约化研究群与运算符的交换模。让 M 是一个可交换的群，其运算符是加法编写的，所有的外法则都是乘法编写的。令Ω \Omega Ω Ω \Omega Ω Ω \Omega Ω Ω \Omega Ω
( s , x ) ↦ s . x
(s, x) \mapsto s.x
( s , x ) ↦ s . x 定义在 M 上，以 Mo(Ω \Omega Ω Ω \Omega Ω x ∈ M x \in M x ∈ M e . x = x e.x = x e . x = x n ≥ 1 n \geq 1 n ≥ 1 s = t u s = tu s = t u n − 1 n - 1 n − 1 t ∈ Ω t \in \Omega t ∈ Ω s . x = t . ( u . x ) s.x = t.(u.x) s . x = t . ( u . x ) Ω \Omega Ω s ′ s' s ′ s . ( s ′ . x ) = ( s s ′ ) . x s.(s'.x) = (ss').x s . ( s ′ . x ) = ( s s ′ ) . x
然后如上所述的应用方法，在M上得到左Z ( M o ( Ω ) ) \mathbf{Z}^{(Mo(\Omega))} Z ( M o ( Ω ))
引用了 4 引用了 I § 5. 。 群运行于集合 , 命题 6 , III § 7. 。外观代数 , E II § 4. 集合家庭的并和交
7. 免费代数
令I为集合；设 M(I)（分别为 Mo(I)、分别为 N ( I ) \mathbf{N}^{(I)} N ( I ) N ( I ) \mathbf{N}^{(I)} N ( I )
被以下引用 3 被以下引用 习题 6, § 1 , 习题 13, § 1 , LIE II § 3. 。包容代数免费李代数
我们将用 Lib_A(I)（分别为 Libas_A(I)、分别为 Libasc_A(I)）来表示 A 上的 I 的自由代数（分别为自由结合代数、分别为自由交换代数）。将 I 的典范映射组合成 M(I)（分别为 Mo(I)、分别为 N ( I ) \mathbf{N}^{(I)} N ( I ) N ( I ) \mathbf{N}^{(I)} N ( I ) A ≠ { 0 } A \neq \{0\} A = { 0 } X i X_i X i i ∈ I i \in I i ∈ I X i X_i X i
由于 Mo(I) 和 N ( I ) \mathbf{N}^{(I)} N ( I ) α ↦ α e \alpha \mapsto \alpha e α ↦ α e
令 F m F_m F m F m F_m F m N ( I ) \mathbf{N}^{(I)} N ( I ) g g g g ( i ) = f ( i ) g(i) = f(i) g ( i ) = f ( i ) i ∈ I i \in I i ∈ I
8. 生成元和关系的代数的定义
设F是A上的代数，并且( x i ) i ∈ I (x_i)_{i \in I} ( x i ) i ∈ I f : L i b A ( I ) → F f : \mathrm{Lib}_A(I) \to F f : Lib A ( I ) → F f ( X i ) = x i f(X_i) = x_i f ( X i ) = x i i ∈ I i \in I i ∈ I f f f ( x i ) i ∈ I (x_i)_{i \in I} ( x i ) i ∈ I
如果U ∈ L i b A ( I ) U \in \mathrm{Lib}_A(I) U ∈ Lib A ( I ) f ( U ) f(U) f ( U ) x i x_i x i X i X_i X i x i x_i x i X i X_i X i U ( ( x i ) i ∈ I ) U((x_i)_{i \in I}) U (( x i ) i ∈ I ) U ( ( X i ) i ∈ I ) = U U((X_i)_{i \in I}) = U U (( X i ) i ∈ I ) = U λ \lambda λ F ′ F' F ′
λ ( U ( ( x i ) i ∈ I ) ) = U ( ( λ ( x i ) ) i ∈ I ) .
\lambda(U((x_i)_{i \in I})) = U((\lambda(x_i))_{i \in I}).
λ ( U (( x i ) i ∈ I )) = U (( λ ( x i ) ) i ∈ I ) . 特别考虑F = L i b A ( J ) F = \mathrm{Lib}_A(J) F = Lib A ( J ) L i b A ( J ) \mathrm{Lib}_A(J) Lib A ( J ) ( H i ) i ∈ I (H_i)_{i \in I} ( H i ) i ∈ I F ′ F' F ′ ( y j ′ ) j ∈ J (y_j')_{j \in J} ( y j ′ ) j ∈ J
U ( ( H i ) i ∈ I ) ) ( ( y j ′ ) j ∈ J ) = U ( ( H i ( ( y j ′ ) j ∈ J ) ) ) i ∈ I ) .
U((H_i)_{i \in I}))((y_j')_{j \in J}) = U((H_i((y_j')_{j \in J})))_{i \in I}).
U (( H i ) i ∈ I )) (( y j ′ ) j ∈ J ) = U (( H i (( y j ′ ) j ∈ J )) ) i ∈ I ) . 在上面的记号中，L i b A ( I ) \mathrm{Lib}_A(I) Lib A ( I ) U ( ( x i ) i ∈ I ) = 0 U((x_i)_{i \in I}) = 0 U (( x i ) i ∈ I ) = 0 f ( U ) = 0 f(U) = 0 f ( U ) = 0 ( x i ) i ∈ I (x_i)_{i \in I} ( x i ) i ∈ I K e r ( f ) \mathrm{Ker}(f) Ker ( f ) ( x i ) (x_i) ( x i )
设( R j ) j ∈ J (R_j)_{j \in J} ( R j ) j ∈ J L i b A ( I ) \mathrm{Lib}_A(I) Lib A ( I ) ( x i ) i ∈ I (x_i)_{i \in I} ( x i ) i ∈ I R j R_j R j L i b A ( I ) \mathrm{Lib}_A(I) Lib A ( I ) ( x i ) i ∈ I (x_i)_{i \in I} ( x i ) i ∈ I ( ( x i ) i ∈ I , ( R j ) j ∈ J ) ((x_i)_{i \in I}, (R_j)_{j \in J}) (( x i ) i ∈ I , ( R j ) j ∈ J ) x i x_i x i 生成元 ，R j R_j R j
现在考虑任何集合 I 和( R j ) j ∈ J (R_j)_{j \in J} ( R j ) j ∈ J L i b A ( I ) \mathrm{Lib}_A(I) Lib A ( I ) ( R j ) (R_j) ( R j ) L i b A ( I ) \mathrm{Lib}_A(I) Lib A ( I ) ( R j ) j ∈ J (R_j)_{j \in J} ( R j ) j ∈ J X ‾ i \overline{X}_i X i X i X_i X i
( ( X ‾ i ) i ∈ I , ( R j ) j ∈ J )
((\overline{X}_i)_{i \in I}, (R_j)_{j \in J})
(( X i ) i ∈ I , ( R j ) j ∈ J ) ( x i ) i ∈ I (x_i)_{i \in I} ( x i ) i ∈ I R j ( ( x i ) i ∈ I ) = 0 R_j((x_i)_{i \in I}) = 0 R j (( x i ) i ∈ I ) = 0 j ∈ J j \in J j ∈ J g : E → F g : E \to F g : E → F g ( X ‾ i ) = x i g(\overline{X}_i) = x_i g ( X i ) = x i i ∈ I i \in I i ∈ I ( ( x i ) i ∈ I , ( R j ) j ∈ J ) ((x_i)_{i \in I}, (R_j)_{j \in J}) (( x i ) i ∈ I , ( R j ) j ∈ J ) g g g 这些注证明了以下滥用语言的合理性：而不是说“( ( x i ) i ∈ I , ( R j ) j ∈ J ) ((x_i)_{i \in I}, (R_j)_{j \in J}) (( x i ) i ∈ I , ( R j ) j ∈ J ) F是由生成元x i x_i x i R j ( ( x i ) i ∈ I ) = 0 R_j((x_i)_{i \in I}) = 0 R j (( x i ) i ∈ I ) = 0 ”。当R j R_j R j P j − Q j P_j - Q_j P j − Q j F是由x i x_i x i P j ( ( x i ) ) = Q j ( ( x i ) ) P_j((x_i)) = Q_j((x_i)) P j (( x i )) = Q j (( x i )) 生成的代数”。
设H为集合；我们可以说，L i b A ( H ) \mathrm{Lib}_A(H) Lib A ( H ) ( x h ) h ∈ H (x_h)_{h \in H} ( x h ) h ∈ H S ( ( x h ) h ∈ H ) = 0 S((x_h)_{h \in H}) = 0 S (( x h ) h ∈ H ) = 0
(1)取H = { 1 , 2 , 3 } H = \{1, 2, 3\} H = { 1 , 2 , 3 }
( X 1 X 2 ) X 3 − X 1 ( X 2 X 3 )
(X_1 X_2) X_3 - X_1 (X_2 X_3)
( X 1 X 2 ) X 3 − X 1 ( X 2 X 3 ) 作为通用关系词的是联想词代数。承认
X 1 X 2 − X 2 X 1 X_1 X_2 - X_2 X_1 X 1 X 2 − X 2 X 1 为万能关系子的代数是交换代数。 *代数承认通用关系器
X 1 X 1 X_1 X_1 X 1 X 1 并且
( X 1 X 2 ) X 3 + ( X 2 X 3 ) X 1 + ( X 3 X 1 ) X 2
(X_1 X_2) X_3 + (X_2 X_3) X_1 + (X_3 X_1) X_2
( X 1 X 2 ) X 3 + ( X 2 X 3 ) X 1 + ( X 3 X 1 ) X 2 作为通用关系器是李代数.
令I为集合；给定 Lib_A(H) 的元素的族 ( S k ) k ∈ K (S_k)_{k \in K} ( S k ) k ∈ K S k ( ( U h ) ) h ∈ H S_k((U_h))_{h \in H} S k (( U h ) ) h ∈ H ( U h ) h ∈ H (U_h)_{h \in H} ( U h ) h ∈ H ( R j ) j ∈ J (R_j)_{j \in J} ( R j ) j ∈ J ( R j ) j ∈ J (R_j)_{j \in J} ( R j ) j ∈ J u : L i b A ( I ) → F u : \mathrm{Lib}_A(I) \to F u : Lib A ( I ) → F S k ( ( U h ) h ∈ H ) S_k((U_h)_{h \in H}) S k (( U h ) h ∈ H ) k ∈ K k \in K k ∈ K ( U h ) h ∈ H (U_h)_{h \in H} ( U h ) h ∈ H S k S_k S k k ∈ K k \in K k ∈ K ( x i ) i ∈ I (x_i)_{i \in I} ( x i ) i ∈ I S k S_k S k u ′ : L i b A ( I ) → F ′ u' : \mathrm{Lib}_A(I) \to F' u ′ : Lib A ( I ) → F ′ u ′ ( X i ) = x i u'(X_i) = x_i u ′ ( X i ) = x i i ∈ I i \in I i ∈ I ⊂ \subset ⊂ u ′ u' u ′ u ′ = h ∘ u u' = h \circ u u ′ = h ∘ u h : F → F ′ h : F \to F' h : F → F ′ h ( X ‾ i ) = x i h(\overline{X}_i) = x_i h ( X i ) = x i i ∈ I i \in I i ∈ I ( S k ) k ∈ K (S_k)_{k \in K} ( S k ) k ∈ K ( u h ) h ∈ H (u_h)_{h \in H} ( u h ) h ∈ H S k ( ( u h ) ) = 0 S_k((u_h)) = 0 S k (( u h )) = 0
例* 2. 设L'为I生成的通用代数，并服从于L'的每个三口之家元素的恒等式( u v ) w − u ( v w ) = 0 (uv)w - u(vw) = 0 ( uv ) w − u ( v w ) = 0 i ∈ I i \in I i ∈ I g ( X ‾ i ) = X i g(\overline{X}_i) = X_i g ( X i ) = X i
与上述类似的考虑可以应用于结合律代数（分别是交换结合律代数），同时考虑以下注。当上下文充分表明代数被认为是统一的代数时，通过滥用语言，代数F F F ( x i ) i ∈ I (x_i)_{i \in I} ( x i ) i ∈ I x i x_i x i i ∈ I i \in I i ∈ I F F F F F F F F F A A A ( x i ) i ∈ I (x_i)_{i \in I} ( x i ) i ∈ I F F F f : L i b a s A ( I ) → F f : \mathrm{Libas}_A(I) \to F f : Libas A ( I ) → F f ( X i ) = x i f(X_i) = x_i f ( X i ) = x i i ∈ I i \in I i ∈ I U ∈ L i b a s A ( I ) U \in \mathrm{Libas}_A(I) U ∈ Libas A ( I ) f ( U ) f(U) f ( U ) F F F U U U x i x_i x i X i X_i X i U ( ( x i ) i ∈ I ) U((x_i)_{i \in I}) U (( x i ) i ∈ I ) L i b A ( I ) \mathrm{Lib}_A(I) Lib A ( I ) L i b a s A ( I ) \mathrm{Libas}_A(I) Libas A ( I ) 由关系族相关的生成的系统 I 定义的通用单位结合律代数 ( R j ) j ∈ J (R_j)_{j \in J} ( R j ) j ∈ J ( R j ) (R_j) ( R j ) L i b a s A ( I ) \mathrm{Libas}_A(I) Libas A ( I ) I I I L i b a s c A ( I ) \mathrm{Libasc}_A(I) Libasc A ( I ) L i b a s A ( I ) \mathrm{Libas}_A(I) Libas A ( I )
引用了 1 引用了 命题 7
令 L ′ L' L ′ I I I L ′ L' L ′ u v − v u = 0 uv - vu = 0 uv − v u = 0 L ′ L' L ′ L i b a s c A ( I ) \mathrm{Libasc}_A(I) Libasc A ( I )
9. 多项式代数
令B B B A A A ( x i ) i ∈ I (x_i)_{i \in I} ( x i ) i ∈ I B B B x i x_i x i i ∈ I i \in I i ∈ I B B B A [ ( x i ) i ∈ I ] B A[(x_i)_{i \in I}]_B A [( x i ) i ∈ I ] B A [ ( x i ) i ∈ I ] A[(x_i)_{i \in I}] A [( x i ) i ∈ I ] I I I L i b a s c A ( I ) \mathrm{Libasc}_A(I) Libasc A ( I ) A [ ( X i ) i ∈ I ] A[(X_i)_{i \in I}] A [( X i ) i ∈ I ] A [ X i ] i ∈ I A[X_i]_{i \in I} A [ X i ] i ∈ I A [ ( X i ) i ∈ I ] A[(X_i)_{i \in I}] A [( X i ) i ∈ I ] X i X_i X i i ∈ I i \in I i ∈ I 系数为 A A A A [ ( X i ) i ∈ I ] A[(X_i)_{i \in I}] A [( X i ) i ∈ I ] X i X_i X i I I I J J J L i b a s c A ( J ) \mathrm{Libasc}_A(J) Libasc A ( J ) i ∈ J i \in J i ∈ J X i X_i X i L i b a s c A ( I ) \mathrm{Libasc}_A(I) Libasc A ( I ) I = { 1 , 2 , … , n } I = \{1, 2, \ldots, n\} I = { 1 , 2 , … , n } A [ X 1 , X 2 , … , X n ] A[X_1, X_2, \ldots, X_n] A [ X 1 , X 2 , … , X n ] A [ ( X i ) i ∈ I ] A[(X_i)_{i \in I}] A [( X i ) i ∈ I ]
如果I I I I ′ I' I ′ L i b a s c A ( I ) \mathrm{Libasc}_A(I) Libasc A ( I ) L i b a s c A ( I ′ ) \mathrm{Libasc}_A(I') Libasc A ( I ′ ) A [ X ] A[X] A [ X ] X X X A [ X , Y ] A[X, Y] A [ X , Y ] A [ X , Y , Z ] A[X, Y, Z] A [ X , Y , Z ] A ≠ { 0 } A \neq \{0\} A = { 0 } A [ X ] A[X] A [ X ] A [ Y ] A[Y] A [ Y ] A [ X , Y , Z ] A[X, Y, Z] A [ X , Y , Z ]
元素
X ν = ∏ i ∈ I X i ν ( i ) ,
X^\nu = \prod_{i \in I} X_i^{\nu(i)},
X ν = i ∈ I ∏ X i ν ( i ) , ν \nu ν A [ ( X i ) i ∈ I ] A[(X_i)_{i \in I}] A [( X i ) i ∈ I ] N ( I ) \mathbf{N}^{(I)} N ( I ) X i X_i X i ∣ ν ∣ = ∑ i ∈ I ν ( i ) |\nu| = \sum_{i \in I} \nu(i) ∣ ν ∣ = ∑ i ∈ I ν ( i ) X ν X^\nu X ν A [ ( X i ) i ∈ I ] A[(X_i)_{i \in I}] A [( X i ) i ∈ I ] A A A A [ ( X i ) i ∈ I ] A[(X_i)_{i \in I}] A [( X i ) i ∈ I ] u u u u = ∑ ν ∈ N ( I ) α ν X ν
u = \sum_{\nu \in \mathbf{N}^{(I)}} \alpha_\nu X^\nu
u = ν ∈ N ( I ) ∑ α ν X ν α ν ∈ A \alpha_\nu \in A α ν ∈ A α ν \alpha_\nu α ν ν ∈ N ( I ) \nu \in \mathbf{N}^{(I)} ν ∈ N ( I ) u u u α ν X ν \alpha_\nu X^\nu α ν X ν u u u α ν X ν \alpha_\nu X^\nu α ν X ν X ν X^\nu X ν α 0 X 0 \alpha_0 X^0 α 0 X 0 α 0 ∈ A \alpha_0 \in A α 0 ∈ A u u u J J J I I I u u u A [ ( X i ) i ∈ J ] A[(X_i)_{i \in J}] A [( X i ) i ∈ J ] α ν = 0 \alpha_\nu = 0 α ν = 0 ν ∉ N ( J ) \nu \notin \mathbf{N}^{(J)} ν ∈ / N ( J ) A [ ( X i ) i ∈ I ] A[(X_i)_{i \in I}] A [( X i ) i ∈ I ] A [ ( X i ) i ∈ J ] A[(X_i)_{i \in J}] A [( X i ) i ∈ J ] J J J I I I α ν = 0 \alpha_\nu = 0 α ν = 0 ∣ ν ∣ > n |\nu| > n ∣ ν ∣ > n u u u 次数 ≤ n \leq n ≤ n α ν = 0 \alpha_\nu = 0 α ν = 0 u u u X ν X^\nu X ν α 0 = 0 \alpha_0 = 0 α 0 = 0 u u u 对于每个*非零*多项式u = ∑ α ν X ν u = \sum \alpha_\nu X^\nu u = ∑ α ν X ν u u u ν \nu ν ∣ ν ∣ |\nu| ∣ ν ∣ α ν ≠ 0 \alpha_\nu \neq 0 α ν = 0
令F F F A A A ( x i ) i ∈ I (x_i)_{i \in I} ( x i ) i ∈ I F F F x i x_i x i F F F F ′ F' F ′ x i x_i x i u ∈ A [ ( X i ) i ∈ I ] u \in A[(X_i)_{i \in I}] u ∈ A [( X i ) i ∈ I ] X i X_i X i F F F u ( ( x i ) i ∈ I ) u((x_i)_{i \in I}) u (( x i ) i ∈ I ) F ′ F' F ′ F F F h : u → u ( ( x i ) i ∈ I ) h: u \to u((x_i)_{i \in I}) h : u → u (( x i ) i ∈ I ) A [ ( X i ) i ∈ I ] A[(X_i)_{i \in I}] A [( X i ) i ∈ I ] F F F h h h A [ ( X i ) i ∈ I ] A[(X_i)_{i \in I}] A [( X i ) i ∈ I ] ( x i ) (x_i) ( x i ) x i x_i x i A A A h h h F ′ F' F ′ A [ ( x i ) i ∈ I ] A[(x_i)_{i \in I}] A [( x i ) i ∈ I ] F F F a a a x i x_i x i A A A
0 ⟶ a ⟶ A [ ( X i ) i ∈ I ] → h A [ ( x i ) i ∈ I ] ⟶ 0.
0 \longrightarrow a \longrightarrow A[(X_i)_{i \in I}] \xrightarrow{h} A[(x_i)_{i \in I}] \longrightarrow 0.
0 ⟶ a ⟶ A [( X i ) i ∈ I ] h A [( x i ) i ∈ I ] ⟶ 0.
令A [ ( X i ) i ∈ I ] A[(X_i)_{i \in I}] A [( X i ) i ∈ I ] I I I J J J K K K I I I J J J A ′ = A [ ( X j ) j ∈ J ] A' = A[(X_j)_{j \in J}] A ′ = A [( X j ) j ∈ J ] X k ′ X'_k X k ′ k ∈ K k \in K k ∈ K
被以下引用 1 被以下引用 推论 1（希尔伯特）, VIII § 1
显然，A[(X_i)_{i \in I}] 是由 X_k 对于 k \in K 生成的 A' 代数。另一方面，作为 X_k (k \in K) 与 A' 中的系数之间的多项式关系器，可以唯一地写为 ∑ v h v ( ( X j ) j ∈ J ) X v \sum_v h_v((X_j)_{j \in J}) X^v ∑ v h v (( X j ) j ∈ J ) X v N ( K ) \mathbf{N}^{(K)} N ( K )
命题 8 中描述的同构通常用于 A[(X_i)_{i \in I}] 的认同元素，其中多项式的系数为 A' = A[(X_j)_{j \in J}]。如果 u 是 A[(X_i)_{i \in I}] 的元素 ≠ 0 \neq 0 = 0
10. 幺半群总数代数
A 上幺半群 S 的代数（如A-模）是有限支撑的族( α s ) s ∈ S (\alpha_s)_{s \in S} ( α s ) s ∈ S A S A^S A S ( α s ) ( β s ) = ( γ s ) (\alpha_s)(\beta_s) = (\gamma_s) ( α s ) ( β s ) = ( γ s ) s ∈ S s \in S s ∈ S
γ s = ∑ t u = s α t β u
\gamma_s = \sum_{tu = s} \alpha_t \beta_u
γ s = t u = s ∑ α t β u （cf.no. 6，公式（35））。 (38) 右侧的和是有意义的，因为( α s ) (\alpha_s) ( α s ) ( β s ) (\beta_s) ( β s ) ( α t β u ) ( t , u ) ∈ S × S (\alpha_t\beta_u)_{(t,u)\in S\times S} ( α t β u ) ( t , u ) ∈ S × S S S S A S A^S A S ( α s ) , β s ) (\alpha_s),\ \beta_s) ( α s ) , β s )
(D) 对于所有 s ∈ S s\in S s ∈ S ( t , u ) (t,u) ( t , u ) 在 S × S S\times S S × S 使得 t u = s tu=s t u = s
假设S S S A S A^S A S A S A^S A S A S A^S A S α , β , γ \alpha,\beta,\gamma α , β , γ
∑ u v w = t α u β v γ w = ∑ r w = t ( ( ∑ u v = r α u β v ) γ w ) = ∑ u s = t ( α u ( ∑ v w = s β v γ w ) ) .
\sum_{uvw=t} \alpha_u\beta_v\gamma_w = \sum_{rw=t} \left( \left( \sum_{uv=r} \alpha_u\beta_v \right) \gamma_w \right) = \sum_{us=t} \left( \alpha_u \left( \sum_{vw=s} \beta_v\gamma_w \right) \right).
uv w = t ∑ α u β v γ w = r w = t ∑ ( ( uv = r ∑ α u β v ) γ w ) = u s = t ∑ ( α u ( v w = s ∑ β v γ w ) ) . 乘法和 A-模结构在 A S A^S A S A S A^S A S A S A^S A S
直接的，幺半群 S 在 A 上的 代数 A ( S ) A^{(S)} A ( S ) 子代数 （当 S 有限时与后者相同）。 滥用语言 ，A 上 S 的总代数中的每一个元素( ξ s ) s ∈ S (\xi_s)_{s\in S} ( ξ s ) s ∈ S ∑ s ∈ S ξ s e s \sum_{s\in S} \xi_s e_s ∑ s ∈ S ξ s e s ∑ s ∈ S ξ s s \sum_{s\in S} \xi_s s ∑ s ∈ S ξ s s ≠ 0 \neq 0 = 0
如果 S 是可交换的，那么它的总代数 A S A^S A S A T A^T A T ρ : A → B \rho : A \to B ρ : A → B ρ S : A S → B S \rho^S : A^S \to B^S ρ S : A S → B S A ( S ) → B ( S ) A^{(S)} \to B^{(S)} A ( S ) → B ( S )
11. 正式幂级数超过交换环
对于每个集合 I，加法幺半群N ( I ) N^{(I)} N ( I ) s = ( n i ) i ∈ I s = (n_i)_{i\in I} s = ( n i ) i ∈ I n i = 0 n_i = 0 n i = 0 i i i s = t + u s = t + u s = t + u t = ( p i ) i ∈ I t = (p_i)_{i\in I} t = ( p i ) i ∈ I u = ( q i ) i ∈ I u = (q_i)_{i\in I} u = ( q i ) i ∈ I p i + q i = n i p_i + q_i = n_i p i + q i = n i i i i i ∉ H i \notin H i ∈ / H p i = q i = 0 p_i = q_i = 0 p i = q i = 0 p i ≤ n i , q i ≤ n i p_i \leq n_i, q_i \leq n_i p i ≤ n i , q i ≤ n i i ∈ H i \in H i ∈ H N ( I ) N^{(I)} N ( I ) ∏ i ∈ H ( n i + 1 ) \prod_{i\in H} (n_i + 1) ∏ i ∈ H ( n i + 1 ) ( t , u ) (t, u) ( t , u ) t + u = s t + u = s t + u = s
因此，我们可以考虑幺半群N ( I ) \mathbf{N}^{(I)} N ( I ) A A A A [ X i ] i ∈ I A[\mathbf{X}_i]_{i \in I} A [ X i ] i ∈ I X i ( i ∈ I ) X_i \ (i \in I) X i ( i ∈ I ) A A A A [ [ X i ] ] i ∈ I A[[\mathbf{X}_i]]_{i \in I} A [[ X i ] ] i ∈ I X i ( i ∈ I ) X_i \ (i \in I) X i ( i ∈ I ) A A A ( α v ) v ∈ N ( I ) (\alpha_v)_{v \in \mathbf{N}^{(I)}} ( α v ) v ∈ N ( I ) ∑ v ∈ N ( I ) α v X v \sum_{v \in \mathbf{N}^{(I)}} \alpha_v X^v ∑ v ∈ N ( I ) α v X v α v \alpha_v α v α v X v \alpha_v X^v α v X v X i X_i X i ≠ 0 \neq 0 = 0
显然，代数同构A [ [ X i ] ] i ∈ I 1 → A [ [ X i ] ] i ∈ I 2 A[[\mathbf{X}_i]]_{i \in I_1} \to A[[\mathbf{X}_i]]_{i \in I_2} A [[ X i ] ] i ∈ I 1 → A [[ X i ] ] i ∈ I 2 σ : I 1 → I 2 \sigma : I_1 \to I_2 σ : I 1 → I 2 ∑ ( n i ) α ( n i ) ⋅ ∏ i ∈ I 1 X i n i \sum_{(n_i)} \alpha_{(n_i)} \cdot \prod_{i \in I_1} X_i^{n_i} ∑ ( n i ) α ( n i ) ⋅ ∏ i ∈ I 1 X i n i ∑ ( n i ) α ( n i ) ⋅ ∏ i ∈ I 1 X σ ( i ) n i \sum_{(n_i)} \alpha_{(n_i)} \cdot \prod_{i \in I_1} X_{\sigma(i)}^{n_i} ∑ ( n i ) α ( n i ) ⋅ ∏ i ∈ I 1 X σ ( i ) n i
设J J J I I I A [ [ X i ] ] i ∈ J A[[\mathbf{X}_i]]_{i \in J} A [[ X i ] ] i ∈ J ∑ ( n i ) α ( n i ) ⋅ ∏ i ∈ I X i n i \sum_{(n_i)} \alpha_{(n_i)} \cdot \prod_{i \in I} X_i^{n_i} ∑ ( n i ) α ( n i ) ⋅ ∏ i ∈ I X i n i A [ [ X i ] ] i ∈ I A[[\mathbf{X}_i]]_{i \in I} A [[ X i ] ] i ∈ I α ( n i ) = 0 \alpha_{(n_i)} = 0 α ( n i ) = 0 ( n i ) ∈ N ( I ) (n_i) \in \mathbf{N}^{(I)} ( n i ) ∈ N ( I ) n i ≠ 0 n_i \neq 0 n i = 0 i ∈ I − J i \in I - J i ∈ I − J K = I − J K = I - J K = I − J A [ [ X i ] ] i ∈ I A[[\mathbf{X}_i]]_{i \in I} A [[ X i ] ] i ∈ I ( A [ [ X j ] ] j ∈ J ) [ [ X k ] ] k ∈ K (A[[\mathbf{X}_j]]_{j \in J})[[\mathbf{X}_k]]_{k \in K} ( A [[ X j ] ] j ∈ J ) [[ X k ] ] k ∈ K ∑ ( n i ) α ( n i ) ⋅ ∏ i ∈ I X i n i \sum_{(n_i)} \alpha_{(n_i)} \cdot \prod_{i \in I} X_i^{n_i} ∑ ( n i ) α ( n i ) ⋅ ∏ i ∈ I X i n i ∑ ( m k ) β ( m k ) ⋅ ∏ k ∈ K X k m k \sum_{(m_k)} \beta_{(m_k)} \cdot \prod_{k \in K} X_k^{m_k} ∑ ( m k ) β ( m k ) ⋅ ∏ k ∈ K X k m k
β ( m k ) = ∑ ( p j ) γ ( p j ) ⋅ ∏ j ∈ J X j p j
\beta_{(m_k)} = \sum_{(p_j)} \gamma_{(p_j)} \cdot \prod_{j \in J} X_j^{p_j}
β ( m k ) = ( p j ) ∑ γ ( p j ) ⋅ j ∈ J ∏ X j p j γ ( p j ) = α ( n i ) \gamma_{(p_j)} = \alpha_{(n_i)} γ ( p j ) = α ( n i ) ( n i ) (n_i) ( n i ) n i = p i n_i = p_i n i = p i i ∈ J i \in J i ∈ J n i = m i n_i = m_i n i = m i i ∈ K i \in K i ∈ K 给定一个正式的幂级数u = ∑ v α v X v u = \sum_v \alpha_v X^v u = ∑ v α v X v α v X v \alpha_v X^v α v X v ∣ v ∣ = p |v| = p ∣ v ∣ = p p p p u u u u p u_p u p p p p u u u u u u p p p I I I u p u_p u p p p p u 0 u_0 u 0 A A A u u u u u u v v v w = u v w = uv w = uv
w p = ∑ r = 0 p u r v p − r
w_p = \sum_{r=0}^p u_r v_{p-r}
w p = r = 0 ∑ p u r v p − r p ≥ 0 p \geq 0 p ≥ 0 对于每个正式的幂级数u ≠ 0 u \neq 0 u = 0 p ≥ 0 p \geq 0 p ≥ 0 u p ≠ 0 u_p \neq 0 u p = 0 u u u ω ( u ) \omega(u) ω ( u ) u u u v v v ≠ 0 \neq 0 = 0
ω ( u + v ) ≥ inf ( ω ( u ) , ω ( v ) ) if u + v ≠ 0 ω ( u v ) ≥ ω ( u ) + ω ( v ) if u v ≠ 0.
\omega(u + v) \geq \inf(\omega(u), \omega(v)) \quad \text{if } u + v \neq 0 \\
\omega(uv) \geq \omega(u) + \omega(v) \quad \text{if } uv \neq 0.
ω ( u + v ) ≥ inf ( ω ( u ) , ω ( v )) if u + v = 0 ω ( uv ) ≥ ω ( u ) + ω ( v ) if uv = 0. 进一步，如果ω ( u ) ≠ ω ( v ) \omega(u) \neq \omega(v) ω ( u ) = ω ( v ) u + v ≠ 0 u + v \neq 0 u + v = 0
注 0 的阶 未定义 。 By an abuse of language, it is a convention to say that "f is a formal power series of order ≥ p \geq p ≥ p > p > p > p n < p n < p n < p n ≤ p n \leq p n ≤ p > p > p > p every integer p ≥ 0 p \geq 0 p ≥ 0
令J为I的子集，并让A [ [ X i ] ] i ∈ I A[[X_i]]_{i \in I} A [[ X i ] ] i ∈ I B [ [ X k ] ] k ∈ K B[[X_k]]_{k \in K} B [[ X k ] ] k ∈ K K = I − J K = I - J K = I − J B = A [ [ X j ] ] j ∈ J B = A[[X_j]]_{j \in J} B = A [[ X j ] ] j ∈ J B [ [ X k ] ] k ∈ K B[[X_k]]_{k \in K} B [[ X k ] ] k ∈ K u ∈ A [ [ X i ] ] i ∈ I u \in A[[X_i]]_{i \in I} u ∈ A [[ X i ] ] i ∈ I α ( n i ) ⋅ ∏ i ∈ I X i n i \alpha_{(n_i)} \cdot \prod_{i \in I} X_i^{n_i} α ( n i ) ⋅ ∏ i ∈ I X i n i X i X_i X i i ∈ K i \in K i ∈ K ∑ i ∈ K n i = p \sum_{i \in K} n_i = p ∑ i ∈ K n i = p B [ [ X k ] ] k ∈ K B[[X_k]]_{k \in K} B [[ X k ] ] k ∈ K p p p u u u i ∈ K i \in K i ∈ K X i X_i X i u ≠ 0 u \neq 0 u = 0 i ∈ K i \in K i ∈ K X i X_i X i ω K ( u ) \omega_K(u) ω K ( u ) p ≥ 0 p \geq 0 p ≥ 0 u u u i ∈ K i \in K i ∈ K X i X_i X i p p p ≠ 0 \neq 0 = 0 ω \omega ω ω K \omega_K ω K
习题
参见习题为 § 2 。