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。
§ 8。 范数和痕迹
在本段中，K 表示域。
1. 记起
令 A 为 K 上的有限次数 n n n x ∈ A x \in A x ∈ A L x L_x L x a ↦ x a a \mapsto xa a ↦ x a L x L_x L x T r A / K ( x ) \mathrm{Tr}_{A/K}(x) Tr A / K ( x ) L x L_x L x N A / K ( x ) N_{A/K}(x) N A / K ( x ) ( x 1 , . . . , x n ) (x_1, ..., x_n) ( x 1 , ... , x n ) ( T r A / K ( x i x j ) ) 1 ≤ i , j ≤ n (\mathrm{Tr}_{A/K}(x_i x_j))_{1 \leq i,j \leq n} ( Tr A / K ( x i x j ) ) 1 ≤ i , j ≤ n D A / K ( x 1 , . . . , x n ) D_{A/K}(x_1, ..., x_n) D A / K ( x 1 , ... , x n )
令 K' 为 K 的扩张，并令 A ′ = A ( K ′ ) A' = A_{(K')} A ′ = A ( K ′ )
(1)
T r A ′ / K ′ ( 1 ⊗ x ) = T r A / K ( x ) .1 , N A ′ / K ′ ( 1 ⊗ x ) = N A / K ( x ) .1
\mathrm{Tr}_{A'/K'}(1 \otimes x) = \mathrm{Tr}_{A/K}(x) . 1 , \quad N_{A'/K'}(1 \otimes x) = N_{A/K}(x) . 1
Tr A ′ / K ′ ( 1 ⊗ x ) = Tr A / K ( x ) .1 , N A ′ / K ′ ( 1 ⊗ x ) = N A / K ( x ) .1 x ∈ A x \in A x ∈ A ( x 1 , . . . , x n ) (x_1, ..., x_n) ( x 1 , ... , x n ) (2)
D A ′ / K ′ ( 1 ⊗ x 1 , . . . , 1 ⊗ x n ) = D A / K ( x 1 , . . . , x n ) .1 ,
D_{A'/K'}(1 \otimes x_1, ..., 1 \otimes x_n) = D_{A/K}(x_1, ..., x_n) . 1 ,
D A ′ / K ′ ( 1 ⊗ x 1 , ... , 1 ⊗ x n ) = D A / K ( x 1 , ... , x n ) .1 , 2. 范数和 etale 代数中的痕迹
设 A 是 K 上（有限）次数 n n n u 1 , . . . , u n u_1, ..., u_n u 1 , ... , u n
a) A A A L L L u i u_i u i L L L u : A ( L ) → L n u : A_{(L)} \to L^n u : A ( L ) → L n
u ( 1 ⊗ x ) = ( u 1 ( x ) , … , u n ( x ) ) for all x ∈ A .
u(1 \otimes x) = (u_1(x), \ldots, u_n(x)) \quad \text{for all } x \in A .
u ( 1 ⊗ x ) = ( u 1 ( x ) , … , u n ( x )) for all x ∈ A . 此外，K K K L L L
我们在下面修复L , u 1 , … , u n L, u_1, \ldots, u_n L , u 1 , … , u n x ∈ A x \in A x ∈ A
Tr A / K ( x ) ⋅ 1 = ∑ i = 1 n u i ( x ) , N A / K ( x ) ⋅ 1 = ∏ i = 1 n u i ( x ) .
\text{Tr}_{A/K}(x) \cdot 1 = \sum_{i=1}^n u_i(x) , \quad \text{N}_{A/K}(x) \cdot 1 = \prod_{i=1}^n u_i(x) .
Tr A / K ( x ) ⋅ 1 = i = 1 ∑ n u i ( x ) , N A / K ( x ) ⋅ 1 = i = 1 ∏ n u i ( x ) . Let v v v l ⊗ x l \otimes x l ⊗ x A ( L ) A_{(L)} A ( L ) A ( L ) A_{(L)} A ( L ) u − 1 u^{-1} u − 1 L n L^n L n v v v u 1 ( x ) , … , u n ( x ) u_1(x), \ldots, u_n(x) u 1 ( x ) , … , u n ( x )
T r A ( L ) / L ( 1 ∘ x ) ⋅ 1 = ∑ r = 1 n u i ( x ) , whence Tr A / K ( x ) ⋅ 1 = ∑ i = 1 n u i ( x ) by (1) ;
Tr_{A_{(L)}/L}(1 \circ x) \cdot 1 = \sum_{r=1}^n u_i(x) , \quad \text{whence} \quad \text{Tr}_{A/K}(x) \cdot 1 = \sum_{i=1}^n u_i(x) \text{ by (1)} ;
T r A ( L ) / L ( 1 ∘ x ) ⋅ 1 = r = 1 ∑ n u i ( x ) , whence Tr A / K ( x ) ⋅ 1 = i = 1 ∑ n u i ( x ) by (1) ; 范数的情况也类似处理。
进一步令( x 1 , … , x n ) (x_1, \ldots, x_n) ( x 1 , … , x n ) A A A
( u i ( x j ) ) 1 ≤ i , j ≤ n
(u_i(x_j))_{1 \leq i,j \leq n}
( u i ( x j ) ) 1 ≤ i , j ≤ n 令( t i j ) = ′ U . U (t_{ij}) = 'U . U ( t ij ) = ′ U . U
Tr A / K ( x i x j ) ⋅ 1 = ∑ k = 1 n u k ( x i x j ) = ∑ k = 1 n u k ( x i ) u k ( x j ) = t i j ;
\text{Tr}_{A/K}(x_i x_j) \cdot 1 = \sum_{k=1}^n u_k(x_i x_j) = \sum_{k=1}^n u_k(x_i) u_k(x_j) = t_{ij} ;
Tr A / K ( x i x j ) ⋅ 1 = k = 1 ∑ n u k ( x i x j ) = k = 1 ∑ n u k ( x i ) u k ( x j ) = t ij ; 经过到行列式，我们得到
D A / K ( x 1 , … , x r ) ⋅ 1 = [ det u i ( x j ) ] 2 .
D_{A/K}(x_1, \ldots, x_r) \cdot 1 = [\det u_i(x_j)]^2 .
D A / K ( x 1 , … , x r ) ⋅ 1 = [ det u i ( x j ) ] 2 .
— 令 A A A K K K ： a) 代数 A A A b) A A A 对于A A A x ≠ 0 x \neq 0 x = 0 A A A y y y Tr A / K ( x y ) ≠ 0 \text{Tr}_{A/K}(xy) \neq 0 Tr A / K ( x y ) = 0 *此外，当满足这些条件时，A A A
引用了 1 · 被以下引用 2 引用了 定理 1, V § 6
被以下引用 注 (1), AC V § 2 , 推论
我们将证明当A A A A A A A A A ( x 1 , … , x n ) (x_1, \ldots, x_n) ( x 1 , … , x n ) K K K a ) ⇒ b ) a) \Rightarrow b) a ) ⇒ b )
∑ i = 1 n λ i u i ( x j ) = 0 ( for 1 ≤ j ≤ n )
\sum_{i=1}^n \lambda_i u_i(x_j) = 0 \quad (\text{for } 1 \leq j \leq n)
i = 1 ∑ n λ i u i ( x j ) = 0 ( for 1 ≤ j ≤ n ) L 中只有解 λ 1 = ⋯ = A , = 0 \lambda_1 = \cdots = A, = 0 λ 1 = ⋯ = A , = 0 ∑ i = 1 n λ i u i ( x ) = 0 \sum_{i=1}^n \lambda_i u_i(x) = 0 ∑ i = 1 n λ i u i ( x ) = 0 x ∈ A x \in A x ∈ A λ i = 0 \lambda_i = 0 λ i = 0 1 ≤ i ≤ n 1 \leq i \leq n 1 ≤ i ≤ n
b)和c)的等价是以下一般引理的推论：
令 V 为 K 和 B 上有限维的向量空间，在 * V x V 上为双线性 形式。设( v 1 , … , v n ) (v_1, \ldots, v_n) ( v 1 , … , v n ) 是*V*的基*超过*K*和*Δ = det B ( v i , v j ) \Delta = \det B(v_i, v_j) Δ = det B ( v i , v j ) A ≠ 0 A \neq 0 A = 0 x ≠ 0 x \neq 0 x = 0 B ( x , y ) ≠ 0 B(x, y) \neq 0 B ( x , y ) = 0
被以下引用 1 被以下引用 命题 3, VIII § 19
我们有 A ≠ 0 A \neq 0 A = 0
∑ i = 1 n λ i B ( v i , v j ) = 0 ( 1 ≤ j ≤ n )
\sum_{i=1}^n \lambda_i B(v_i, v_j) = 0 \quad (1 \leq j \leq n)
i = 1 ∑ n λ i B ( v i , v j ) = 0 ( 1 ≤ j ≤ n ) K 中只有解 A , = ⋯ = A , = 0 A, = \cdots = A, = 0 A , = ⋯ = A , = 0 x = ∑ i = 1 n λ i v i x = \sum_{i=1}^n \lambda_i v_i x = ∑ i = 1 n λ i v i 1 ≤ j ≤ n 1 \leq j \leq n 1 ≤ j ≤ n B ( x , v j ) = 0 B(x, v_j) = 0 B ( x , v j ) = 0 ( v 1 , … , v n ) (v_1, \ldots, v_n) ( v 1 , … , v n ) B ( x , y ) = 0 B(x, y) = 0 B ( x , y ) = 0 y ∈ V y \in V y ∈ V
让我们证明条件 c) 意味着 A 被约简。设 x 为 A 的幂零元 ：对于任何 y ∈ A y \in A y ∈ A L x y L_{xy} L x y T r ( x y ) = 0 \mathrm{Tr}(xy) = 0 Tr ( x y ) = 0 y ∈ A y \in A y ∈ A x = 0 x = 0 x = 0
令V为K上有限维的向量空间和V的u u u T r ( u ) = 0 \mathrm{Tr}(u) = 0 Tr ( u ) = 0
引用了 1 引用了 定理 4, V § 6
对于每个整数 n ≥ 0 n \geq 0 n ≥ 0 V n V_n V n u n u^n u n u u u r ≥ 0 r \geq 0 r ≥ 0 V 0 = V , V r = 0 V_0 = V, V_r = 0 V 0 = V , V r = 0 V i ≠ V i + 1 V_i \neq V_{i+1} V i = V i + 1 0 ≤ i < r − 1 0 \leq i < r - 1 0 ≤ i < r − 1 d i d_i d i V i − 1 V_{i-1} V i − 1 1 ≤ i ≤ r 1 \leq i \leq r 1 ≤ i ≤ r ( x 1 , … , x d ) (x_1, \ldots, x_d) ( x 1 , … , x d ) x j x_j x j d − d i < j ≤ d d - d_i < j \leq d d − d i < j ≤ d V i − 1 V_{i-1} V i − 1 1 ≤ i ≤ r 1 \leq i \leq r 1 ≤ i ≤ r u ( V i − 1 ) ⊂ V i u(V_{i-1}) \subset V_i u ( V i − 1 ) ⊂ V i ( x 1 , … , x d ) (x_1, \ldots, x_d) ( x 1 , … , x d ) u u u T r ( u ) = 0 \mathrm{Tr}(u) = 0 Tr ( u ) = 0
最后让我们证明 b) 蕴涵 a)。令( x 1 , … , x n ) (x_1, \ldots, x_n) ( x 1 , … , x n ) D A / K ( x 1 , … , x n ) ≠ 0 D_{A/K}(x_1, \ldots, x_n) \neq 0 D A / K ( x 1 , … , x n ) = 0 1 ≤ i ≤ n 1 \leq i \leq n 1 ≤ i ≤ n x i ′ = 1 ⊗ x i x'_i = 1 \otimes x_i x i ′ = 1 ⊗ x i D A ′ / K ′ ( x 1 ′ , … , x n ′ ) ≠ 0 D_{A'/K'}(x'_1, \ldots, x'_n) \neq 0 D A ′ / K ′ ( x 1 ′ , … , x n ′ ) = 0
— 令 E 为 K 的扩张 有限 次数。为了使 E 可分离，E 中存在 a 使得 T r E / K ( a ) ≠ 0 \mathrm{Tr}_{E/K}(a) \neq 0 Tr E / K ( a ) = 0
引用了 1 · 被以下引用 2 引用了 命题 1
被以下引用 习题 9, § 11 , 命题 5, VIII § 20
根据命题1，条件是必要的。相反，假设存在a ∈ E a \in E a ∈ E T r E / K ( a ) ≠ 0 \mathrm{Tr}_{E/K}(a) \neq 0 Tr E / K ( a ) = 0 x ≠ 0 x \neq 0 x = 0 y = a x ′ y = a x' y = a x ′ T r E / K ( x y ) ≠ 0 \mathrm{Tr}_{E/K}(xy) \neq 0 Tr E / K ( x y ) = 0
3. 范数以及有限次数扩张的痕迹
代数（III，p. 548）中的传递性公式在扩张或有限次数的情况下意味着以下命题。
— 令 F 为 K 的有限次数的扩张，E 为 F 的子扩展。那么对于每个 x ∈ F x \in F x ∈ F
Tr F / K ( x ) = Tr E / K ( Tr F / E ( x ) ) N F / K ( x ) = N E / K ( N F / E ( x ) ) .
\text{Tr}_{F/K}(x) = \text{Tr}_{E/K}(\text{Tr}_{F/E}(x)) \\
\text{N}_{F/K}(x) = \text{N}_{E/K}(\text{N}_{F/E}(x)) .
Tr F / K ( x ) = Tr E / K ( Tr F / E ( x )) N F / K ( x ) = N E / K ( N F / E ( x )) .
— 将m = [ F : E ] m = [F : E] m = [ F : E ] x ∈ E x \in E x ∈ E
Tr F / K ( x ) = m ⋅ Tr E / K ( x ) N F / K ( x ) = N E / K ( x ) m .
\text{Tr}_{F/K}(x) = m \cdot \text{Tr}_{E/K}(x) \\
\text{N}_{F/K}(x) = \text{N}_{E/K}(x)^m .
Tr F / K ( x ) = m ⋅ Tr E / K ( x ) N F / K ( x ) = N E / K ( x ) m . 被以下引用 1 被以下引用 命题 3
— 令 E 为 K 的有限次数 n 的扩张，以及 E 的 x 元素，K 上的次数 d。将 f ( X ) = X d + ∑ i = 1 d a i X d − i f(X) = X^d + \sum_{i=1}^d a_i X^{d-i} f ( X ) = X d + ∑ i = 1 d a i X d − i
Tr E / K ( x ) = − n d a 1 N E / K ( x ) = ( ( ( − 1 ) d a d ) n / d = ( − 1 ) n a d n / d .
\text{Tr}_{E/K}(x) = - \frac{n}{d} a_1 \\
\text{N}_{E/K}(x) = (((-1)^d a_d)^{n/d} = (-1)^n a_d^{n/d} .
Tr E / K ( x ) = − d n a 1 N E / K ( x ) = ((( − 1 ) d a d ) n / d = ( − 1 ) n a d n / d . 引用了 1 · 被以下引用 1 引用了 推论
被以下引用 引理 3
令 R 为交换环、f ( X ) = X d + ∑ i = 1 d a i X d − i f(X) = X^d + \sum_{i=1}^d a_i X^{d-i} f ( X ) = X d + ∑ i = 1 d a i X d − i R [ X ] \mathbf{R}[X] R [ X ] R [ X ] / ( f ) \mathbf{R}[X]/(f) R [ X ] / ( f ) Tr A / R ( x ) = − a 1 \text{Tr}_{A/R}(x) = -a_1 Tr A / R ( x ) = − a 1 N A / R ( x ) = ( − 1 ) d a d \text{N}_{A/R}(x) = (-1)^d a_d N A / R ( x ) = ( − 1 ) d a d
引用了 1 引用了 命题 3
由科。 (IV, p. 11) 序列 ( 1 , x , . . . , x d − 1 ) (1, x, ..., x^{d-1}) ( 1 , x , ... , x d − 1 )
x ⋅ 1 = x , x ⋅ x = x 2 , … , x ⋅ x d − 2 = x d − 1 ,
x \cdot 1 = x , \quad x \cdot x = x^2 , \ldots , x \cdot x^{d-2} = x^{d-1} ,
x ⋅ 1 = x , x ⋅ x = x 2 , … , x ⋅ x d − 2 = x d − 1 , x ⋅ x d − 1 = − a d ⋅ 1 − a d − 1 ⋅ x − … − a 1 ⋅ x d − 1
x \cdot x^{d-1} = -a_d \cdot 1 - a_{d-1} \cdot x - \ldots - a_1 \cdot x^{d-1}
x ⋅ x d − 1 = − a d ⋅ 1 − a d − 1 ⋅ x − … − a 1 ⋅ x d − 1 矩阵用x相对于A A A ( 1 , x , . . . , x d − 1 ) (1, x, ..., x^{d-1}) ( 1 , x , ... , x d − 1 ) d = 5 d = 5 d = 5
( 0 0 0 0 − a 1 0 0 0 − a 0 1 0 0 − a 3 0 0 1 0 − a 2 0 0 0 1 − a 1 )
\begin{pmatrix}
0 & 0 & 0 & 0 & -a \\
1 & 0 & 0 & 0 & -a \\
0 & 1 & 0 & 0 & -a_3 \\
0 & 0 & 1 & 0 & -a_2 \\
0 & 0 & 0 & 1 & -a_1
\end{pmatrix}
0 1 0 0 0 0 0 1 0 0 0 0 0 1 0 0 0 0 0 1 − a − a − a 3 − a 2 − a 1 这个矩阵的痕迹明显是− a -a − a
( − 1 ) d − 1 ( − a d ) = ( − 1 ) d a d .
(-1)^{d-1} (-a_d) = (-1)^d a_d .
( − 1 ) d − 1 ( − a d ) = ( − 1 ) d a d . 对于这个编号的其余部分，我们用 K 的有限次数的 E 扩张和 E 的 x x x x x x
a) 可分扩张 的情况：假设E在K上与次数n n n Ω \Omega Ω σ 1 , … , σ n \sigma_1, \ldots, \sigma_n σ 1 , … , σ n n n n K K K V V V
Tr E / K ( x ) = ∑ i = 1 n σ i ( x ) , N E / K ( x ) = ∏ i = 1 n σ i ( x ) .
\text{Tr}_{E/K}(x) = \sum_{i=1}^n \sigma_i(x) , \quad N_{E/K}(x) = \prod_{i=1}^n \sigma_i(x) .
Tr E / K ( x ) = i = 1 ∑ n σ i ( x ) , N E / K ( x ) = i = 1 ∏ n σ i ( x ) . b) p 根式扩展的情况 ：假设 K 具有特征 p > 0 p > 0 p > 0 e ≥ 0 e \geq 0 e ≥ 0 [ E : K ] = p e [E:K] = p^e [ E : K ] = p e V V V f f f x x x x x x X p f − x p f X^{p^f} - x^{p^f} X p f − x p f V V V N E / K ( x ) = ( x p f ) p e / p f N_{E/K}(x) = (x^{p^f})^{p^e/p^f} N E / K ( x ) = ( x p f ) p e / p f
N E / K ( x ) = x p e = x [ E : K ]
N_{E/K}(x) = x^{p^e} = x^{[E:K]}
N E / K ( x ) = x p e = x [ E : K ] 对于迹，我们找到Tr E / K ( x ) = − p e − f a \text{Tr}_{E/K}(x) = -p^{e-f} a Tr E / K ( x ) = − p e − f a a a a X p f − x p f X^{p^f} - x^{p^f} X p f − x p f X p f − 1 X^{p^f-1} X p f − 1
Tr E / K ( x ) = p e ⋅ x = [ E : K ] x = { x if [ E : K ] = 1 0 if [ E : K ] > 1.
\text{Tr}_{E/K}(x) = p^e \cdot x = [E:K] x = \begin{cases} x & \text{if } [E:K] = 1 \\ 0 & \text{if } [E:K] > 1 . \end{cases}
Tr E / K ( x ) = p e ⋅ x = [ E : K ] x = { x 0 if [ E : K ] = 1 if [ E : K ] > 1. c) 一般情况 ：我们可以总结范数的计算和跟踪如下命题：
令 p 为 K 的有限次数的 K 和 E 扩张的特征指数。令 σ 1 , … , σ n \sigma_1, \ldots, \sigma_n σ 1 , … , σ n Ω \Omega Ω p e = [ E : K ] i p^e = [E:K]_i p e = [ E : K ] i x ∈ E x \in E x ∈ E Ω \Omega Ω
被以下引用 1 被以下引用 命题 5, AC V § 2
Tr E / K ( x ) = p e ⋅ ∑ i = 1 n σ i ( x ) , N E / K ( x ) = ( ∏ i = 1 n σ i ( x ) ) p e .
\text{Tr}_{E/K}(x) = p^e \cdot \sum_{i=1}^n \sigma_i(x) , \quad N_{E/K}(x) = \left( \prod_{i=1}^n \sigma_i(x) \right)^{p^e} .
Tr E / K ( x ) = p e ⋅ i = 1 ∑ n σ i ( x ) , N E / K ( x ) = ( i = 1 ∏ n σ i ( x ) ) p e . 令E为E中K的相对可分闭包；则E s E_s E s n n n σ 1 , … , σ n \sigma_1, \ldots, \sigma_n σ 1 , … , σ n e ^e e V V V
习题
参见习题为 § 8 。