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。
§ 9。 范数和痕迹
在本段中，K K K A A A K K K A A A K K K K K K
1. 范数以及与模相关的痕迹
因此a a a K K K Tr M / K ( a ) \operatorname{Tr}_{M/K}(a) Tr M / K ( a ) N M / K ( a ) N_{M/K}(a) N M / K ( a ) a a a K [ X ] K[X] K [ X ] P c M / K ( a ; X ) \mathrm{Pc}_{M/K}(a; X) Pc M / K ( a ; X ) K K K
根据自同态（II, § 4, no. 3和§ 8, no. 1）的迹和行列式的性质，我们得到关系
Tr M ( a + a ′ ) = Tr M ( a ) + Tr M ( a ′ )
\operatorname{Tr}_M(a + a') = \operatorname{Tr}_M(a) + \operatorname{Tr}_M(a')
Tr M ( a + a ′ ) = Tr M ( a ) + Tr M ( a ′ ) (2) T r M ( a a ′ ) = T r M ( a ′ a ) \mathrm{Tr}_M(aa') = \mathrm{Tr}_M(a'a) Tr M ( a a ′ ) = Tr M ( a ′ a ) N M ( a a ′ ) = N M ( a ) N M ( a ′ ) N_M(aa') = N_M(a)N_M(a') N M ( a a ′ ) = N M ( a ) N M ( a ′ )
对于A A A a , a ′ a, a' a , a ′
就该基而言，令( e i ) 1 ≤ i ≤ n (e_i)_{1 \leq i \leq n} ( e i ) 1 ≤ i ≤ n K K K M M M a M a_M a M ( m i j ( a ) ) (m_{ij}(a)) ( m ij ( a )) m i j m_{ij} m ij K K K A A A
(4) T r M ( a ) = ∑ i = 1 n m i i ( a ) \mathrm{Tr}_M(a) = \sum_{i=1}^n m_{ii}(a) Tr M ( a ) = ∑ i = 1 n m ii ( a ) N M ( a ) = det ( m i j ( a ) ) N_M(a) = \det(m_{ij}(a)) N M ( a ) = det ( m ij ( a )) P c ( a ; X ) = det ( δ i j X − m i j ( a ) ) . \mathrm{Pc}(a; X) = \det(\delta_{ij}X - m_{ij}(a)). Pc ( a ; X ) = det ( δ ij X − m ij ( a )) .
由计算行列式 (§ 8, no. 11, 公式 (50)) 的方法得出：
(7) P c M ( a ; X ) = X n + c 1 X n − 1 + ⋯ + c n \mathrm{Pc}_M(a; X) = X^n + c_1 X^{n-1} + \cdots + c_n Pc M ( a ; X ) = X n + c 1 X n − 1 + ⋯ + c n
哪里
(8) c 1 = − T r M ( a ) , c n = ( − 1 ) n N M ( a ) . c_1 = -\mathrm{Tr}_M(a), \quad c_n = (-1)^n N_M(a). c 1 = − Tr M ( a ) , c n = ( − 1 ) n N M ( a ) .
对于λ ∈ K \lambda \in K λ ∈ K
(9) T r M ( λ ) = n . λ , N M ( λ ) = λ n , P c M ( λ ; X ) = ( X − λ ) n . \mathrm{Tr}_M(\lambda) = n.\lambda, \quad N_M(\lambda) = \lambda^n, \quad \mathrm{Pc}_M(\lambda; X) = (X - \lambda)^n. Tr M ( λ ) = n . λ , N M ( λ ) = λ n , Pc M ( λ ; X ) = ( X − λ ) n .
设K ′ K' K ′ K K K M ′ = K ′ ⊗ K M M' = K' \otimes_K M M ′ = K ′ ⊗ K M A ′ = K ′ ⊗ K A A' = K' \otimes_K A A ′ = K ′ ⊗ K A M ′ M' M ′ A ′ A' A ′ K ′ K' K ′ M ′ M' M ′ 1 ⊗ e i 1 \otimes e_i 1 ⊗ e i 1 ≤ i ≤ n 1 \leq i \leq n 1 ≤ i ≤ n a M a_M a M ( e i ) (e_i) ( e i ) ( 1 ⊗ a ) M ′ (1 \otimes a)_M' ( 1 ⊗ a ) M ′ ( e i ′ ) (e'_i) ( e i ′ )
(12) T r M ′ ( 1 ⊗ a ) = T r M ( a ) .1 , N M ′ ( 1 ⊗ a ) = N M ( a ) .1 \mathrm{Tr}_{M'}(1 \otimes a) = \mathrm{Tr}_M(a).1, \quad N_{M'}(1 \otimes a) = N_M(a).1 Tr M ′ ( 1 ⊗ a ) = Tr M ( a ) .1 , N M ′ ( 1 ⊗ a ) = N M ( a ) .1 P c M ′ ( 1 ⊗ a ; X ) = P c M ( a ; X ) .1 \mathrm{Pc}_{M'}(1 \otimes a; X) = \mathrm{Pc}_M(a; X).1 Pc M ′ ( 1 ⊗ a ; X ) = Pc M ( a ; X ) .1
对于所有 a ∈ A a \in A a ∈ A K ′ K' K ′ K ′ = K [ X ] K' = K[X] K ′ = K [ X ]
(13) P c M / K ( a ; X ) = N M [ X ] / K [ X ] ( X − a ) . \mathrm{Pc}_{M/K}(a; X) = N_{M[X]/K[X]}(X - a). Pc M / K ( a ; X ) = N M [ X ] / K [ X ] ( X − a ) .
2. 范数的性质以及与模相关的踪迹
如果M M M M ′ M' M ′ A A A K K K a ∈ A a \in A a ∈ A
(14) T r M ′ ( a ) = T r M ( a ) , N M ′ ( a ) = N M ( a ) , P c M ′ ( a ; X ) = P c M ( a ; X ) \mathrm{Tr}_{M'}(a) = \mathrm{Tr}_M(a), \quad N_{M'}(a) = N_M(a), \quad \mathrm{Pc}_{M'}(a; X) = \mathrm{Pc}_M(a; X) Tr M ′ ( a ) = Tr M ( a ) , N M ′ ( a ) = N M ( a ) , Pc M ′ ( a ; X ) = Pc M ( a ; X )
因为如果f f f M M M M ′ M' M ′ a M a_M a M M M M B B B K K K a M ′ a_{M'} a M ′ M ′ M' M ′ f ( B ) f(B) f ( B )
Tr M ( a ) = ∑ i = 1 r Tr P i ( a ) , N M ( a ) = ∏ i = 1 r N P i ( a )
\operatorname{Tr}_M(a) = \sum_{i=1}^r \operatorname{Tr}_{P_i}(a), \qquad N_M(a) = \prod_{i=1}^r N_{P_i}(a)
Tr M ( a ) = i = 1 ∑ r Tr P i ( a ) , N M ( a ) = i = 1 ∏ r N P i ( a ) (15)
P C M ( a ; X ) = ∏ i = 1 r P C P i ( a ; X ) .
P_{C_M}(a; X) = \prod_{i=1}^r P_{C_{P_i}}(a; X).
P C M ( a ; X ) = i = 1 ∏ r P C P i ( a ; X ) . 令B i ′ B'_i B i ′ P i P_i P i K K K B i ′ B'_i B i ′ M i M_i M i B i B_i B i K K K M i − 1 M_{i-1} M i − 1 K K K M i M_i M i B i B_i B i 1 ≤ i ≤ r 1 \leq i \leq r 1 ≤ i ≤ r B B B K K K M M M X t t X_{tt} X tt a P i a_{P_i} a P i B i ′ B'_i B i ′ a M a_M a M B B B
( X r r X r , r − 1 ⋯ X r , 1 0 X r − 1 , r − 1 ⋯ X r − 1 , 1 ⋮ ⋮ ⋱ ⋮ 0 0 ⋯ X 11 )
\begin{pmatrix}
X_{rr} & X_{r,r-1} & \cdots & X_{r,1} \\
0 & X_{r-1,r-1} & \cdots & X_{r-1,1} \\
\vdots & \vdots & \ddots & \vdots \\
0 & 0 & \cdots & X_{11}
\end{pmatrix}
X r r 0 ⋮ 0 X r , r − 1 X r − 1 , r − 1 ⋮ 0 ⋯ ⋯ ⋱ ⋯ X r , 1 X r − 1 , 1 ⋮ X 11 命题由no. 1的公式(4)、(5)和(6)以及§ 8, no. 6的公式(31)得出。
Tr M ⊗ M ′ ( a ⊗ a ′ ) = Tr M ( a ) Tr M ′ ( a ′ )
\operatorname{Tr}_{M \otimes M'}(a \otimes a') = \operatorname{Tr}_M(a) \operatorname{Tr}_{M'}(a')
Tr M ⊗ M ′ ( a ⊗ a ′ ) = Tr M ( a ) Tr M ′ ( a ′ ) (16)
N M ⊗ M ′ ( a ⊗ a ′ ) = ( N M ( a ) ) n ′ ( N M ′ ( a ′ ) ) n .
N_{M \otimes M'}(a \otimes a') = (N_M(a))^{n'} (N_{M'}(a'))^n.
N M ⊗ M ′ ( a ⊗ a ′ ) = ( N M ( a ) ) n ′ ( N M ′ ( a ′ ) ) n . (17)
公式 (16) 源自II, § 4, no. 4、公式 (26)，公式 (17) 源自§ 8, no. 6、公式 (33)。
3. 范数并追踪代数
令 A A A K K K K K K a ∈ A a \in A a ∈ A a a a A A A K K K K K K A A A x ↦ a x x \mapsto ax x ↦ a x
引用了 2 · 被以下引用 2 引用了 命题 1 , 命题 2
被以下引用 定义 2, IV § 6 , 注 5, IV § 6
a ∈ A a \in A a ∈ A A A A K K K T r A / K ( a ) \mathrm{Tr}_{A/K}(a) Tr A / K ( a ) N A / K ( a ) N_{A/K}(a) N A / K ( a ) P c A / K ( a ; X ) \mathrm{Pc}_{A/K}(a; X) Pc A / K ( a ; X ) K K K A A A a ∈ A a \in A a ∈ A a a a A A A A s A_s A s
假设 A A A K K K A 1 × A 2 × ⋯ × A m A_1 \times A_2 \times \cdots \times A_m A 1 × A 2 × ⋯ × A m K K K
a = ( a 1 , … , a m ) ∈ A ,
a = (a_1, \ldots, a_m) \in A,
a = ( a 1 , … , a m ) ∈ A ,
T r A / K ( a ) = ∑ i = 1 m T r A i / K ( a i ) , N A / K ( a ) = ∏ i = 1 m N A i / K ( a i )
\mathrm{Tr}_{A/K}(a) = \sum_{i=1}^m \mathrm{Tr}_{A_i/K}(a_i), \qquad N_{A/K}(a) = \prod_{i=1}^m N_{A_i/K}(a_i)
Tr A / K ( a ) = i = 1 ∑ m Tr A i / K ( a i ) , N A / K ( a ) = i = 1 ∏ m N A i / K ( a i ) P c A / K ( a ; X ) = ∏ i = 1 m P c A i / K ( a i ; X ) .
\mathrm{Pc}_{A/K}(a; X) = \prod_{i=1}^m \mathrm{Pc}_{A_i/K}(a_i; X).
Pc A / K ( a ; X ) = i = 1 ∏ m Pc A i / K ( a i ; X ) . 类似地，no. 2的命题 2表明，如果A A A A ′ A' A ′ K K K n , n ′ n, n' n , n ′ K K K a ∈ A , a ′ ∈ A ′ a \in A, a' \in A' a ∈ A , a ′ ∈ A ′
T r A ⊗ A ′ ( a ⊗ a ′ ) = T r A ( a ) T r A ′ ( a ′ )
\mathrm{Tr}_{A \otimes A'}(a \otimes a') = \mathrm{Tr}_A(a) \, \mathrm{Tr}_{A'}(a')
Tr A ⊗ A ′ ( a ⊗ a ′ ) = Tr A ( a ) Tr A ′ ( a ′ )
N A ⊗ A ′ ( a ⊗ a ′ ) = ( N A ( a ) ) n ′ ( N A ′ ( a ′ ) ) n .
N_{A \otimes A'}(a \otimes a') = (N_A(a))^{n'} (N_{A'}(a'))^n.
N A ⊗ A ′ ( a ⊗ a ′ ) = ( N A ( a ) ) n ′ ( N A ′ ( a ′ ) ) n . Finally let A A A K K K K K K h h h K K K K ′ K' K ′ A ′ = A ( K ′ ) A' = A_{(K')} A ′ = A ( K ′ ) K ′ K' K ′ A A A h h h a ∈ A a \in A a ∈ A
T r A ′ / K ′ ( 1 ⊗ a ) = h ( T r A / K ( a ) ) , N A ′ / K ′ ( 1 ⊗ a ) = h ( N A / K ( a ) )
\mathrm{Tr}_{A'/K'}(1 \otimes a) = h(\mathrm{Tr}_{A/K}(a)), \qquad N_{A'/K'}(1 \otimes a) = h(N_{A/K}(a))
Tr A ′ / K ′ ( 1 ⊗ a ) = h ( Tr A / K ( a )) , N A ′ / K ′ ( 1 ⊗ a ) = h ( N A / K ( a )) P c A ′ / K ′ ( 1 ⊗ a ; X ) = h ˉ ( P c A / K ( a ; X ) )
\mathrm{Pc}_{A'/K'}(1 \otimes a; X) = \bar{h}(\mathrm{Pc}_{A/K}(a; X))
Pc A ′ / K ′ ( 1 ⊗ a ; X ) = h ˉ ( Pc A / K ( a ; X )) 其中h ˉ \bar{h} h ˉ K [ X ] → K ′ [ X ] K[X] \to K'[X] K [ X ] → K ′ [ X ] h h h K ′ = K [ X ] K' = K[X] K ′ = K [ X ] A [ X ] = A ⊗ K K [ X ] A[X] = A \otimes_K K[X] A [ X ] = A ⊗ K K [ X ]
P c A / K ( a ; X ) = N A [ X ] / K [ X ] ( X − a ) .
\mathrm{Pc}_{A/K}(a; X) = N_{A[X]/K[X]}(X - a).
Pc A / K ( a ; X ) = N A [ X ] / K [ X ] ( X − a ) . 更一般地，如果 K ′ K' K ′ K K K A ′ = A ⊗ K K ′ A' = A \otimes_K K' A ′ = A ⊗ K K ′ x ∈ A ′ x \in A' x ∈ A ′
P c A / K ( a ; x ) = N A ′ / K ′ ( x − a ) .
\mathrm{Pc}_{A/K}(a; x) = N_{A'/K'}(x - a).
Pc A / K ( a ; x ) = N A ′ / K ′ ( x − a ) .
(1) Let A A A K K K ( α , β ) (\alpha, \beta) ( α , β ) ( e 1 , e 2 ) (e_1, e_2) ( e 1 , e 2 ) ( α , β ) (\alpha, \beta) ( α , β ) x = ξ e 1 + η e 2 x = \xi e_1 + \eta e_2 x = ξ e 1 + η e 2 T r A / K ( x ) = 2 ξ + β η \mathrm{Tr}_{A/K}(x) = 2\xi + \beta \eta Tr A / K ( x ) = 2 ξ + β η N A / K ( x ) = ξ 2 + β ξ η − α η 2 N_{A/K}(x) = \xi^2 + \beta \xi \eta - \alpha \eta^2 N A / K ( x ) = ξ 2 + β ξ η − α η 2 x x x
(2) 设 A A A K K K T r A / K ( x ) = 2 T ( x ) \mathrm{Tr}_{A/\mathbf{K}}(x) = 2T(x) Tr A / K ( x ) = 2 T ( x ) N A / K ( x ) = ( N ( x ) ) 2 N_{A/\mathbf{K}}(x) = (N(x))^2 N A / K ( x ) = ( N ( x ) ) 2
(3) 令A(II, § 10, no. 3)的A = M n ( K ) A = \mathbf{M}_n(\mathbf{K}) A = M n ( K ) ( E i j ) (E_{ij}) ( E ij ) Y ↦ X Y Y \mapsto XY Y ↦ X Y X = ∑ i , j ξ i j E i j X = \sum_{i,j} \xi_{ij} E_{ij} X = ∑ i , j ξ ij E ij n 2 n^2 n 2
( X 0 … 0 0 X … 0 … … … … 0 0 … X )
\begin{pmatrix}
X & 0 & \ldots & 0 \\
0 & X & \ldots & 0 \\
\ldots & \ldots & \ldots & \ldots \\
0 & 0 & \ldots & X
\end{pmatrix}
X 0 … 0 0 X … 0 … … … … 0 0 … X 从而T r A / K ( X ) = n ⋅ T r ( X ) \mathrm{Tr}_{A/\mathbf{K}}(X) = n \cdot \mathrm{Tr}(X) Tr A / K ( X ) = n ⋅ Tr ( X ) N A / K ( X ) = ( det ( X ) ) n N_{A/\mathbf{K}}(X) = (\det(X))^n N A / K ( X ) = ( det ( X ) ) n
引用了 2 引用了 III § 2. 。 例 / 代数 , II § 10. 。 矩阵
4. 范数的性质和代数中的痕迹
如果a a a A A A a ′ a' a ′
N A / K ( a ) N A / K ( a ′ ) = N A / K ( a a ′ ) = N A / K ( 1 ) = 1
N_{A/\mathbf{K}}(a) N_{A/\mathbf{K}}(a') = N_{A/\mathbf{K}}(aa') = N_{A/\mathbf{K}}(1) = 1
N A / K ( a ) N A / K ( a ′ ) = N A / K ( a a ′ ) = N A / K ( 1 ) = 1 由 no. 1 的公式 (3) 获得。相反，如果 N A / K ( a ) N_{A/\mathbf{K}}(a) N A / K ( a ) h : x ↦ a x h : x \mapsto ax h : x ↦ a x a ′ ∈ A a' \in A a ′ ∈ A a a ′ = 1 aa' = 1 a a ′ = 1 h ( a ′ a − 1 ) = a a ′ a − a = ( a a ′ − 1 ) a = 0 h(a'a - 1) = aa'a - a = (aa' - 1)a = 0 h ( a ′ a − 1 ) = a a ′ a − a = ( a a ′ − 1 ) a = 0 a a ′ = 1 aa' = 1 a a ′ = 1 h h h a ′ a' a ′ a a a
设 A 为 K-代数，承认有限的基。对于所有 a ∈ A a \in A a ∈ A P c A / K ( a ; a ) = 0 \mathrm{Pc}_{A/\mathbf{K}}(a; a) = 0 Pc A / K ( a ; a ) = 0
引用了 1 引用了 命题 20（“凯利-汉密尔顿定理”）, III § 8
这紧随 Cayley-Hamilton 定理（§ 8, no. 11、命题 20）而来。
设A为K-代数，m为A的双向理想。假设A 0 = A / m A_0 = A/m A 0 = A / m r > 0 r > 0 r > 0 m r = { 0 } m^r = \{0\} m r = { 0 } m i − 1 / m i m^{i-1}/m^i m i − 1 / m i A 0 A_0 A 0 s i s_i s i 1 ≤ i ≤ r 1 \leq i \leq r 1 ≤ i ≤ r s = s 1 + ⋯ + s r s = s_1 + \cdots + s_r s = s 1 + ⋯ + s r a ∈ A a \in A a ∈ A a 0 a_0 a 0 a a a n s ns n s a ∈ A a \in A a ∈ A
引用了 2 引用了 命题 25, II § 1 , 命题 1
T r A ( a ) = s ⋅ T r A 0 ( a 0 ) , N A ( a ) = ( N A 0 ( a 0 ) ) s
\mathrm{Tr}_A(a) = s \cdot \mathrm{Tr}_{A_0}(a_0), \qquad N_A(a) = (N_{A_0}(a_0))^s
Tr A ( a ) = s ⋅ Tr A 0 ( a 0 ) , N A ( a ) = ( N A 0 ( a 0 ) ) s P c A ( a ; X ) = ( P c A 0 ( a 0 ; X ) ) s .
\mathrm{Pc}_A(a; X) = (\mathrm{Pc}_{A_0}(a_0; X))^s.
Pc A ( a ; X ) = ( Pc A 0 ( a 0 ; X ) ) s . 由于II, § 1, no. 13、命题25，m i − 1 / m i m^{i-1}/m^i m i − 1 / m i n s i ns_i n s i P i = m i − 1 / m i P_i = m^{i-1}/m^i P i = m i − 1 / m i
这首先表明 A A A n ( s 1 + ⋯ + s r ) = n s n(s_1 + \cdots + s_r) = ns n ( s 1 + ⋯ + s r ) = n s K K K A A A P i P_i P i s i s_i s i A A A A 0 A_0 A 0 N P i ( a ) = N A 0 ( a ) s i N_{P_i}(a) = N_{A_0}(a)^{s_i} N P i ( a ) = N A 0 ( a ) s i
N A ( a ) = N A 0 ( a ) s .
N_A(a) = N_{A_0}(a)^s.
N A ( a ) = N A 0 ( a ) s . 在此，通过将
A 0 A_0 A 0 视为左
A A A 模来定义公式
N A 0 ( a ) N_{A_0}(a) N A 0 ( a ) ，并且它等于
A 0 A_0 A 0 的
K K K 线性映射
x ↦ a x x \mapsto ax x ↦ a x 到自身的行列式；但是，如
a x = a 0 x ax = a_0x a x = a 0 x 代表
x ∈ A 0 x \in A_0 x ∈ A 0 ，
N A 0 ( a ) = N A 0 ( a 0 ) N_{A_0}(a) = N_{A_0}(a_0) N A 0 ( a ) = N A 0 ( a 0 ) ，它完成证明或公式（23）的范数。其他两个的显示方式类似。
设A A A K K K K K K V V V A A A A A A V V V K K K V V V A A A u u u u K u_K u K u u u K K K V V V
Tr ( u K ) = Tr A / K ( Tr ( u ) ) , det ( u K ) = N A / K ( det ( u ) ) Pc ( u K ; X ) = N A [ X ] / K [ X ] ( Pc ( u ; X ) ) .
\begin{align*}
\operatorname{Tr}(u_K) &= \operatorname{Tr}_{A/K}(\operatorname{Tr}(u)), \qquad \det(u_K) = N_{A/K}(\det(u)) \\
\operatorname{Pc}(u_K; X) &= N_{A[X]/K[X]}(\operatorname{Pc}(u; X)).
\end{align*}
Tr ( u K ) Pc ( u K ; X ) = Tr A / K ( Tr ( u )) , det ( u K ) = N A / K ( det ( u )) = N A [ X ] / K [ X ] ( Pc ( u ; X )) . (24)
引用了 2 · 被以下引用 3 引用了 命题 25, II § 1 , III § 8. 。决定因素
被以下引用 引理 1 , 推论 , 命题 1, VIII § 19
设( a i ) 1 ≤ i ≤ m (a_i)_{1 \leq i \leq m} ( a i ) 1 ≤ i ≤ m A A A K K K ( e j ) 1 ≤ j ≤ n (e_j)_{1 \leq j \leq n} ( e j ) 1 ≤ j ≤ n V V V A A A ( a i e j ) (a_i e_j) ( a i e j ) V V V K K K A [ X ] A[X] A [ X ] A [ X ] ⊗ A V A[X] \otimes_A V A [ X ] ⊗ A V X − u ˉ X - \bar{u} X − u ˉ
令 X i j X_{ij} X ij 1 ≤ i ≤ n , 1 ≤ j ≤ n 1 \leq i \leq n,\ 1 \leq j \leq n 1 ≤ i ≤ n , 1 ≤ j ≤ n n 2 n^2 n 2 X X X ( X i j ) (X_{ij}) ( X ij ) n n n D ( X 11 , … , X n n ) ∈ Z [ X 11 , … , X n n ] D(X_{11}, \ldots, X_{nn}) \in \mathbf{Z}[X_{11}, \ldots, X_{nn}] D ( X 11 , … , X nn ) ∈ Z [ X 11 , … , X nn ] det ( X ) \det(X) det ( X ) A A A M i j M_{ij} M ij 1 ≤ i ≤ n , 1 ≤ j ≤ n 1 \leq i \leq n,\ 1 \leq j \leq n 1 ≤ i ≤ n , 1 ≤ j ≤ n n 2 n^2 n 2 m m m A A A M M M m n mn mn A A A
M = ( M 11 M 12 ⋯ M 1 n M 21 M 22 ⋯ M 2 n ⋯ ⋯ ⋯ ⋯ M n 1 M n 2 ⋯ M n n )
M = \begin{pmatrix}
M_{11} & M_{12} & \cdots & M_{1n} \\
M_{21} & M_{22} & \cdots & M_{2n} \\
\cdots & \cdots & \cdots & \cdots \\
M_{n1} & M_{n2} & \cdots & M_{nn}
\end{pmatrix}
M = M 11 M 21 ⋯ M n 1 M 12 M 22 ⋯ M n 2 ⋯ ⋯ ⋯ ⋯ M 1 n M 2 n ⋯ M nn 那么
M M M 的行列式等于平方矩阵的行列式
D ( M 11 , … , M n n )
D(M_{11}, \ldots, M_{nn})
D ( M 11 , … , M nn ) 阶
m m m 。
引用了 3 引用了 II § 10. 。 矩阵 , III § 8. 。决定因素 , 命题 6
我们在n n n n = 0 n = 0 n = 0 n = 1 n = 1 n = 1
设 Z Z Z N i j N_{ij} N ij M i j + δ i j Z I m M_{ij} + \delta_{ij} Z I_m M ij + δ ij Z I m δ i j \delta_{ij} δ ij D i j ( X 11 , … , X n n ) D^{ij}(X_{11}, \ldots, X_{nn}) D ij ( X 11 , … , X nn ) X X X X i j X_{ij} X ij
(25) X j i D k i ( X 11 , … , X n n ) = δ j k D ( X 11 , … , X n n )
\text{(25)} \quad X_{ji} D^{ki}(X_{11}, \ldots, X_{nn}) = \delta_{jk} D(X_{11}, \ldots, X_{nn})
(25) X j i D k i ( X 11 , … , X nn ) = δ j k D ( X 11 , … , X nn ) （§ 8, no. 6，公式（28））。我们写 N i j ′ = D i j ( N 11 , … , N n n ) N'_{ij} = D^{ij}(N_{11}, \ldots, N_{nn}) N ij ′ = D ij ( N 11 , … , N nn ) m m m A [ Z ] \mathbf{A}[Z] A [ Z ] N . U N . U N . U
U = ( N 11 ′ 0 ⋯ 0 N 12 ′ I m ⋯ 0 ⋯ ⋯ ⋯ ⋯ N 1 n ′ 0 ⋯ I m ) ,
U = \begin{pmatrix}
N'_{11} & 0 & \cdots & 0 \\
N'_{12} & I_m & \cdots & 0 \\
\cdots & \cdots & \cdots & \cdots \\
N'_{1n} & 0 & \cdots & I_m
\end{pmatrix},
U = N 11 ′ N 12 ′ ⋯ N 1 n ′ 0 I m ⋯ 0 ⋯ ⋯ ⋯ ⋯ 0 0 ⋯ I m , N = ( N 11 N 12 ⋯ N 1 n N 21 N 22 ⋯ N 2 n ⋯ ⋯ ⋯ ⋯ N n 1 N n 2 ⋯ N n n )
N = \begin{pmatrix}
N_{11} & N_{12} & \cdots & N_{1n} \\
N_{21} & N_{22} & \cdots & N_{2n} \\
\cdots & \cdots & \cdots & \cdots \\
N_{n1} & N_{n2} & \cdots & N_{nn}
\end{pmatrix}
N = N 11 N 21 ⋯ N n 1 N 12 N 22 ⋯ N n 2 ⋯ ⋯ ⋯ ⋯ N 1 n N 2 n ⋯ N nn 在块（II, § 10, no. 5）中执行乘积并使用公式（25），我们得到
N . U = ( P N 12 ⋯ N 1 n 0 N 22 ⋯ N 2 n ⋯ ⋯ ⋯ ⋯ 0 N n 2 ⋯ N n n )
N . U = \begin{pmatrix}
P & N_{12} & \cdots & N_{1n} \\
0 & N_{22} & \cdots & N_{2n} \\
\cdots & \cdots & \cdots & \cdots \\
0 & N_{n2} & \cdots & N_{nn}
\end{pmatrix}
N . U = P 0 ⋯ 0 N 12 N 22 ⋯ N n 2 ⋯ ⋯ ⋯ ⋯ N 1 n N 2 n ⋯ N nn 我们写了P = D ( N 11 , … , N n n ) P = D(N_{11}, \ldots, N_{nn}) P = D ( N 11 , … , N nn )
Q = ( N 22 ⋯ N 2 n ⋯ ⋯ ⋯ N n 2 ⋯ N n n )
Q = \begin{pmatrix}
N_{22} & \cdots & N_{2n} \\
\cdots & \cdots & \cdots \\
N_{n2} & \cdots & N_{nn}
\end{pmatrix}
Q = N 22 ⋯ N n 2 ⋯ ⋯ ⋯ N 2 n ⋯ N nn 即阶 m ( n − 1 ) m(n-1) m ( n − 1 ) N N N U U U P P P Q Q Q U = det N 11 ′ U = \det N'_{11} U = det N 11 ′ Q = det ( D 11 ( N 11 , … , N n n ) ) = det N 11 ′ Q = \det(D^{11}(N_{11}, \ldots, N_{nn})) = \det N'_{11} Q = det ( D 11 ( N 11 , … , N nn )) = det N 11 ′ N i j N_{ij} N ij Q Q Q m ( n − 1 ) m(n-1) m ( n − 1 ) A [ Z ] \mathbf{A}[Z] A [ Z ] Z m ( n − 1 ) Z^{m(n-1)} Z m ( n − 1 ) A [ Z ] \mathbf{A}[Z] A [ Z ] Q Q Q A [ Z ] \mathbf{A}[Z] A [ Z ] N = det ( D ( N 11 , … , N n n ) ) N = \det(D(N_{11}, \ldots, N_{nn})) N = det ( D ( N 11 , … , N nn )) Z Z Z M = det ( D ( M 11 , … , M n n ) ) M = \det(D(M_{11}, \ldots, M_{nn})) M = det ( D ( M 11 , … , M nn ))
显示了这个引理，K-模V就是K-模A e j Ae_j A e j 1 ≤ j ≤ n 1 \leq j \leq n 1 ≤ j ≤ n u ( e j ) = ∑ k = 1 n c j k e k u(e_j) = \sum_{k=1}^n c_{jk} e_k u ( e j ) = ∑ k = 1 n c j k e k x e j ∈ A e j xe_j \in Ae_j x e j ∈ A e j x ∈ A x \in A x ∈ A u ( x e j ) u(xe_j) u ( x e j ) A e k Ae_k A e k x c j k e k xc_{jk} e_k x c j k e k u k u_k u k ( a i e j ) (a_i e_j) ( a i e j ) ( M j k ) (M_{jk}) ( M j k ) ( a i e j ) 1 ≤ i ≤ m (a_i e_j)_{1 \leq i \leq m} ( a i e j ) 1 ≤ i ≤ m ( a i e k ) 1 ≤ i ≤ m (a_i e_k)_{1 \leq i \leq m} ( a i e k ) 1 ≤ i ≤ m M j k M_{jk} M j k A e j Ae_j A e j x e j ↦ x c j k e k xe_j \mapsto xc_{jk} e_k x e j ↦ x c j k e k A e k Ae_k A e k t ∈ A t \in A t ∈ A M ( t ) M(t) M ( t ) ( a i ) 1 ≤ i ≤ m (a_i)_{1 \leq i \leq m} ( a i ) 1 ≤ i ≤ m x ↦ x t x \mapsto xt x ↦ x t M j k = M ( c j k ) M_{jk} = M(c_{jk}) M j k = M ( c j k ) t ↦ M ( t ) t \mapsto M(t) t ↦ M ( t ) M j k M_{jk} M j k 可与 1 另一个 置换。然后
det u K = det ( D ( M 11 , … , M n n ) )
\det u_K = \det(D(M_{11}, \ldots, M_{nn}))
det u K = det ( D ( M 11 , … , M nn )) 由引理 1。但由于
t ↦ M ( t ) t \mapsto M(t) t ↦ M ( t ) 是环同态，所以
D ( M 11 , … , M n n ) D(M_{11}, \ldots, M_{nn}) D ( M 11 , … , M nn ) 是A相对于基
( a i ) (a_i) ( a i ) 的K-自同态
x ↦ x . det ( c j k ) x \mapsto x.\det(c_{jk}) x ↦ x . det ( c j k ) 的矩阵；因此，通过定义，它的行列式是
N A / K ( det ( u ) ) N_{A/K}(\det(u)) N A / K ( det ( u )) ，这证明了（24）的第二个公式。另一方面，
Tr ( u K ) = ∑ j = 1 n Tr ( M j j ) = ∑ j = 1 n Tr A / K ( c j j ) = Tr A / K ( ∑ j = 1 n c j j ) = Tr A / K ( Tr ( u ) )
\operatorname{Tr}(u_K) = \sum_{j=1}^n \operatorname{Tr}(M_{jj}) = \sum_{j=1}^n \operatorname{Tr}_{A/K}(c_{jj}) = \operatorname{Tr}_{A/K}\left( \sum_{j=1}^n c_{jj} \right) = \operatorname{Tr}_{A/K}(\operatorname{Tr}(u))
Tr ( u K ) = j = 1 ∑ n Tr ( M j j ) = j = 1 ∑ n Tr A / K ( c j j ) = Tr A / K ( j = 1 ∑ n c j j ) = Tr A / K ( Tr ( u )) 命题 6 中的证明已完成。
设 A 交换K-代数在 K 上承认有限基，并且 B A-代数在 A 上承认有限基。然后 B 在 K 上承认有限基，并且，对于所有 b ∈ B b \in B b ∈ B
Tr B / K ( b ) = Tr A / K ( Tr B / A ( b ) ) , N B / K ( b ) = N A / K ( N B / A ( b ) ) P C B / K ( b ; X ) = N A [ X ] / K [ X ] ( P C B / A ( b ; X ) ) .
\begin{align*}
\operatorname{Tr}_{B/K}(b) &= \operatorname{Tr}_{A/K}(\operatorname{Tr}_{B/A}(b)), \quad N_{B/K}(b) = N_{A/K}(N_{B/A}(b)) \\
P_{C_{B/K}}(b; X) &= N_{A[X]/K[X]}(P_{C_{B/A}}(b; X)).
\end{align*}
Tr B / K ( b ) P C B / K ( b ; X ) = Tr A / K ( Tr B / A ( b )) , N B / K ( b ) = N A / K ( N B / A ( b )) = N A [ X ] / K [ X ] ( P C B / A ( b ; X )) . (26) 这紧接着命题 6、设定
V = B V = B V = B 和
u ( x ) = b x u(x) = bx u ( x ) = b x 。
引用了 1 引用了 命题 6
5. AN 代数的判别式
设 A 为 K-代数，承认 n 元素的有限基。 A 的 n 元素的序列 ( x 1 , … , x n ) (x_1, \ldots, x_n) ( x 1 , … , x n ) D A / K ( x 1 , … , x n ) D_{A/\mathbf{K}}(x_1, \ldots, x_n) D A / K ( x 1 , … , x n )
( Tr A / K ( x i x j ) ) 1 ≤ i ≤ n , 1 ≤ j ≤ n .
(\operatorname{Tr}_{A/\mathbf{K}}(x_i x_j))_{1 \leq i \leq n, 1 \leq j \leq n}.
( Tr A / K ( x i x j ) ) 1 ≤ i ≤ n , 1 ≤ j ≤ n . 首先考虑 A 右 K 的基
( e i ) 1 ≤ i ≤ n (e_i)_{1 \leq i \leq n} ( e i ) 1 ≤ i ≤ n 并写
e i e j = ∑ k = 1 n c i j k e k with c i j k ∈ K .
e_i e_j = \sum_{k=1}^n c_{ij k} e_k \quad \text{with} \quad c_{ij k} \in \mathbf{K}.
e i e j = k = 1 ∑ n c ij k e k with c ij k ∈ K . 然后
Tr A / K ( e i ) = ∑ s = 1 n c i s s \operatorname{Tr}_{A/\mathbf{K}}(e_i) = \sum_{s=1}^n c_{iss} Tr A / K ( e i ) = ∑ s = 1 n c i ss , 从而
Tr A / K ( e i e j ) = ∑ k , s c i j k c k s s \operatorname{Tr}_{A/\mathbf{K}}(e_i e_j) = \sum_{k,s} c_{ij k} c_{kss} Tr A / K ( e i e j ) = ∑ k , s c ij k c k ss 因此
D A / K ( e 1 , … , e n ) = det ( ( ∑ k , s c i j k c k s s ) 1 ≤ i ≤ n , 1 ≤ j ≤ n ) .
D_{A/\mathbf{K}}(e_1, \ldots, e_n) = \det \left( \left( \sum_{k,s} c_{ij k} c_{kss} \right)_{1 \leq i \leq n, 1 \leq j \leq n} \right).
D A / K ( e 1 , … , e n ) = det k , s ∑ c ij k c k ss 1 ≤ i ≤ n , 1 ≤ j ≤ n . 现在令
( x i ) 1 ≤ i ≤ n , ( x i ′ ) 1 ≤ i ≤ n (x_i)_{1 \leq i \leq n}, (x'_i)_{1 \leq i \leq n} ( x i ) 1 ≤ i ≤ n , ( x i ′ ) 1 ≤ i ≤ n 为A的n元素的两个序列，并假设存在阶n、
M = ( m i j ) M = (m_{ij}) M = ( m ij ) 的平方矩阵，系数为K，使得
x i = ∑ j = 1 n m i j x j ′ x_i = \sum_{j=1}^n m_{ij} x'_j x i = ∑ j = 1 n m ij x j ′ 对应于
1 ≤ i ≤ n 1 \leq i \leq n 1 ≤ i ≤ n 。我们写
T = ( Tr A / K ( x i x j ) ) 1 ≤ i ≤ n , 1 ≤ j ≤ n , T ′ = ( Tr A / K ( x i ′ x j ′ ) ) 1 ≤ i ≤ n , 1 ≤ j ≤ n .
T = (\operatorname{Tr}_{A/\mathbf{K}}(x_i x_j))_{1 \leq i \leq n, 1 \leq j \leq n}, \qquad T' = (\operatorname{Tr}_{A/\mathbf{K}}(x'_i x'_j))_{1 \leq i \leq n, 1 \leq j \leq n}.
T = ( Tr A / K ( x i x j ) ) 1 ≤ i ≤ n , 1 ≤ j ≤ n , T ′ = ( Tr A / K ( x i ′ x j ′ ) ) 1 ≤ i ≤ n , 1 ≤ j ≤ n . 然后
Tr A / K ( x i x j ) = ∑ p , q m i p m j q Tr A / K ( x p ′ x q ′ ) \operatorname{Tr}_{A/\mathbf{K}}(x_i x_j) = \sum_{p,q} m_{ip} m_{jq} \operatorname{Tr}_{A/\mathbf{K}}(x'_p x'_q) Tr A / K ( x i x j ) = ∑ p , q m i p m j q Tr A / K ( x p ′ x q ′ ) 、从而
T = M ⋅ T ′ ⋅ t M T = M \cdot T' \cdot {}^t M T = M ⋅ T ′ ⋅ t M ；因此，行列式乘法规则给出
det T = det M ⋅ det T ′ ⋅ det t M = ( det M ) 2 det T ′
\det T = \det M \cdot \det T' \cdot \det {}^t M = (\det M)^2 \det T'
det T = det M ⋅ det T ′ ⋅ det t M = ( det M ) 2 det T ′ 终于从而
D A / K ( x 1 , … , x n ) = ( det M ) 2 D A / K ( x 1 ′ , … , x n ′ ) .
D_{A/\mathbf{K}}(x_1, \ldots, x_n) = (\det M)^2 D_{A/\mathbf{K}}(x'_1, \ldots, x'_n).
D A / K ( x 1 , … , x n ) = ( det M ) 2 D A / K ( x 1 ′ , … , x n ′ ) . 被以下引用 1 被以下引用 习题 8, § 1
上述公式特别表明，A的两个基对K的判别式相差K的可逆元的平方，因此生成是K的相同（主要）理想。这个理想Δ A / K \Delta_{A/K} Δ A / K ± 1 \pm 1 ± 1
(1) 如果 A 是型 ( α , β ) (\alpha, \beta) ( α , β ) Tr ( e 1 ) = 2 , Tr ( e 2 ) = β \operatorname{Tr}(e_1) = 2, \operatorname{Tr}(e_2) = \beta Tr ( e 1 ) = 2 , Tr ( e 2 ) = β
Tr ( e 2 2 ) = α Tr ( e 1 ) + β Tr ( e 2 ) = 2 α + β 2 ,
\operatorname{Tr}(e_2^2) = \alpha \operatorname{Tr}(e_1) + \beta \operatorname{Tr}(e_2) = 2\alpha + \beta^2,
Tr ( e 2 2 ) = α Tr ( e 1 ) + β Tr ( e 2 ) = 2 α + β 2 , 从而D A / K ( e 1 , e 2 ) = β 2 + 4 α D_{A/K}(e_1, e_2) = \beta^2 + 4\alpha D A / K ( e 1 , e 2 ) = β 2 + 4 α
(2) 令A = K [ X ] / K [ X ] P A = K[X]/K[X]P A = K [ X ] / K [ X ] P P ( X ) = X 3 + p X + q P(X) = X^3 + pX + q P ( X ) = X 3 + pX + q x 3 = − p x − q x^3 = -px - q x 3 = − p x − q x 2 x^2 x 2 Tr ( 1 ) = 3 , Tr ( x ) = 0 , Tr ( x 2 ) = − 2 p \operatorname{Tr}(1) = 3, \operatorname{Tr}(x) = 0, \operatorname{Tr}(x^2) = -2p Tr ( 1 ) = 3 , Tr ( x ) = 0 , Tr ( x 2 ) = − 2 p x 3 = − p x − q , Tr ( x 3 ) = − 3 q x^3 = -px - q, \operatorname{Tr}(x^3) = -3q x 3 = − p x − q , Tr ( x 3 ) = − 3 q Tr ( x 4 ) = 2 p 2 \operatorname{Tr}(x^4) = 2p^2 Tr ( x 4 ) = 2 p 2 D A / K ( 1 , x , x 2 ) = − 4 p 3 − 27 q 2 D_{A/K}(1, x, x^2) = -4p^3 - 27q^2 D A / K ( 1 , x , x 2 ) = − 4 p 3 − 27 q 2
(3) 令A为K上型( α , β , γ ) (\alpha, \beta, \gamma) ( α , β , γ ) ( α , β , γ ) (\alpha, \beta, \gamma) ( α , β , γ ) ( 1 , i , j , k ) (1, i, j, k) ( 1 , i , j , k ) Tr ( 1 ) = 4 , Tr ( i ) = 2 β , Tr ( j ) = Tr ( k ) = 0 \operatorname{Tr}(1) = 4, \operatorname{Tr}(i) = 2\beta, \operatorname{Tr}(j) = \operatorname{Tr}(k) = 0 Tr ( 1 ) = 4 , Tr ( i ) = 2 β , Tr ( j ) = Tr ( k ) = 0
D A / K ( 1 , i , j , k ) = − 16 γ 2 ( β 2 + 4 α ) 2 .
D_{A/K}(1, i, j, k) = -16\gamma^2(\beta^2 + 4\alpha)^2.
D A / K ( 1 , i , j , k ) = − 16 γ 2 ( β 2 + 4 α ) 2 . (4) 令 A = M n ( K ) A = M_n(K) A = M n ( K ) ( E i j ) 1 ≤ i ≤ n , 1 ≤ j ≤ n (E_{ij})_{1 \leq i \leq n, 1 \leq j \leq n} ( E ij ) 1 ≤ i ≤ n , 1 ≤ j ≤ n j ≠ i j \neq i j = i Tr A / K ( E i i ) = n \operatorname{Tr}_{A/K}(E_{ii}) = n Tr A / K ( E ii ) = n Tr A / K ( E i j ) = 0 \operatorname{Tr}_{A/K}(E_{ij}) = 0 Tr A / K ( E ij ) = 0 n 2 n^2 n 2 ( Tr ( E i j E h k ) ) (\operatorname{Tr}(E_{ij}E_{hk})) ( Tr ( E ij E hk )) n . P n.P n . P D A / K ( ( E i j ) ) = ± n n 2 D_{A/K}((E_{ij})) = \pm n^{n^2} D A / K (( E ij )) = ± n n 2
引用了 3 引用了 III § 2. 。 例 / 代数 , III § 3. 。评分代数 , II § 10. 。 矩阵
习题
参见习题为 § 9 。