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. 扩张代数估值扩张
1. 衍生指标。 剩余类类次数
令K \mathbf{K} K L \mathbf{L} L K \mathbf{K} K A ′ \mathbf{A}' A ′ L \mathbf{L} L A = K ∩ A ′ \mathbf{A} = \mathbf{K} \cap \mathbf{A}' A = K ∩ A ′ K \mathbf{K} K
m ( A ) = m ( A ′ ) ∩ K .
m(\mathbf{A}) = m(\mathbf{A}') \cap \mathbf{K}.
m ( A ) = m ( A ′ ) ∩ K . v ′ v' v ′ A ′ \mathbf{A}' A ′ v ′ v' v ′ K \mathbf{K} K v v v A \mathbf{A} A K \mathbf{K} K v v v Γ v \Gamma_v Γ v v ′ v' v ′ Γ v ′ \Gamma_{v'} Γ v ′
指标* [ p v : Γ v ′ ] [\mathfrak{p}_v : \Gamma_{v'}] [ p v : Γ v ′ ] 被称为 v ′ v' v ′ 超过 v v v （或超过 K \mathbf{K} K e ( v ′ / v ) e(v'/v) e ( v ′ / v ) （或 e ( A ′ / A ) e(A'/A) e ( A ′ / A ) e ( L / K ) e(L/K) e ( L / K )
这个指标是自然数或+ ∞ +\infty + ∞ v 0 ′ v'_0 v 0 ′ v ′ v' v ′ e ( v ′ / v ) e(v'/v) e ( v ′ / v ) v 0 ′ v'_0 v 0 ′ v v v e ( v ′ / v ) = 1 e(v'/v) = 1 e ( v ′ / v ) = 1 v ′ v' v ′ v v v
另一方面，v v v κ ( A ) \kappa(\mathbf{A}) κ ( A ) v ′ v' v ′ κ ( A ′ ) \kappa(\mathbf{A}') κ ( A ′ )
次数* [ κ ( A ′ ) : κ ( A ) ] [\kappa(\mathbf{A}') : \kappa(\mathbf{A})] [ κ ( A ′ ) : κ ( A )] 称为 v ′ v' v ′ 超过 v v v （或超过 K \mathbf{K} K f ( v ′ / v ) f(v'/v) f ( v ′ / v ) （或 f ( A ′ / A ) f(A'/A) f ( A ′ / A ) 或有时 f ( L / K ) f(L/K) f ( L / K )
这个次数是自然数或+ ∞ +\infty + ∞
设*K , K ′ , K ′ ′ \mathbf{K}, \mathbf{K}', \mathbf{K}'' K , K ′ , K ′′ K ⊂ K ′ ⊂ K ′ ′ K \subset K' \subset K'' K ⊂ K ′ ⊂ K ′′ v ′ ′ v'' v ′′ K ′ ′ K'' K ′′ v v v v ′ v' v ′ K \mathbf{K} K K ′ \mathbf{K}' K ′ 然后还有关系:
( 1 ) e ( v ′ ′ / v ) = e ( v ′ ′ / v ′ ) e ( v ′ / v ) , f ( v ′ ′ / v ) = f ( v ′ ′ / v ′ ) f ( v ′ / v ) . (1)\quad e(v''/v) = e(v''/v') e(v'/v), \qquad f(v''/v) = f(v''/v') f(v'/v). ( 1 ) e ( v ′′ / v ) = e ( v ′′ / v ′ ) e ( v ′ / v ) , f ( v ′′ / v ) = f ( v ′′ / v ′ ) f ( v ′ / v ) . 被以下引用 4 被以下引用 定理 1 , 定理 2 , 推论 4 , 引理 5, VII § 4
令K K K L L L n n n K K K v ′ v' v ′ L L L v v v K K K
e ( v ′ / v ) f ( v ′ / v ) ≤ n
e(v'/v) f(v'/v) \leq n
e ( v ′ / v ) f ( v ′ / v ) ≤ n 被以下引用 8 被以下引用 推论 1, VI § 10 , 习题 11, § 8 , 习题 8, § 8 , 推论 3 , 命题 1 , 命题 10 , 命题 2 , 推论 4
持有；特别是e ( v ′ / v ) e(v'/v) e ( v ′ / v ) f ( v ′ / v ) f(v'/v) f ( v ′ / v )
让我们分别取自然数、r r r s s s e ( v ′ / v ) e(v'/v) e ( v ′ / v ) f ( v ′ / v ) f(v'/v) f ( v ′ / v ) r s ≤ n rs \leq n r s ≤ n r r r L L L 1 ≤ i ≤ r 1 \leq i \leq r 1 ≤ i ≤ r x i x_i x i i ≠ j i \neq j i = j v ′ ( x i ) ≢ v ′ ( x j ) v'(x_i) \not\equiv v'(x_j) v ′ ( x i ) ≡ v ′ ( x j ) Γ v \Gamma_v Γ v s s s v ′ v' v ′ A ′ A' A ′ y k y_k y k 1 ≤ k ≤ s 1 \leq k \leq s 1 ≤ k ≤ s κ ( A ′ ) \kappa(A') κ ( A ′ ) y ˉ k \bar{y}_k y ˉ k κ ( A ) \kappa(A) κ ( A ) v ′ ( y k ) = 0 v'(y_k) = 0 v ′ ( y k ) = 0 k k k r s rs r s x i y k x_i y_k x i y k K K K r s ≤ n rs \leq n r s ≤ n
假设存在形式的非平凡线性关系
∑ i , k a i k x i y k = 0 ( a i k ∈ K ) .
\sum_{i,k} a_{ik} x_i y_k = 0 \quad (a_{ik} \in K).
i , k ∑ a ik x i y k = 0 ( a ik ∈ K ) . 让我们选择指标j , m j, m j , m
v ′ ( a j m x j y m ) ≤ v ′ ( a i k x i y k )
v'(a_{jm} x_j y_m) \leq v'(a_{ik} x_i y_k)
v ′ ( a j m x j y m ) ≤ v ′ ( a ik x i y k ) 每有序对 ( i , k ) (i, k) ( i , k ) a j m ≠ 0 a_{jm} \neq 0 a j m = 0 i ≠ j i \neq j i = j v ′ ( a i k x i y k ) = v ′ ( a j m x j y m ) v'(a_{ik} x_i y_k) = v'(a_{jm} x_j y_m) v ′ ( a ik x i y k ) = v ′ ( a j m x j y m )
v ′ ( x i ) − v ′ ( x j ) = v ′ ( a j m ) − v ′ ( a i k ) ∈ Γ v ,
v'(x_i) - v'(x_j) = v'(a_{jm}) - v'(a_{ik}) \in \Gamma_v,
v ′ ( x i ) − v ′ ( x j ) = v ′ ( a j m ) − v ′ ( a ik ) ∈ Γ v , 与x i x_i x i ( a j m x j ) − 1 (a_{jm} x_j)^{-1} ( a j m x j ) − 1
∑ k b k y k + z = 0 , where b k = a j k x j a j m x j ∈ A ′ , z ∈ A ′
\sum_k b_k y_k + z = 0, \text{ where } b_k = \frac{a_{jk} x_j}{a_{jm} x_j} \in A', \quad z \in A'
k ∑ b k y k + z = 0 , where b k = a j m x j a j k x j ∈ A ′ , z ∈ A ′ 和v ′ ( b k ) ≥ 0 , v ′ ( z ) > 0 v'(b_k) \geq 0, v'(z) > 0 v ′ ( b k ) ≥ 0 , v ′ ( z ) > 0 κ ( A ′ ) \kappa(A') κ ( A ′ ) ∑ k b ˉ k y ˉ k = 0 \sum_k \bar{b}_k \bar{y}_k = 0 ∑ k b ˉ k y ˉ k = 0 b m = 1 b_m = 1 b m = 1 y k y_k y k
令K K K L L L K K K v ′ v' v ′ L L L v v v K K K A A A A ′ A' A ′ v v v v ′ v' v ′ Γ v ′ / Γ v \Gamma_{v'}/\Gamma_v Γ v ′ / Γ v κ ( A ′ ) \kappa(A') κ ( A ′ ) κ ( A ) \kappa(A) κ ( A )
引用了 1 · 被以下引用 4 引用了 引理 2
被以下引用 定理 1, VI § 10 , 推论 1 , 推论 1 , 命题 9
令( L α ) (L_\alpha) ( L α ) L L L Γ α = v ′ ( L α ∗ ) \Gamma_\alpha = v'(L_\alpha^*) Γ α = v ′ ( L α ∗ ) Γ v ′ \Gamma_{v'} Γ v ′ Γ α \Gamma_\alpha Γ α Γ α / Γ v \Gamma_\alpha/\Gamma_v Γ α / Γ v Γ v ′ / Γ v \Gamma_{v'}/\Gamma_v Γ v ′ / Γ v κ ( A ′ ) \kappa(A') κ ( A ′ ) κ ( A ) \kappa(A) κ ( A )
v ′ v' v ′ v v v
引用了 1 引用了 命题 1
令G ′ G' G ′ G G G G ′ \mathscr{G}' G ′ G \mathscr{G} G G G G G ′ G' G ′ G G G H ′ ↦ H ′ ∩ G H' \mapsto H' \cap G H ′ ↦ H ′ ∩ G G ′ \mathscr{G}' G ′ G \mathscr{G} G G ′ / G G'/G G ′ / G
引用了 1 · 被以下引用 1 引用了 VI § 4. 估值的高度
被以下引用 定理 1
显然H ′ ∈ G ′ H' \in \mathscr{G}' H ′ ∈ G ′ H ′ ∩ G ∈ G H' \cap G \in \mathscr{G} H ′ ∩ G ∈ G H ∈ G H \in \mathscr{G} H ∈ G H ′ H' H ′ x ′ ∈ G ′ x' \in G' x ′ ∈ G ′ h ∈ H h \in H h ∈ H − h ≤ x ′ ≤ h -h \leq x' \leq h − h ≤ x ′ ≤ h H ′ H' H ′ G ′ G' G ′ H ′ ∩ G = H H' \cap G = H H ′ ∩ G = H H H H H ′ ↦ H ′ ∩ G H' \mapsto H' \cap G H ′ ↦ H ′ ∩ G G ′ / G G'/G G ′ / G H 1 ′ H'_1 H 1 ′ H 2 ′ H'_2 H 2 ′ G ′ G' G ′ H 1 ′ ∩ G = H 2 ′ ∩ G H'_1 \cap G = H'_2 \cap G H 1 ′ ∩ G = H 2 ′ ∩ G H 1 ′ ⊃ H 2 ′ H'_1 \supset H'_2 H 1 ′ ⊃ H 2 ′ H 1 ′ / H 2 ′ H'_1/H'_2 H 1 ′ / H 2 ′ H 1 ′ / ( H 1 ′ ∩ G ) H'_1/(H'_1 \cap G) H 1 ′ / ( H 1 ′ ∩ G ) G ′ / G G'/G G ′ / G H 1 ′ / H 2 ′ H'_1/H'_2 H 1 ′ / H 2 ′
假设L L L K K K v ′ v' v ′ v v v v ′ v' v ′ Γ v ′ \Gamma_{v'} Γ v ′ Z \mathbf{Z} Z Z \mathbf{Z} Z v v v Γ v \Gamma_v Γ v Z \mathbf{Z} Z Γ v ′ / Γ v \Gamma_{v'}/\Gamma_v Γ v ′ / Γ v Γ v ′ \Gamma_{v'} Γ v ′ Z \mathbf{Z} Z
引用了 2 · 被以下引用 3 引用了 引理 2 , 推论 2
被以下引用 命题 5 , 推论 4 , 命题 12, VII § 1
2. 扩张评估和完成
假设K K K v v v K K K L L L K K K L L L ( v i ′ ) i ∈ I (v'_i)_{i \in I} ( v i ′ ) i ∈ I v v v L L L v v v v i ′ v'_i v i ′ v v v L L L
设K K K v v v K K K K ^ \hat{K} K ^ K K K v v v v ^ \hat{v} v ^ v v v K ^ \hat{K} K ^ L L L n n n K K K v ′ v' v ′ L L L v v v L ^ v ′ \hat{L}_{v'} L ^ v ′ L L L v ′ v' v ′ 8 ′ 8' 8 ′ v ′ v' v ′ L ^ v ′ \hat{L}_{v'} L ^ v ′ K ^ \hat{K} K ^ L ^ v ′ \hat{L}_{v'} L ^ v ′ K K K
e ( v ^ ′ ∣ v ^ ) = e ( v ′ ∣ v ) , f ( v ^ ′ ∣ v ^ ) = f ( v ′ ∣ v ) ,
e(\hat{v}'|\hat{v}) = e(v'|v), \quad f(\hat{v}'|\hat{v}) = f(v'|v),
e ( v ^ ′ ∣ v ^ ) = e ( v ′ ∣ v ) , f ( v ^ ′ ∣ v ^ ) = f ( v ′ ∣ v ) , [ L ^ v ′ : K ^ ] ≤ n ,
[\hat{L}_{v'} : \hat{K}] \leq n,
[ L ^ v ′ : K ^ ] ≤ n , e ( v ′ ∣ v ) f ( v ′ ∣ v ) ⪯ [ L ^ v ′ : K ^ ] .
e(v'|v)f(v'|v) \preceq [\hat{L}_{v'} : \hat{K}].
e ( v ′ ∣ v ) f ( v ′ ∣ v ) ⪯ [ L ^ v ′ : K ^ ] . (b)
对L L L v v v v 1 ′ , … , v s ′ v'_1, \ldots, v'_s v 1 ′ , … , v s ′ L L L v的成对独立评估，使得对L扩展v的每个评估都依赖于v_i'之一；令 L_i 为域 L，其中拓扑由 v_i' 定义，
L ^ i \hat{L}_i L ^ i 完备的；我们写
n i = [ L ^ i : K ^ ] n_i = [\hat{L}_i : \hat{K}] n i = [ L ^ i : K ^ ] 。然后典范映射
引用了 7 · 被以下引用 5 引用了 推论, VI § 5 , 命题 5, VI § 5 , 引理 2 , 定理 1（估值的近似值定理）, VI § 7 , 命题 5, VI § 3 , 命题 3 , 命题 4
被以下引用 § 8. 扩张代数估值扩张 , 注 , 命题 11 , 定理 1 , 推论 2
ϕ : K ^ ⊗ K L → ∏ i = 1 s L ^ i
\phi : \hat{K} \otimes_K L \to \prod_{i=1}^s \hat{L}_i
ϕ : K ^ ⊗ K L → i = 1 ∏ s L ^ i （通过连续性延伸对角线映射 L → ∏ i = 1 s L i L \to \prod_{i=1}^s L_i L → ∏ i = 1 s L i K ^ ⊗ K L \hat{K} \otimes_K L K ^ ⊗ K L
∑ i = 1 s n i ≤ n .
\sum_{i=1}^s n_i \leq n.
i = 1 ∑ s n i ≤ n . 我们首先证明(a)。假设 v 不是不恰当的。由于 v 和 v ^ \hat{v} v ^ v ′ v' v ′ v ^ ′ \hat{v}' v ^ ′ L ^ v ′ \hat{L}_{v'} L ^ v ′ K ^ \hat{K} K ^ L ^ v ′ \hat{L}_{v'} L ^ v ′
我们现在转到(b)。我们仍然可以假设 v 不是不恰当的。令 ( v 1 ′ , … , v r ′ ) (v_1', \ldots, v_r') ( v 1 ′ , … , v r ′ ) ∏ i = 1 r L i \prod_{i=1}^r L_i ∏ i = 1 r L i ∏ i = 1 r L i \prod_{i=1}^r L_i ∏ i = 1 r L i ∏ i = 1 r L ^ i \prod_{i=1}^r \hat{L}_i ∏ i = 1 r L ^ i ∏ i = 1 r L ^ i \prod_{i=1}^r \hat{L}_i ∏ i = 1 r L ^ i K ^ ⊗ K L \hat{K} \otimes_K L K ^ ⊗ K L ∏ i = 1 r L ^ i \prod_{i=1}^r \hat{L}_i ∏ i = 1 r L ^ i K ^ \hat{K} K ^ ∏ i = 1 r L ^ i \prod_{i=1}^r \hat{L}_i ∏ i = 1 r L ^ i K ^ \hat{K} K ^ K ^ ⊗ K L \hat{K} \otimes_K L K ^ ⊗ K L ∏ i = 1 r L ^ i \prod_{i=1}^r \hat{L}_i ∏ i = 1 r L ^ i K ^ ⊗ K L \hat{K} \otimes_K L K ^ ⊗ K L ∑ i = 1 r n i ≤ n \sum_{i=1}^r n_i \leq n ∑ i = 1 r n i ≤ n
我们现在采用( v 1 ′ , … , v s ′ ) (v_1', \ldots, v_s') ( v 1 ′ , … , v s ′ )
ϕ : K ^ ⊗ K L → ∏ i = 1 s L ^ i
\phi : \hat{K} \otimes_K L \to \prod_{i=1}^s \hat{L}_i
ϕ : K ^ ⊗ K L → i = 1 ∏ s L ^ i 是满射且关系 (7) 已被证明。仍有待验证ϕ \phi ϕ K ^ ⊗ K L \hat{K} \otimes_K L K ^ ⊗ K L t t t ∏ i = 1 s L ^ i \prod_{i=1}^s \hat{L}_i ∏ i = 1 s L ^ i t ⊂ n t \subset n t ⊂ n K ^ ⊗ K L \hat{K} \otimes_K L K ^ ⊗ K L L ( m ) = ( K ^ ⊗ K L ) / m L(m) = (\hat{K} \otimes_K L)/m L ( m ) = ( K ^ ⊗ K L ) / m K ^ \hat{K} K ^
章VIII, § 8、命题1)。存在对 L(m) 延伸 ϑ \vartheta ϑ v ′ v' v ′ v v v [ L ( m ) : K ^ ] [L(m): \hat{K}] [ L ( m ) : K ^ ] L ( m ) L(m) L ( m ) § 5 § 5 §5 L ( m ) L(m) L ( m ) K \mathbf{K} K L ( m ) L(m) L ( m ) L ( m ) L(m) L ( m ) L ^ v ′ \hat{L}_{v'} L ^ v ′ K ^ ⊗ K L \hat{K} \otimes_K L K ^ ⊗ K L L ^ v ′ \hat{L}_{v'} L ^ v ′ i i i v ′ v' v ′ v i ′ v'_i v i ′ L v ′ = L i L_{v'} = L_i L v ′ = L i § 7 § 7 §7 n ⊂ m n \subset m n ⊂ m n ⊂ r n \subset r n ⊂ r
如果 K 相对于 v 是完整的并且 v 不是不恰当的，则扩展 v 的 L 的两个估值是相关的。
被以下引用 1 被以下引用 推论 2
这是从K ^ ⊗ K L = L \hat{K} \otimes_K L = L K ^ ⊗ K L = L
ϕ : K ^ ⊗ K L → ∏ i = 1 s L ^ i
\phi : \hat{K} \otimes_K L \to \prod_{i=1}^s \hat{L}_i
ϕ : K ^ ⊗ K L → i = 1 ∏ s L ^ i 是同构 。
K ^ ⊗ K L \hat{K} \otimes_K L K ^ ⊗ K L 代数 , 章 VIII, § 7, no. 3, 定理 1)。
3. 关系∑ i e i f i ≤ n \sum_i e_i f_i \leq n ∑ i e i f i ≤ n
令 K 为域，v v v n n n ( v 1 ′ , … , v r ′ ) (v'_1, \ldots, v'_r) ( v 1 ′ , … , v r ′ ) v v v v v v
∑ i ∈ I e ( v i ′ ∣ v ) f ( v i ′ ∣ v ) ≤ n
\sum_{i \in I} e(v'_i|v) f(v'_i|v) \leq n
i ∈ I ∑ e ( v i ′ ∣ v ) f ( v i ′ ∣ v ) ≤ n 命题 2 (公式 (6) 和 (7))。我们将看到这个结果在一般情况下是正确的。准确地说：
令 K 为域，v 为 K 的估值，L 次数 n 的 K 的有限扩张。然后：
引用了 8 · 被以下引用 2 引用了 命题 3, VI § 7 , 命题 2 , 命题 4, VI § 7 , 引理 1 , VI § 4. 估值的高度 , 推论, VI § 4 , 命题 2, I § 1 , 引理 3
被以下引用 推论 , 命题 12, VII § 1
(a) v 到 L 的扩张的每个完整系统 ( v i ′ ) i ∈ I (v'_i)_{i \in I} ( v i ′ ) i ∈ I
(b) ∑ i ∈ I e ( v i ′ ∣ v ) f ( v i ′ ∣ v ) ≤ n \sum_{i \in I} e(v'_i|v) f(v'_i|v) \leq n ∑ i ∈ I e ( v i ′ ∣ v ) f ( v i ′ ∣ v ) ≤ n Card ( I ) ≤ n \operatorname{Card}(I) \leq n Card ( I ) ≤ n
(c) 环和 v i ′ v'_i v i ′
由于如果v v v v v v ( v 1 ′ , … , v s ′ ) (v'_1, \ldots, v'_s) ( v 1 ′ , … , v s ′ ) v v v
关系∑ i e i f i ≤ n \sum_i e_i f_i \leq n ∑ i e i f i ≤ n
其中两个是等价的。我们首先要证明∑ i = 1 s e ( v i ′ / v ) f ( v i ′ / v ) ≤ n \sum_{i=1}^s e(v'_i/v)f(v'_i/v) \leq n ∑ i = 1 s e ( v i ′ / v ) f ( v i ′ / v ) ≤ n
我们通过归纳对 s 进行论证，因此假设不等式已针对 0, 1, ..., s - 1 估值的情况成立。我们区分两种情况。
(1) 假设至少存在两个独立的估值v i ′ v'_i v i ′ [ 1 , s ] [1, s] [ 1 , s ] [ 1 , s ] = I 1 ∪ … ∪ I t [1, s] = I_1 \cup \ldots \cup I_t [ 1 , s ] = I 1 ∪ … ∪ I t v i ′ v'_i v i ′ v j ′ v'_j v j ′ i i i j j j I k I_k I k k k k ( I k ) < s (I_k) < s ( I k ) < s
我们每个I k I_k I k i ( k ) i(k) i ( k ) L ^ i ( k ) \hat{L}_{i(k)} L ^ i ( k ) L L L v i ( k ) ′ v'_{i(k)} v i ( k ) ′ n ( k ) = [ L ^ i ( k ) : K ^ ] n(k) = [\hat{L}_{i(k)} : \hat{K}] n ( k ) = [ L ^ i ( k ) : K ^ ] L L L i ∈ I k i \in I_k i ∈ I k v i ′ v'_i v i ′ v i ( k ) ′ v'_{i(k)} v i ( k ) ′ L ^ i ( k ) \hat{L}_{i(k)} L ^ i ( k ) v ˉ i ′ \bar{v}'_i v ˉ i ′ K ^ \hat{K} K ^ i ∈ I i \in I i ∈ I v i ′ v'_i v i ′ v ˉ i ′ \bar{v}'_i v ˉ i ′ ( K ^ , L ^ i ( k ) ) (\hat{K}, \hat{L}_{i(k)}) ( K ^ , L ^ i ( k ) ) ∑ i ∈ I k e ( v i ′ / v ) f ( v i ′ / v ) ≤ n ( k ) \sum_{i \in I_k} e(v'_i/v)f(v'_i/v) \leq n(k) ∑ i ∈ I k e ( v i ′ / v ) f ( v i ′ / v ) ≤ n ( k ) ∑ k = 1 t n ( k ) ≤ n \sum_{k=1}^t n(k) \leq n ∑ k = 1 t n ( k ) ≤ n ∑ i = 1 s e ( v i ′ / v ) f ( v i ′ / v ) ≤ n \sum_{i=1}^s e(v'_i/v)f(v'_i/v) \leq n ∑ i = 1 s e ( v i ′ / v ) f ( v i ′ / v ) ≤ n
(2) 现在我们讨论任意两个 v i ′ v'_i v i ′ A i ′ A'_i A i ′ v i ′ v'_i v i ′ 1 ≤ i ≤ s 1 \leq i \leq s 1 ≤ i ≤ s v v v A A A i i i A i ′ ∩ K = A A'_i \cap K = A A i ′ ∩ K = A B ′ B' B ′ A 1 ′ , … , A s ′ A'_1, \ldots, A'_s A 1 ′ , … , A s ′ L L L B = B ′ ∩ K B = B' \cap K B = B ′ ∩ K B ⊇ A B \supseteq A B ⊇ A B B B K K K w w w B ′ B' B ′ w ′ w' w ′ w w w κ ( B ′ ) \kappa(B') κ ( B ′ ) f ( w ′ / w ) f(w'/w) f ( w ′ / w ) κ ( B ) \kappa(B) κ ( B ) κ ( B ′ ) \kappa(B') κ ( B ′ ) A i ′ A'_i A i ′ A A A A ˉ i ′ , A ˉ \bar{A}'_i, \bar{A} A ˉ i ′ , A ˉ A ˉ \bar{A} A ˉ κ ( B ) \kappa(B) κ ( B ) v ˉ \bar{v} v ˉ A ˉ i ′ \bar{A}'_i A ˉ i ′ κ ( B ′ ) \kappa(B') κ ( B ′ ) v ˉ i ′ \bar{v}'_i v ˉ i ′ v ˉ \bar{v} v ˉ A i ′ A'_i A i ′ B ′ B' B ′ A ˉ i ′ \bar{A}'_i A ˉ i ′ κ ( B ′ ) \kappa(B') κ ( B ′ ) v ˉ i ′ \bar{v}'_i v ˉ i ′
∑ i = 1 s e ( v ˉ i ′ / v ˉ ) f ( v ˉ i ′ / v ˉ ) ≤ [ κ ( B ′ ) : κ ( B ) ] = f ( w ′ / w )
\sum_{i=1}^s e(\bar{v}'_i/\bar{v})f(\bar{v}'_i/\bar{v}) \leq [\kappa(B') : \kappa(B)] = f(w'/w)
i = 1 ∑ s e ( v ˉ i ′ / v ˉ ) f ( v ˉ i ′ / v ˉ ) ≤ [ κ ( B ′ ) : κ ( B )] = f ( w ′ / w ) 因此
∑ i = 1 s e ( w ′ / w ) e ( v ˉ i ′ / v ˉ ) f ( v ˉ i ′ / v ˉ ) ≤ e ( w ′ / w ) f ( w ′ / w ) ≤ n (no. 1, Lemma 1).
\sum_{i=1}^s e(w'/w)e(\bar{v}'_i/\bar{v})f(\bar{v}'_i/\bar{v}) \leq e(w'/w)f(w'/w) \leq n \quad \text{(no. 1, Lemma 1).}
i = 1 ∑ s e ( w ′ / w ) e ( v ˉ i ′ / v ˉ ) f ( v ˉ i ′ / v ˉ ) ≤ e ( w ′ / w ) f ( w ′ / w ) ≤ n (no. 1, Lemma 1). 因此，如果我们证明，（a）和（b）的证明将完成
f ( v ˉ i ′ / v ˉ ) = f ( v ˉ i ′ / v ) , e ( w ′ / w ) e ( v ˉ i ′ / v ˉ ) = e ( v i ′ / v ) .
f(\bar{v}'_i/\bar{v}) = f(\bar{v}'_i/v), \quad e(w'/w)e(\bar{v}'_i/\bar{v}) = e(v'_i/v).
f ( v ˉ i ′ / v ˉ ) = f ( v ˉ i ′ / v ) , e ( w ′ / w ) e ( v ˉ i ′ / v ˉ ) = e ( v i ′ / v ) . 为此，我们注v v v v ˉ \bar{v} v ˉ v i ′ v'_i v i ′ v ˉ i ′ \bar{v}'_i v ˉ i ′
0 → Γ v ˉ → Γ v → Γ w → 0 ↓ ↓ ↓ 0 → Γ v ˉ i ′ → Γ v i ′ → Γ w ′ → 0
\begin{array}{ccccccccc}
0 & \to & \Gamma_{\bar{v}} & \to & \Gamma_v & \to & \Gamma_w & \to & 0 \\
& & \downarrow & & \downarrow & & \downarrow & & \\
0 & \to & \Gamma_{\bar{v}_i'} & \to & \Gamma_{v_i'} & \to & \Gamma_{w'} & \to & 0
\end{array}
0 0 → → Γ v ˉ ↓ Γ v ˉ i ′ → → Γ v ↓ Γ v i ′ → → Γ w ↓ Γ w ′ → → 0 0 我们推断有正合序列
0 → Γ v ˉ i ′ / Γ v ˉ → Γ v i ′ / Γ v → Γ w ′ / Γ w → 0
0 \to \Gamma_{\bar{v}_i'}/\Gamma_{\bar{v}} \to \Gamma_{v_i'}/\Gamma_v \to \Gamma_{w'}/\Gamma_w \to 0
0 → Γ v ˉ i ′ / Γ v ˉ → Γ v i ′ / Γ v → Γ w ′ / Γ w → 0 由章I, § 1, no. 4，命题2，证明第二个公式(8)。
要完成定理 1 的证明，还需要证明(c)。如果v i ′ v_i' v i ′ v j ′ v_j' v j ′ Γ v i ′ \Gamma_{v_i'} Γ v i ′ Γ v j ′ / H \Gamma_{v_j'}/H Γ v j ′ / H H H H
Γ v → Γ v j → Γ v j ′ / H = Γ v i ′
\Gamma_v \to \Gamma_{v_j} \to \Gamma_{v_j'}/H = \Gamma_{v_i'}
Γ v → Γ v j → Γ v j ′ / H = Γ v i ′ 是单射，H ∩ Γ v = { 0 } H \cap \Gamma_v = \{0\} H ∩ Γ v = { 0 } H = { 0 } H = \{0\} H = { 0 } v i ′ v_i' v i ′ v j ′ v_j' v j ′ i = j i = j i = j
4. 初始后果指标
设 G 是有限指标的 G 的全序交换群和 H 子群。 The number of major subsets of G consisting of strictly positive elements and containing all the elements > O of H is called the initial index of H in G and denoted by ε ( G , H ) \varepsilon(G, H) ε ( G , H )
被以下引用 1 被以下引用 命题 3
由于以下命题，这个初始指标是自然数：
在定义 4的假设下，如果G的严格正元素的集合不存在最小元素，则对于所有H来说ε ( G , H ) = 1 \varepsilon(G, H) = 1 ε ( G , H ) = 1 G ′ G' G ′
ε ( G , H ) = ( G : ( G ∩ H ) ) .
\varepsilon(G, H) = (G : (G \cap H)).
ε ( G , H ) = ( G : ( G ∩ H )) . 引用了 1 · 被以下引用 1 引用了 定义 4
被以下引用 命题 2
在第一种情况下，令 x 在 G 中为元素 > 0。 y ∈ G y \in G y ∈ G 0 < y < x 0 < y < x 0 < y < x 0 < z < x 0 < z < x 0 < z < x
在第二种情况下，令 x x x n n n n x ∈ H nx \in H n x ∈ H n = ( G : ( G ∩ H ) ) n = (G : (G \cap H)) n = ( G : ( G ∩ H )) z ∈ G z \in G z ∈ G M ( y ) M(y) M ( y ) y ≤ z ( y ∈ G ) y \leq z \ (y \in G) y ≤ z ( y ∈ G ) M ( x ) , M ( 2 x ) , … , M ( n x ) M(x), M(2x), \ldots, M(nx) M ( x ) , M ( 2 x ) , … , M ( n x )
初始指标 ε ( G , H ) \varepsilon(G, H) ε ( G , H ) ( G : H ) (G : H) ( G : H ) 并且等于它，如果 G G G 同构于 Z \mathbf{Z} Z
被以下引用 1 被以下引用 推论
特别是ε ( G , H ) ≤ ( G : H ) \varepsilon(G, H) \leq (G : H) ε ( G , H ) ≤ ( G : H )
设* K K K 为域、L K K K w w w 对 L L L v v v 其限制到 K K K 和 Γ w \Gamma_w Γ w 和 Γ v \Gamma_v Γ v Γ v \Gamma_v Γ v 在 Γ w \Gamma_w Γ w 中的初始指标称为 w w w 相对于 v v v （或相对于 K K K ε ( w / v ) \varepsilon(w/v) ε ( w / v )
被以下引用 1 被以下引用 命题 4
从上述推论，在离散估值的情况下，ε ( w / v ) \varepsilon(w/v) ε ( w / v ) e ( w / v ) e(w/v) e ( w / v )
[ A ′ / m A ′ : A / m ] = ε ( w / v ) f ( w / v ) .
[A'/mA': A/m] = \varepsilon(w/v) f(w/v).
[ A ′ / m A ′ : A / m ] = ε ( w / v ) f ( w / v ) . A ′ A' A ′ m A ′ mA' m A ′ A ′ A' A ′ Γ w \Gamma_w Γ w Γ v \Gamma_v Γ v ε ( w / v ) \varepsilon(w/v) ε ( w / v ) A ′ / m A ′ A'/mA' A ′ / m A ′ A ′ A' A ′ A ′ / m ′ A'/m' A ′ / m ′ A A A f ( w / v ) f(w/v) f ( w / v ) A ′ / m A ′ A'/mA' A ′ / m A ′ A ′ A' A ′ ε ( w / v ) \varepsilon(w/v) ε ( w / v ) A A A ε ( w / v ) f ( w / v ) \varepsilon(w/v) f(w/v) ε ( w / v ) f ( w / v ) A / m A/m A / m
5. 关系∑ i e i f i = n \sum_i e_i f_i = n ∑ i e i f i = n
设* K K K 为域，v 对 K K K A A A 其环，m 为理想值，L 为 K K K 次数 的有限扩张* n n n B B B A A A 在 L L L 和 ( v i ′ ) 1 ≤ i ≤ s (v'_i)_{1 \leq i \leq s} ( v i ′ ) 1 ≤ i ≤ s 完备的 v v v 到 L L L
引用了 4 · 被以下引用 2 引用了 II § 2. 环和模分数 , 命题 8, II § 3 , 推论 3 , 命题 4
被以下引用 命题 9 , 定理 2
[ B / m B : A / m ] = ∑ i = 1 s ε ( v i ′ / v ) f ( v i ′ / v ) .
[B/mB : A/m] = \sum_{i=1}^s \varepsilon(v'_i/v) f(v'_i/v).
[ B / m B : A / m ] = i = 1 ∑ s ε ( v i ′ / v ) f ( v i ′ / v ) . 设A i A_i A i v i ′ v'_i v i ′ A i = B m i A_i = B_{m_i} A i = B m i m i m_i m i B B B q i q_i q i m B mB m B m i m_i m i B / m B → ∏ i = 1 s B / q i B/mB \to \prod_{i=1}^s B/q_i B / m B → ∏ i = 1 s B / q i m i m_i m i q i q_i q i B / q i B/q_i B / q i ( B / q i ) m i (B/q_i)_{m_i} ( B / q i ) m i
B m i / m B m i = A i / m A i .
B_{m_i}/mB_{m_i} = A_i/mA_i.
B m i / m B m i = A i / m A i . 因此，凭借 no. 4 的命题 4 得到典范同构
B / m B → ∏ i = 1 s A i / m A i B/mB \to \prod_{i=1}^s A_i/mA_i B / m B → ∏ i = 1 s A i / m A i 、从而。
使用相同的假设和记号，
[ B / m B : A / m ] = ∑ i = 1 s ε ( v i ′ / v ) f ( v i ′ / v ) ≤ ∑ i = 1 s ε ( v i ′ / v ) f ( v i ′ / v ) ≤ n .
[B/mB : A/m] = \sum_{i=1}^s \varepsilon(v'_i/v) f(v'_i/v) \leq \sum_{i=1}^s \varepsilon(v'_i/v) f(v'_i/v) \leq n.
[ B / m B : A / m ] = i = 1 ∑ s ε ( v i ′ / v ) f ( v i ′ / v ) ≤ i = 1 ∑ s ε ( v i ′ / v ) f ( v i ′ / v ) ≤ n . 我们知道
ε ( v i ′ / v ) ≤ ε ( v i ′ / v ) \varepsilon(v'_i/v) \leq \varepsilon(v'_i/v) ε ( v i ′ / v ) ≤ ε ( v i ′ / v ) （no. 4，命题3的推论）和
∑ i = 1 s ε ( v i ′ / v ) f ( v i ′ / v ) ≤ n \sum_{i=1}^s \varepsilon(v'_i/v) f(v'_i/v) \leq n ∑ i = 1 s ε ( v i ′ / v ) f ( v i ′ / v ) ≤ n （no. 3，定理1）。
引用了 2 · 被以下引用 1 引用了 推论 , 定理 1
被以下引用 定理 2
使用命题 5 的假设和记号，以下条件是等价的： (a) B 为有限生成A-模； (b) B 为自由A-模； （三）[ B / m B : A / m ] = n ; [B/mB : A/m] = n; [ B / m B : A / m ] = n ; ∑ i = 1 n ε ( v i ′ / v ) f ( v i ′ / v ) = n \sum_{i=1}^n \varepsilon(v'_i/v) f(v'_i/v) = n ∑ i = 1 n ε ( v i ′ / v ) f ( v i ′ / v ) = n ε ( v i ′ / v ) = ε ( v i ′ / v ) \varepsilon(v'_i/v) = \varepsilon(v'_i/v) ε ( v i ′ / v ) = ε ( v i ′ / v )
引用了 7 · 被以下引用 4 引用了 命题 5 , VI § 1. 估值环 , VI § 3. 估值 , 引理 1 , 推论 , 推论 1, II § 3 , 引理 1, VI § 3
被以下引用 命题 11 , 命题 12 , 命题 9 , VII § 4. 模过度封闭诺特域
(a)和(b)的等价由引理1、§ 3, no. 6得出。显然 (b) 意味着 (c) (代数 , 章 11, § 1, no. 5, 公式 (19))。 (c) 和 (d) 的等价由命题 5 的推论得出。 仍有待证明 (c) 蕴含 (b)。
一般来说，如果M是A-模，我们将在A / m A/m A / m M / m M M/mM M / m M dim ( V ( B ) ) = n \dim(V(B)) = n dim ( V ( B )) = n x 1 , … , x n x_1, \ldots, x_n x 1 , … , x n ⊂ B \subset B ⊂ B B = L B = L B = L y ∈ B y \in B y ∈ B M = L + A y M = L + Ay M = L + A y L → M → B L \to M \to B L → M → B V ( L ) → V ( M ) → V ( B ) V(L) \to V(M) \to V(B) V ( L ) → V ( M ) → V ( B ) ≤ n \leq n ≤ n V ( L ) → V ( B ) V(L) \to V(B) V ( L ) → V ( B ) V ( B ) V(B) V ( B ) V ( L ) V(L) V ( L ) V ( M ) V(M) V ( M ) V ( L ) → V ( M ) V(L) \to V(M) V ( L ) → V ( M ) L → M L \to M L → M L = M , y ∈ L L = M, y \in L L = M , y ∈ L B = L B = L B = L
∑ i = 1 s e ( v i ′ / v ) f ( v i ′ / v ) = n .
\sum_{i=1}^{s} e(v_i'/v) f(v_i'/v) = n.
i = 1 ∑ s e ( v i ′ / v ) f ( v i ′ / v ) = n . 那么 A 的积分闭包 B 就是秩 n n n
adic 拓扑（其中 m = m ( A ) m = m(A) m = m ( A ) A ′ / m A ′ A'/mA' A ′ / m A ′ A / m A/m A / m e ( v ′ / v ) f ( v ′ / v ) = n e(v'/v) f(v'/v) = n e ( v ′ / v ) f ( v ′ / v ) = n
ϕ : K ^ ⊗ K L → ∏ i = 1 s L ^ i
\phi : \hat{K} \otimes_K L \to \prod_{i=1}^s \hat{L}_i
ϕ : K ^ ⊗ K L → i = 1 ∏ s L ^ i （no. 2，命题2）。 *是双射的。对于所有x ∈ L x \in L x ∈ L P C L / K ( x ; X ) P_{C_{L/K}}(x; X) P C L / K ( x ; X ) P C L ^ i / K ˉ ( x ; X ) P_{C_{\hat{L}_i/\bar{K}}}(x; X) P C L ^ i / K ˉ ( x ; X ) 1 ≤ i ≤ s 1 \leq i \leq s 1 ≤ i ≤ s
{ Tr L / K ( x ) = ∑ i = 1 s Tr L ^ i / K ˉ ( x ) N L / K ( x ) = ∏ i = 1 s N L ^ i / K ˉ ( x ) v ( N L / K ( x ) ) = ∑ i = 1 s n i v i ′ ( x ) .
\begin{cases}
\operatorname{Tr}_{L/K}(x) = \sum_{i=1}^s \operatorname{Tr}_{\hat{L}_i/\bar{K}}(x) \\
N_{L/K}(x) = \prod_{i=1}^s N_{\hat{L}_i/\bar{K}}(x) \\
v(N_{L/K}(x)) = \sum_{i=1}^s n_i v_i'(x).
\end{cases}
⎩ ⎨ ⎧ Tr L / K ( x ) = ∑ i = 1 s Tr L ^ i / K ˉ ( x ) N L / K ( x ) = ∏ i = 1 s N L ^ i / K ˉ ( x ) v ( N L / K ( x )) = ∑ i = 1 s n i v i ′ ( x ) . （（9）中的最后一个关系是有意义的，因为我们显然可以假设v i ′ v_i' v i ′ R \mathbf{R} R
由于没有两个v i ′ v'_i v i ′ e ( v i ′ / v ) f ( v i ′ / v ) ≤ n i e(v'_i/v) f(v'_i/v) \leq n_i e ( v i ′ / v ) f ( v i ′ / v ) ≤ n i i i i ∑ i = 1 s n i ≤ n \sum_{i=1}^s n_i \leq n ∑ i = 1 s n i ≤ n ∑ i = 1 s e ( v i ′ / v ) f ( v i ′ / v ) = n \sum_{i=1}^s e(v'_i/v) f(v'_i/v) = n ∑ i = 1 s e ( v i ′ / v ) f ( v i ′ / v ) = n K ^ ⊗ K L \hat{K} \otimes_K L K ^ ⊗ K L ϕ \phi ϕ z ↦ z ( 1 ⊗ x ) z \mapsto z(1 \otimes x) z ↦ z ( 1 ⊗ x ) x ∈ L x \in L x ∈ L ∏ i = 1 s L ^ i \prod_{i=1}^s \hat{L}_i ∏ i = 1 s L ^ i x x x L L L L ^ i \hat{L}_i L ^ i x x x K ^ \hat{K} K ^ L ^ i \hat{L}_i L ^ i K ^ \hat{K} K ^ ν ^ \hat{\nu} ν ^ ν ^ \hat{\nu} ν ^ w w w K ^ \hat{K} K ^ σ \sigma σ w ( σ ( x ) ) = v i ′ ( x ) w(\sigma(x)) = v'_i(x) w ( σ ( x )) = v i ′ ( x )
ϑ ( N L ^ i / K ^ ( x ) ) = n i v i ′ ( x )
\vartheta(N_{\hat{L}_i/\hat{K}}(x)) = n_i v'_i(x)
ϑ ( N L ^ i / K ^ ( x )) = n i v i ′ ( x ) (代数, 章 VIII, § 12, no. 2, 公式 (15))，证明了(9)的公式。
在推论3的假设下，如果L是K的可分扩张，则每个L ^ i \hat{L}_i L ^ i K ^ \hat{K} K ^ G \mathscr{G} G G i \mathscr{G}_i G i v i ′ v'_i v i ′ L ^ i \hat{L}_i L ^ i K ^ \hat{K} K ^ G i \mathscr{G}_i G i
引用了 10 引用了 定义 2, V § 2 , 推论 3 , 命题 6, V § 2 , 推论 1, VI § 7 , 命题 10, A V § 7 , VI § 3. 估值 , 引理 2 , 推论 2 , 引理 1 , 命题 4
显然L ^ i = K ^ ( L ) \hat{L}_i = \hat{K}(L) L ^ i = K ^ ( L ) L ^ i \hat{L}_i L ^ i K ^ \hat{K} K ^ σ ∈ G i \sigma \in \mathscr{G}_i σ ∈ G i v i ′ v'_i v i ′ v i ′ v'_i v i ′ v i ′ = v i ′ ∘ σ v'_i = v'_i \circ \sigma v i ′ = v i ′ ∘ σ G i \mathscr{G}_i G i σ \sigma σ L ^ i \hat{L}_i L ^ i K ^ \hat{K} K ^ σ ^ \hat{\sigma} σ ^ K ^ \hat{K} K ^ L ^ i \hat{L}_i L ^ i C a r d ( G i ) \mathrm{Card}(\mathscr{G}_i) Card ( G i ) v i ′ v'_i v i ′ G \mathscr{G} G s = ( G i ) s = (\mathscr{G}_i) s = ( G i )
C a r d ( G i ) = n / s ≤ n .
\mathrm{Card}(\mathscr{G}_i) = n/s \leq n.
Card ( G i ) = n / s ≤ n . 另一方面，
n = s n n = sn n = s n ，由推论 3 得出；这证明
L ^ i \hat{L}_i L ^ i 是
K ^ \hat{K} K ^ 的伽罗瓦扩张，并且扩张通过自同构
σ ∈ G i \sigma \in \mathscr{G}_i σ ∈ G i 的连续性是
L ^ i \hat{L}_i L ^ i 唯一的
K ^ \hat{K} K ^ 自同构。
注(2) 。上述结果中的部分扩展适用于对*不一定可交换*域K（cf.§ 3, no. 1）进行估值的情况。令L为K的扩张域，令v ′ v' v ′ v v v v ′ v' v ′ v v v e ( v ′ / v ) e(v'/v) e ( v ′ / v ) κ ( A ) \kappa(A) κ ( A ) κ ( A ′ ) \kappa(A') κ ( A ′ ) v ′ v' v ′ v v v f ( v ′ / v ) f(v'/v) f ( v ′ / v ) κ ( A ) \kappa(A) κ ( A ) κ ( A ′ ) \kappa(A') κ ( A ′ ) + ∞ +\infty + ∞ n n n v v v u ′ u' u ′ n n n
首先，L上的v ′ v' v ′ v v v § 5 § 5 §5 § 6 § 6 §6 m = m ( A ) m = m(A) m = m ( A ) A ′ / m A ′ A'/mA' A ′ / m A ′ e ( v ′ / v ) f ( v ′ / v ) e(v'/v) f(v'/v) e ( v ′ / v ) f ( v ′ / v ) ( A / m ) (A/m) ( A / m ) e = e ( v ′ / v ) e = e(v'/v) e = e ( v ′ / v ) v ( K ∗ ) = Z v(K^*) = Z v ( K ∗ ) = Z v ′ ( L ∗ ) = e − 1 Z v'(L^*) = e^{-1}Z v ′ ( L ∗ ) = e − 1 Z u ′ u' u ′ v ′ ( u ′ ) = e − 1 v'(u') = e^{-1} v ′ ( u ′ ) = e − 1 u u u v ( u ) = 1 v(u) = 1 v ( u ) = 1 u = z u ′ e u = z{u'}^e u = z u ′ e z ∈ L z \in L z ∈ L v ′ ( z ) = 0 v'(z) = 0 v ′ ( z ) = 0 u u u m A ′ = u ′ e A ′ = A ′ u ′ e mA' = {u'}^eA' = A'{u'}^e m A ′ = u ′ e A ′ = A ′ u ′ e 0 ≤ k ≤ e − 1 0 \leq k \leq e - 1 0 ≤ k ≤ e − 1 A ′ u ′ k / A ′ u ′ k + 1 A'{u'}^k/A'{u'}^{k+1} A ′ u ′ k / A ′ u ′ k + 1 f ( v ′ / v ) f(v'/v) f ( v ′ / v ) ( A / m ) (A/m) ( A / m ) t ↦ t u ′ k t \mapsto t{u'}^k t ↦ t u ′ k A ′ u ′ k A'{u'}^k A ′ u ′ k A ′ u ′ A'u' A ′ u ′ A ′ u ′ k + 1 A'{u'}^{k+1} A ′ u ′ k + 1 A ′ / A ′ u ′ A'/A'u' A ′ / A ′ u ′ ( A / m ) (A/m) ( A / m ) A ′ u ′ k / A ′ u ′ k + 1 A'{u'}^k/A'{u'}^{k+1} A ′ u ′ k / A ′ u ′ k + 1 f ( v ′ / v ) f(v'/v) f ( v ′ / v ) u ′ u' u ′ A ′ A' A ′ § 3 § 3 §3
6. 代数估值环扩张
令 K 为域，v v v A ′ A' A ′ A ′ A' A ′ v v v M ′ \mathfrak{M}' M ′ V ↦ m ( V ) ∩ A ′ V \mapsto m(V) \cap A' V ↦ m ( V ) ∩ A ′ M ′ \mathfrak{M}' M ′ m ′ ↦ A m ′ ′ m' \mapsto A'_{m'} m ′ ↦ A m ′ ′
引用了 2 · 被以下引用 2 引用了 命题 1, V § 2 , 命题 6, VI § 3
被以下引用 习题 8, § 8 , 命题 7
A ′ A' A ′ m ′ m' m ′ m ′ ∩ A m' \cap A m ′ ∩ A A m ′ ′ A'_{m'} A m ′ ′ v v v § 1 § 1 §1 V = A m ′ ′ V = A'_{m'} V = A m ′ ′ V ∩ K = A m ′ ′ ∩ K V \cap K = A'_{m'} \cap K V ∩ K = A m ′ ′ ∩ K α \alpha α A α ′ = A ′ ∩ K α A'_\alpha = A' \cap K_\alpha A α ′ = A ′ ∩ K α A α ′ A'_\alpha A α ′ v v v V α V_\alpha V α A α ′ A'_\alpha A α ′ 1 ≤ i ≤ n 1 \leq i \leq n 1 ≤ i ≤ n ( A α ′ ) i ∣ m α (A'_\alpha)_{i|m_\alpha} ( A α ′ ) i ∣ m α m i α ′ m'_{i\alpha} m i α ′ A α ′ A'_\alpha A α ′ m ′ ∩ A α ′ m' \cap A'_\alpha m ′ ∩ A α ′ m i α ′ m'_{i\alpha} m i α ′ V ∩ K α V \cap K_\alpha V ∩ K α ( A α ′ ) m i α ′ ⊂ A m ′ ′ (A'_\alpha)_{m'_{i\alpha}} \subset A'_{m'} ( A α ′ ) m i α ′ ⊂ A m ′ ′ V = A m ′ ′ V = A'_{m'} V = A m ′ ′ V ∈ V V \in \mathfrak{V} V ∈ V A ′ ⊂ V A' \subset V A ′ ⊂ V m ′ = m ( V ) ∩ A ′ m' = m(V) \cap A' m ′ = m ( V ) ∩ A ′ m ′ ∩ A = m m' \cap A = m m ′ ∩ A = m m ′ m' m ′ A ′ A' A ′ V = A m ′ ′ V = A'_{m'} V = A m ′ ′
设 K 为域、L K 和 f 的拟伽罗瓦扩展以及 L 的 f ′ f' f ′ f ′ f' f ′ f ′ = f ∘ s f' = f \circ s f ′ = f ∘ s
引用了 3 · 被以下引用 2 引用了 推论 1, V § 2 , 推论 3, VI § 1 , 命题 6
被以下引用 推论 1 , 推论 2
令 A 为 K 公共的限制位置的环和 f ′ f' f ′ f ′ f' f ′ A ′ A' A ′ f ∘ s f \circ s f ∘ s A ′ A' A ′ m ′ m' m ′ m ′ ∩ A m' \cap A m ′ ∩ A m ′ m' m ′ A ′ A' A ′ f ′ f' f ′ f ∘ s f \circ s f ∘ s A m ′ ′ A'_{m'} A m ′ ′ A m ′ ′ A'_{m'} A m ′ ′ A m ′ ′ A'_{m'} A m ′ ′ f ′ f' f ′ f ∘ s f \circ s f ∘ s
设f ′ f' f ′ f ′ ′ f'' f ′′ v ′ v' v ′ v ′ ′ v'' v ′′ f ′ ′ = f ′ ∘ s f'' = f' \circ s f ′′ = f ′ ∘ s v ′ ′ v'' v ′′ v ′ ∘ s v' \circ s v ′ ∘ s
设 K 为域，f 为 K 的位（分别为 v 对 K 的估价），L 为 K 的根数扩张。那么 f（分别为 v）到 L 的所有扩张都是等价的。
引用了 2 引用了 命题 7 , 推论 1
L是准伽罗瓦扩展，其唯一的自同构是恒等式。因此，推论 2 由命题 7 得出（分别推论 1）。
令 K 为域，v 对 K 的估值，L K 的次数 n 的有限准伽罗瓦扩展，( v i ′ ) 1 ≤ i ≤ g (v'_i)_{1 \leq i \leq g} ( v i ′ ) 1 ≤ i ≤ g e ( v i ′ / v ) e(v'_i/v) e ( v i ′ / v ) f ( v i ′ / v ) f(v'_i/v) f ( v i ′ / v ) f \mathfrak{f} f e f g ≤ n ef g \leq n e f g ≤ n e f g = n ef g = n e f g = n
这紧接着定理 1 (no. 3) 和 2 (no. 5)。
7. 绝对值的扩张
设 K 为域，L K 的代数扩张，f \mathfrak{f} f f \mathfrak{f} f
引用了 4 · 被以下引用 1 引用了 推论 2 , 命题 1 , 命题 5 , 定理 2
被以下引用 命题 10
首先假设存在对 K K K v v v f ( x ) = e − v ( x ) f(x) = e^{-v(x)} f ( x ) = e − v ( x ) L L L v ′ v' v ′ K K K v v v § 3 § 3 §3 v ′ v' v ′ x ↦ e − v ′ ( x ) x \mapsto e^{-v'(x)} x ↦ e − v ′ ( x ) K K K f f f f s f^s f s s > 0 s > 0 s > 0 § 3 § 3 §3
x ↦ e − v ′ ( x ) / s
x \mapsto e^{-v'(x)/s}
x ↦ e − v ′ ( x ) / s 是 L L L f f f
现在假设 f f f K K K C \mathbf{C} C f ( x ) = ∣ x ∣ s f(x) = |x|^s f ( x ) = ∣ x ∣ s 0 ≤ s ≤ 1 0 \leq s \leq 1 0 ≤ s ≤ 1 § 6 § 6 §6 C \mathbf{C} C L L L C \mathbf{C} C x ↦ ∣ x ∣ s x \mapsto |x|^s x ↦ ∣ x ∣ s f f f
令 K K K f f f K K K K K K f f f L L L K K K f f f L L L f ′ f' f ′ L L L n n n
引用了 3 引用了 引理 2 , 命题 4 , 命题 9
f ′ ( x ) = ( f ( N L / K ( x ) ) ) 1 / n
f'(x) = (f(N_{L/K}(x)))^{1/n}
f ′ ( x ) = ( f ( N L / K ( x )) ) 1/ n 对于所有人 x ∈ L x \in L x ∈ L
f ′ f' f ′ L L L L L L § 6 § 6 §6 f ′ f' f ′ K K K L L L n n n N L / K ( x ) = ∏ i = 1 n x i N_{L/K}(x) = \prod_{i=1}^n x_i N L / K ( x ) = ∏ i = 1 n x i x i x_i x i x x x K K K § 12 § 12 §12 f ′ f' f ′ i i i f ′ ( x i ) = f ′ ( x ) f'(x_i) = f'(x) f ′ ( x i ) = f ′ ( x )
令 K K K f f f K K K K ^ \hat{K} K ^ K K K f f f f ^ \hat{f} f ^ K ^ \hat{K} K ^ L L L K K K n n n
(a) 令f ′ f' f ′ L L L f f f L ^ f ′ \hat{L}_{f'} L ^ f ′ L L L f ′ f' f ′ K ^ \hat{K} K ^ L ^ f ′ \hat{L}_{f'} L ^ f ′ K K K [ L ^ f ′ : K ] ≤ n [\hat{L}_{f'} : K] \leq n [ L ^ f ′ : K ] ≤ n
(b) 在L L L f f f f 1 ′ , … , f s ′ f'_1, \ldots, f'_s f 1 ′ , … , f s ′ 表示，并且相对于 f i ′ f'_i f i ′ 由 L ^ i \hat{L}_i L ^ i L L L
K ^ ⊗ K L → ∏ i = 1 s L ^ i
\hat{K} \otimes_K L \to \prod_{i=1}^s \hat{L}_i
K ^ ⊗ K L → i = 1 ∏ s L ^ i 是同构并且
引用了 6 · 被以下引用 1 引用了 命题 2 , 推论, VI § 5 , 定理 1（估值的近似值定理）, VI § 7 , 定理 2（绝对值值的近似值定理）, VI § 7 , 命题 5, TG IX § 3 , 定理 2
被以下引用 命题 12
∑ i = 1 s [ L ^ i : K ^ ] = n .
\sum_{i=1}^s [\hat{L}_i : \hat{K}] = n.
i = 1 ∑ s [ L ^ i : K ^ ] = n . (10)
证明与命题 2 (no. 2) 中的类似断言相同。参考文献
§ 7, no. 2、定理1； § 5, no. 2，命题 4 的推论应替换为以下内容
§ 7, no. 3、定理2；拓扑向量空间、章 I, § 2, no. 3，推论 1 至定理 2。
观察两个扩张到L，其中定义相同的拓扑是相等的（一般拓扑，章IX, § 3, no. 2，命题5）。最后，由于 f f f K ^ ⊗ K L \hat{K} \otimes_K L K ^ ⊗ K L
注 (1) 命题 11 (b) 表明，K 上K ^ \hat{K} K ^ L ^ i \hat{L}_i L ^ i K ^ \hat{K} K ^ L ^ i \hat{L}_i L ^ i R \mathbf{R} R C \mathbf{C} C § 6 § 6 §6 K ^ \hat{K} K ^ C \mathbf{C} C i i i L ^ i \hat{L}_i L ^ i f i ′ f_i' f i ′ 等于 n 。如果K ^ \hat{K} K ^ R \mathbf{R} R K = Q K = \mathbf{Q} K = Q r 1 r_1 r 1 r 2 r_2 r 2 i i i L ^ i \hat{L}_i L ^ i R \mathbf{R} R C \mathbf{C} C
r 1 + 2 r 2 = n .
r_1 + 2r_2 = n.
r 1 + 2 r 2 = n .
设 K 为域，f 是 K 上的绝对值，L K 的准伽罗瓦扩展，以及 f ′ f' f ′ f ′ ′ f'' f ′′ f ′ ′ = f ′ ∘ s f'' = f' \circ s f ′′ = f ′ ∘ s
引用了 4 引用了 推论 2 , 推论 1 , 定理 2 , 命题 11
如果f f f f ′ ′ f'' f ′′ f ′ ∘ s f' \circ s f ′ ∘ s a > 0 a > 0 a > 0 f ′ ′ ( x ) = ( f ′ ( s ( x ) ) ) a f''(x) = (f'(s(x)))^a f ′′ ( x ) = ( f ′ ( s ( x )) ) a x ∈ L x \in L x ∈ L f f f x ∈ K ∗ x \in K^* x ∈ K ∗ f ( x ) ≠ 1 f(x) \neq 1 f ( x ) = 1 a = 1 a = 1 a = 1 f f f f ′ f' f ′ f ′ ′ f'' f ′′
如果f f f u ′ , u ′ ′ u', u'' u ′ , u ′′ C \mathbf{C} C a ′ > 0 , a ′ ′ > 0 a' > 0, a'' > 0 a ′ > 0 , a ′′ > 0 f ′ ( x ) = ∣ u ′ ( x ) ∣ a ′ f'(x) = |u'(x)|^{a'} f ′ ( x ) = ∣ u ′ ( x ) ∣ a ′
f ′ ′ ( x ) = ∣ u ′ ′ ( x ) ∣ a ′ ′
f''(x) = |u''(x)|^{a''}
f ′′ ( x ) = ∣ u ′′ ( x ) ∣ a ′′ 对于所有x ∈ L x \in L x ∈ L § 6 § 6 §6 x = 2 x = 2 x = 2 a ′ = a ′ ′ a' = a'' a ′ = a ′′ u ′ u' u ′ u ′ ′ u'' u ′′ u 1 u_1 u 1 K ^ \hat{K} K ^ u 2 u_2 u 2 R \mathbf{R} R C \mathbf{C} C u 2 ∘ u ˉ 1 − 1 u_2 \circ \bar{u}_1^{-1} u 2 ∘ u ˉ 1 − 1 R \mathbf{R} R C \mathbf{C} C c : ζ → ζ ˉ c : \zeta \to \bar{\zeta} c : ζ → ζ ˉ u ′ u' u ′ c ∘ u ′ c \circ u' c ∘ u ′ u ′ u' u ′ u ′ ′ u'' u ′′ C \mathbf{C} C u ′ u' u ′ u ′ ′ u'' u ′′ C \mathbf{C} C u ′ ′ = u ′ ∘ s u'' = u' \circ s u ′′ = u ′ ∘ s a ′ = a ′ ′ a' = a'' a ′ = a ′′ f ′ ′ = f ′ ∘ s f'' = f' \circ s f ′′ = f ′ ∘ s
注 (3)。如果K ^ \hat{K} K ^ R \mathbf{R} R L L L L ^ t \hat{L}_t L ^ t r 1 = n r_1 = n r 1 = n r 2 = 0 r_2 = 0 r 2 = 0 r 1 = 0 r_1 = 0 r 1 = 0 2 r 2 = n 2r_2 = n 2 r 2 = n
习题
请参阅习题为 § 8 。