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。 环
1. Rings
(AN I) 在加法 A \mathbf{A} A 下是可交换的群 。 (AN II) 乘法具有结合律并拥有单位元素 。 (AN III) 乘法相对于加法 是可分配的。 环 A \mathbf{A} A 如果其乘法是可交换的，则称其是可交换的 。
其中，( x , y ) ↦ x + y (x, y) \mapsto x + y ( x , y ) ↦ x + y ( x , y ) ↦ x y (x, y) \mapsto xy ( x , y ) ↦ x y 0 0 0 1 1 1 − x -x − x x x x
( 1 ) x + ( y + z ) = ( x + y ) + z (associativity of addition) ( 2 ) x + y = y + x (commutativity of addition) ( 3 ) 0 + x = x + 0 = x (zero) ( 4 ) x + ( − x ) = ( − x ) + x = 0 (negative) ( 5 ) x ( y z ) = ( x y ) z (associativity of multiplication) ( 6 ) x .1 = 1. x = x (unit element) ( 7 ) ( x + y ) . z = x z + y z ( 8 ) x . ( y + z ) = x y + x z (distributivity)
\begin{array}{ll}
(1) & x + (y + z) = (x + y) + z \quad \text{(associativity of addition)} \\
(2) & x + y = y + x \quad \text{(commutativity of addition)} \\
(3) & 0 + x = x + 0 = x \quad \text{(zero)} \\
(4) & x + (-x) = (-x) + x = 0 \quad \text{(negative)} \\
(5) & x(yz) = (xy)z \quad \text{(associativity of multiplication)} \\
(6) & x.1 = 1.x = x \quad \text{(unit element)} \\
(7) & (x + y).z = xz + yz \\
(8) & x.(y + z) = xy + xz \quad \text{(distributivity)}
\end{array}
( 1 ) ( 2 ) ( 3 ) ( 4 ) ( 5 ) ( 6 ) ( 7 ) ( 8 ) x + ( y + z ) = ( x + y ) + z (associativity of addition) x + y = y + x (commutativity of addition) 0 + x = x + 0 = x (zero) x + ( − x ) = ( − x ) + x = 0 (negative) x ( y z ) = ( x y ) z (associativity of multiplication) x .1 = 1. x = x (unit element) ( x + y ) . z = x z + y z x . ( y + z ) = x y + x z (distributivity) 最后，如果x y = y x xy = yx x y = y x A \mathbf{A} A x , y x, y x , y A \mathbf{A} A
单独使用加法 A \mathbf{A} A A \mathbf{A} A 加法群 。对于所有x ∈ A x \in \mathbf{A} x ∈ A γ x \gamma_x γ x δ x \delta_x δ x γ x ( y ) = x y \gamma_x(y) = xy γ x ( y ) = x y δ x ( y ) = y x \delta_x(y) = yx δ x ( y ) = y x γ x \gamma_x γ x δ x \delta_x δ x A \mathbf{A} A
( 9 ) x .0 = 0. x = 0 ( 10 ) x . ( − y ) = ( − x ) . y = − x y ;
\begin{align*}
(9) & \quad x.0 = 0.x = 0 \\
(10) & \quad x.(-y) = (-x).y = -xy;
\end{align*}
( 9 ) ( 10 ) x .0 = 0. x = 0 x . ( − y ) = ( − x ) . y = − x y ; ( − x ) ( − y ) = − ( ( − x ) . y ) = − ( − x y ) (-x)(-y) = -((-x).y) = -(-xy) ( − x ) ( − y ) = − (( − x ) . y ) = − ( − x y ) ( 11 ) ( − x ) ( − y ) = x y . (11) \quad (-x)(-y) = xy. ( 11 ) ( − x ) ( − y ) = x y . − x = ( − 1 ) x = x ( − 1 ) -x = (-1)x = x(-1) − x = ( − 1 ) x = x ( − 1 ) ( − 1 ) ( − 1 ) = 1 (-1)(-1) = 1 ( − 1 ) ( − 1 ) = 1 从 (11) 可以得出 n n n
( 12 ) ( − x ) n = { x n if n is even − x n if n is odd. (12) \quad (-x)^n = \begin{cases} x^n & \text{if } n \text{ is even} \\ -x^n & \text{if } n \text{ is odd.} \end{cases} ( 12 ) ( − x ) n = { x n − x n if n is even if n is odd. 当我们说到环A \mathbf{A} A A \mathbf{A} A x , y ∈ A x, y \in \mathbf{A} x , y ∈ A y y y A \mathbf{A} A A \mathbf{A} A x y − 1 xy^{-1} x y − 1 x / y x/y x / y A \mathbf{A} A A \mathbf{A} A A ∗ \mathbf{A}^* A ∗
设x , y x, y x , y A \mathbf{A} A y ′ ∈ A y' \in \mathbf{A} y ′ ∈ A x = y ′ y x = y'y x = y ′ y x = y y ′ x = yy' x = y y ′ x x x y y y y y y
左）x 的除子。当A是可交换的时，不需要区分“左”和“右”。
与上述术语一致，每一个元素y ∈ A y \in \mathbf{A} y ∈ A y y y x y = 0 xy = 0 x y = 0 y x = 0 yx = 0 y x = 0 x ≠ 0 x \neq 0 x = 0
让x ∈ A x \in \mathbf{A} x ∈ A n > 0 n > 0 n > 0 x n = 0 x^n = 0 x n = 0 x x x 1 − x 1 - x 1 − x
1 + x + x 2 + ⋯ + x n − 1 .
1 + x + x^2 + \cdots + x^{n-1}.
1 + x + x 2 + ⋯ + x n − 1 . 由于 A 是加法下的交换律群，n ∈ Z n \in \mathbf{Z} n ∈ Z x ∈ A x \in \mathbf{A} x ∈ A n x nx n x γ x \gamma_x γ x δ x \delta_x δ x γ x ( n y ) = n γ x ( y ) \gamma_x(ny) = n\gamma_x(y) γ x ( n y ) = n γ x ( y ) δ y ( n x ) = n δ y ( x ) \delta_y(nx) = n\delta_y(x) δ y ( n x ) = n δ y ( x )
x . ( n y ) = ( n x ) . y = n . ( x y ) .
x.(ny) = (nx).y = n.(xy).
x . ( n y ) = ( n x ) . y = n . ( x y ) . n x = ( n .1 ) x nx = (n.1)x n x = ( n .1 ) x 集合加法和乘法满足公理为环，但保证乘法下存在单位元素，称为伪环。
2. 推论分配律
乘法的分配律相对于加法允许我们得到 § 3, no. 4 的应用命题 1，这给出
∏ i = 1 n ( ∑ λ ∈ L i x i , λ ) = ∑ α 1 , … , α n ∏ i = 1 n x i , α i
\prod_{i=1}^n \left( \sum_{\lambda \in L_i} x_{i,\lambda} \right) = \sum_{\alpha_1, \ldots, \alpha_n} \prod_{i=1}^n x_{i, \alpha_i}
i = 1 ∏ n ( λ ∈ L i ∑ x i , λ ) = α 1 , … , α n ∑ i = 1 ∏ n x i , α i ( α 1 , … , α n ) (\alpha_1, \ldots, \alpha_n) ( α 1 , … , α n ) L 1 × ⋯ × L n L_1 \times \cdots \times L_n L 1 × ⋯ × L n i = 1 , … , n i = 1, \ldots, n i = 1 , … , n ( x i , λ ) λ ∈ L i (x_{i,\lambda})_{\lambda \in L_i} ( x i , λ ) λ ∈ L i
令 A 为交换环和 ( x λ ) λ ∈ L (x_\lambda)_{\lambda \in L} ( x λ ) λ ∈ L β = ( β λ ) λ ∈ L \beta = (\beta_\lambda)_{\lambda \in L} β = ( β λ ) λ ∈ L ∣ β ∣ = ∑ λ ∈ L β λ |\beta| = \sum_{\lambda \in L} \beta_\lambda ∣ β ∣ = ∑ λ ∈ L β λ
( ∑ λ ∈ L x λ ) n = ∑ ∣ β ∣ = n n ! ∏ λ ∈ L β λ ! ∏ λ ∈ L x λ β λ .
\left( \sum_{\lambda \in L} x_\lambda \right)^n = \sum_{|\beta|=n} \frac{n!}{\prod_{\lambda \in L} \beta_\lambda!} \prod_{\lambda \in L} x_\lambda^{\beta_\lambda}.
( λ ∈ L ∑ x λ ) n = ∣ β ∣ = n ∑ ∏ λ ∈ L β λ ! n ! λ ∈ L ∏ x λ β λ . 我们将公式 (13) 与
L i = L L_i = L L i = L 一起应用，将
x i , λ = x λ x_{i,\lambda} = x_\lambda x i , λ = x λ 应用于
1 ≤ i ≤ n 1 \leq i \leq n 1 ≤ i ≤ n 。然后
( ∑ λ ∈ L x λ ) n = ∑ α 1 , … , α n x α 1 ⋯ x α n ,
\left( \sum_{\lambda \in L} x_\lambda \right)^n = \sum_{\alpha_1, \ldots, \alpha_n} x_{\alpha_1} \cdots x_{\alpha_n},
( λ ∈ L ∑ x λ ) n = α 1 , … , α n ∑ x α 1 ⋯ x α n , 和延伸至所有序列
α = ( α 1 , … , α n ) ∈ L n \alpha = (\alpha_1, \ldots, \alpha_n) \in L^n α = ( α 1 , … , α n ) ∈ L n 。
引用了 1 引用了 I § 5. 。 群运行于集合
设α \alpha α L n L^n L n λ ∈ L \lambda \in L λ ∈ L U λ α U_\lambda^\alpha U λ α i i i 1 ≤ i ≤ n 1 \leq i \leq n 1 ≤ i ≤ n α i = λ \alpha_i = \lambda α i = λ Φ ( α ) = ( U λ α ) λ ∈ L \Phi(\alpha) = (U_\lambda^\alpha)_{\lambda \in L} Φ ( α ) = ( U λ α ) λ ∈ L Φ \Phi Φ L n L^n L n L L L { 1 , 2 , … , n } \{1, 2, \ldots, n\} { 1 , 2 , … , n } β ∈ N L \beta \in \mathbf{N}^L β ∈ N L ∣ β ∣ = n |\beta| = n ∣ β ∣ = n L β n L_\beta^n L β n α ∈ L n \alpha \in L^n α ∈ L n C a r d U β α = β λ \mathrm{Card}\ U_\beta^\alpha = \beta_\lambda Card U β α = β λ λ ∈ L \lambda \in L λ ∈ L ( L β n ) ∣ β ∣ = n (L_\beta^n)_{|\beta|=n} ( L β n ) ∣ β ∣ = n L n L^n L n
C a r d L β n = n ! ∏ λ ∈ L β λ !
\mathrm{Card}\ L_\beta^n = \frac{n!}{\prod_{\lambda \in L} \beta_\lambda!}
Card L β n = ∏ λ ∈ L β λ ! n ! (§ 5, no. 5)。
最后，对于α ∈ L β n \alpha \in L_\beta^n α ∈ L β n
x α 1 ⋯ x α n = ∏ λ ∈ L ∏ i ∈ U λ α x α i = ∏ λ ∈ L ∏ i ∈ U λ α x λ = ∏ λ ∈ L x λ ,
x_{\alpha_1} \cdots x_{\alpha_n} = \prod_{\lambda \in L} \prod_{i \in U_\lambda^\alpha} x_{\alpha_i} = \prod_{\lambda \in L} \prod_{i \in U_\lambda^\alpha} x_\lambda = \prod_{\lambda \in L} x_\lambda,
x α 1 ⋯ x α n = λ ∈ L ∏ i ∈ U λ α ∏ x α i = λ ∈ L ∏ i ∈ U λ α ∏ x λ = λ ∈ L ∏ x λ , whence
∑ ( α 1 , … , α n ) x α 1 ⋯ x α n = ∑ ∣ β ∣ = n ∑ α ∈ L β n x α 1 ⋯ x α n = ∑ ∣ β ∣ = n ∑ α ∈ L β n ∏ λ ∈ L x λ β λ = ∑ ∣ β ∣ = n n ! ∏ λ ∈ L β λ ! x λ β λ
\sum_{(\alpha_1, \ldots, \alpha_n)} x_{\alpha_1} \cdots x_{\alpha_n} = \sum_{|\beta|=n} \sum_{\alpha \in L_\beta^n} x_{\alpha_1} \cdots x_{\alpha_n}
= \sum_{|\beta|=n} \sum_{\alpha \in L_\beta^n} \prod_{\lambda \in L} x_\lambda^{\beta_\lambda}
= \sum_{|\beta|=n} \frac{n!}{\prod_{\lambda \in L} \beta_\lambda!} x_\lambda^{\beta_\lambda}
( α 1 , … , α n ) ∑ x α 1 ⋯ x α n = ∣ β ∣ = n ∑ α ∈ L β n ∑ x α 1 ⋯ x α n = ∣ β ∣ = n ∑ α ∈ L β n ∑ λ ∈ L ∏ x λ β λ = ∣ β ∣ = n ∑ ∏ λ ∈ L β λ ! n ! x λ β λ 公式 (14) 因此从 (15) 得出。
( x + y ) n = ∑ p = 0 n ( n p ) x p y n − p .
(x + y)^n = \sum_{p=0}^n \binom{n}{p} x^p y^{n-p}.
( x + y ) n = p = 0 ∑ n ( p n ) x p y n − p . 将公式 (14) 应用于 L = { 1 , 2 } L = \{1, 2\} L = { 1 , 2 } x 1 = x x_1 = x x 1 = x x 2 = y x_2 = y x 2 = y
( x + y ) n = ∑ p + q = n n ! p ! q ! x p y q ,
(x + y)^n = \sum_{p+q=n} \frac{n!}{p! q!} x^p y^q,
( x + y ) n = p + q = n ∑ p ! q ! n ! x p y q , 和用 p + q = n p + q = n p + q = n p , q p, q p , q
设A为A的交换环、X、集合、u = ( u x ) x ∈ X \mathbf{u} = (u_x)_{x \in X} u = ( u x ) x ∈ X v = ( v x ) x ∈ X \mathbf{v} = (v_x)_{x \in X} v = ( v x ) x ∈ X u + v \mathbf{u} + \mathbf{v} u + v ( u x + v x ) x ∈ X (u_x + v_x)_{x \in X} ( u x + v x ) x ∈ X λ ∈ N ( X ) \lambda \in \mathbf{N}^{(X)} λ ∈ N ( X ) λ ! = ∏ x ∈ X λ ( x ) ! \lambda! = \prod_{x \in X} \lambda(x)! λ ! = ∏ x ∈ X λ ( x )! α ∈ N ( X ) \alpha \in \mathbf{N}^{(X)} α ∈ N ( X )
引用了 2 · 被以下引用 2 引用了 I § 7. 。免费幺半群，免费群 , 推论 1（二项式公式）
被以下引用 推论, IV § 1 , IV § 4. 。正式幂级数
( u + v ) α = ∑ β + γ = α α ! β ! γ ! u β v γ .
(\mathbf{u} + \mathbf{v})^\alpha = \sum_{\beta + \gamma = \alpha} \frac{\alpha!}{\beta! \gamma!} \mathbf{u}^\beta \mathbf{v}^\gamma.
( u + v ) α = β + γ = α ∑ β ! γ ! α ! u β v γ . 对于x ∈ X x \in X x ∈ X
( u x + v x ) α ( x ) = ∑ m + n = α ( x ) α ( x ) ! m ! n ! u x m v x n
(u_x + v_x)^{\alpha(x)} = \sum_{m+n=\alpha(x)} \frac{\alpha(x)!}{m! n!} u_x^m v_x^n
( u x + v x ) α ( x ) = m + n = α ( x ) ∑ m ! n ! α ( x )! u x m v x n 由推论 1。将这些方程中的乘积代入 x ∈ X x \in X x ∈ X
设A A A x 1 , … , x n x_1, \ldots, x_n x 1 , … , x n A A A I = { 1 , 2 , … , n } I = \{1, 2, \ldots, n\} I = { 1 , 2 , … , n } H ⊂ I H \subset I H ⊂ I x H = ∑ i ∈ H x i x_H = \sum_{i \in H} x_i x H = ∑ i ∈ H x i
被以下引用 4 被以下引用 习题 7, § 8 , 注, III § 6 , 命题 3, IV § 5 , 引理 1, LIE VIII § 8
( − 1 ) n ∑ σ ∈ S n x σ ( 1 ) ⋯ x σ ( n ) = ∑ H ⊂ I ( − 1 ) C a r d H ( x H ) n .
(-1)^n \sum_{\sigma \in S_n} x_{\sigma(1)} \cdots x_{\sigma(n)} = \sum_{H \subset I} (-1)^{\mathrm{Card}\, H}(x_H)^n.
( − 1 ) n σ ∈ S n ∑ x σ ( 1 ) ⋯ x σ ( n ) = H ⊂ I ∑ ( − 1 ) Card H ( x H ) n . *特别是，如果 A A A
( − 1 ) n n ! x 1 x 2 ⋯ x n = ∑ H ⊂ I ( − 1 ) C a r d H ( x H ) n .
(-1)^n n! x_1 x_2 \cdots x_n = \sum_{H \subset I} (-1)^{\mathrm{Card}\, H}(x_H)^n.
( − 1 ) n n ! x 1 x 2 ⋯ x n = H ⊂ I ∑ ( − 1 ) Card H ( x H ) n . 设C C C I I I { 0 , 1 } \{0, 1\} { 0 , 1 } H ⊂ I H \subset I H ⊂ I P ( I ) \mathscr{P}(I) P ( I ) C C C
∑ a ∈ C ( − 1 ) a ( 1 ) + ⋯ + a ( n ) ( ∑ i ∈ I a ( i ) x i ) n
\sum_{a \in C} (-1)^{a(1)+\cdots+a(n)} \left( \sum_{i \in I} a(i) x_i \right)^n
a ∈ C ∑ ( − 1 ) a ( 1 ) + ⋯ + a ( n ) ( i ∈ I ∑ a ( i ) x i ) n = ∑ a ∈ C ( − 1 ) a ( 1 ) + ⋯ + a ( n ) ∑ ( i 1 , … , i n ) ∈ I n a ( i 1 ) … a ( i n ) x i 1 … x i n
= \sum_{a \in C} (-1)^{a(1)+\cdots+a(n)} \sum_{(i_1, \ldots, i_n) \in I^n} a(i_1) \ldots a(i_n) x_{i_1} \ldots x_{i_n}
= a ∈ C ∑ ( − 1 ) a ( 1 ) + ⋯ + a ( n ) ( i 1 , … , i n ) ∈ I n ∑ a ( i 1 ) … a ( i n ) x i 1 … x i n = ∑ ( i 1 , … , i n ) ∈ I n c i 1 … i n x i 1 … x i n
= \sum_{(i_1, \ldots, i_n) \in I^n} c_{i_1 \ldots i_n} x_{i_1} \ldots x_{i_n}
= ( i 1 , … , i n ) ∈ I n ∑ c i 1 … i n x i 1 … x i n 哪里
c i 1 … i n = ∑ a ∈ C ( − 1 ) a ( 1 ) + ⋯ + a ( n ) a ( i 1 ) … a ( i n ) .
c_{i_1 \ldots i_n} = \sum_{a \in C} (-1)^{a(1)+\cdots+a(n)} a(i_1) \ldots a(i_n).
c i 1 … i n = a ∈ C ∑ ( − 1 ) a ( 1 ) + ⋯ + a ( n ) a ( i 1 ) … a ( i n ) . (1) 假设( i 1 , … , i n ) (i_1, \ldots, i_n) ( i 1 , … , i n ) I I I i 1 , … , i n i_1, \ldots, i_n i 1 , … , i n j ∈ I j \in I j ∈ I C ′ C' C ′ a ∈ C a \in C a ∈ C a ( j ) = 0 a(j) = 0 a ( j ) = 0 a ∈ C ′ a \in C' a ∈ C ′ a ∗ a^* a ∗ a a a { j } \{j\} { j } a ∗ ( 1 ) + ⋯ + a ∗ ( n ) = a ( 1 ) + ⋯ + a ( n ) + 1 a^*(1) + \cdots + a^*(n) = a(1) + \cdots + a(n) + 1 a ∗ ( 1 ) + ⋯ + a ∗ ( n ) = a ( 1 ) + ⋯ + a ( n ) + 1
c i 1 … i n = ∑ a ∈ C ′ ( − 1 ) a ( 1 ) + ⋯ + a ( n ) a ( i 1 ) … a ( i n ) + ( − 1 ) a ∗ ( 1 ) + ⋯ + a ∗ ( n ) a ∗ ( i 1 ) … a ∗ ( i n )
c_{i_1 \ldots i_n} = \sum_{a \in C'} (-1)^{a(1)+\cdots+a(n)} a(i_1) \ldots a(i_n) + (-1)^{a^*(1)+\cdots+a^*(n)} a^*(i_1) \ldots a^*(i_n)
c i 1 … i n = a ∈ C ′ ∑ ( − 1 ) a ( 1 ) + ⋯ + a ( n ) a ( i 1 ) … a ( i n ) + ( − 1 ) a ∗ ( 1 ) + ⋯ + a ∗ ( n ) a ∗ ( i 1 ) … a ∗ ( i n ) = ∑ a ∈ C ′ ( ( − 1 ) a ( 1 ) + ⋯ + ( n ) + ( − 1 ) a ( 1 ) + ⋯ + a ( n ) + 1 ) a ( i 1 ) … a ( i n ) = 0.
= \sum_{a \in C'} ((-1)^{a(1)+\cdots+(n)} + (-1)^{a(1)+\cdots+a(n)+1}) a(i_1) \ldots a(i_n) = 0.
= a ∈ C ′ ∑ (( − 1 ) a ( 1 ) + ⋯ + ( n ) + ( − 1 ) a ( 1 ) + ⋯ + a ( n ) + 1 ) a ( i 1 ) … a ( i n ) = 0. (2) 假设存在σ ∈ S n \sigma \in S_n σ ∈ S n i 1 = σ ( 1 ) , … , i n = σ ( n ) i_1 = \sigma(1), \ldots, i_n = \sigma(n) i 1 = σ ( 1 ) , … , i n = σ ( n ) a ( i 1 ) … a ( i n ) = 0 a(i_1) \ldots a(i_n) = 0 a ( i 1 ) … a ( i n ) = 0 a a a c i 1 … i n = ( − 1 ) n c_{i_1 \ldots i_n} = (-1)^n c i 1 … i n = ( − 1 ) n
3. 例 / 环
I。 零环。设 A 为环。对于A中的0 = 1 0 = 1 0 = 1 0 = 1 0 = 1 0 = 1 x ∈ A x \in A x ∈ A x = x .1 = x .0 = 0 x = x.1 = x.0 = 0 x = x .1 = x .0 = 0
II。有理整数的环。由于§ 2, no. 5中定义了加法，§ 2, no. 6中定义了乘法，Z \mathbf{Z} Z − x -x − x
III。实值函数的环。令I为实数的集合R \mathbf{R} R f f f g g g f + g f + g f + g f . g f.g f . g ( f + g ) ( t ) = f ( t ) + g ( t ) , ( f g ) ( t ) = f ( t ) g ( t ) ( t ∈ I ) . (f + g)(t) = f(t) + g(t), \quad (fg)(t) = f(t)g(t) \quad (t \in I). ( f + g ) ( t ) = f ( t ) + g ( t ) , ( f g ) ( t ) = f ( t ) g ( t ) ( t ∈ I ) .
IV。卷积伪环。令 E 为 R \mathbf{R} R ( f g ) ( t ) = ∫ − ∞ ∞ f ( s ) g ( t − s ) d s (fg)(t) = \int_{-\infty}^{\infty} f(s)g(t-s)\ ds ( f g ) ( t ) = ∫ − ∞ ∞ f ( s ) g ( t − s ) d s
V。 环 A 的逆环。设 A 为环。与 A 相同的加法和乘法 ( x , y ) ↦ y x (x, y) \mapsto yx ( x , y ) ↦ y x A 0 A^0 A 0
VI。 群的交换律自同态环。设 G 是可加写的交换律群。令E表示G的自同态的集合。给定E中的f f f g g g f + g f + g f + g f g fg f g
( f + g ) ( x ) = f ( x ) + g ( x ) , ( f g ) ( x ) = f ( g ( x ) ) ( x ∈ G ) . (f + g)(x) = f(x) + g(x), \quad (fg)(x) = f(g(x)) \quad (x \in G). ( f + g ) ( x ) = f ( x ) + g ( x ) , ( f g ) ( x ) = f ( g ( x )) ( x ∈ G ) . f + g f + g f + g f g = f ∘ g fg = f \circ g f g = f ∘ g I d G \mathrm{Id}_G Id G f , g f, g f , g h h h ϕ = f . ( g + h ) \phi = f.(g + h) ϕ = f . ( g + h ) x ∈ G x \in G x ∈ G ϕ ( x ) = f ( ( g + h ) ( x ) ) = f ( g ( x ) + h ( x ) ) = f ( g ( x ) ) + f ( h ( x ) ) \phi(x) = f((g + h)(x)) = f(g(x) + h(x)) = f(g(x)) + f(h(x)) ϕ ( x ) = f (( g + h ) ( x )) = f ( g ( x ) + h ( x )) = f ( g ( x )) + f ( h ( x )) f f f ϕ = f g + f h \phi = fg + fh ϕ = f g + f h ( g + h ) f = g f + h f . (g + h)f = gf + hf. ( g + h ) f = g f + h f . 因此 E 是一个（一般而言不可交换的）环，称为 G* 的 *自同态环。
VII。 零平方的伪环。 如果 x y = 0 xy = 0 x y = 0 x , y ∈ A x, y \in A x , y ∈ A ( x , y ) ↦ 0 (x, y) \mapsto 0 ( x , y ) ↦ 0 G = { 0 } G = \{0\} G = { 0 }
4. Ring homomorphisms
设A和B为两个环。 A 到 B 的态射或同态是 A 到 B 的任意映射 f 满足关系：
f ( x + y ) = f ( x ) + f ( y ) , f ( x y ) = f ( x ) ⋅ f ( y ) , f ( 1 ) = 1 ,
f(x + y) = f(x) + f(y), \quad f(xy) = f(x) \cdot f(y), \quad f(1) = 1,
f ( x + y ) = f ( x ) + f ( y ) , f ( x y ) = f ( x ) ⋅ f ( y ) , f ( 1 ) = 1 , 对于所有 x , y x, y x , y
两个环同态中的合成是环同态。设A和B是A到B的两个环和f f f f f f f − 1 f^{-1} f − 1
设f : A → B f : A \to B f : A → B f f f x ∈ A x \in A x ∈ A f ( 0 ) = 0 f(0) = 0 f ( 0 ) = 0 f ( − x ) = − f ( x ) f(-x) = -f(x) f ( − x ) = − f ( x ) f f f f f f
(1) 设A为环。立即可见，Z \mathbf{Z} Z n ↦ n ⋅ 1 n \mapsto n \cdot 1 n ↦ n ⋅ 1 Z \mathbf{Z} Z Z \mathbf{Z} Z Z \mathbf{Z} Z
特别地，取A为加法群Z \mathbf{Z} Z Z \mathbf{Z} Z n ↦ n ⋅ 1 n \mapsto n \cdot 1 n ↦ n ⋅ 1 Z \mathbf{Z} Z Z \mathbf{Z} Z
(2) 设a a a x ↦ a x a − 1 x \mapsto axa^{-1} x ↦ a x a − 1
a ( x + y ) a − 1 = a x a − 1 + a y a − 1 ,
a(x + y)a^{-1} = axa^{-1} + aya^{-1},
a ( x + y ) a − 1 = a x a − 1 + a y a − 1 , a ( x y ) a − 1 = ( a x a − 1 ) ( a y a − 1 ) .
a(xy)a^{-1} = (axa^{-1})(aya^{-1}).
a ( x y ) a − 1 = ( a x a − 1 ) ( a y a − 1 ) . 它是双射的，因为关系x ′ = a x a − 1 x' = axa^{-1} x ′ = a x a − 1 x = a − 1 x ′ a x = a^{-1}x'a x = a − 1 x ′ a a a a 内自同构 。
引用了 1 引用了 I § 2. 。 单位元素；可取消元素； 可逆元
5. Subrings
设A A A A A A A A A B B B A A A A A A
上面的条件可以写成如下
0 ∈ B , B + B ⊂ B , − B ⊂ B , B . B ⊂ B , 1 ∈ B .
0 \in B, \quad B + B \subset B, \quad -B \subset B, \quad B.B \subset B, \quad 1 \in B.
0 ∈ B , B + B ⊂ B , − B ⊂ B , B . B ⊂ B , 1 ∈ B . 如果B B B A A A A A A B B B A A A
(1) 加法群 Z \mathbf{Z} Z Z \mathbf{Z} Z Z \mathbf{Z} Z Z \mathbf{Z} Z
(2) 设A A A ( A t ) t ∈ I (A_t)_{t \in I} ( A t ) t ∈ I A A A ⋂ t ∈ I A t \bigcap_{t \in I} A_t ⋂ t ∈ I A t A A A A A A X X X A A A A A A X X X
(3) 设X X X A A A A A A X X X A A A A A A A A A
(4) 设G G G Ω \Omega Ω G G G Ω \Omega Ω α ↦ f α \alpha \mapsto f_\alpha α ↦ f α E E E G G G G G G F F F F F F ϕ \phi ϕ G G G ϕ ⋅ f α = f α ⋅ ϕ \phi \cdot f_\alpha = f_\alpha \cdot \phi ϕ ⋅ f α = f α ⋅ ϕ α ∈ Ω \alpha \in \Omega α ∈ Ω F F F E E E F F F G G G F 1 F_1 F 1 f α f_\alpha f α E E E F F F F 1 F_1 F 1 E E E
引用了 1 引用了 II § 1. 。 模
6. 理想
设A A A A A A a ∈ A , x ∈ a a \in A, x \in a a ∈ A , x ∈ a a x ∈ a ax \in a a x ∈ a x a ∈ a xa \in a x a ∈ a A A A a a a a a a A A A A A A
左理想的定义可以表示为关系
0 ∈ a , a + a ⊂ a , A . a ⊂ a
0 \in a, \quad a + a \subset a, \quad A.a \subset a
0 ∈ a , a + a ⊂ a , A . a ⊂ a 关系 − a ⊂ a -a \subset a − a ⊂ a ( − 1 ) . x = − x (-1).x = -x ( − 1 ) . x = − x A . a ⊂ a A.a \subset a A . a ⊂ a x ∈ A x \in A x ∈ A γ x \gamma_x γ x A A A a ↦ x a a \mapsto xa a ↦ x a A A A x ↦ γ x x \mapsto \gamma_x x ↦ γ x A A A A + A^+ A + A A A A A A A + A^+ A +
环A A A A 0 A^0 A 0
(1) 设A A A A A A A A A ( 0 ) (0) ( 0 ) { 0 } \{0\} { 0 }
(2) 对于A A A a a a a a a A . a A.a A . a a . A a.A a . A a a a A A A A . a = a . A A.a = a.A A . a = a . A a a a ( a ) (a) ( a ) ( a ) = A (a) = A ( a ) = A a a a
(3) 设M M M A A A x ∈ A x \in A x ∈ A x y = 0 xy = 0 x y = 0 y ∈ M y \in M y ∈ M A A A M M M M M M
(4) A A A X X X A A A X X X X X X
令a a a A A A 1 ∉ a , a ≠ A 1 \notin a, a \neq A 1 ∈ / a , a = A
令A A A a a a A A A
在其他字中，如果a ≠ A a \neq A a = A A A A a a a a a a A A A a a a
将 A A A A A A A + A^+ A + A A A A + A^+ A + A + A^+ A + P = { 1 } P = \{1\} P = { 1 }
令A A A ( x λ ) λ ∈ L (x_\lambda)_{\lambda \in L} ( x λ ) λ ∈ L A A A a a a b b b ∑ λ ∈ L a λ x λ \sum_{\lambda \in L} a_\lambda x_\lambda ∑ λ ∈ L a λ x λ ( a λ ) λ ∈ L (a_\lambda)_{\lambda \in L} ( a λ ) λ ∈ L A A A ∑ λ ∈ L a λ x λ b λ \sum_{\lambda \in L} a_\lambda x_\lambda b_\lambda ∑ λ ∈ L a λ x λ b λ ( a λ ) λ ∈ L , ( b λ ) λ ∈ L (a_\lambda)_{\lambda \in L}, (b_\lambda)_{\lambda \in L} ( a λ ) λ ∈ L , ( b λ ) λ ∈ L A A A a a a b b b x λ x_\lambda x λ A A A
公式
0 = ∑ λ ∈ L 0. x λ
0 = \sum_{\lambda \in L} 0.x_\lambda
0 = λ ∈ L ∑ 0. x λ ∑ λ ∈ L a λ x λ + ∑ λ ∈ L a λ ′ x λ = ∑ λ ∈ L ( a λ + a λ ′ ) x λ
\sum_{\lambda \in L} a_\lambda x_\lambda + \sum_{\lambda \in L} a'_\lambda x_\lambda = \sum_{\lambda \in L} (a_\lambda + a'_\lambda)x_\lambda
λ ∈ L ∑ a λ x λ + λ ∈ L ∑ a λ ′ x λ = λ ∈ L ∑ ( a λ + a λ ′ ) x λ a . ∑ λ ∈ L a λ x λ = ∑ λ ∈ L ( a a λ ) x λ
a.\sum_{\lambda \in L} a_\lambda x_\lambda = \sum_{\lambda \in L} (aa_\lambda)x_\lambda
a . λ ∈ L ∑ a λ x λ = λ ∈ L ∑ ( a a λ ) x λ 证明a a a a ′ a' a ′ x λ ∈ a ′ x_\lambda \in a' x λ ∈ a ′ λ ∈ L \lambda \in L λ ∈ L ( a λ ) λ ∈ L (a_\lambda)_{\lambda \in L} ( a λ ) λ ∈ L A A A a λ x λ ∈ a ′ a_\lambda x_\lambda \in a' a λ x λ ∈ a ′ λ ∈ L \lambda \in L λ ∈ L ∑ λ ∈ L a λ x λ ∈ a ′ \sum_{\lambda \in L} a_\lambda x_\lambda \in a' ∑ λ ∈ L a λ x λ ∈ a ′ a ⊂ a ′ a \subset a' a ⊂ a ′ a a a x λ x_\lambda x λ A A A b b b
设A A A ( a λ ) λ ∈ L (a_\lambda)_{\lambda \in L} ( a λ ) λ ∈ L A A A ⋃ λ ∈ L a λ \bigcup_{\lambda \in L} a_\lambda ⋃ λ ∈ L a λ ∑ λ ∈ L y λ \sum_{\lambda \in L} y_\lambda ∑ λ ∈ L y λ ( y λ ) λ ∈ L (y_\lambda)_{\lambda \in L} ( y λ ) λ ∈ L y λ ∈ a λ y_\lambda \in a_\lambda y λ ∈ a λ λ ∈ L \lambda \in L λ ∈ L
引用了 1 引用了 II § 1. 。 模
对于所有 λ ∈ L \lambda \in L λ ∈ L a a a ∑ λ ∈ L y λ \sum_{\lambda \in L} y_\lambda ∑ λ ∈ L y λ y λ ∈ a λ y_\lambda \in a_\lambda y λ ∈ a λ ∑ λ ∈ L x λ + ∑ λ ∈ L y λ = ∑ λ ∈ L ( x λ + y λ ) \sum_{\lambda \in L} x_\lambda + \sum_{\lambda \in L} y_\lambda = \sum_{\lambda \in L} (x_\lambda + y_\lambda) ∑ λ ∈ L x λ + ∑ λ ∈ L y λ = ∑ λ ∈ L ( x λ + y λ ) a ⋅ ∑ λ ∈ L x λ = ∑ λ ∈ L a x λ a \cdot \sum_{\lambda \in L} x_\lambda = \sum_{\lambda \in L} a x_\lambda a ⋅ ∑ λ ∈ L x λ = ∑ λ ∈ L a x λ a a a A A A λ ∈ L \lambda \in L λ ∈ L x ∈ a λ x \in a_\lambda x ∈ a λ μ ≠ λ \mu \neq \lambda μ = λ y λ = x y_\lambda = x y λ = x y μ = 0 y_\mu = 0 y μ = 0 x = ∑ λ ∈ L y λ x = \sum_{\lambda \in L} y_\lambda x = ∑ λ ∈ L y λ x ∈ a x \in a x ∈ a a λ ⊂ a a_\lambda \subset a a λ ⊂ a λ ∈ L \lambda \in L λ ∈ L a ′ a' a ′ a λ a_\lambda a λ a a a a a a ⋃ λ ∈ L a λ \bigcup_{\lambda \in L} a_\lambda ⋃ λ ∈ L a λ
由⋃ λ ∈ L a λ \bigcup_{\lambda \in L} a_\lambda ⋃ λ ∈ L a λ a a a a λ a_\lambda a λ ∑ λ ∈ L a λ \sum_{\lambda \in L} a_\lambda ∑ λ ∈ L a λ a 1 + a 2 a_1 + a_2 a 1 + a 2 a 1 + a 2 a_1 + a_2 a 1 + a 2 a 1 ∈ a 1 a_1 \in a_1 a 1 ∈ a 1 a 2 ∈ a 2 a_2 \in a_2 a 2 ∈ a 2
7. 商环
设A A A a a a A A A A A A x x x y y y a a a x ≡ y x \equiv y x ≡ y a a a x ≡ y ( a ) x \equiv y(a) x ≡ y ( a ) x − y ∈ a x - y \in a x − y ∈ a A A A x ≡ y ( a ) x \equiv y(a) x ≡ y ( a ) x ′ ≡ y ′ ( a ) x' \equiv y'(a) x ′ ≡ y ′ ( a ) x + x ′ ≡ y + y ′ ( a ) x + x' \equiv y + y'(a) x + x ′ ≡ y + y ′ ( a ) a a a x x ′ ≡ x y ′ ( a ) xx' \equiv xy'(a) x x ′ ≡ x y ′ ( a ) x y ′ ≡ y y ′ ( a ) xy' \equiv yy'(a) x y ′ ≡ y y ′ ( a ) a a a x x ′ ≡ y y ′ ( a ) xx' \equiv yy'(a) x x ′ ≡ y y ′ ( a ) R R R A A A x ≡ 0 x \equiv 0 x ≡ 0 a a a R R R x ≡ y x \equiv y x ≡ y R R R x ≡ y x \equiv y x ≡ y a a a
令A A A a a a A A A A / a A/a A / a A A A x ≡ y ( a ) x \equiv y(a) x ≡ y ( a ) A A A § 1 § 1 §1 A / a A/a A / a
(a) 在加法下，A / a A/a A / a A A A a a a
(b) 在乘法下，A / a A/a A / a § 2 § 2 §2
(c) 令ξ , η , ζ \xi, \eta, \zeta ξ , η , ζ A / a A/a A / a π : A → A / a \pi : A \to A/a π : A → A / a x , y , z x, y, z x , y , z A A A π ( x ) = ξ , π ( y ) = η \pi(x) = \xi, \pi(y) = \eta π ( x ) = ξ , π ( y ) = η π ( z ) = ζ \pi(z) = \zeta π ( z ) = ζ
ξ ( η + ζ ) = π ( x ) π ( y + z ) = π ( x ( y + z ) ) = π ( x y + x z ) = π ( x ) π ( y ) + π ( x ) π ( z ) = ξ η + ξ ζ
\xi(\eta + \zeta) = \pi(x)\pi(y + z) = \pi(x(y + z)) = \pi(xy + xz)
= \pi(x)\pi(y) + \pi(x)\pi(z) = \xi\eta + \xi\zeta
ξ ( η + ζ ) = π ( x ) π ( y + z ) = π ( x ( y + z )) = π ( x y + x z ) = π ( x ) π ( y ) + π ( x ) π ( z ) = ξ η + ξ ζ ( ξ + η ) ζ = ξ ζ + η ζ (\xi + \eta)\zeta = \xi\zeta + \eta\zeta ( ξ + η ) ζ = ξ ζ + η ζ
设A A A a a a A A A A A A a a a A / a A/a A / a A A A x ≡ y ( a ) x \equiv y(a) x ≡ y ( a ) A A A
环A / { 0 } A/\{0\} A / { 0 } A A A A / A A/A A / A
令A A A a a a A A A A A A A / a A/a A / a π \pi π B B B A A A f f f B B B f ( a ) = { 0 } f(a) = \{0\} f ( a ) = { 0 } A / a A/a A / a B B B f ˉ \bar{f} f ˉ f = f ˉ ∘ π f = \bar{f} \circ \pi f = f ˉ ∘ π
引用了 1 · 被以下引用 8 引用了 命题 5, I § 4
被以下引用 命题 1, AC I § 3 , 命题 8, AC I § 3 , AC II § 1. 基本理想 , 定义 1, AC II § 1 , 推论 2, AC II § 2 , 命题 8, AC II § 2 , 引理 1, AC VI § 1 , 定理 3
由构造、π ( x + y ) = π ( x ) + π ( y ) \pi(x + y) = \pi(x) + \pi(y) π ( x + y ) = π ( x ) + π ( y ) π ( x y ) = π ( x ) π ( y ) \pi(xy) = \pi(x)\pi(y) π ( x y ) = π ( x ) π ( y ) A A A x , y x, y x , y π ( 1 ) \pi(1) π ( 1 ) A / a A/a A / a ε \varepsilon ε
设A + A^+ A + A A A B + B^+ B + B B B f f f A + A^+ A + B + B^+ B + A + A^+ A + a a a A + / a A^+/a A + / a f ˉ \bar{f} f ˉ B + B^+ B + f = f ˉ ∘ π f = \bar{f} \circ \pi f = f ˉ ∘ π ξ , η \xi, \eta ξ , η A / a A/a A / a x , y x, y x , y A A A π ( x ) = ξ \pi(x) = \xi π ( x ) = ξ π ( y ) = η \pi(y) = \eta π ( y ) = η ξ η = π ( x y ) \xi \eta = \pi(xy) ξ η = π ( x y )
f ˉ ( ξ η ) = f ˉ ( π ( x y ) ) = f ( x y ) = f ( x ) f ( y ) = f ˉ ( ξ ) f ˉ ( η )
\bar{f}(\xi \eta) = \bar{f}(\pi(xy)) = f(xy) = f(x)f(y) = \bar{f}(\xi)\bar{f}(\eta)
f ˉ ( ξ η ) = f ˉ ( π ( x y )) = f ( x y ) = f ( x ) f ( y ) = f ˉ ( ξ ) f ˉ ( η ) 和f ˉ ( ε ) = f ( π ( 1 ) ) = f ( 1 ) \bar{f}(\varepsilon) = f(\pi(1)) = f(1) f ˉ ( ε ) = f ( π ( 1 )) = f ( 1 ) f ˉ \bar{f} f ˉ
设A A A B B B A A A f f f B B B f f f a a a B B B f f f B ′ = f ( B ) B' = f(B) B ′ = f ( B ) B B B π : A → A / a \pi : A \to A/a π : A → A / a i : B ′ → B i : B' \to B i : B ′ → B A / a A/a A / a B ′ B' B ′ f ˉ \bar{f} f ˉ f = i ∘ f ˉ ∘ π f = i \circ \bar{f} \circ \pi f = i ∘ f ˉ ∘ π f ˉ \bar{f} f ˉ
引用了 2 · 被以下引用 1 引用了 定理 2 , E II § 6. 等价关系
被以下引用 命题 5
由于f f f A A A B B B a a a A A A x ∈ a x \in a x ∈ a a ∈ A a \in A a ∈ A f ( a x ) = f ( a ) f ( x ) = 0 f(ax) = f(a)f(x) = 0 f ( a x ) = f ( a ) f ( x ) = 0 a x ∈ a ax \in a a x ∈ a x a ∈ a xa \in a x a ∈ a a a a A A A f f f a a a A / a A/a A / a f ˉ \bar{f} f ˉ B ′ B' B ′ f = i ∘ f ˉ ∘ π f = i \circ \bar{f} \circ \pi f = i ∘ f ˉ ∘ π f ˉ \bar{f} f ˉ f ˉ \bar{f} f ˉ
8. 子环和商环的理想
令A A A A ′ A' A ′ A A A f f f A ′ A' A ′ f f f a a a B ′ B' B ′ A ′ A' A ′ B = f − 1 ( B ′ ) B = f^{-1}(B') B = f − 1 ( B ′ ) A A A a a a f f f B / a B/a B / a B ′ B' B ′ f ( B ) = B ′ f(B) = B' f ( B ) = B ′ f ∣ B f|_B f ∣ B b ′ b' b ′ A ′ A' A ′ b = f − 1 ( b ′ ) b = f^{-1}(b') b = f − 1 ( b ′ ) A A A a a a
引用了 1 · 被以下引用 1 引用了 定理 3
被以下引用 推论
(c) *如果b ′ b' b ′ A ′ A' A ′ A ′ → A ′ / b ′ A' \to A'/b' A ′ → A ′ / b ′ f : A → A ′ f : A \to A' f : A → A ′ f ˉ \bar{f} f ˉ A / b A/b A / b A ′ / b ′ A'/b' A ′ / b ′ f f f f ˉ \bar{f} f ˉ A / b A/b A / b A ′ / b ′ A'/b' A ′ / b ′
(d) *假设 f f f Φ \Phi Φ A A A a a a Φ ′ \Phi' Φ ′ A ′ A' A ′ B → f ( B ) B \to f(B) B → f ( B ) B ′ ↦ f ˉ ( B ′ ) − 1 B' \mapsto \bar{f}(B')^{-1} B ′ ↦ f ˉ ( B ′ ) − 1 Φ \Phi Φ Φ ′ \Phi' Φ ′ Φ ′ \Phi' Φ ′ Φ \Phi Φ
(a) 和 (b) 是显而易见的，除了 (a) 的最后一个断言是从 no. 7、定理 3 得出的。
(c) 中考虑的复合态射 g : A → A ′ → A ′ / b ′ g : A \to A' \to A'/b' g : A → A ′ → A ′ / b ′ b b b f ˉ \bar{f} f ˉ A / b A/b A / b A ′ / b ′ A'/b' A ′ / b ′ f f f g g g f ˉ \bar{f} f ˉ
假设 f f f θ : B ′ ↦ f ˉ ( B ′ ) − 1 \theta : B' \mapsto \bar{f}(B')^{-1} θ : B ′ ↦ f ˉ ( B ′ ) − 1 Φ ′ \Phi' Φ ′ Φ \Phi Φ η : B ↦ f ( B ) \eta : B \mapsto f(B) η : B ↦ f ( B ) Φ \Phi Φ Φ ′ \Phi' Φ ′ θ ∘ η = I d Φ , η ∘ θ = I d Φ ′ , \theta \circ \eta = \mathrm{Id}_{\Phi}, \eta \circ \theta = \mathrm{Id}_{\Phi'}, θ ∘ η = Id Φ , η ∘ θ = Id Φ ′ ,
令A A A a a a A A A A / a A/a A / a b / a b/a b / a b b b A A A a a a b b b A / b A/b A / b ( A / a ) / ( b / a ) (A/a)/(b/a) ( A / a ) / ( b / a ) A → A / a → ( A / a ) / ( b / a ) A \to A/a \to (A/a)/(b/a) A → A / a → ( A / a ) / ( b / a )
引用了 1 引用了 命题 5
它足以将应用命题 5 转换为A A A A / a A/a A / a
9. 乘法理想
令A A A a a a b b b A A A x 1 y 1 + ⋯ + x n y n x_1 y_1 + \cdots + x_n y_n x 1 y 1 + ⋯ + x n y n n ≥ 0 , x i ∈ a n \geq 0, x_i \in a n ≥ 0 , x i ∈ a y i ∈ b y_i \in b y i ∈ b 1 ≤ i ≤ n 1 \leq i \leq n 1 ≤ i ≤ n A A A a b ab ab a a a b b b A A A A A A a , b , c a, b, c a , b , c A A A a ( b + c ) = a b + a c , ( b + c ) a = b a + c a a(b+c) = ab + ac, (b+c)a = ba + ca a ( b + c ) = ab + a c , ( b + c ) a = ba + c a A A A
a b ⊂ a A ⊂ a ab \subset aA \subset a ab ⊂ a A ⊂ a a b ⊂ A b ⊂ b ab \subset Ab \subset b ab ⊂ A b ⊂ b
a b ⊂ a ∩ b .
ab \subset a \cap b.
ab ⊂ a ∩ b . (21)
设a , b 1 , … , b n a, b_1, \ldots, b_n a , b 1 , … , b n A A A A = a + b i A = a + b_i A = a + b i i i i A = a + b 1 b 2 … b n = a + ( b 1 ∩ b 2 ∩ ⋯ ∩ b n ) A = a + b_1 b_2 \ldots b_n = a + (b_1 \cap b_2 \cap \cdots \cap b_n) A = a + b 1 b 2 … b n = a + ( b 1 ∩ b 2 ∩ ⋯ ∩ b n )
被以下引用 4 被以下引用 推论, AC III § 2 , 习题 2, § 8 , 命题 7 , 命题 9
由(21)足以证明A = a + b 1 b 2 … b n A = a + b_1 b_2 \ldots b_n A = a + b 1 b 2 … b n n = 2 n = 2 n = 2 a , a ′ ∈ a , b 1 ∈ b 1 , b 2 ∈ b 2 a, a' \in a, b_1 \in b_1, b_2 \in b_2 a , a ′ ∈ a , b 1 ∈ b 1 , b 2 ∈ b 2 1 = a + b 1 = a ′ + b 2 1 = a + b_1 = a' + b_2 1 = a + b 1 = a ′ + b 2
1 = a ′ + ( a + b 1 ) b 2 = ( a ′ + a b 2 ) + b 1 b 2 ∈ a + b 1 b 2 ,
1 = a' + (a + b_1) b_2 = (a' + a b_2) + b_1 b_2 \in a + b_1 b_2,
1 = a ′ + ( a + b 1 ) b 2 = ( a ′ + a b 2 ) + b 1 b 2 ∈ a + b 1 b 2 , 从而
A = a + b 1 b 2 A = a + b_1 b_2 A = a + b 1 b 2 。
令b 1 , … , b n b_1, \ldots, b_n b 1 , … , b n A A A b i + b j = A b_i + b_j = A b i + b j = A i ≠ j i \neq j i = j b 1 ∩ b 2 ∩ ⋯ ∩ b n = ∑ σ ∈ S n b σ ( 1 ) b σ ( 2 ) ⋯ b σ ( n ) b_1 \cap b_2 \cap \cdots \cap b_n = \sum_{\sigma \in S_n} b_{\sigma(1)} b_{\sigma(2)} \cdots b_{\sigma(n)} b 1 ∩ b 2 ∩ ⋯ ∩ b n = ∑ σ ∈ S n b σ ( 1 ) b σ ( 2 ) ⋯ b σ ( n ) A A A b 1 ∩ b 2 ∩ ⋯ ∩ b n = b 1 b 2 … b n b_1 \cap b_2 \cap \cdots \cap b_n = b_1 b_2 \ldots b_n b 1 ∩ b 2 ∩ ⋯ ∩ b n = b 1 b 2 … b n
引用了 2 · 被以下引用 2 引用了 习题 2, § 8 , 命题 6
被以下引用 命题 10 , 命题 9
假设首先n = 2 n = 2 n = 2 a 1 ∈ b 1 , a 2 ∈ b 2 a_1 \in b_1, a_2 \in b_2 a 1 ∈ b 1 , a 2 ∈ b 2 a 1 + a 2 = 1 a_1 + a_2 = 1 a 1 + a 2 = 1 x ∈ b 1 ∩ b 2 x \in b_1 \cap b_2 x ∈ b 1 ∩ b 2 x = x ( a 1 + a 2 ) = x a 1 + x a 2 ∈ b 2 b 1 + b 1 b 2 x = x(a_1 + a_2) = x a_1 + x a_2 \in b_2 b_1 + b_1 b_2 x = x ( a 1 + a 2 ) = x a 1 + x a 2 ∈ b 2 b 1 + b 1 b 2 b 1 ∩ b 2 = b 1 b 2 + b 2 b 1 b_1 \cap b_2 = b_1 b_2 + b_2 b_1 b 1 ∩ b 2 = b 1 b 2 + b 2 b 1
假设现在对于所有整数 < n < n < n b n + ( b 1 b 2 … b n − 1 ) = A b_n + (b_1 b_2 \ldots b_{n-1}) = A b n + ( b 1 b 2 … b n − 1 ) = A
b 1 ∩ b 2 ∩ ⋯ ∩ b n = ( b 1 ∩ b 2 ∩ ⋯ ∩ b n − 1 ) b n + b n ( b 1 ∩ b 2 ∩ ⋯ ∩ b n − 1 )
b_1 \cap b_2 \cap \cdots \cap b_n = (b_1 \cap b_2 \cap \cdots \cap b_{n-1}) b_n + b_n (b_1 \cap b_2 \cap \cdots \cap b_{n-1})
b 1 ∩ b 2 ∩ ⋯ ∩ b n = ( b 1 ∩ b 2 ∩ ⋯ ∩ b n − 1 ) b n + b n ( b 1 ∩ b 2 ∩ ⋯ ∩ b n − 1 ) = ( ∑ σ ∈ S n − 1 b σ ( 1 ) b σ ( 2 ) ⋯ b σ ( n − 1 ) ) b n
= \left( \sum_{\sigma \in S_{n-1}} b_{\sigma(1)} b_{\sigma(2)} \cdots b_{\sigma(n-1)} \right) b_n
= σ ∈ S n − 1 ∑ b σ ( 1 ) b σ ( 2 ) ⋯ b σ ( n − 1 ) b n + b n ( ∑ σ ∈ S n − 1 b σ ( 1 ) b σ ( 2 ) ⋯ b σ ( n − 1 ) )
+ b_n \left( \sum_{\sigma \in S_{n-1}} b_{\sigma(1)} b_{\sigma(2)} \cdots b_{\sigma(n-1)} \right)
+ b n σ ∈ S n − 1 ∑ b σ ( 1 ) b σ ( 2 ) ⋯ b σ ( n − 1 ) ⊂ ∑ τ ∈ S n b τ ( 1 ) b τ ( 2 ) ⋯ b τ ( n ) ⊂ b 1 ∩ b 2 ∩ ⋯ ∩ b n .
\subset \sum_{\tau \in S_n} b_{\tau(1)} b_{\tau(2)} \cdots b_{\tau(n)} \subset b_1 \cap b_2 \cap \cdots \cap b_n.
⊂ τ ∈ S n ∑ b τ ( 1 ) b τ ( 2 ) ⋯ b τ ( n ) ⊂ b 1 ∩ b 2 ∩ ⋯ ∩ b n .
10. 乘积 / 环
设( A i ) i ∈ I (A_i)_{i \in I} ( A i ) i ∈ I A A A ∏ i ∈ I A i \prod_{i \in I} A_i ∏ i ∈ I A i A A A
( x i ) + ( y i ) = ( x i + y i ) , ( x i ) ( y i ) = ( x i y i ) .
(x_i) + (y_i) = (x_i + y_i), \quad (x_i)(y_i) = (x_i y_i).
( x i ) + ( y i ) = ( x i + y i ) , ( x i ) ( y i ) = ( x i y i ) . A A A A i A_i A i 0 = ( 0 i ) i ∈ I 0 = (0_i)_{i \in I} 0 = ( 0 i ) i ∈ I 0 i 0_i 0 i A i A_i A i 1 = ( 1 i ) i ∈ I 1 = (1_i)_{i \in I} 1 = ( 1 i ) i ∈ I 1 i 1_i 1 i A i A_i A i A i A_i A i A A A C i C_i C i A i A_i A i A A A ∏ i ∈ I C i \prod_{i \in I} C_i ∏ i ∈ I C i 对于所有i ∈ I i \in I i ∈ I A A A A i A_i A i p r i \mathrm{pr}_i pr i B B B f i : B → A i f_i : B \to A_i f i : B → A i f : B → A f : B \to A f : B → A f i = p r i ∘ f f_i = \mathrm{pr}_i \circ f f i = pr i ∘ f i ∈ I i \in I i ∈ I f ( b ) = ( f i ( b ) ) i ∈ I f(b) = (f_i(b))_{i \in I} f ( b ) = ( f i ( b ) ) i ∈ I
对于所有i ∈ I i \in I i ∈ I a i a_i a i A i A_i A i a = ∏ i ∈ I a i a = \prod_{i \in I} a_i a = ∏ i ∈ I a i A A A
假设a i a_i a i i ∈ I i \in I i ∈ I f i f_i f i A i A_i A i A i / a i A_i / a_i A i / a i ∏ i ∈ I A i \prod_{i \in I} A_i ∏ i ∈ I A i f : ( x i ) i ∈ I ↦ ( f i ( x i ) ) i ∈ I f : (x_i)_{i \in I} \mapsto (f_i(x_i))_{i \in I} f : ( x i ) i ∈ I ↦ ( f i ( x i ) ) i ∈ I ∏ i ∈ I ( A i / a i ) \prod_{i \in I} (A_i / a_i) ∏ i ∈ I ( A i / a i ) ∏ i ∈ I a i \prod_{i \in I} a_i ∏ i ∈ I a i ( ∏ i ∈ I A i ) / ( ∏ i ∈ I a i ) \left( \prod_{i \in I} A_i \right) / \left( \prod_{i \in I} a_i \right) ( ∏ i ∈ I A i ) / ( ∏ i ∈ I a i ) ∏ i ∈ I ( A i / a i ) \prod_{i \in I} (A_i / a_i) ∏ i ∈ I ( A i / a i )
令( I λ ) λ ∈ L (I_\lambda)_{\lambda \in L} ( I λ ) λ ∈ L I I I ∏ i ∈ I A i \prod_{i \in I} A_i ∏ i ∈ I A i ∏ λ ∈ L ( ∏ i ∈ I λ A i ) \prod_{\lambda \in L} \left( \prod_{i \in I_\lambda} A_i \right) ∏ λ ∈ L ( ∏ i ∈ I λ A i )
让J ⊂ I J \subset I J ⊂ I e J e_J e J A A A ( x i ) i ∈ I (x_i)_{i \in I} ( x i ) i ∈ I i ∈ J i \in J i ∈ J x i = 1 i x_i = 1_i x i = 1 i i ∈ I − J i \in I - J i ∈ I − J x i = 0 i x_i = 0_i x i = 0 i e J e_J e J A A A § 1 § 1 §1
e I = 1 ; e ∅ = 0 ; e J ∩ K = e J e K for J ⊂ I , K ⊂ I ; e J ∪ K = e J + e K for J ⊂ I , K ⊂ I , J ∩ K = ∅ ; ∑ λ e J λ = 1 if ( J λ ) is a finite partition of I .
\begin{align*}
e_I &= 1; \\
e_\varnothing &= 0; \\
e_{J \cap K} &= e_J e_K & \text{for } J \subset I, K \subset I; \\
e_{J \cup K} &= e_J + e_K & \text{for } J \subset I, K \subset I, J \cap K = \varnothing; \\
\sum_\lambda e_{J_\lambda} &= 1 & \text{if } (J_\lambda) \text{ is a finite partition of } I.
\end{align*}
e I e ∅ e J ∩ K e J ∪ K λ ∑ e J λ = 1 ; = 0 ; = e J e K = e J + e K = 1 for J ⊂ I , K ⊂ I ; for J ⊂ I , K ⊂ I , J ∩ K = ∅ ; if ( J λ ) is a finite partition of I . 让A J = ∏ i ∈ J A i A_J = \prod_{i \in J} A_i A J = ∏ i ∈ J A i η J \eta_J η J A A A A J A_J A J x = ( x i ) i ∈ J ∈ A J x = (x_i)_{i \in J} \in A_J x = ( x i ) i ∈ J ∈ A J ε J ( x ) \varepsilon_J(x) ε J ( x ) A A A ( y i ) i ∈ I (y_i)_{i \in I} ( y i ) i ∈ I i ∈ J i \in J i ∈ J y i = x i y_i = x_i y i = x i i ∈ I − J i \in I - J i ∈ I − J y i = 0 i y_i = 0_i y i = 0 i η J \eta_J η J A J A_J A J A A A ε J \varepsilon_J ε J A A A A J A_J A J
A J → ε J A → η I − J A I − J
A_J \xrightarrow{\varepsilon_J} A \xrightarrow{\eta_{I-J}} A_{I-J}
A J ε J A η I − J A I − J η I − J \eta_{I-J} η I − J a J a_J a J ε J \varepsilon_J ε J ε J ( x x ′ ) = ε J ( x ) ε J ( x ′ ) \varepsilon_J(xx') = \varepsilon_J(x)\varepsilon_J(x') ε J ( x x ′ ) = ε J ( x ) ε J ( x ′ ) x , x ′ ∈ A J x, x' \in A_J x , x ′ ∈ A J ε J \varepsilon_J ε J ε J ( 1 ) = e J \varepsilon_J(1) = e_J ε J ( 1 ) = e J a J = e J A = A e J a_J = e_J A = A e_J a J = e J A = A e J
让e ( i ) = e i e_{(i)} = e_i e ( i ) = e i a i = a ( i ) = e i A = A e i a_i = a_{(i)} = e_i A = A e_i a i = a ( i ) = e i A = A e i i ∈ I i \in I i ∈ I e i 2 = e i e_i^2 = e_i e i 2 = e i
e i e j = e j e i = 0
e_i e_j = e_j e_i = 0
e i e j = e j e i = 0 i ≠ j i \neq j i = j I I I ∑ i ∈ I e i = 1 \sum_{i \in I} e_i = 1 ∑ i ∈ I e i = 1 A A A a i a_i a i x ∈ A x \in A x ∈ A a i a_i a i x e i x e_i x e i
11. 直接分解 / 环
设A A A ( b i ) i ∈ I (b_i)_{i \in I} ( b i ) i ∈ I A A A
x ↦ ( ϕ i ( x ) ) i ∈ I ,
x \mapsto (\phi_i(x))_{i \in I},
x ↦ ( ϕ i ( x ) ) i ∈ I , 其中ϕ i \phi_i ϕ i A A A A / b i A/b_i A / b i A A A ∏ i ∈ I ( A / b i ) \prod_{i \in I} (A/b_i) ∏ i ∈ I ( A / b i )
令A A A ( b 1 , … , b n ) (b_1, \ldots, b_n) ( b 1 , … , b n ) A A A b i + b j = A b_i + b_j = A b i + b j = A i ≠ j i \neq j i = j A A A ∏ i = 1 n ( A / b i ) \prod_{i=1}^n (A/b_i) ∏ i = 1 n ( A / b i ) ⋂ i = 1 n b i = ∑ σ ∈ S n b σ ( 1 ) b σ ( 2 ) ⋯ b σ ( n ) . \bigcap_{i=1}^n b_i = \sum_{\sigma \in S_n} b_{\sigma(1)} b_{\sigma(2)} \cdots b_{\sigma(n)}. ⋂ i = 1 n b i = ∑ σ ∈ S n b σ ( 1 ) b σ ( 2 ) ⋯ b σ ( n ) .
引用了 2 · 被以下引用 5 引用了 命题 6 , 命题 7
被以下引用 命题 10 , 命题 11 , 推论 1, V § 15 , 命题 4, VII § 1 , 命题 3, VIII § 8
显然核是⋂ i = 1 n b i \bigcap_{i=1}^n b_i ⋂ i = 1 n b i A A A ( x i ) 1 ≤ i ≤ n (x_i)_{1 \leq i \leq n} ( x i ) 1 ≤ i ≤ n x ∈ A x \in A x ∈ A 1 ≤ i ≤ n 1 \leq i \leq n 1 ≤ i ≤ n x ≡ x i ( b i ) x \equiv x_i (b_i) x ≡ x i ( b i ) I I I n n n n ≤ 1 n \leq 1 n ≤ 1 y ∈ A y \in A y ∈ A y ≡ x i ( b i ) y \equiv x_i (b_i) y ≡ x i ( b i ) 1 ≤ i ≤ n − 1 1 \leq i \leq n-1 1 ≤ i ≤ n − 1 y + z y + z y + z z ∈ A z \in A z ∈ A x x x i < n i < n i < n z ≡ 0 ( b i ) z \equiv 0 (b_i) z ≡ 0 ( b i ) z ∈ b = ⋂ i = 1 n − 1 b i z \in b = \bigcap_{i=1}^{n-1} b_i z ∈ b = ⋂ i = 1 n − 1 b i z ≡ x n − y ( b n ) z \equiv x_n - y (b_n) z ≡ x n − y ( b n ) b n + b = A b_n + b = A b n + b = A z z z
令A A A A A A ( b i ) i ∈ I (b_i)_{i \in I} ( b i ) i ∈ I A A A ∏ i ∈ I ( A / b i ) \prod_{i \in I} (A/b_i) ∏ i ∈ I ( A / b i ) A A A
被以下引用 1 被以下引用 VII § 2. 。 挠元模超过主理想整环
令A A A A ′ A' A ′ ( b i ) i ∈ I (b_i)_{i \in I} ( b i ) i ∈ I A A A
(a) 家庭( b i ) i ∈ I (b_i)_{i \in I} ( b i ) i ∈ I A A A A ′ A' A ′ ( e i ) i ∈ I (e_i)_{i \in I} ( e i ) i ∈ I e i e j = 0 e_i e_j = 0 e i e j = 0 i ≠ j i \neq j i = j 1 = ∑ i ∈ I e i 1 = \sum_{i \in I} e_i 1 = ∑ i ∈ I e i b i = A ( 1 − e i ) b_i = A(1 - e_i) b i = A ( 1 − e i ) i ∈ I i \in I i ∈ I b i + b j = A b_i + b_j = A b i + b j = A i ≠ j i \neq j i = j ⋂ i ∈ I b i = { 0 } \bigcap_{i \in I} b_i = \{0\} ⋂ i ∈ I b i = { 0 } b i + b j = A b_i + b_j = A b i + b j = A i ≠ j i \neq j i = j ∏ i ∈ I b i = { 0 } \prod_{i \in I} b_i = \{0\} ∏ i ∈ I b i = { 0 } I I I A ′ A' A ′ ( b i ′ ) i ∈ I (b'_i)_{i \in I} ( b i ′ ) i ∈ I b i = A b i ′ b_i = A b'_i b i = A b i ′ i ∈ I i \in I i ∈ I
引用了 2 · 被以下引用 3 引用了 命题 7 , 命题 9
被以下引用 注 , 引理 8, VIII § 16 , 推论, VIII § 8
(a) ⇒ \Rightarrow ⇒ A A A ∏ i ∈ I ( A / b i ) \prod_{i \in I} (A/b_i) ∏ i ∈ I ( A / b i ) b i b_i b i p r i \mathrm{pr}_i pr i e i e_i e i
(b) ⇒ \Rightarrow ⇒ e i e_i e i i ≠ j i \neq j i = j 1 − e i ∈ b i , e i = e i ( 1 − e j ) ∈ b j 1 - e_i \in b_i, \ e_i = e_i(1 - e_j) \in b_j 1 − e i ∈ b i , e i = e i ( 1 − e j ) ∈ b j 1 ∈ b i + b j 1 \in b_i + b_j 1 ∈ b i + b j A = b i + b j A = b_i + b_j A = b i + b j ( x i ) i ∈ I (x_i)_{i \in I} ( x i ) i ∈ I e i e_i e i
∏ i ∈ I x i ( 1 − e i ) = ( ∏ i ∈ I x i ) ( ∏ i ∈ I ( 1 − e i ) ) = ( ∏ i ∈ I x i ) ( 1 − ∏ i ∈ I e i ) = 0
\prod_{i \in I} x_i (1 - e_i) = \left( \prod_{i \in I} x_i \right) \left( \prod_{i \in I} (1 - e_i) \right) = \left( \prod_{i \in I} x_i \right) \left( 1 - \prod_{i \in I} e_i \right) = 0
i ∈ I ∏ x i ( 1 − e i ) = ( i ∈ I ∏ x i ) ( i ∈ I ∏ ( 1 − e i ) ) = ( i ∈ I ∏ x i ) ( 1 − i ∈ I ∏ e i ) = 0 因此∏ i ∈ I b i = { 0 } \prod_{i \in I} b_i = \{0\} ∏ i ∈ I b i = { 0 }
(d) ⇒ \Rightarrow ⇒
(c) ⇒ \Rightarrow ⇒
因此条件 (a), (b), (c) 和 (d) 是等价的。假设他们持有。到 (b) ⇒ \Rightarrow ⇒ b i ′ = A ′ ( 1 − e i ) b'_i = A'(1 - e_i) b i ′ = A ′ ( 1 − e i ) A ′ A' A ′ b i = A ( 1 − e i ) = A b i ′ b_i = A(1 - e_i) = Ab'_i b i = A ( 1 − e i ) = A b i ′ i ∈ I i \in I i ∈ I
最后，假设条件 (e) 成立。通过(a)⇒ \Rightarrow ⇒ A ′ A' A ′ ( e i ) i ∈ I (e_i)_{i \in I} ( e i ) i ∈ I e i e j = 0 e_i e_j = 0 e i e j = 0 i ≠ j i \neq j i = j 1 = ∑ i ∈ I e i 1 = \sum_{i \in I} e_i 1 = ∑ i ∈ I e i b i ′ = A ′ ( 1 − e i ) b'_i = A'(1 - e_i) b i ′ = A ′ ( 1 − e i ) i ∈ I i \in I i ∈ I b i = A b i ′ = A ( 1 − e i ) b_i = Ab'_i = A(1 - e_i) b i = A b i ′ = A ( 1 − e i ) i ∈ I i \in I i ∈ I
理想和商环*Z \mathbf{Z} Z
Z \mathbf{Z} Z Z \mathbf{Z} Z n . Z n.\mathbf{Z} n . Z n ⩾ 0 n \geqslant 0 n ⩾ 0 n ⩾ 0 n \geqslant 0 n ⩾ 0 n . Z n.\mathbf{Z} n . Z ( n ) (n) ( n ) Z \mathbf{Z} Z n Z n\mathbf{Z} n Z n ⩾ 0 n \geqslant 0 n ⩾ 0 ( 1 ) (1) ( 1 ) Z \mathbf{Z} Z ( 0 ) (0) ( 0 ) Z \mathbf{Z} Z { 0 } \{0\} { 0 } n Z n\mathbf{Z} n Z n > 1 n > 1 n > 1 m ⩾ 1 m \geqslant 1 m ⩾ 1 n ⩾ 1 n \geqslant 1 n ⩾ 1 m Z ⊃ n Z m\mathbf{Z} \supset n\mathbf{Z} m Z ⊃ n Z n ∈ m . Z n \in m.\mathbf{Z} n ∈ m . Z m m m n n n n Z n\mathbf{Z} n Z n n n n n n m > 1 m > 1 m > 1 Z \mathbf{Z} Z p Z p\mathbf{Z} p Z p p p
令m m m n n n ⩾ 1 \geqslant 1 ⩾ 1 m Z + n Z m\mathbf{Z} + n\mathbf{Z} m Z + n Z d ⩾ 1 d \geqslant 1 d ⩾ 1 d Z = m Z + n Z d\mathbf{Z} = m\mathbf{Z} + n\mathbf{Z} d Z = m Z + n Z r Z ⊃ d Z r\mathbf{Z} \supset d\mathbf{Z} r Z ⊃ d Z r Z ⊃ m Z r\mathbf{Z} \supset m\mathbf{Z} r Z ⊃ m Z r Z ⊃ n Z r\mathbf{Z} \supset n\mathbf{Z} r Z ⊃ n Z ⩾ 1 \geqslant 1 ⩾ 1 最大的公共的除子 （缩写为 g.c.d.）。作为d Z = m Z + n Z d\mathbf{Z} = m\mathbf{Z} + n\mathbf{Z} d Z = m Z + n Z d = m x + n y d = mx + ny d = m x + n y m x + n y = 1 mx + ny = 1 m x + n y = 1
理想m Z m\mathbf{Z} m Z n Z n\mathbf{Z} n Z m n mn mn r Z r\mathbf{Z} r Z r ⩾ 1 r \geqslant 1 r ⩾ 1 the least of the integers ⩾ 1 \geqslant 1 ⩾ 1 the least common multiple (l.c.m.) of m and n.
对于x 1 , … , y r ∈ Z x_1, \ldots, y_r \in \mathbf{Z} x 1 , … , y r ∈ Z m Z m\mathbf{Z} m Z n Z n\mathbf{Z} n Z ∑ i = 1 r m x i n y i = m n ( ∑ i = 1 r x i y i ) \sum_{i=1}^{r} mx_i ny_i = mn \left( \sum_{i=1}^{r} x_i y_i \right) ∑ i = 1 r m x i n y i = mn ( ∑ i = 1 r x i y i ) m n Z mn\mathbf{Z} mn Z
对于每个整数n ⩾ 1 n \geqslant 1 n ⩾ 1 Z / n Z \mathbf{Z}/n\mathbf{Z} Z / n Z n − 1 n - 1 n − 1 n = 1 n = 1 n = 1
引用了 1 引用了 定义 16, I § 4
设n 1 , … , n d n_1, \ldots, n_d n 1 , … , n d ⩾ 1 \geqslant 1 ⩾ 1 n = n 1 … n r n = n_1 \ldots n_r n = n 1 … n r Z \mathbf{Z} Z ∏ i = 1 r Z / n i Z \prod_{i=1}^{r} \mathbf{Z}/n_i\mathbf{Z} ∏ i = 1 r Z / n i Z n Z n\mathbf{Z} n Z Z / n Z \mathbf{Z}/n\mathbf{Z} Z / n Z ∏ i = 1 r Z / n i Z \prod_{i=1}^{r} \mathbf{Z}/n_i\mathbf{Z} ∏ i = 1 r Z / n i Z
引用了 2 · 被以下引用 1 引用了 命题 9 , VII § 1. 。 主理想整环
被以下引用 V § 12. 。 有限域
让a i = n i Z a_i = n_i\mathbf{Z} a i = n i Z i = 1 , … , r i = 1, \ldots, r i = 1 , … , r a i + a j = Z a_i + a_j = \mathbf{Z} a i + a j = Z i ≠ j i \neq j i = j
上述结果以及有关分解到素数因子的结果，将推广到致力于研究主理想整环的章 VII, § 1 和致力于研究阶乘域的 交换代数 、章 VII, § 3 中。
12. 环 / 分数
设 A 为 A 的交换环，S 子集。设 A S A_S A S ε : A → A S \varepsilon : A \to A_S ε : A → A S A S A_S A S A S A_S A S ε \varepsilon ε
引用了 1 · 被以下引用 1 引用了 I § 2. 。 单位元素；可取消元素； 可逆元
被以下引用 定义 8
假设已经为 A S A_S A S
让x , y ∈ A S x, y \in A_S x , y ∈ A S S ′ S' S ′ S S S A A A a , b ∈ A a, b \in A a , b ∈ A p , q ∈ S ′ p, q \in S' p , q ∈ S ′ x = a / p , y = b / q x = a/p, y = b/q x = a / p , y = b / q
x = ε ( a q ) ε ( p q ) − 1 , y = ε ( b p ) ε ( p q ) − 1 ,
x = \varepsilon(aq)\varepsilon(pq)^{-1}, \quad y = \varepsilon(bp)\varepsilon(pq)^{-1},
x = ε ( a q ) ε ( pq ) − 1 , y = ε ( b p ) ε ( pq ) − 1 , 从而
(23)
x + y = ( ε ( a q ) + ε ( b p ) ) ε ( p q ) − 1 = ε ( a q + b p ) ε ( p q ) − 1 = ( a q + b p ) / p q .
\begin{align*}
x + y &= (\varepsilon(aq) + \varepsilon(bp))\varepsilon(pq)^{-1} \\
&= \varepsilon(aq + bp)\varepsilon(pq)^{-1} \\
&= (aq + bp)/pq.
\end{align*}
x + y = ( ε ( a q ) + ε ( b p )) ε ( pq ) − 1 = ε ( a q + b p ) ε ( pq ) − 1 = ( a q + b p ) / pq . 这就证明了唯一性或加法。
我们现在*定义* 加法于A S A_S A S x + y = ( a q + b p ) / p q x + y = (aq + bp)/pq x + y = ( a q + b p ) / pq a , b , p , q a, b, p, q a , b , p , q a ′ , b ′ ∈ A , p ′ , q ′ ∈ S ′ a', b' \in A, p', q' \in S' a ′ , b ′ ∈ A , p ′ , q ′ ∈ S ′ x = a ′ / p ′ , y = b ′ / q ′ x = a'/p', y = b'/q' x = a ′ / p ′ , y = b ′ / q ′ S ′ S' S ′ s s s t t t a p ′ s = a ′ p s , b q ′ t = b ′ q t ap's = a'ps, bq't = b'qt a p ′ s = a ′ p s , b q ′ t = b ′ q t
( a q + b p ) ( p ′ q ′ ) ( s t ) = ( a ′ q ′ + b ′ p ′ ) ( p q ) ( s t )
(aq + bp)(p'q')(st) = (a'q' + b'p')(pq)(st)
( a q + b p ) ( p ′ q ′ ) ( s t ) = ( a ′ q ′ + b ′ p ′ ) ( pq ) ( s t ) 因此
( a q + b p ) / p q = ( a ′ q ′ + b ′ p ′ ) / p ′ q ′ .
(aq + bp)/pq = (a'q' + b'p')/p'q'.
( a q + b p ) / pq = ( a ′ q ′ + b ′ p ′ ) / p ′ q ′ . 很容易验证A S A_S A S 0 / 1 0/1 0/1 ( − a ) / p (-a)/p ( − a ) / p a / p a/p a / p x , y , z ∈ A S x, y, z \in A_S x , y , z ∈ A S x ( y + z ) = x y + x z x(y + z) = xy + xz x ( y + z ) = x y + x z a , b ∈ A a, b \in A a , b ∈ A
ε ( a + b ) = ( a + b ) / 1 = a / 1 + b / 1 = ε ( a ) + ε ( b )
\varepsilon(a + b) = (a + b)/1 = a/1 + b/1 = \varepsilon(a) + \varepsilon(b)
ε ( a + b ) = ( a + b ) /1 = a /1 + b /1 = ε ( a ) + ε ( b ) 因此 ε \varepsilon ε
在定理 4 中定义的环称为分数的环，与S S S S S S A [ S − 1 ] A[S^{-1}] A [ S − 1 ]
引用了 2 引用了 定理 4 , 命题 6, I § 2
A [ S − 1 ] A[S^{-1}] A [ S − 1 ] 0 / 1 0/1 0/1 A [ S − 1 ] A[S^{-1}] A [ S − 1 ] 1 / 1 1/1 1/1
我们将回到*交换代数*、章II, § 2中A [ S − 1 ] A[S^{-1}] A [ S − 1 ]
若S S S A A A A [ S − 1 ] A[S^{-1}] A [ S − 1 ] A A A ε \varepsilon ε A A A A [ S − 1 ] A[S^{-1}] A [ S − 1 ]
设A A A A A A S S S B B B A A A f f f B B B f ( S ) f(S) f ( S ) A [ S − 1 ] A[S^{-1}] A [ S − 1 ] f ˉ \bar{f} f ˉ B B B f = f ˉ ∘ ε f = \bar{f} \circ \varepsilon f = f ˉ ∘ ε
引用了 1 · 被以下引用 1 引用了 定理 1, I § 2
被以下引用 命题 3, I § 9
我们知道（§ 2, no. 4，定理 1），乘法幺半群A [ S − 1 ] A[S^{-1}] A [ S − 1 ] B B B f ˉ \bar{f} f ˉ f = f ˉ ∘ ε f = \bar{f} \circ \varepsilon f = f ˉ ∘ ε a , b ∈ A , p , q ∈ S ′ a, b \in A, p, q \in S' a , b ∈ A , p , q ∈ S ′ S S S A A A f ( A ) f(A) f ( A )
f ˉ ( a / p + b / q ) = f ˉ ( ( a q + b p ) / p q ) = f ( a q + b p ) f ( p q ) − 1 = ( f ( a ) f ( q ) + f ( b ) f ( p ) ) f ( p ) − 1 f ( q ) − 1 = f ( a ) f ( p ) − 1 + f ( b ) f ( q ) − 1 = f ˉ ( a / p ) + f ˉ ( b / q ) .
\bar{f}(a/p + b/q) = \bar{f}((aq + bp)/pq) = f(aq + bp)f(pq)^{-1}
= (f(a)f(q) + f(b)f(p))f(p)^{-1}f(q)^{-1}
= f(a)f(p)^{-1} + f(b)f(q)^{-1}
= \bar{f}(a/p) + \bar{f}(b/q).
f ˉ ( a / p + b / q ) = f ˉ (( a q + b p ) / pq ) = f ( a q + b p ) f ( pq ) − 1 = ( f ( a ) f ( q ) + f ( b ) f ( p )) f ( p ) − 1 f ( q ) − 1 = f ( a ) f ( p ) − 1 + f ( b ) f ( q ) − 1 = f ˉ ( a / p ) + f ˉ ( b / q ) . 因此
f ˉ \bar{f} f ˉ 是环同态。
习题
参见习题为 § 8 。