环的基本定义
定义 (环)
一个非空集合 R R R + + + ⋅ \cdot ⋅ ( R , + , ⋅ ) (R, +, \cdot) ( R , + , ⋅ ) 环 (Ring) ：
( R , + ) (R, +) ( R , + ) 阿贝尔群 （Abelian Group）。乘法满足结合律 : ∀ α , β , γ ∈ R , ( α ⋅ β ) ⋅ γ = α ⋅ ( β ⋅ γ ) \forall \alpha, \beta, \gamma \in R, \quad (\alpha \cdot \beta) \cdot \gamma = \alpha \cdot (\beta \cdot \gamma) ∀ α , β , γ ∈ R , ( α ⋅ β ) ⋅ γ = α ⋅ ( β ⋅ γ )
乘法对加法满足分配律 : ∀ α , β , γ ∈ R \forall \alpha, \beta, \gamma \in R ∀ α , β , γ ∈ R
α ⋅ ( β + γ ) = α ⋅ β + α ⋅ γ \alpha \cdot (\beta + \gamma) = \alpha \cdot \beta + \alpha \cdot \gamma α ⋅ ( β + γ ) = α ⋅ β + α ⋅ γ ( α + β ) ⋅ γ = α ⋅ γ + β ⋅ γ (\alpha + \beta) \cdot \gamma = \alpha \cdot \gamma + \beta \cdot \gamma ( α + β ) ⋅ γ = α ⋅ γ + β ⋅ γ
特殊的环
根据环是否满足一些额外的性质，我们可以定义出几类重要的环：
含幺环 (Ring with Unity) : 如果环 R R R 1 1 1 ∀ α ∈ R , 1 ⋅ α = α ⋅ 1 = α \forall \alpha \in R, 1\cdot \alpha = \alpha \cdot 1 = \alpha ∀ α ∈ R , 1 ⋅ α = α ⋅ 1 = α R R R 交换环 (Commutative Ring) : 如果环 R R R ∀ α , β ∈ R , α β = β α \forall \alpha, \beta \in R, \alpha \beta = \beta \alpha ∀ α , β ∈ R , α β = β α R R R 单位 (Unit) : 在含幺环 R R R α ∈ R \alpha \in R α ∈ R ∃ β ∈ R \exists \beta \in R ∃ β ∈ R α β = β α = 1 \alpha \beta = \beta \alpha = 1 α β = β α = 1 α \alpha α 单位 。环中所有单位构成的集合关于乘法是一个群。零因子 (Zero Divisor) : 对于环 R R R α ≠ 0 \alpha \neq 0 α = 0 β ≠ 0 \beta \neq 0 β = 0 α β = 0 \alpha \beta = 0 α β = 0 β α = 0 \beta \alpha = 0 β α = 0 α \alpha α 零因子 。整环 (Integral Domain) : 如果一个环 R R R 含幺 、可交换 、无零因子 ，则称 R R R 域 (Field) : 如果一个含幺环 R R R 1 ≠ 0 1 \neq 0 1 = 0 R R R
说明 : 域一定是整环。因为域中所有非零元素关于乘法构成群，所以任意两个非零元素之积必然非零，所以域没有零因子。
常见的例子
整数环 ( Z , + , ⋅ ) (\mathbb{Z}, +, \cdot) ( Z , + , ⋅ ) : 这是最典型的环，它是一个整环 ，但不是域（因为除了 ± 1 \pm 1 ± 1 模 n 剩余类环 ( Z n , + , ⋅ ) (\mathbb{Z}_n, +, \cdot) ( Z n , + , ⋅ ) : 这是一个交换含幺环。
当 n n n Z 4 \mathbb{Z}_4 Z 4 不是 整环。因为存在零因子，如在 Z 4 \mathbb{Z}_4 Z 4 2 ˉ ⋅ 2 ˉ = 4 ˉ = 0 ˉ \bar{2} \cdot \bar{2} = \bar{4} = \bar{0} 2 ˉ ⋅ 2 ˉ = 4 ˉ = 0 ˉ
当 n n n p p p ( Z p , + , ⋅ ) (\mathbb{Z}_p, +, \cdot) ( Z p , + , ⋅ ) 域 。因为对于任意 a ˉ ∈ Z p \bar{a} \in \mathbb{Z}_p a ˉ ∈ Z p a ˉ ≠ 0 ˉ \bar{a} \neq \bar{0} a ˉ = 0 ˉ gcd ( a , p ) = 1 \operatorname{gcd}(a,p)=1 gcd ( a , p ) = 1 x , y x, y x , y a x + p y = 1 ax+py=1 a x + p y = 1 p p p a ˉ x ˉ = 1 ˉ \bar{a}\bar{x}=\bar{1} a ˉ x ˉ = 1 ˉ a ˉ \bar{a} a ˉ
n 阶实方阵环 ( M n , + , ⋅ ) (M_n, +, \cdot) ( M n , + , ⋅ ) : 这是一个含幺环（幺元为单位矩阵 I n I_n I n n ≥ 2 n \ge 2 n ≥ 2 不是 交换环，也不是 整环（存在零因子）。
子环与同态
子环 (Subring) : 设 ( R , + , ⋅ ) (R, +, \cdot) ( R , + , ⋅ ) S S S R R R ( S , + , ⋅ ) (S, +, \cdot) ( S , + , ⋅ ) S S S R R R 子环 。子域 (Subfield) : 设 ( E , + , ⋅ ) (E, +, \cdot) ( E , + , ⋅ ) F F F E E E ( F , + , ⋅ ) (F, +, \cdot) ( F , + , ⋅ ) F F F E E E 子域 ，记为 F < E F < E F < E E E E F F F 扩域 (Extension Field) 。同态 (Homomorphism) : 对于两个环 ( R , + , ⋅ ) (R, +, \cdot) ( R , + , ⋅ ) ( S , ⊕ , ⊙ ) (S, \oplus, \odot) ( S , ⊕ , ⊙ ) f : R → S f: R \to S f : R → S ∀ α , β ∈ R \forall \alpha, \beta \in R ∀ α , β ∈ R
f ( α + β ) = f ( α ) ⊕ f ( β ) f(\alpha + \beta) = f(\alpha) \oplus f(\beta) f ( α + β ) = f ( α ) ⊕ f ( β ) f ( α ⋅ β ) = f ( α ) ⊙ f ( β ) f(\alpha \cdot \beta) = f(\alpha) \odot f(\beta) f ( α ⋅ β ) = f ( α ) ⊙ f ( β ) f f f 环同态 。
同构 (Isomorphism) : 如果同态映射 f f f f f f 环同构 ，称环 R R R S S S
环的基本算术性质
对于任意环 ( R , + , ⋅ ) (R, +, \cdot) ( R , + , ⋅ )
与零元相乘 : ∀ α ∈ R \forall \alpha \in R ∀ α ∈ R 0 ⋅ α = α ⋅ 0 = 0 0 \cdot \alpha = \alpha \cdot 0 = 0 0 ⋅ α = α ⋅ 0 = 0
证明 : 0 ⋅ α = ( 0 + 0 ) ⋅ α = 0 ⋅ α + 0 ⋅ α 0 \cdot \alpha = (0+0) \cdot \alpha = 0 \cdot \alpha + 0 \cdot \alpha 0 ⋅ α = ( 0 + 0 ) ⋅ α = 0 ⋅ α + 0 ⋅ α ( R , + ) (R, +) ( R , + ) 0 ⋅ α 0 \cdot \alpha 0 ⋅ α 0 = 0 ⋅ α 0 = 0 \cdot \alpha 0 = 0 ⋅ α
与逆元相乘 : ∀ α , β ∈ R \forall \alpha, \beta \in R ∀ α , β ∈ R α ( − β ) = ( − α ) β = − ( α β ) \alpha (-\beta) = (-\alpha) \beta = -(\alpha \beta) α ( − β ) = ( − α ) β = − ( α β ) ( − α ) ( − β ) = α β (-\alpha)(-\beta) = \alpha \beta ( − α ) ( − β ) = α β
证明 : 因为 α β + α ( − β ) = α ( β + ( − β ) ) = α ⋅ 0 = 0 \alpha \beta + \alpha(-\beta) = \alpha(\beta + (-\beta)) = \alpha \cdot 0 = 0 α β + α ( − β ) = α ( β + ( − β )) = α ⋅ 0 = 0 α ( − β ) \alpha(-\beta) α ( − β ) α β \alpha\beta α β α ( − β ) = − ( α β ) \alpha(-\beta) = -(\alpha\beta) α ( − β ) = − ( α β ) ( − α ) β = − ( α β ) (-\alpha)\beta = -(\alpha\beta) ( − α ) β = − ( α β ) 进而，( − α ) ( − β ) = − ( α ( − β ) ) = − ( − ( α β ) ) = α β (-\alpha)(-\beta) = -(\alpha(-\beta)) = -(-(\alpha\beta)) = \alpha\beta ( − α ) ( − β ) = − ( α ( − β )) = − ( − ( α β )) = α β
广义分配律 : ( ∑ i = 1 n α i ) ( ∑ j = 1 m β j ) = ∑ i = 1 n ∑ j = 1 m α i β j \displaystyle \left(\sum_{i=1}^{n} \alpha_{i}\right) \left(\sum_{j=1}^{m} \beta_j\right) = \sum_{i=1}^{n}\sum_{j=1}^{m} \alpha_i \beta_j ( i = 1 ∑ n α i ) ( j = 1 ∑ m β j ) = i = 1 ∑ n j = 1 ∑ m α i β j
说明 : 这是环的分配律通过数学归纳法得到的直接推广。
整数与环元素相乘 : 对 n ∈ Z n \in \mathbb{Z} n ∈ Z α , β ∈ R \alpha, \beta \in R α , β ∈ R ( n α ) β = α ( n β ) = n ( α β ) (n \alpha)\beta = \alpha(n \beta) = n(\alpha \beta) ( n α ) β = α ( n β ) = n ( α β )
说明 : 这里的 n α n\alpha n α n n n α \alpha α n n n − n -n − n − α -\alpha − α ( 2 α ) β = ( α + α ) β = α β + α β = 2 ( α β ) (2\alpha)\beta = (\alpha+\alpha)\beta = \alpha\beta + \alpha\beta = 2(\alpha\beta) ( 2 α ) β = ( α + α ) β = α β + α β = 2 ( α β )
理想 (Ideal)
核心思想 : “理想” 在环论中的地位，等价于 “正规子群” 在群论中的地位。正规子群是为了构造商群，而理想则是为了构造商环 (Quotient Ring) 。
我们知道，对于环 ( R , + ) (R, +) ( R , + ) ( S , + ) (S, +) ( S , + ) ( R / S , + ) (R/S, +) ( R / S , + ) a + S a+S a + S
为了让这个商集也成为一个环，我们需要定义陪集间的乘法。最自然的定义是：
( a + S ) ⋅ ( b + S ) = a b + S (a+S) \cdot (b+S) = ab + S
( a + S ) ⋅ ( b + S ) = ab + S
即 a ˉ ⋅ b ˉ = a b ‾ \bar{a} \cdot \bar{b} = \overline{ab} a ˉ ⋅ b ˉ = ab
定理 : 对于环 ( R , + , ⋅ ) (R, +, \cdot) ( R , + , ⋅ ) S S S ( R / S , + , ⋅ ) (R / S, +, \cdot) ( R / S , + , ⋅ ) S S S 吸收律 ：
∀ a ∈ S , ∀ r ∈ R 都有 a r ∈ S 且 r a ∈ S \forall a \in S, \forall r \in R \quad \text{都有} \quad ar \in S \text{ 且 } ra \in S
∀ a ∈ S , ∀ r ∈ R 都有 a r ∈ S 且 r a ∈ S
证明 :
必要性 (⇒ \Rightarrow ⇒ :( R / S , + , ⋅ ) (R/S, +, \cdot) ( R / S , + , ⋅ ) 0 ˉ = 0 + S = S \bar{0} = 0+S = S 0 ˉ = 0 + S = S a ∈ S a \in S a ∈ S r ∈ R r \in R r ∈ R a ∈ S a \in S a ∈ S a a a S S S a ˉ = S = 0 ˉ \bar{a} = S = \bar{0} a ˉ = S = 0 ˉ a r ‾ = a ˉ ⋅ r ˉ = 0 ˉ ⋅ r ˉ = 0 ˉ = S \overline{ar} = \bar{a} \cdot \bar{r} = \bar{0} \cdot \bar{r} = \bar{0} = S a r = a ˉ ⋅ r ˉ = 0 ˉ ⋅ r ˉ = 0 ˉ = S a r ‾ = S \overline{ar} = S a r = S a r ∈ S ar \in S a r ∈ S r a ∈ S ra \in S r a ∈ S
充分性 (⇐ \Leftarrow ⇐ :S S S
乘法良定义 :α ′ ∈ α ˉ \alpha' \in \bar{\alpha} α ′ ∈ α ˉ β ′ ∈ β ˉ \beta' \in \bar{\beta} β ′ ∈ β ˉ α ′ = α + s 1 \alpha' = \alpha + s_1 α ′ = α + s 1 β ′ = β + s 2 \beta' = \beta + s_2 β ′ = β + s 2 s 1 , s 2 ∈ S s_1, s_2 \in S s 1 , s 2 ∈ S α ′ β ′ ‾ = α β ‾ \overline{\alpha'\beta'} = \overline{\alpha\beta} α ′ β ′ = α β α ′ β ′ − α β ∈ S \alpha'\beta' - \alpha\beta \in S α ′ β ′ − α β ∈ S α ′ β ′ − α β = ( α + s 1 ) ( β + s 2 ) − α β = ( α β + α s 2 + s 1 β + s 1 s 2 ) − α β = α s 2 + s 1 β + s 1 s 2 \begin{aligned}
\alpha'\beta' - \alpha\beta &= (\alpha+s_1)(\beta+s_2) - \alpha\beta \\
&= (\alpha\beta + \alpha s_2 + s_1 \beta + s_1 s_2) - \alpha\beta \\
&= \alpha s_2 + s_1 \beta + s_1 s_2
\end{aligned}
α ′ β ′ − α β = ( α + s 1 ) ( β + s 2 ) − α β = ( α β + α s 2 + s 1 β + s 1 s 2 ) − α β = α s 2 + s 1 β + s 1 s 2
根据吸收律：因为 s 1 , s 2 ∈ S s_1, s_2 \in S s 1 , s 2 ∈ S α , β ∈ R \alpha, \beta \in R α , β ∈ R
α s 2 ∈ S \alpha s_2 \in S α s 2 ∈ S s 1 β ∈ S s_1 \beta \in S s 1 β ∈ S s 1 s 2 ∈ S s_1 s_2 \in S s 1 s 2 ∈ S S S S S S S
结合律与分配律 : 这些性质由 R R R
结合律 : ( α ˉ β ˉ ) γ ˉ = α β ‾ γ ˉ = ( α β ) γ ‾ = α ( β γ ) ‾ = α ˉ β γ ‾ = α ˉ ( β ˉ γ ˉ ) (\bar{\alpha}\bar{\beta})\bar{\gamma} = \overline{\alpha\beta}\bar{\gamma} = \overline{(\alpha\beta)\gamma} = \overline{\alpha(\beta\gamma)} = \bar{\alpha}\overline{\beta\gamma} = \bar{\alpha}(\bar{\beta}\bar{\gamma}) ( α ˉ β ˉ ) γ ˉ = α β γ ˉ = ( α β ) γ = α ( β γ ) = α ˉ β γ = α ˉ ( β ˉ γ ˉ ) 分配律 : α ˉ ( β ˉ + γ ˉ ) = α ˉ ( β + γ ‾ ) = α ( β + γ ) ‾ = α β + α γ ‾ = α β ‾ + α γ ‾ = α ˉ β ˉ + α ˉ γ ˉ \bar{\alpha}(\bar{\beta} + \bar{\gamma}) = \bar{\alpha}(\overline{\beta+\gamma}) = \overline{\alpha(\beta+\gamma)} = \overline{\alpha\beta + \alpha\gamma} = \overline{\alpha\beta} + \overline{\alpha\gamma} = \bar{\alpha}\bar{\beta} + \bar{\alpha}\bar{\gamma} α ˉ ( β ˉ + γ ˉ ) = α ˉ ( β + γ ) = α ( β + γ ) = α β + α γ = α β + α γ = α ˉ β ˉ + α ˉ γ ˉ
定义 (理想)
对于环 ( R , + , ⋅ ) (R, +, \cdot) ( R , + , ⋅ ) I I I
∀ a ∈ I , ∀ r ∈ R 都有 a r ∈ I 且 r a ∈ I \forall a \in I, \forall r \in R \quad \text{都有} \quad ar \in I \text{ 且 } ra \in I
∀ a ∈ I , ∀ r ∈ R 都有 a r ∈ I 且 r a ∈ I
那么称 I I I R R R 双边理想 (Two-sided Ideal) ，简称理想 。
理想的例子
平凡理想 : 对于任何环 R R R { 0 } \{0\} { 0 } R R R 整数环的理想 : 在整数环 ( Z , + , ⋅ ) (\mathbb{Z}, +, \cdot) ( Z , + , ⋅ ) n Z = { . . . , − 2 n , − n , 0 , n , 2 n , . . . } n\mathbb{Z} = \{..., -2n, -n, 0, n, 2n, ...\} n Z = { ... , − 2 n , − n , 0 , n , 2 n , ... }
环的特征 (Characteristic)
定义 (特征)
对于一个环 ( R , + , ⋅ ) (R, +, \cdot) ( R , + , ⋅ ) m m m 任意 r ∈ R r \in R r ∈ R m ⋅ r = r + r + ⋯ + r ⏟ m times = 0 m \cdot r = \underbrace{r + r + \dots + r}_{m \text{ times}} = 0 m ⋅ r = m times r + r + ⋯ + r = 0 m m m R R R 特征 ，记为 c h a r ( R ) = m char(R) = m c ha r ( R ) = m
如果这样的正整数不存在，则记 c h a r ( R ) = 0 char(R) = 0 c ha r ( R ) = 0
等价定义 : 对于一个含幺环 R R R m m m m ⋅ 1 = 0 m \cdot 1 = 0 m ⋅ 1 = 0
说明 : 这个定义与前者等价。若 m ⋅ 1 = 0 m \cdot 1 = 0 m ⋅ 1 = 0 r ∈ R r \in R r ∈ R m ⋅ r = m ⋅ ( 1 ⋅ r ) = ( m ⋅ 1 ) ⋅ r = 0 ⋅ r = 0 m \cdot r = m \cdot (1 \cdot r) = (m \cdot 1) \cdot r = 0 \cdot r = 0 m ⋅ r = m ⋅ ( 1 ⋅ r ) = ( m ⋅ 1 ) ⋅ r = 0 ⋅ r = 0 r r r m ⋅ r = 0 m \cdot r = 0 m ⋅ r = 0 r = 1 r=1 r = 1 m ⋅ 1 = 0 m \cdot 1 = 0 m ⋅ 1 = 0
附注 : 除非是零环 R = { 0 } R=\{0\} R = { 0 } c h a r ( R ) ≠ 1 char(R) \neq 1 c ha r ( R ) = 1
例子
整数环 ( Z , + , ⋅ ) (\mathbb{Z}, +, \cdot) ( Z , + , ⋅ ) c h a r ( Z ) = 0 char(\mathbb{Z}) = 0 c ha r ( Z ) = 0
剩余类环 ( Z n , + , ⋅ ) (\mathbb{Z}_n, +, \cdot) ( Z n , + , ⋅ ) c h a r ( Z n ) = n char(\mathbb{Z}_n) = n c ha r ( Z n ) = n
重要结论
定理 ：在一个交换环 R R R c h a r ( R ) = p char(R)=p c ha r ( R ) = p ∀ α , β ∈ R \forall \alpha, \beta \in R ∀ α , β ∈ R ( α + β ) p = α p + β p (\alpha + \beta)^{p} = \alpha ^{p} + \beta ^{p} ( α + β ) p = α p + β p
证明 :( α + β ) p = ∑ k = 0 p ( p k ) α p − k β k = α p + ( p 1 ) α p − 1 β + ⋯ + ( p p − 1 ) α β p − 1 + β p (\alpha + \beta)^{p} = \sum_{k=0}^{p} \binom{p}{k} \alpha^{p-k} \beta^k = \alpha^p + \binom{p}{1}\alpha^{p-1}\beta + \dots + \binom{p}{p-1}\alpha\beta^{p-1} + \beta^p ( α + β ) p = ∑ k = 0 p ( k p ) α p − k β k = α p + ( 1 p ) α p − 1 β + ⋯ + ( p − 1 p ) α β p − 1 + β p p p p 1 ≤ k ≤ p − 1 1 \le k \le p-1 1 ≤ k ≤ p − 1 ( p k ) = p ! k ! ( p − k ) ! \binom{p}{k} = \frac{p!}{k!(p-k)!} ( k p ) = k ! ( p − k )! p ! p p p c h a r ( R ) = p char(R)=p c ha r ( R ) = p x ∈ R x \in R x ∈ R p ⋅ x = 0 p \cdot x = 0 p ⋅ x = 0 ( p k ) α p − k β k \binom{p}{k}\alpha^{p-k}\beta^k ( k p ) α p − k β k 0 0 0 ( α + β ) p = α p + β p (\alpha + \beta)^{p} = \alpha^p + \beta^p ( α + β ) p = α p + β p
定理 : 一个非零整环 的特征，如果不是 0 0 0
证明 :R R R c h a r ( R ) = m ≠ 0 char(R) = m \neq 0 c ha r ( R ) = m = 0 m m m m = s t m=st m = s t 1 < s , t < m 1 < s, t < m 1 < s , t < m m ⋅ 1 = 0 m \cdot 1 = 0 m ⋅ 1 = 0 ( s t ) ⋅ 1 = 0 (st) \cdot 1 = 0 ( s t ) ⋅ 1 = 0 ( s ⋅ 1 ) ( t ⋅ 1 ) = 0 (s \cdot 1)(t \cdot 1) = 0 ( s ⋅ 1 ) ( t ⋅ 1 ) = 0 R R R ( s ⋅ 1 ) ( t ⋅ 1 ) = 0 (s \cdot 1)(t \cdot 1) = 0 ( s ⋅ 1 ) ( t ⋅ 1 ) = 0 s ⋅ 1 = 0 s \cdot 1 = 0 s ⋅ 1 = 0 t ⋅ 1 = 0 t \cdot 1 = 0 t ⋅ 1 = 0 m m m k ⋅ 1 = 0 k \cdot 1 = 0 k ⋅ 1 = 0 最小 正整数的定义相矛盾，因为 s s s t t t m m m m m m
整环上的整除理论
在域中，所有非零元素都是 unit，任何非零元素都能“整除”其他任何元素，因此讨论整除没有太大意义。我们主要在整环上讨论这个话题。
设 ( R , + , ⋅ ) (R, +, \cdot) ( R , + , ⋅ ) α , β ∈ R \alpha, \beta \in R α , β ∈ R
因子 (Divisor) : 若存在 r ∈ R r \in R r ∈ R β = α r \beta = \alpha r β = α r α \alpha α 整除 β \beta β α ∣ β \alpha \mid \beta α ∣ β α \alpha α β \beta β 因子 ，β \beta β α \alpha α 倍数 。真因子 (Proper Divisor) : 如果 α ∣ β \alpha \mid \beta α ∣ β α \alpha α r r r α \alpha α β \beta β 真因子 。相伴 (Associate) : 如果 α = β u \alpha = \beta u α = β u u u u α \alpha α β \beta β 相伴 。不可约元 (Irreducible Element) : 对于一个非零、非单位的元素 β \beta β β = a b \beta = ab β = ab a a a b b b β \beta β 不可约元 。素元 (Prime Element) : 对于一个非零、非单位的元素 β \beta β β ∣ a b \beta \mid ab β ∣ ab β ∣ a \beta \mid a β ∣ a β ∣ b \beta \mid b β ∣ b β \beta β 素元 。
关系 : 在整环中，素元一定是不可约元 。
证明 :p p p p p p p = a b p = ab p = ab a a a b b b p = a b p=ab p = ab p ∣ a b p \mid ab p ∣ ab p p p p ∣ a p \mid a p ∣ a p ∣ b p \mid b p ∣ b p ∣ a p \mid a p ∣ a c ∈ R c \in R c ∈ R a = p c a = pc a = p c p = ( p c ) b = p ( c b ) p = (pc)b = p(cb) p = ( p c ) b = p ( c b ) p ≠ 0 p \neq 0 p = 0 1 = c b 1 = cb 1 = c b c c c b b b b b b p ∣ b p \mid b p ∣ b a a a p p p
然而，不可约元不一定是素元。 这也是某些环（如 Z [ − 5 ] \mathbb{Z}[\sqrt{-5}] Z [ − 5 ]
例子: Z [ − 10 ] \mathbb{Z}[\sqrt{-10}] Z [ − 10 ]
考虑环 Z [ − 10 ] = { a + b − 10 ∣ a , b ∈ Z } \mathbb{Z}[\sqrt{-10}] = \{ a+b \sqrt{-10} \mid a, b \in \mathbb{Z} \} Z [ − 10 ] = { a + b − 10 ∣ a , b ∈ Z } 范数 (Norm) : N ( a + b − 10 ) = a 2 + 10 b 2 N(a+b\sqrt{-10}) = a^2 + 10b^2 N ( a + b − 10 ) = a 2 + 10 b 2 N ( α β ) = N ( α ) N ( β ) N(\alpha\beta) = N(\alpha)N(\beta) N ( α β ) = N ( α ) N ( β ) α \alpha α ⟺ N ( α ) = 1 \iff N(\alpha)=1 ⟺ N ( α ) = 1
在这个环中，数字 14 14 14
14 = 2 ⋅ 7 = ( 2 + − 10 ) ( 2 − − 10 ) 14 = 2 \cdot 7 = (2+\sqrt{-10})(2-\sqrt{-10})
14 = 2 ⋅ 7 = ( 2 + − 10 ) ( 2 − − 10 )
我们可以用范数来验证这些因子都是不可约元 :
N ( 2 ) = 2 2 = 4 N(2) = 2^2 = 4 N ( 2 ) = 2 2 = 4 N ( 7 ) = 7 2 = 49 N(7) = 7^2 = 49 N ( 7 ) = 7 2 = 49 N ( 2 ± − 10 ) = 2 2 + 10 ( 1 2 ) = 14 N(2\pm\sqrt{-10}) = 2^2 + 10(1^2) = 14 N ( 2 ± − 10 ) = 2 2 + 10 ( 1 2 ) = 14
以 2 2 2 2 = α β 2 = \alpha\beta 2 = α β N ( 2 ) = N ( α ) N ( β ) = 4 N(2) = N(\alpha)N(\beta) = 4 N ( 2 ) = N ( α ) N ( β ) = 4 α , β \alpha, \beta α , β N ( α ) ≠ 1 , N ( β ) ≠ 1 N(\alpha) \neq 1, N(\beta) \neq 1 N ( α ) = 1 , N ( β ) = 1 N ( α ) = N ( β ) = 2 N(\alpha)=N(\beta)=2 N ( α ) = N ( β ) = 2 a 2 + 10 b 2 = 2 a^2+10b^2=2 a 2 + 10 b 2 = 2 2 2 2 7 7 7 2 ± − 10 2\pm\sqrt{-10} 2 ± − 10
现在我们来考察 2 2 2 2 ∣ ( 2 + − 10 ) ( 2 − − 10 ) 2 \mid (2+\sqrt{-10})(2-\sqrt{-10}) 2 ∣ ( 2 + − 10 ) ( 2 − − 10 ) 2 2 2 ( 2 + − 10 ) (2+\sqrt{-10}) ( 2 + − 10 ) ( 2 − − 10 ) (2-\sqrt{-10}) ( 2 − − 10 ) 2 ∣ ( 2 + − 10 ) 2 \mid (2+\sqrt{-10}) 2 ∣ ( 2 + − 10 ) a + b − 10 ∈ Z [ − 10 ] a+b\sqrt{-10} \in \mathbb{Z}[\sqrt{-10}] a + b − 10 ∈ Z [ − 10 ]
2 ( a + b − 10 ) = 2 + − 10 ⟹ 2 a + 2 b − 10 = 2 + − 10 2(a+b\sqrt{-10}) = 2+\sqrt{-10} \implies 2a + 2b\sqrt{-10} = 2+\sqrt{-10}
2 ( a + b − 10 ) = 2 + − 10 ⟹ 2 a + 2 b − 10 = 2 + − 10
比较系数可知 2 a = 2 , 2 b = 1 2a=2, 2b=1 2 a = 2 , 2 b = 1 a = 1 , b = 1 / 2 a=1, b=1/2 a = 1 , b = 1/2 b b b 2 2 2 ( 2 + − 10 ) (2+\sqrt{-10}) ( 2 + − 10 ) ( 2 − − 10 ) (2-\sqrt{-10}) ( 2 − − 10 )
结论 : 在环 Z [ − 10 ] \mathbb{Z}[\sqrt{-10}] Z [ − 10 ] 2 2 2 不可约元 ，但不是 一个素元 。这也解释了为什么这个环中的元素没有唯一的因子分解。
