Fragment 1 数域
/Define/
定义1 设 K K K C C C K K K K K K K K K
例1 Q ( 2 ) = { a + b 2 ∣ a , b ∈ Q } \mathbb{Q}(\sqrt{2}) = \{ a + b\sqrt{2} \mid a, b \in \mathbb{Q} \} Q ( 2 ) = { a + b 2 ∣ a , b ∈ Q }
/proof/
先证明 a + b 2 = 0 ⇒ a = b = 0 a + b\sqrt{2} = 0 \Rightarrow a = b = 0 a + b 2 = 0 ⇒ a = b = 0
( a + b 2 ) ± ( c + d 2 ) = ( a ± c ) ± ( b ± d ) 2 ∈ Q ( 2 ) ( a + b 2 ) ⋅ ( c + d 2 ) = ( a c + 2 b d ) + ( a d + b c ) 2 ∈ Q ( 2 ) c + d 2 ≠ 0 ⇒ c ≠ 0 or d ≠ 0 ⇒ c − d 2 ≠ 0 a + b 2 c + d 2 = ( a + b 2 ) ( c − d 2 ) ( c + d 2 ) ( c − d 2 ) = a c − 2 b d c 2 − 2 d 2 + b c − a d c 2 − 2 d 2 2 ∈ Q ( 2 ) (a + b\sqrt{2}) \pm (c + d\sqrt{2}) = (a \pm c) \pm (b \pm d)\sqrt{2} \in \mathbb{Q}(\sqrt{2}) \\
(a + b\sqrt{2}) \cdot (c + d\sqrt{2}) = (ac + 2bd) + (ad + bc)\sqrt{2} \in \mathbb{Q}(\sqrt{2}) \\
c + d\sqrt{2} \neq 0 \Rightarrow c \neq 0 \text{ or } d \neq 0 \Rightarrow c - d\sqrt{2} \neq 0 \\
\frac{a + b\sqrt{2}}{c + d\sqrt{2}} = \frac{(a + b\sqrt{2})(c - d\sqrt{2})}{(c + d\sqrt{2})(c - d\sqrt{2})} = \frac{ac - 2bd}{c^2 - 2d^2} + \frac{bc - ad}{c^2 - 2d^2}\sqrt{2} \in \mathbb{Q}(\sqrt{2})
( a + b 2 ) ± ( c + d 2 ) = ( a ± c ) ± ( b ± d ) 2 ∈ Q ( 2 ) ( a + b 2 ) ⋅ ( c + d 2 ) = ( a c + 2 b d ) + ( a d + b c ) 2 ∈ Q ( 2 ) c + d 2 = 0 ⇒ c = 0 or d = 0 ⇒ c − d 2 = 0 c + d 2 a + b 2 = ( c + d 2 ) ( c − d 2 ) ( a + b 2 ) ( c − d 2 ) = c 2 − 2 d 2 a c − 2 b d + c 2 − 2 d 2 b c − a d 2 ∈ Q ( 2 )
例2 $ \mathbb{Q}(\sqrt{3}) = { a + b\sqrt{3} + c\sqrt{4} \mid a, b, c \in \mathbb{Q} } $ 是数域。
/proof/
先证明 a + b 3 + c 4 = 0 ⇒ a 2 + 2 b 2 + 4 c 2 − 2 a b c = 0 a + b\sqrt{3} + c\sqrt{4} = 0 \Rightarrow a^2 + 2b^2 + 4c^2 - 2abc = 0 a + b 3 + c 4 = 0 ⇒ a 2 + 2 b 2 + 4 c 2 − 2 ab c = 0
⇒ a = b = c = 0 \Rightarrow a = b = c = 0
⇒ a = b = c = 0
推广1： Q ( 2 ) = { a 0 + a 1 2 + ⋯ + a n 2 n ∣ a i ∈ Q , 0 ≤ i ≤ n } \mathbb{Q}(\sqrt{2}) = \{ a_0 + a_1\sqrt{2} + \cdots + a_n\sqrt{2^n} \mid a_i \in \mathbb{Q}, 0 \leq i \leq n \} Q ( 2 ) = { a 0 + a 1 2 + ⋯ + a n 2 n ∣ a i ∈ Q , 0 ≤ i ≤ n }
推广2： P P P Q ( p ) = { a + b p ∣ a , b ∈ Q } \mathbb{Q}(\sqrt{p}) = \{ a + b\sqrt{p} \mid a, b \in \mathbb{Q} \} Q ( p ) = { a + b p ∣ a , b ∈ Q }
例3 { a 0 + a 1 π + ⋯ + a n π n ∣ a i ∈ Q } \left\{ a_0 + a_1\pi + \cdots + a_n\pi^n \mid a_i \in \mathbb{Q} \right\} { a 0 + a 1 π + ⋯ + a n π n ∣ a i ∈ Q }
/proof/
π \pi π ⟺ d e f \overset{def}\iff ⟺ d e f
则 b 0 = b 1 = ⋯ = b m = 0 b_0 = b_1 = \cdots = b_m = 0 b 0 = b 1 = ⋯ = b m = 0
α ∈ C \alpha \in \mathbb{C} α ∈ C α \alpha α ⟺ d e f ∃ f ( x ) = a n x n + a n − 1 x n − 1 + ⋯ + a 1 x + a 0 \overset{def}\iff \exist f(x) = a_nx^n + a_{n-1}x^{n-1} + \cdots + a_1x + a_0 ⟺ d e f ∃ f ( x ) = a n x n + a n − 1 x n − 1 + ⋯ + a 1 x + a 0
s t . f ( α ) = 0 st. f(\alpha) = 0 s t . f ( α ) = 0 α \alpha α f ( x ) f(x) f ( x )
例4 S = { a + b 2 ∣ a , b ∈ Z } S = \{ a + b\sqrt{2} \mid a, b \in \mathbb{Z} \} S = { a + b 2 ∣ a , b ∈ Z }
/proof/
用反证法。设 S S S Z ⊆ S \mathbb{Z} \subseteq S Z ⊆ S
1 ∈ S ⇒ 2 ∈ S ⇒ 1 2 ∈ S 1 \in S \Rightarrow 2 \in S \Rightarrow \frac{1}{2} \in S 1 ∈ S ⇒ 2 ∈ S ⇒ 2 1 ∈ S 1 2 = a + b 2 \frac{1}{2} = a + b\sqrt{2} 2 1 = a + b 2 a , b ∈ Z a, b \in \mathbb{Z} a , b ∈ Z
若 $b \neq 0 $ ，则 $ \sqrt{2} = \frac{a - \frac{1}{2}}{b} \in \mathbb{Q}$ 矛盾！
例5 S = { a 2 ∣ a ∈ R } S = \{ a\sqrt{2} \mid a \in \mathbb{R} \} S = { a 2 ∣ a ∈ R }
/proof/
反证法。设 S S S 2 ∈ S ⇒ 2 = 2 ⋅ 2 ∈ S \sqrt{2} \in S \Rightarrow 2 = \sqrt{2} \cdot \sqrt{2} \in S 2 ∈ S ⇒ 2 = 2 ⋅ 2 ∈ S
即 2 = a 2 a ∈ Q ⇒ a = 2 ∉ Q 2 = a\sqrt{2}\quad a \in \mathbb{Q} \Rightarrow a = \sqrt{2} \notin \mathbb{Q} 2 = a 2 a ∈ Q ⇒ a = 2 ∈ / Q
/Theorem/ 定理2：任一数域 K K K Q \mathbb{Q} Q Q \mathbb{Q} Q
/proof/
∀ a ∈ K ⇒ 0 = a − a ∈ K \forall a \in K \Rightarrow 0 = a - a \in K ∀ a ∈ K ⇒ 0 = a − a ∈ K
由定义（数域至少有2个不同元素）再取 K K K b b b 1 = b b ∈ K 1 = \frac{b}{b} \in K 1 = b b ∈ K
{ ∀ m ∈ Z + , m = ( 1 + ⋯ + 1 ) ∈ K , − m = 0 − m ∈ K } ⇒ Z ⊆ K \left\{ \forall m \in \mathbb{Z}^+ ,\quad m = (1 + \cdots + 1) \in K,\quad
-m = 0 - m \in K \right\} \Rightarrow \mathbb{Z} \subseteq K\\
{ ∀ m ∈ Z + , m = ( 1 + ⋯ + 1 ) ∈ K , − m = 0 − m ∈ K } ⇒ Z ⊆ K
任取 m n ∈ Q \frac{m}{n} \in \mathbb{Q} n m ∈ Q n ∈ Z + n \in \mathbb{Z}^+ n ∈ Z + m ∈ Z + m \in \mathbb{Z}^+ m ∈ Z + n ∈ K n \in K n ∈ K m ∈ K ⇒ m n ∈ K m \in K \Rightarrow \frac{m}{n} \in K m ∈ K ⇒ n m ∈ K Q ⊆ K \mathbb{Q} \subseteq K Q ⊆ K
