近世代數--環同态--環的擴張定理
部落客是初學近世代數(群環域),本意是想整理一些較難了解的定理、算法,加深記憶也友善日後查找;如果有錯,歡迎指正。
我整理成一個系列:近世代數,友善檢索。
域的擴張定理用來将一已知的環擴大為某一具有特定性質的環。
S ˉ 、 R \bar{S}、R Sˉ、R是環, S ˉ ∩ R = ∅ , φ ˉ : S ˉ → R \bar{S}\cap R=\empty,\bar{\varphi}:\bar{S}\rightarrow R Sˉ∩R=∅,φˉ:Sˉ→R是單同态,
則
- ∃ S , S \exists S,S ∃S,S是環, S ≅ R , φ : S → R S\cong R,\varphi:S\rightarrow R S≅R,φ:S→R是同構,
- S ′ ≤ S , S'\le S, S′≤S,
- 且 φ ∣ s ˉ = φ ˉ \varphi|\bar{s}=\bar{\varphi} φ∣sˉ=φˉ
證明:把已知環 S ′ S' S′擴大為環 S S S
- 構造環 S S S:
- S = ( R − φ ˉ ( S ˉ ) ) ∪ S ˉ S=(R-\bar{\varphi}(\bar{S}))\cup \bar{S} S=(R−φˉ(Sˉ))∪Sˉ
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構造映射 φ : S → R \varphi:S\rightarrow R φ:S→R φ ( x ) = { φ ˉ ( x ) , x ∈ S ˉ x , x ∉ S ˉ \varphi(x)=\left\{ \begin{aligned} \bar{\varphi}(x),x\in \bar{S}\\ x,x\notin \bar{S} \end{aligned} \right. φ(x)={φˉ(x),x∈Sˉx,x∈/Sˉ
R ∩ S ˉ = ∅ → R − φ ˉ ( S ˉ ) ∩ S ˉ = ∅ → ∣ R − φ ˉ ( S ˉ ) ∩ S ˉ ∣ = ∣ R − φ ˉ ( S ˉ ) ∣ + ∣ S ˉ ∣ R\cap \bar{S}=\empty\\\rightarrow R-\bar{\varphi}(\bar{S})\cap \bar{S}=\empty\\\rightarrow |R-\bar{\varphi}(\bar{S})\cap \bar{S}|=|R-\bar{\varphi}(\bar{S})|+|\bar{S}| R∩Sˉ=∅→R−φˉ(Sˉ)∩Sˉ=∅→∣R−φˉ(Sˉ)∩Sˉ∣=∣R−φˉ(Sˉ)∣+∣Sˉ∣
x ∉ S ˉ → φ ( x ) = x = R − φ ˉ ( S ˉ ) x\notin \bar{S}\rightarrow \varphi(x)=x=R-\bar{\varphi}(\bar{S}) x∈/Sˉ→φ(x)=x=R−φˉ(Sˉ)
x ∈ S ˉ → φ ( x ) = φ ˉ ( x ) = φ ˉ ( S ˉ ) x\in \bar{S}\rightarrow \varphi(x)=\bar{\varphi}(x)=\bar{\varphi}(\bar{S}) x∈Sˉ→φ(x)=φˉ(x)=φˉ(Sˉ)
→ ∀ x ∈ S , φ ( x ) = ( R − φ ˉ ( S ˉ ) ) + φ ˉ ( S ˉ ) = R \rightarrow\forall x\in S,\varphi(x)=(R-\bar{\varphi}(\bar{S}))+\bar{\varphi}(\bar{S})=R →∀x∈S,φ(x)=(R−φˉ(Sˉ))+φˉ(Sˉ)=R
φ : S → R , φ ( x ) = R , → φ \varphi:S\rightarrow R,\varphi(x)=R,\rightarrow \varphi φ:S→R,φ(x)=R,→φ是滿映射;
φ ˉ : S ˉ → R \bar{\varphi}:\bar{S}\rightarrow R φˉ:Sˉ→R是單同态,恒等映射 f ( x ) = x f(x)=x f(x)=x是單同态, → φ \rightarrow \varphi →φ是單映射;
→ φ \rightarrow \varphi →φ是雙射,且 φ ∣ S ˉ = φ ˉ \varphi|\bar{S}=\bar{\varphi} φ∣Sˉ=φˉ
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加法運算、乘法運算:
∀ x , y ∈ S , x + y = φ − 1 ( φ ( x ) + φ ( y ) ) x ⋅ y = φ − 1 ( φ ( x ) ⋅ φ ( y ) ) \forall x,y\in S,\\x+y=\varphi^{-1}(\varphi(x)+\varphi(y))\\x·y=\varphi^{-1}(\varphi(x)·\varphi(y)) ∀x,y∈S,x+y=φ−1(φ(x)+φ(y))x⋅y=φ−1(φ(x)⋅φ(y))
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φ \varphi φ是同态
φ ( x + y ) = φ ( x ) + φ ( y ) φ ( x ⋅ y ) = φ ( x ) ⋅ φ ( y ) \varphi(x+y)=\varphi(x)+\varphi(y)\\\varphi(x·y)=\varphi(x)·\varphi(y) φ(x+y)=φ(x)+φ(y)φ(x⋅y)=φ(x)⋅φ(y)
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S ˉ ≤ S \bar{S}\le S Sˉ≤S
∀ x , y ∈ S ˉ , x + S y = φ − 1 ( φ ( x ) + R φ ( y ) ) = φ − 1 ( φ ˉ ( x ) + φ ˉ ( y ) ) = φ − 1 ( φ ˉ ( x + S ˉ ( y ) ) = φ − 1 ( φ ( x + S ˉ y ) ) = x + S ˉ y \forall x,y\in \bar{S},\\x+_{S}y\\=\varphi^{-1}(\varphi(x)+_{R}\varphi(y))\\=\varphi^{-1}(\bar{\varphi}(x)+\bar{\varphi}(y))\\=\varphi^{-1}(\bar{\varphi}(x+_{\bar{S}}(y))\\=\varphi^{-1}(\varphi(x+_{\bar{S}}y))\\=x+_{\bar{S}}y ∀x,y∈Sˉ,x+Sy=φ−1(φ(x)+Rφ(y))=φ−1(φˉ(x)+φˉ(y))=φ−1(φˉ(x+Sˉ(y))=φ−1(φ(x+Sˉy))=x+Sˉy
同理, x ⋅ S y = x ⋅ S ˉ y x·_{S}y=x·_{\bar{S}}y x⋅Sy=x⋅Sˉy
是以, S S S的代數運算在 S ˉ \bar{S} Sˉ上的限制就是 S ˉ \bar{S} Sˉ的代數運算, S ˉ ≤ S \bar{S}\le S Sˉ≤S